Important Questions

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 13 Amines Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 13 Amines

Question 1.
What are amines?
Answer:
Amines : The alkyl or aryl derivatives of ammonia in which one, two or all the three hydrogen atoms attached to nitrogen are replaced by same or different alkyl or aryl groups are called amines. OR Amines are nitrogen-containing organic compounds having basic character.

Example : methyl amine : CH₃ – NH₂

Question 2.
Classify the following amines as primary, secondary and tertiary.
Answer:

Question 3.
Mention the functional group in :
(1) Primary amine
(2) Secondary amine
(3) Tertiary amine.
Answer:
(1) A primary amine has a functional group – NH₂ (amino group).
Example : ethylamine, C₂H₅ – NH₂
(2) A secondary amine has a functional group – NH – (imino group).
Example :
(3) A tertiary amine has a functional group (tertiary nitrogen atom)

Example :

Question 4.
Write common and IUPAC names of following compounds :
Answer:
(A) Primary amines :


(B) Secondary amine :

(C) Tertiary Aimines :
.

Question 5.
Give the structures of the following :
Answer:

Question 6.
Give the IUPAC names of the following amines :
Answer:

Question 7.
Write the IUPAC names of the following amines :
Answer:

Question 8.
Give the structures and IUPAC names of the following amines :
Answer:

Question 9.
Classify the following amines as primary, secondary and tertiary and write the IUPAC names.
Answer:

Question 10.
Write the structures and classify the following amines as primary, secondary, tertiary amines.
Answer:

Question 11.
Write the common and IUPAC name of a tertiary amine in which one methyl, one ethyl and one w-propyl group is attached to nitrogen.
Answer:

Question 12.
How will you prepare ethanamine from ethyl iodide?
Answer:
When ethyl iodide is heated with excess of alcoholic ammonia, under pressure at 373 K ethanamine is obtained as a major product.

Question 13.
How is a nitroalkane converted to a primary amine?
OR
What is the action of LiAlH4/ether on (i) 1-Nitropropane (ii) 2-MethyI-l-nitropropane?
Answer:
When a nitroalkane is refluxed with tin (or iron) and concentrated HCl it gives corresponding primary amine.

For example, (1) nitromethane on reduction by refluxing with Sn and concentrated HCl gives methylamine.

(2) 1-Nitropropane on reduction with Sn and concentrated HCl gives propan-1-amine.

(3) Niirobenzcnc on reducion with tin and concentrated HCI or by using H₂/Pd in ethanol gives anilinc.

(4) When nitropropane is reduced in the presence of LiAlH₄ in ether, n-propyl amine is obtained.

(5) When 2-methyl-1-nitropropane is reduced in the presence of LiAlH₄ in ether, 2-methyl propan-1-amine is obtained.

Question 14.
How will you prepare aniline from nitrobenzene?
OR
How is aniline prepared from nitro compounds?
Answer:
Nitrobenzene is reduced to aniline by passing hydrogen gas in the presence of finely divided nickel, palladium or platinum.

Question 15.
Identify the compounds A and B in the following reactions

Answer:

Question 16.
How will you obtain a primary amine from an alkyl cyanide (nitrile)?
OR
Write a short note on Mendius reduction.
Answer:
Alkyl cyanides (nitriles) on reduction by sodium and ethyl alcohol form corresponding primary amines. This reaction is called Mendius reduction.

For example; propionitrile on reduction by sodium and ethanol gives n-propyl amine (Propan-1-amine).

Methyl cyanide or acetonitrile on reduction by sodium and ethanol gives ethanaminc.

Question 17.
How will you prepare ethylamine from acetonitrile?
OR
How is ethanamine prepared from methyl cyanide?
OR
What is the action of a mixture of sodium and alcohol on acetonitrile?
Answer:
Methyl cyanide or acetonitrile on reduction by sodium and ethyl alcohol forms ethanamine. The reaction is called Mendius reduction.

Question 18.
How will convert phenyl acetonitrile to β-phenylethylamine?
Answer:
When phenyl acetonitrile is reduced in the presence of sodium and ethanol, β-phenyl ethylamine is obtained.

Question 19.
How will you obtain primary amine from an acid amide?
Answer:
Acid amides on reduction with lithium aluminium hydride or sodium, ethanol form corresponding primary amines.

For example : Acetamide on reduction with lithium aluminium hydride or sodium, ethanol gives ethylamines.

Question 20.
Explain Hoffmann degradation of amides.
Write a note on Hoffmann bromamide degradation.
Answer:
The conversion of amides into amines in the presence of bromine and alkali is known as Hoffmann degradation of amides. An important characteristic of this reaction is that an amine with one carbon less than those in the amide is formed. Thus, decreasing the length of carbon chain. This reaction is an example of molecular rearrangement and involves the migration of an alkyl or aryl group from the carbonyl carbon to the adjacent nitrogen atom. For example,

(1) When propanamide is treated with bromine and aqueous or alcoholic sodium hydroxide, ethanamine is obtained which has one carbon atom less.

(2) When benzamide is treated with bromine and aqueous or alcoholic sodium hydroxide, aniline is obtained.

Question 21.
How will you obtain methyl amine from acetamide?
Answer:
When acetamide is treated with bromine and aq or alcoholic solution of KOH, methyl amine is obtained, which has one cabon atom less.

Question 22.
How will you convert the following?

(1) Ethyl bromide to ethylamine.
Answer:

(2) Propionitrile to n-propyl amine.
Answer:

(3) Acetonitrile to ethylamine.
Answer:

(4) Phenyl acetonitrile to β-phenylethyl amine.
Answer:

(5) Acetamide to ethylamine.
Answer:

(6) Nitropropane to propan-l-amine.
Answer:

(7) Nitrobenzene to Aniline.
Answer:

(8) Benzamide to aniline.
Answer:

Question 23.
How will you prepare propan-l-amine from (1) butane nitrile (2) 1-nitropropane (3) propanamide (4) butanamide?
Answer:
(1) From butane nitrile :
When butane nitrile is reduced by sodium and ethanol, it gives propan-l-amine.

(2) From 1-nitropropane :
When 1-nitropropane is reduced in the presence of tin and cone, hydrochloric acid, propan-l-amine is obtained.

(3) From propanamide :
When propanamide is reduced in the presence of lithium aluminium hydride, propan-l-amine is obtained.

(4) From butanamide :
When butanamide is treated with bromine and aq. KOH, propan-l-amine is obtained.

Question 24.
Write a reaction to, convert acetic acid into methyl amine.
Answer:

Question 25.
Primary and secondary amines have boiling points higher than the tertiary amines. Explain why?
Answer:
(1) The N – H bond in amines is polar in nature because of electronegativities of nitrogen (3.0) and hydrogen (2.1) are different.
(2) Due to the polar nature of N – H bond, primary and secondary have strong intermolecular hydrogen bonding. Tertiary amines do not have intermolecular hydrogen bonding as there is no hydrogen atom on nitrogen of tertiary amine. Thus, intermolecular forces of attraction are strongest in primary and secondary amines and weakest in to tertiary amines. Hence, primary and secondary amines have boiling points higher than the tertiary amines.

Question 26.
Amines have boiling points higher than the hydrocarbon but lower than the alcohols of comparable masses. Explain, why?
Answer:
Amines are polar than alkanes but less polar than alcohols. Primary and secondary amines form intermolecular hydrogen bonds. This hydrogen bonding leads to an associated structure. The association is more in primary amines than that in secondary amines as there are two hydrogen atoms attached to the nitrogen atom. However, tertiary amines do not form intermolecular hydrogen bonds because they do not contain any hydrogen atoms attached to the nitrogen atom. Hence, amines have higher boiling points than the hydrocarbons but lower boiling points than the alcohols of comparable masses.

Compound	Molar mass	Boiling points (K)
nC2H5CH(CH3)2	72	300
nC4H9NH2	73	350.8
nC4H9OH	74	391

Question 27.
Arrange the following compounds in the decreasing order of their solubility in water.
(a) Ethyl amine, diethyl amine and triethyl amine.
Answer:
Diethyl amine > triethyl amine > ethyl amine
(The reason that ethyl group has greater +1 effect than methyl group)

(b) Ethyl amine, n-propyl amine and n-butyl amine.
Answer: n-butyl amine < n-propyl amine < ethyl amine

(c) n-Butane, n -butyl alcohol and n-butyl amine
Answer:
n-butyl alcohol < n-butyl amine < n-butane

Question 28.
Arrange the following compounds in the decreasing order of their boiling points.
(a) Ethane, ethyl amine and ethyl alcohol.
Answer:
Ethyl alcohol < ethyl amine < ethane

(b) Ethyl amine, n-propyl amine and n-butyl amine.
Answer:
n-butyl amine < n-propyl amine < ethyl amine

(c) n-propyl amine, ethyl methyl amine and trimethyl amine.
Answer:
n-propyl amine < ethyl methyl amine < trimethyl amine.

(d) Ethyl alcohol, dimethyl amine and ethyl amine.
Answer:
Ethyl alcohol < ethyl amine < dimethyl amine.

Question 29.
Explain the basic nature of amines with a suitable examples.
OR
Explain why amines are basic.

Question 38.
Tertiary amine (R₃N) or 3° amine is weaker base than secondary amine R₂NH or 2° amine. Explain.
Answer:


The increase in basic strength from 1° amine to 2° amine is explained on the basis of increased stabilization of conjugate acids by +1 effect of the increased number of the alkyl group. However, decreased basic strength of 3° implies that the conjugate acid of 3° amine is less stabilized and is weak base though the +1 effect of three alkyl groups in R₃NH is large.

R₂NH is best stabilized by solvation while the stabilization by solvation is very poor in R₃NH. Hence (R₃N) or tertiary amine or 3° amine is weaker base than secondary amine (R₂NH) or 2° amine.

Question 30.
Primary or aliphatic amine is a stronger base than ammonia. Explain.
Answer:
(1) The alkyl group in primary amines has +I effect i.e. (electron releasing).

The alkyl group tends to increase the electron density on the nitrogen atom. As a result, amines can donate the lone pair of electrons on nitrogen more easily than ammonia.

(2) The amine being a base, can donate a pair of electrons to an acid. The alkyl group with +I effect will disperse the positive charge on the cation more than ammonia.

Due to +I effect of alkyl group cation formed by primary amine is more stable compared to cation formed from ammonia. Also it is seen that observed increasing basic strength from ammonia to primary amine is explained on the basis of increased stabilization of conjugate acids by +I effect for the presence of alkyl (R) groups. Hence, primary or aliphatic amine is a stronger base than ammonia.

Question 31.
Aniline is less basic than ammonia. Explain.
Answer:
The less basic character of aniline can be explained on the basis of resonance shown by aniline.

Due to resonance, the nitrogen atom of amino group in aniline acquires a positive charge, hence, lone pair of electrons is less available for protonation as compared to that of ammonia. Aniline is resonance stabilized by five resonance structures. On the other hand, aniline in aqueous medium, accepts a proton does not have lone pair of electrons on nitrogen to produce a very low concentration of anilium ion and anilium ion shows only two resonance structures and therefore less stabilized than anline.

Thus, aniline is more stable than anilium ion. Hence aniline accepts proton less readily or less basic in nature than ammonia.

Question 32.
Explain the order of basicity in ammonia and aliphatic amines.
Answer:
Since nitrogen atom in ammonia molecule has a lone pair of electrons, it is a Lewis base.
Greater the availability of an electron pair, more is the basic character.

Since alkyl group (R -) is an electron releasing group with (+I) inductive effect, alkyl amines act as a stronger base than ammonia.

The decreasing order of basicity is –

The availability of a lone pair of electrons on a nitrogen atom in amines is influenced by steric factor due to crowding of alkyl groups which affects solvation along with inductive effect of alkyl groups.

Due to high energy of solvation of \(\mathrm{NH}_{4}^{+}\) ions, they acquire higher stability in aqueous solutions.

The presence of alkyl groups in secondary and tertiary amines, due to steric hindrance decrease the solvation energy.

This effect is more in tertiary amines making the tertiary ammonium ions (R₃NH⁺) unstable as compared to secondary ammonium ion (R₂N⁺H₂).
Hence the cumulative effect on the order of basicity of amines is, secondary amine > primary amine > tertiary amine > ammonia (NH₃).

Question 33.
Arrange the following amines in the decreasing order of their basic nature.
(a) Aniline, propan-l-amine and N-methylethanamine.
Answer:
N-methylethanamine < propan-l-amine < aniline

(b) Benzene-1, 4-diamine, ammonia and 4-aminobenzoic acid.
Answer:
Ammonia < benzene-1, 4-diamine < 4-aminobenzoic acid

(c) N-Methylaniline, phenylmethylamine and N-phenylaniline.
Answer:
N-Methylaniline < N-phenylaniline < phenylmethylamine

Question 34.
Arrange the following amines in the increasing order of their pKb values.
(a) Aniline, N-methylaniline and cyclohexalamine.
Answer:

(b) Phenylmethylamine, 2-aminotoluene and 2-fluoroaniline.
Answer:

(c) Aniline, 4-methoxyaniline and 4-nitroaniline.
Answer:

Question 35.
Arrange the following compounds in the decreasing order of their basic nature in the gaseous phase.
Ammonia, N-methylhexanamine, propan-1-amine and N, N-dimethylethanolamine.
Answer:
Propan-1-amine < N-methylethanamine < N,N-dimethylmethanamine < ammonia

Question 36.
Explain laboratory test for amines.
Answer:
(1) All amines are basic compounds. Aqueous solution of water soluble amines turns red litmus blue.

(2) When water insoluble amine is dissolved in aqueous HCl, forms water soluble substituted ammonium chloride, further a substituted ammonium chloride on reaction with excess aqueous NaOH regenerates the original insoluble amine.

(3) Diazotization reaction/ Orange dye test: In a sample of aromatic primary amine, 1-2 mL of cone. HCl is added. The aqueous solution of NaNO₂ is added with cooling. This solution is transfered to a test tube containing solution of β naphthol in NaOH. Formation of orange dye indicates presence of aromatic primary amino group. (It may be noted that temperature of all the solutions and reaction mixtures is maintained near 0 °C throughout the reaction).

Question 37.
Explain Hofmann’s exhaustive alkylation.
OR
Explain Hofmann’s exhaustive methylation of amines.
Answer:
Hofmann’s Exhaustive alkylation : When a primary amine is heated with excess of primary alkyl halide it gives a mixture of secondary amine, tertiary amine along with tetraalkylammonium halide

If excess of alkyl halide is used, tetraalkyl ammonium halide is obtained as major product. The reaction is known as exhaustive alkylation of amines.

Hofmann’s Exhaustive Methylation : The process of converting a primary, secondary or tertiary amine into quaternary ammonium halide by heating them with excess of methyl iodide, is called exhaustive methylation or Hoffmann’s exhaustive methylation.

Thus when methyl amine is heated with excess of methyl iodide it forms dimethylamine (secondary amine), then trimethylamine (a tertiary amine) and finally of quaternary ammonium iodide. The reaction is carried out in the presence of mild base NaHCO₃, to neutralize the large quantity of HI formed.

Question 38.
Predict the products of exhaustive methylation of following compounds.
(1) Ethylamine.
Answer:
A primary amine, ethylamine (CH₃ – CH₂ – NH₂) on exhaustive methylation, i.e., on heating with excess methyl iodide, forms secondary amine, tertiary amine and finally a quaternary ammonium salt, ethyl-trimethyl ammonium iodide.

(2) Benzylamine.
Answer:
Benzylamine C6H5CH2NH2 on exhaustive methylation i.e., on heating with excess methyl iodide forms benzylmethyl amine, benzyldimethyl ammonium chloride and finally benzyltrimethyl ammonium iodide.

Question 39.
Explain Hofmann elimination.
OR
Write a note on Hoffmann elimination.
Answer:
When tetra alkyl ammonium halide is heated with moist silver hydroxide, a quaternary ammonium hydroxide is obtained. Quaternary ammonium hydroxides are deliquescent crystalline solids and are basic in nature. Quaternary ammonium hydroxides on strong heating undergo ^-elimination to give tertiaryamine, alkenes and water, the reaction is called Hofmann elimination. The major product is least substituted alkene.

Question 40.
Write the bond line formula of the alkene which is obtained as major product from the following amines, on heating with excess of methyl iodide followed by strong heating with moist silver oxide.

Answer:

Answer:

Answer:

Question 41.
Compound X with a molecular formula C₅H₁₃N did not react with nitrous acid, but reacted with one mole of CH3I to form a salt. What is the structure of X?
Answer:
The structure of compound X is ethyl-N-methylethanamine since compound X is tertiary amine. It reacts with one mole of CH₃I to give a quaternary ammonium salt.

Question 42.
What is the action of acetyl chloride on :
(1) ethyl amine (ethanamine)
(2) diethyl amine (N-Ethylethanamine)
(3) triethyl amine?
OR
Write a short note on acylation of amines.
Answer:
The reaction of amines with acetyl chloride is called acetylation of amines.

(1) Acetyl chloride on reaction with ethylamine forms monoacetyl derivative, N-ethylacetamide (or N-acetyl ethylamine).

(2) Diethyl amine on reaction with acetyl chloride forms N-acetyl dimethylamine.

(3) Triethyl amine, being a tertiary amine does not have H atom attached to nitrogen of amine, hence it does not react with acetyl chloride.

Question 43.
What is the action of acetic anhydride on aniline?
Answer:
Aniline on reaction with acetic anhydride forms N-phenyl acetamide.

Question 44.
What is the action of benzoyl chloride on ethanamine?
Answer:
When benzoyl chloride is treated with ethanamine, N-ethyl benzamide is obtained.

Question 45.
What is the action of nitrous acid on ethylamine?
Answer:
Ethyl amine on reaction with nitrous acid in cold forms aliphatic diazonium salt, (unstable intermediate), which decomposes immediately by reaction with solvent water to produce ethyl alcohol and nitrogen gas.

Question 46.
What is the action of nitrous acid on aniline?
Answer:
Aniline reacts with nitrous acid in cold to form diazonium salt which has reasonable solubility at 273 K

Question 47.
How is benzenediazon|um chloride prepared?
Answer:
Benzenediazonium chloride is prepared by the action of nitrous acid on aniline at 273-278 K. Nitrous acid being unstable, is prepared in situ by the reaction between sodium nitrite and dilute hydrochloric acid.

Question 48.
Write resonance stabilized structures of aryl diazonium salt.
Answer:

Question 49.
Write a note on Sandmeyer’s reaction.
OR
How is aryl chloride or aryl bromide or aryl cyanide prepared from diazonium salt?
Answer:
[Replacement by Cl^–, Br^– and -CN : Sandmeyer reaction.] Freshly prepared aromatic diazonium salt on reaction with cuprous chloride gives aryl chloride, on reaction with cuprous bromide gives aryl bromide and on reaction with cuprous cyanide give aryl cyanide. The reaction in which copper (I) salts are used to replace nitrogen in diazonium salt is called Sandmeyer reaction.

Question 50.
How is aryl chloride or aryl bromide prepared by Gattermann reactions?
Answer:
The aryl chloride or bromides can also be prepared by Gattermann reactions in which diazonium salt reacts with

Question 51.
How is aryl iodide obtained from diazonium salt?
Answer:
When diazonium salt is warmed with potassium iodide, aryl iodide is obtained.

Question 52.
Explain the reduction of arene diazonium salt?
OR
How is arene obtained from arene diazonium salt?
OR
What is the action of benzene diazonium chloride on ethanol?
Answer:
Arene diazonium salt on treatment with mild reducing agents like phosphinic acid (hypophosphoric acid) or ethanol, arene is obtained.

Question 53.
How is phenol obtained from arene diazonium salt?
Answer:
When arene diazonium salt is slowly added to a large volume of boiling dilute sulphuric acid, phenol is obtained,

Question 54.
How is aryl fluoride obtained from diazonium salt?
Answer:
When fluoroboric acid is treated with the solution of diazonium salt, a precipitate of diazonium fluoroborate is obtained, which is filtered and dried. When dry diazonium fluoroborate is heated, it decomposes to give aryl fluoride. This reaction is called Balz-Schiemann reaction.

Question 55.
How is nitrobenzene obtained from benzene diazonium fluoroborate?
Answer:
When benzene diazonium fluoroborate is heated with aqueous solution of sodium nitrite in the presence of copper powder, nitrobenzene is obtained.

Benzene diazonium fluorobate can be obtained by reaction of benzene diazonium chloride with HBF₄.

Question 56.
What is meant by a coupling reaction? Explain with suitable examples.
OR
What is the action of benzene diazonium chloride on (a) phenol in alkaline medium (b) aniline?
OR
Write a note on the coupling reaction.
Answer:
Diazonium salts react with certain aromatic compounds having an electron-rich group (e.g.-OH, – NH₂, etc.) to form azo compounds. This reaction is an electrophilic substitution and is called coupling reaction. Azo compounds are brightly coloured and are used as dyes and indicators. Coupling reaction is an electrophilic substitution reaction. Benzene diazonium chloride reacts with alkaline solution of phenol to give p-hydroxy azo benzene (orange dye).

Benzene diazonium chloride reacts with aniline in mild alkaline medium to give p-aminobenzene (yellow dye).

Question 57.
What is the action of p-toluene sulphonyl chloride on ethyl amine and diethyl amine?
Answer:
(1) When ethyl amine is treated with p-toluene sulphonyl chloride, N-ethyl p-toluene sulphonamide is obtained.

(2) When diethyl amine is treated with p-toluene suiphonyl chloride. N.N-dicthyl p-toluene suiphonyl amide is formed.

Question 58.
How will you distinguish between :
(1) Ethylamine, diethyl amine and triethyl amine by using (i) nitrous acid (ii) Hinsberg’s reagent.
(2) Diethyl amine and triethyl amine by using acetic anhydride.
Answer:

Question 59.
Give a chemical test to distinguish between following pairs of compounds.
(i) Ethylamine and diethyl amine :
Answer:
Ethylamine (C₂H₅NH₂) is a primary amine while diethyl amine ( (C₂H₅)₂NH) is a secondary amine. So the two can be distinguished by the following test.

(ii) Ethyl amine and aniline :
Answer:
Ethylamine is an aliphatic amine, while aniline is an aromatic amine. So the two can be distinguished by the following test :

(iii) Aniline and benzyl amine :
Answer:

(iv) Aniline and N-ethylaniline :
Answer:

Question 60.
Compound ‘X’ with a molecular formula C₄H₁₁N did not react with Hinsberg’s reagent, but reacted with one mole of CH₃I to form a salt. What is the structure of ‘X’?
Answer:
The structure of compound ‘X’ is :

Since the compound ‘X’ does not react with NaN02 and HC1 i.e. nitrous acid (HO – N = O), it must be a tertiary amine.

The tertiary amine reacts with one mole of CH3I to give a quaternary ammonium salt.

Question 74.

p-(dimethylamino) azobenzene is yellow dye which was formerly used as a colouring agent in margarine. Write the structures of the reactants used in the preparation of this dye.
Answer:

Question 61.
Convert 3-Methyl aniline into 3-nitrotoluene.
Answer:

Question 62.
How will you bring about following conversions?
(1) N.Methyl aniline into N-methyl benzanilide.
Answer:

(2) 1.4-Dichlorobutane Into hexane-1,6-diamlne.
Answer:

(3) Benzene into 3-bromo aniline.
Answer:

(4) Chlorobenzene into 4-chioroanilinc.
Answer:

(5) 11enaniide into toluene.
Answer:

Question 63.
What is the action of aqueous bromine on aniline?
Answer:
Action of aqueous bromine on aniline : When aniline is treated with bromine water at room temperature, a white precipitate of 2, 4, 6-tri bromoaniline is obtained.

Question 64.
Explain the action of cone, nitric acid (nitrating mixture) on aniline.
Answer:
When aniline is warmed with a mixture of cone, nitric acid and cone, sulphuric acid (a nitrating mixture), a mixture of ortho, meta and para nitroaniline is obtained.

Question 65.
What is the action of acetic anhydride on aniline?
Answer:
When aniline is heated with acetic anhydride, an acetanilide is obtained.

Question 66.
How will you convert aniline to p-nltroanhline? (major product)
Answer:

Question 67.
What is the action of cone, sulphuric acid on aniline?
Answer:
Aniline on treatment with cold sulphuric acid forms anilium hydrogen sulphate which on heating with sulphuric acid at 453 K-475 K gives sulphanilic acid, (p-aminobenzene sulphonic acid) as major product.
Sulphanilic acid exists as a salt; called dipolar ion or zwitter ion. It is produced by the reaction between an acidic group and a basic group present in the same molecule.

Question 68.
How will you convert the following?
(1) Ethylamine to ethyl alcohol.
Answer:

(2) N-Methyl aniline to N-Nitroso-N-methyl aniline.
Answer:

(3) Diethylamine to N-nitrosodiethylamine
Answer:

(4) Triethylamine to triethyl ammonium nitrite.
Answer:

(5) Ethyl amine to N-ethylacetamide.
Answer:

(6) Diethyl amine to N-acetyl diethylamine.
Answer:

(7) Aniline to acetanilide.
Answer:

(8) Aniline to N-ethyl henzamide.
Answer:

(9) Ethylamine to ethyl isocyanide.
Answer:

(10) Aniline to phenyl isocyanide.
Answer:

(11) Aniline to 2,4,6-tribromoaniline.
Answer:

Question 69.
Give a plausible explanation for each of the following statements :
(1) Ethylamine is soluble in water whereas aniline is not.
Answer:
Ethylamine is soluble in water due to intermolecular hydrogen bonding resulting in the formation of C₂H₅NH₃ ion. Whereas in anline the hydrogen bonding with water is negligible due to the phenyl group (C₆H₅) is bulky and has -I effect. Therefore, aniline is nearly insoluble in water.

(2) Butan-1-ol is more soluble in water than butani-amine.
Answer:
Rutan- l-al is more soluble in watcr duc to intermoiccular hydrogen bonding. In alcohols, hydrogen bonding is through oxygen atoms. WIereas hutani-amine is less soluble in water due to the larger hydrocarbon part is hydrophobic in nature. Hence, butan-l-ol is more soluble in water than butani-amine.

(3) Butan-1-amlne has higher boiling point than N-ethylethanamine.
Answer:
Due to the presence of two H-atoms on N-atom in butait- I -amine, they undergo extensive intermolecular H-bonding while in N-cthylethanamine due to the presence of one-H atom on the N-atom, they undergo least intermolecular H-bonding. Hence, butan- l-amine has higher boiling point than-N-ethyl ethanamine.

(4) AnIline Is less basic than ethyl afine.
Answer:
Aniline (K_b4-2 x 10^-10) is less basic than ethyl amine (K_b5.1 x 10^-4). This is because -I effect of phenyl group in aniline as compared to + 1 effect of ethyl group in ethyl amine.

Due to resonance, the lone pair of electrons on the nitrogen atom gets delocalized over the benzene ring and thus less available for protonation. On the other hand, in ethyl anine, delocalization of the lone pair of electrons on the nitrogen atom by resonance is not possible. Further more, the electron density on the nitrogen atom is increased by +1 effect of the ethyl group. Hence, aniline is less basic than ethyl amine.

(5) pK_b value of diethyl amine is less than that of ethyl amine.
Answer:
The basic strength of amines is expressed in terms of pK_b values. Smaller is the value of pK_b more basic is the amine. The pK_b value of ethyl amine is 3.29 and that of diethyl amine is 3.00. Therefore, diethyl amine is more basic than ethyl amine.

(6) Aniline cannot be prepared by Gabriel phthalimide synthesis.
Answer:
In Gabriel-phthalimide synthesis of aniline, potassium phthalimide requires the treatment with chlorobenzene or bromobenzene. Since aryl halides do not undergo nucleophilic substitution reaction. Therefore, chlorobenzene or bromobenzene does not react with potassium phthalimide to give N-phenylphthalimide and hence aniline cannot be prepared by Gabriel phthalimide synthesis.

(7) Gabriel phthalimide synthesis is preferred for the preparation of aliphatic primary amines.
Answer:
In aromatic amines, the lone pair of electrons on the N-atom is delocalized over the benzene ring. As a result electron density on the nitrogen atom decreases. Whereas in aliphatic primary amines, due to +1 effect of alkyl group, electron density on nitrogen atom increases. As the pKh value of aliphatic amines is more than that of aromatic amines, aromatic amines are less basic than primary aliphatic amines. Hence, Gabriel phthalimide synthesis is preferred for the preparation of aliphatic amines.

(8) Arere diazonium salts are relatively more stable than alkyl diazonium salts.
Answer:
Arene diazonium salts are stable due to the dispersal of the positive charge over the benzene ring as shown below :

Alkane diazonium salts are unstable due to their tendency to eliminate a stable molecule of nitrogen to form carbocation. Aromatic diazonium salts have much lower tendency to remove nitrogen than aliphatic diazonium salts. Hence, arene diazonium salts are relatively more stable than alkyl diazonium salts.

(9) Tertiary amines cannot be acylated.
Answer:
Tertiary amines do not react with acetic anhydride or acetyl chloride i.e. they can be acylated because they do not contain a H-atom on the N-atom.

(10) Besides the ortho and para derivatives, considerable amount of meta derivatives is also formed during nitration of aniline.
OR
Although amino group is o- and p-directing in electrophilic substitution reactions, aniline on nitration gives substantial amount of m-nitroaniline.
Answer:
In aromatic amines, -NH₂ is an electron releasing or activating group. It activates the ortho and para positions in the benzene ring towards electrophilic substitution. When aniline is treated with nitrating mixture (cone. HNO₃+ cone. H₂SO₄), a mixture of ortho and para nitroaniline is obtained. However, a substantial amount of m-nitroaniline is also formed. Aniline being a base gets protonated in acidic medium to form anilium cation, which deactivates the ring and the substitution takes place at the meta position.

Question 70.
How will you convert :
(1) Aniline into benzyl alcohol.
Answer:

(2) Aniline into 4-bromoaniline.
Answer:

(3) Aniline into 1,3,5-tribromo benzene.
Answer:

(4) Aniline into 2,4,6-tribromo fluoro benzene.
Answer:

Question 71.
How will you convert :
(1) Propanoic acid into ethanoic acid.
Answer:

(2) Propanoic acid into ethanol
Answer:

(3) Ethanamine into propan-l-amine.
Answer:

(4) Propan-l-amine into ethanamine.
Answer:

(5) Propanoic acid into ethanamine.
Answer:

(6) Ethanamine into propanoic acid.
Answer:

(7) Benzene to aniline.
Answer:

(8) Aniline to Benzene.
Answer:

(9) Aniline into benzoic acid.
Answer:

(10) Benzoic acid into aniline.
Answer:

(11) Aniline into benzamide.
Answer:

(12) 3-Nitrotoluene into 3-methyl aniline.
Answer:

(13) 3-Methyl aniline into 3-Nitrotoluene.
Answer:

Question 72.
An organic compound ‘A’ having molecular formula C₂H₆O evolves hydrogen gas on treatment with sodium metal and on treatment with red phosphorous and iodine gives compound ‘B’. The compound ‘B’ on treatment with alcoholic KCN and on subsequent reduction gives compound ‘C’. The compound ‘C’ on treatment with nitrous acid evolves nitrogen gas. Write the balanced chemical equations for all the reactions involved and identify the compounds ‘A’, ‘B’ and ‘C;.
Answer:
A = C₂H₅OH ethanol
B = C₂H₅I ethyl iodide
C = C₂H₅CH₂NH₂ n-propyl amine
Compound C₂H₆O = C₂H₅OH

Question 73
Identify B, C and D write complete reactions :

Answer:

Question 74.
Identify the compounds B, C and D in the following series of reactions and rewrite the complete equations :

Answer:

Question 75.
Identify the compounds ‘A’ and ‘B’ in the following equation :

Answer:

Question 76.
Answer in one sentence :

(1) Arrange the following compounds in decreasing order of basic strength in their aqueous solutions. NH₃, C₂H₅NH₂, (CH₃)₂NH, (CH₃)₃N
Answer:
The decreasing order of basic strength is – (C₂H₅)₂NH > (C₂H₅)₃N > (C₂H₅)₂NH > NH₃
(The reason that ethyl group has greater +1 effect than methyl group).

(2) Arrange the following compounds in an increasing order of their solubility in water.

Answer:
The solubility increases in order in which molecular mass decreases.

(3) What is Hinsberg’s reagent?
Answer:
Benzenesulphonyl chloride (C₆H₅SO₂Cl) is known as Hinsberg’s reagent.

(4) Name the reaction in which a primary amine is formed from amide.
Answer:
Hoffmann bromamide degradation.

(5) NH₃ is a Lewis base.
Answer:
Since nitrogen in ammonia molecule has a lone pair of electrons, it is a Lewis base.

(6) How many primary amines are possible for the compound C₃H₉N?
Answer:
For the compound C₃H₉N, two primary amines are possible.

(7) State the hybridization of the nitrogen atom in amines.
Answer:
The hybridization of nitrogen atom in amines is sp³.

(8) Arrange the following compounds in an increasing order of basic strength. Aniline, p-nitroaniline, p-toluidine.
Answer:
p-nitroaniline < aniline < p-toluidine.

(9) Which of the two is more basic and why? CH₃NH₂ or NH₃
Answer:
Due to +1 effect of -CH₃ group, electron density on N-atom increases, hence methyl amine is a stronger base than ammonia.

(10) Which of the two is more basic and why? p-toluidine or aniline.
Answer:
p-toluidine is more basic due to the presence of -CH₃ group at para position. Due to +1 effect of -CH₃ group, electron density on nitrogen increases, hence the tendency to donate pair of electrons increases.

Multiple Choice Questions

Question 77.
Select and write the most appropriate answer from the given alternatives for each subquestion :

1. Which of the following is an amine?
(a) C₂H₅N(COCH₃)₂
(b) (C₂H₅)₂N – N = 0
(c) (C₂H₅)₃N
(d) All of these
Answer:
(d) All of these

2. N-methyl-N-ethyl-n-propyl amine is
(a) a primary amine
(b) a secondary amine
(c) a tertiary amine
(d) an alkyl nitrile
Answer:
(c) a tertiary amine

3. Which of the following is a tertiary amine?

Answer:
(d)

4. Tertiary butyl amine is a
(a) primary amine
(b) secondary amine
(c) tertiary amine
(d) quaternary ammonium salt
Answer:
(a) primary amine

5. The IUPAC name of

(a) ethyl propanamine
(b) ethyl butylamine
(c) 2-pentanamine
(d) 3-hexanamine
Answer:
(d) 3-hexanamine

6. The IUPAC name of ethyl dimethyl amine is ……………..
(a) 2-amino propane
(b) N,N-dimethyl ethanolamine
(c) ethyl methanamine
(d) propanamine
Answer:
(b) N,N-dimethyl ethanolamine

7. Isopropyl amine and trimethyl amine are ……………..
(a) acidic in nature
(b) electrophilic compounds
(c) structural isomers
(d) optically active compounds
Answer:
(c) structural isomers

8. N, N-dimethylethanolamine is ……………

Answer:
(b)

9. IUPAC name of diethylmethyl amine is ………………
(a) methyl amino propane
(b) N-Ethyl-N-methylhexanamine
(c) methyl diethanamine
(d) amino pentane
Answer:
(b) N-Ethyl-N-methylhexanamine

10. Ethyl bromide reacts with excess of alcoholic ammonia, the major product is …………..
(a) ethyl amine
(b) diethylamine
(c) triethylamine
(d) tetraethyl ammonium bromide
Answer:
(a) ethyl amine

11. Isopropylamine is obtained by the reduction of
(a) acetoxime
(b) acetaldoxime
(c) formaldoxime
(d) aldoxime
Answer:
(a) acetoxime

12. Which of the following compounds can be converted into amines in the presence of Na and alcohol?
(a) Alkyl nitriles
(b) Aldoxime
(c) Ketoxime
(d) All of these
Answer:
(d) All of these

13. Chloroethane when boiled with excess of aqueous-alcoholic ammonia gives hydrochloric acid and
(a) triethyl amine
(b) trimethyl amine
(c) diethyl amine
(d) ethyl amine
Answer:
(d) ethyl amine

14. How many hydrogen atoms are required for the reduction of 1-nitropropane to n-propyl amine?
(a) Four
(b) Three
(c) Six
(d) Two
Answer:
(c) Six

15. A secondary alkyl halide is heated with excess of ammonia, the major product obtained is
(a) primary amine
(b) secondary amine
(c) tertiary amine
(d) quaternary ammonium salt
Answer:
(a) primary amine

16. The true statement about ethylamine is
(a) it is weaker base than ammonia
(b) it is stronger base than diethyl amine
(c) it is stronger base than triethyl amine
(d) it is stronger base than alkali
Answer:
(c) it is stronger base than triethyl amine

17. The reaction which is given only by primary amines is
(a) acetylation
(b) alkylation
(c) reaction with HNO₂
(d) carbyl amine test
Answer:
(d) carbyl amine test

18. The amine which reacts with NaNO₂ and dil. HCl to give yellow oily compound is
(a) ethylamine
(b) isopropylamine
(c) sec-butylamine
(d) dimethylamine
Answer:
(d) dimethylamine

19. The name of the compound ‘C’ in the following series of reactions, is
(a) propan-l-ol
(b) propan-2-ol
(c) butan-l-ol
(d) butan-2-ol
Answer:
(b) propan-2-ol

20. Triethylamine when treated with nitrous acid gives
(a) an alcohol
(b) a nitrosamine
(c) a monoacetyl derivative
(d) a soluble nitrite salt
Answer:
(d) a soluble nitrite salt

21. Ammes are basic in nature because
(a) of the nitrogen atom contain or lone pair of electrons
(b) they give H+ ions in aqueous medium
(c) they form quaternary ammonium salts when heated with acids
(d) both (a) and (c)
Answer:
(a) of the nitrogen atom contain or lone pair of electrons

22. An aqueous solution of primary amine contains

Answer:
(d)

23. The basic nature of amines in an aqueous solution is in the order of
(a) tert. > sec. > pri.
(b) sec. > pri. > tert.
(b) pri. > sec. > tert.
(d) pri. > tert. > sec.
Answer:
(b) pri. > sec. > tert.

24. In trimethyl ammonium ion, the number of sigma bonds attached to nitrogen are
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(b) 3

25. The number of coordinate bond/bonds in a trialkyl ammonium ion is
(a) one
(b) two
(c) three
(d) four
Answer:
(a) one

26. The number of electrons in the valence shell of nitrogen in methyl amine is
(a) 5
(b) 3
(c) 8
(d) 7
Answer:
(c) 8

27. Ethanamine reacts with excess of acetyl chloride to form
(a) C₂H₅NHCOCH₃
(b) C₂H₅N(CH₃)₂
(c) C₂H₅N(COCH₃)₂
(d) C₂H₅N+H₃Cl
Answer:
(c) C₂H₅N(COCH₃)₂

28. The compound used for acylation of amine is
(a) (CH₃CO)₂O
(b) CH₃COOH
(c) CH₃COCl
(d) both (a) and (c)
Answer:
(d) both (a) and (c)

29. Dimethyl amine reacts with acetyl chloride to give
(a) N-acetyl methyl amine
(b) N-acetyl ethyl amine
(c) N-acetyl dimethyl amine
(d) N-acetyl diethyl amine
Answer:
(c) N-acetyl dimethyl amine

30. Identify ‘A’ in the following reaction :

Answer:
(c)

31. n-propyl alcohol is obtained when HNO₂ is treated with

Answer:
(c)

32. A mixture of CH₃NH₂, (CH₃)₂NH, (CH₃)₃N can be distinguished by using
(a) HCI
(b) HNO₂
(c) HNO₃
(d) H₂SO₄
Answer:
(b) HNO₂

33. In the acetylation reaction the H-atom of an amine is replaced by
(a) a carbonyl group
(b) an alkyl group
(c) an acetyl group
(d) an imino group
Answer:
(c) an acetyl group

34. Amines are basic in nature
(a) as they have a fishy odour
(b) as they form quaternary ammonium salts with alkyl halides
(c) due to the presence of an unshared pair of electrons on the nitrogen atom
(d) all of these
Answer:
(c) due to the presence of an unshared pair of electrons on the nitrogen atom

35. The correct order of increasing basic strength is
(a) NH₃ < CH₃NH₂ < (CH₃)₂NH
(b) CH₃NH₂ < (CH₃)₂NH < NH₃
(c) CH₃NH₂ < NH₃ < (CH₃)₂NH
(d) (CH₃)₂NH < NH₃ < CH₃NH₂
Answer:
(a) NH₃ < CH₃NH₂ < (CH₃)₂NH

36. Which of the following is the strongest base?

Answer:
(d)

37. Identify the weakest base amongst the following :
(a) p-methoxyaniline
(b) o-toluidine
(c) benzene-1, 4-diamine
(d) 4-aminobenzoic acid
Answer:
(d) 4-aminobenzoic acid

38. Amine that cannot be prepared by Gabriel phthalimide synthesis is
(a) aniline
(b) benzyl amine
(c) methyl amine
(d) iso-butyl amine
Answer:
(a) aniline

39. Which of the following exist as Zwitter ion?
(a) Salicylic acid
(b) Suphanilic acid
(c) p-Aminophenol
(d) p-Amino acetophenone
Answer:
(b) Suphanilic acid

40. Reduction of benzene diazonium chloride with Zn/HCl gives
(a) phenyl hydrazine
(b) hydrazine hydrate
(c) aniline
(d) ozo benzene
Answer:
(c) aniline

41. When primary amine reacts with CHCl₃ in alcoholic KOH, the product is
(a) aldehyde
(b) alcohol
(c) cyanide
(d) an isocyanide
Answer:
(d) an isocyanide

42. Which of the following amines cannot be prepared by Gabriel phthalimide synthesis?
(a) sec-Propylamine
(b) tert-Butylamine
(c) 2-Phenylethylamine
(d) N-Methyl benzyl amine
Answer:
(d) N-Methyl benzyl amine

43. Which of the following compounds has highest boiling point?
(a) Ethane
(b) Ethanoic acid
(c) Ethanol
(d) Ethanamine
Answer:
(b) Ethanoic acid

44. Identify the statement about the basic nature of amines.
(a) Alkylamines are weaker bases than ammonia.
(b) Arylamines are stronger bases than alkylamines.
(c) Secondary aliphatic amines are stronger bases than primary aliphatic amines.
(d) Tertiary aliphatic amines are weaker bases than arylamines.
Answer:
(c) Secondary aliphatic amines are stronger bases than primary aliphatic amines.

45. The compounds ‘A’, ‘B’ and ‘C’ react with methyl iodide to give finally quaternary ammonium iodides. Only ‘C’ gives carbylamines test while only ‘A’ forms yellow oily compound on reaction with nitrous acid. The compounds ‘A’, ‘B’ and ‘C’ are respectively.
(a) butan-1-amine, N-ethylethanamine and
N, N-dimethylethanamine.
(b) N-ethylethanamine, N, N-dimethylethanamine and butan-1 – amine.
(c) N, N-dimethylethanamine, N-ethylethanamine and butan-1-amine.
(d) N-ethylethanamine, butan-1-amine and N-ethylethanamine.
Answer:
(b) N-ethylethanamine, N, N-dimethylethanamine and butan-1 – amine.

46. Which of the following amines is most basic in nature?
(a) 2, 4-Dichloroaniline
(b) 2, 4-Dimethylaniline
(c) 2, 4-Dinitroaniline
(d) 2, 4-Dibromoaniline
Answer:
(b) 2, 4-Dimethylaniline

47. How many moles of methyl iodide are required to convert ethylamine, diethylamine and triethylamine into quaternary ammonium salt, respectively?
(a) 1, 2 and 3
(b) 2, 3 and 1
(c) 3, 2 and 1
(d) 3, 1 and 2
Answer:
(c) 3, 2 and 1

48. Which of the following amines does not undergo acetylation?
(a) t-Butylamine
(b) Ethylamine
(c) Diethylamine
(d) Triethylamine
Answer:
(d) Triethylamine

49. n-Propylamine can be prepared by catalytic reduction of
(a) n-propyl cyanide
(b) propionaldoxime
(c) acetoxime
(d) nitroethane
Answer:
(b) propionaldoxime

50. Identify the compound ‘B’ in the following series of reactions :

Answer:
(c)

51. Chloropicrin is used as
(a) antiseptic
(b) antibiotic
(c) insecticide
(d) anaesthetic
Answer:
(c) insecticide

52. Identify the compound B in the following series of reactions.
(a) n-propyl chloride
(b) propanamine
(c) n-propyl alcohol
(d) Isopropyl alcohol
Answer:
(c) n-propyl alcohol

53. Which of the following amines yields foul smelling product with haloform and alcoholic KOH?
(a) Ethyl amine
(b) Diethyl amine
(c) Triethyl amine
(d) Ethyl methyl amine
Answer:
(a) Ethyl amine

54. Which of the following compounds is NOT prepared by the action of alcoholic NH₃ on alkyl halide?
(a) CH₃NH₂
(b) CH₃-CH₂-NH₂
(c) CH₃ – CH₂ – CH₂ – NH₂
(d) (CH₃)₃CNH₂
Answer:
(d) (CH₃)₃CNH₂

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Question 1.
What is a carbonyl group?
Answer:
Carbonyl group : A functional group in which a carbon atom is attached to an oxygen atom by a double bond and remaining two valencies of carbon atom are free is called a carbonyl group and represented as . Carbonyl group is present in aldehydes and ketones.

Question 2.
What are carbonyl compounds?
Answer:
The organic compounds containing a carbonyl group are called carbonyl compounds. For example, acetaldehyde, , acetone, . As carbonyl group is common in aldehydes and ketones, their methods of preparation and properties show similarities.

Question 3.
What are carboxylic compounds?
Answer:
The compounds in which the functional group is – COOH are known as carboxylic compounds. Due to the – OH group bonded to group, carboxylic acids are distinct from aldehydes and ketones.

Question 4.
How are carbonyl compounds classified ?
OR
Name the compounds containing carbonyl group.
Answer:
The carbonyl compounds contain a group . They are classified as follows :

Question 5.
What are aliphatic aldehydes?
Answer:
The compounds in which the – CHO group (formyl group or aldehyde group) is attached directly to sp³ hybridized carbon atom that is saturated carbon atom are called aliphatic aldehydes. (Exception : Formaldehyde, H – CHO is also classified as aliphatic aldehyde though – CHO group is not attached to any carbon).

For example :

Question 6.
What are aromatic aldehydes ?
Answer:
The compounds in which – CHO group is attached directly to an aromatic ring are called aromatic aldehydes.

For example :

Question 7.
Explain the structure of carbonyl functional group.
Answer:

• In the carbonyl functional group, carbon atom is attached to an oxygen atom by a double bond and remaining two valencies of carbon atom are free, and it is represented as .
  
• The carbonyl carbon atom is sp²-hybridised forming coplanar three sigma (σ) bonds with the bond angle 120°.
• One sigma bond is formed with oxygen atom while other two sigma (σ) bonds are formed with hydrogen or carbon atoms.
• The remaining unhybridised 2p_z orbital of carbon atom overlaps with p orbital of oxygen atom colaterally forming a pi (π) bond. Hence, carbon atom is joined to oxygen atom by a double bond of which one is sigma and another is n.
• The oxygen atom in the carbonyl group has two lone pairs of electrons.
• The carbonyl bond is strong, short and polarized.
• The polarity of the carbonyl group is explained on the basis of resonance involving neutral and dipolar structures as shown below :
  

Question 8.
What are aliphatic ketones? How are they classified?
Answer:
Aliphatic ketones : The compounds in which group is attached to two alkyl groups are called aliphatic ketones.

Ketones are classified into two types :

1. Simple or symmetrical ketones and
2. mixed or unsymmetrical ketones.

1. Simple or symmetrical ketone : The ketone in which the carbonyl carbon is attached to two identical alkyl groups is called a simple or symmetrical ketone.

2. Mixed or unsymmetrical ketone : The ketone in which the carbonyl carbon is attached to two different alkyl groups is called a mixed or unsymmetrical ketone.

Question 9.
What are aliphatic carboxylic acids? Give their general formula.
Answer:
The organic compounds in which carboxyl (- COOH) group is bonded to an alkyl group are called aliphatic carboxylic acids or fatty acids. (Exception : Formic acid, H-COOH is also classified as aliphatic carboxylic acid though-COOH group is not attached to any carbon).

For example :

Question 10.
How are carboxylic acids classified ? Give examples. (3 marks)
Answer:
Carboxylic acids are classified according to the presence of number of carboxyl groups into mono-, di-, tri- and polycarboxylic acids.

• Monocarboxylic acids : These carboxylic acids contain one carboxyl group.
  Example :
• Dicarboxylic acids : These contain two carboxyl groups
  Example :
• Tricarboxylic acid : These contain three carboxyl groups
  

Question 11.
What are aromatic carboxylic acids ? Give examples.
Answer:
Aromatic carboxylic acids : These are the compounds in which one or more carboxyl groups (- COOH) are attached directly to the aromatic ring.

For example :

Question 12.
Give examples of common carboxylic acids which are used in daily life.
Answer:
Common carboxylic acids are widely distributed in nature, they are found in both the plants and animals.

• Acetic acid is main constituent of vinegar.
• Butyric acid of butter which is responsible for odour of rancid butter.
• L-lactic acid is present in curd.
• Citric acid is found in citrus fruits.
• Higher carboxylic acids such as palmitic acid, stearic acid and oleic acid are the components of animal fats and vegetable oils.

Nomenclature of Aldehydes :

(A) Common System :

• The names of aldehydes are derived from the common names of acids.
• The suffix ‘-ic acid’ of an acid is replaced by ‘aldehyde’.
• The positions of the substituents in the molecule are indicated by Greek letters α, β, γ, etc.
• starting from the carbon atom attached to the carbonyl group. E.g.
  

(B) I UP AC System :

• The longest carbon atoms chain containing aldehyde carbon atom is selected as a parent hydrocarbon.
• ‘e’ of the alkane is replaced by ‘al’. Alkane → Alkanal
• The position (locant) of aldehyde group need not be mentioned since it is always at the end position.
• The substituents in the alkyl group are prefixed in an alphabetical order by appropriate locants.
• When two – CHO groups are present at the two ends of the chain the ending ‘e’ of alkane is retained and the suffix  ‘-dial’ is added to the name of parent aldehyde.
• In IUPAC nomenclature an alicyclic compound -in which – CHO group is attached directly to the ring is named as a carbaldeliyde. The suffix ‘carbaldehyde’ is added after the full name of parent cycloalkane structure.

(C) Common or trivial names :

(1) The common name of a carboxylic acid is derived from the source from which it was first isolated.
The following table gives common names and the source or origin of name.

(2) In branched carboxylic acids, the position of substituents are indicated by Greek alphabet.

For example :

(D) IUPAC system of nomenclature :

• The longest continuous cha in of carbon atoms including the carbon atom of – COOH group ¡s selected.
• The carboxvlic acid is conside red (IS a derivative of the corresponding parent a/kane.
• The carbon atom of the – COOH group is always at terminal position, hence need not to be indicated while writing IUPAC name.
• The position of the other substitutents are indicated by the appropriate locants in alphabetical order.
• In case of dicarboxylic acids, ‘dioic acid’ is added to parent alkane.
• In an alicyclic compound having a carboxyl group directly attached to alicyclic ring is named as cycloalkane carboxylic acid.

Trivial and IUPAC names of carboxylic acids and aldhydes

Question 13.
Give the common names and IUPAC names of the Miowing aldehydes :
Answer:

Question 14.
Write the structures of the following compounds :
Answer:

Question 15.
What is the IUPAC name of the following compound?

Answer:
IUPAC name : 2-Amino butanoic acid

Question 16.
Write IUPAC name of

Answer:
IUPAC name : Ethanedioic acid

Question 17.
Write the structure and give IUPAC names of following carboxylic acid.

Answer:

Question 18.
Draw the structures of the following compounds :
Answer:

Question 19.
Give IUPAC names of the following carboxylic acids.
Answer:

Question 20.
Write the structures and IUPAC names of all isomers of carboxylic acids having molecular formula C₅H₁₀O₂. HOW many of them are chiral?
Answer:

Nomenclature of ketones :

(A) Common System :

• Ketones are named according to the alkyl groups attached to the carbonyl carbon atom followed by the word ketone.
• The substituents in the alkyl groups are indicated by Greek letters a, f, y, etc. starting from the carbon atom attached to the carbonyl group.

(B) IUPAC System :

• The longest continuous chain containing carbonyl carbon atom is selected as a parent hydrocarbon.
• ‘e’ of the alkane is replaced by ‘one’. Alkane → Alkanone
• The position of carbonyl group is represented by the lowest locant.
• The substituents in the alkyl groups are prefixed in the alphabetical order along with their positions by appropriate locants.
• When two C = O groups are present, then ending ‘e ’ of alkane is retained and the suffix – ‘dione ’ is added to the name of parent ketone indicating the locants of ketonic carbonyl groups.
• In case of polyfunctional ketones, higher priority group is given lower number.
• When ketonic carbonyl is a lower priority group it is named as ‘oxo’, preceded by the locant. In alicyclic ketones, carbonyl carbon is numbered as 1.

Question 21.
Give the common and IUPAC names of the following ketones :
Answer:

Question 22.
Give the structures of the following compounds :
Answer:

Question 23.
Give IUPAC names of the following compounds :
Answer:

Question 24.
Write the structures and give common names and IUPAC names of the carbonyl compounds represented by formula C₅H₁₀O.
Answer:

Question 25.
Write the structure and give IUPAC names of the following compounds :
Answer:

Question 26.
Write the structure of the following compounds :
Answer:

Question 27.
How is an aldehyde obtained from an alcohol ?
Answer:
When a primary alcohol is oxidized with potassium dichromate and dil. H₂SO₄ under controlled conditions, an aldehyde is obtained.

For example, when ethanol is oxidized with potassium dichromate and dil. H₂SO₄ under controlled conditions, acetaldehyde (ethanal) is obtained.

Question 28.
How is ketone obtained from an alcohol?
Answer:
When a secondary alcohol is oxidized with potassium dichromate and dii. H₂SO₄ under controlled conditions, a ketone is obtained.

For example. when 2-propanol is oxidized with potassium dichromate and dii, H₂SO₄ under controlled conditions, accIone (propanone) is obtained.

Question 29.
How are the following compounds obtained from alcohol:
(1) Methanal
(2) Propanal
(3) BuLanal
(4) 3-Methylpentanal?
Answer:

1. Mehanol on controlled oxidation with K₂Cr₂O₇ and dilute H₂SO₄ forms methanal.
   
2. Propan-1-ol on controlled oxidation with K,Cr20 and dilute H₂SO₄ forms propanal.
   
3. Butan..l-ol on controlled oxidation with K₂Cr₂O₇ and dilute H₂SO₄ forms butanal.
   
4. 3-Mcthylpcntan.1-ol on controlled oxidation with K₂Cr₂O₇ and dilute H₂SO₄ gives 3-Methylpentanal.
   

Question 30.
How are the following compounds prepared from alcohol :
(1) Butanone
(2) Pentan-3-one
(3) 2,2-Dimethylpropanal?
Answer:

1. Butan-2-ol on controlled oxidation with K₂Cr₂O₇ and dilute H₂SO₄ forms butanone.
   
2. Pentan-3-ol on oxidation with K₂Cr₂O₇ and dilute H₂SO₄ forms Pentan-3-one.
   
3. 2,2-Dimethylpropan- I -ol on oxidation with K₂Cr₂O₇ or KMnO₄ and dilute H₂SO₄ forms 2,2-Dimethyipropanal.
   

Question 31.
How is an aldehyde obtained from primary alcohol ?
Answer:
When vapours of primary alcohol is passed over heated copper at 573 K, dehydrogenation takes place, an aldehyde is formed.

For example : When vapours of isopropyl alcohol is passed over heated copper at 573 K, acetone (propanone) is obtained.

Question 32.
How is ketone obtained from secondary alcohol?
Answer:
When vapours of secondary alcohol is passed over heated copper at 573 K. dehydrogenation takes place, a ketone is obtained.

For example : When vapours of isopropyl alcohol is passed over heated copper at 573 K. acetone (propanone) is obtained.

Question 33.
How are the following compounds obtained from alkene :
(1) Formaldehyde
(2) Acetaldehyde and
(3) Acetone ?
Answer:
When a stream of ozonised oxygen is passed through a solution of an alkene, in organic solvent, an unstable addition cyclocompound, ozonide is formed which on reduction with zinc dust and water forms an aldehyde or a ketone or a mixture of both.

1. Formaldehyde : Under these conditions ethylene gives formaldehyde.
   
2. AcetuIdeh&: Symmetrically disubstituted alkene like but-2-ene gives acetaldehyde.
   
3. scctnne: Tetrasubshtwed alkene like 2,3-dimethyl but-2-ene gives acetone.
   

Question 34.
Write ozonolysis reaction for
(1) Propylene and
(2) Isobutylene.
Answer:
(1) Propylene on reaction with ozonised oxygen in the organic solvent forms propylene ozonide which on reduction with zinc dust and water forms acetaldehyde and formaldehyde.

Question 35.
How are the following compounds obtained from alkynes :
(1) Acetaldehyde
(2) Acetone?
Answer:

1. Acetaldehyde : On passing acetylene through warm 40% H₂SO₄ in the presence of 1 % H_gSO₄, vinyl alcohol is obtained which tautomerises and forms acetaldehyde. It is a hydration reaction.
   
2. Acetone : On passing propyne through warm 40 % H₂SO₄ in the presence of 1 % H_gSO₄, alkenol is obtained which on tautomerisation form acetone.
   

Question 36.
Predict the products when
(1) dimethyl acetylene
(2) ethyl acetylene and
(3) diethyl acetylene are treated with mercuric sulphate in dilute sulphuric acid.
Answer:
(1) Dimethyl acetylene with 40% H₂SO₄ in the presence of mercuric sulphate (H_gSO₄) forms ethyl methyl ketone by tautomerisation.

(2) Ethyl acetylene with 40% H₂SO₄ in the presence of mercuric sulphate (H_gSO₄) forms Butan-2-one by tautomerisation.

(3) Diethyl acetylene with 40% H₂SO₄ in the presence of mercuric sulphate (H_gSO₄) forms hexan-3-one.

Question 37.
Write the structures of aldehydes and ketones obtained by ozonolysis.

Answer:

Answer:

Question 38.
Predict the structures of ketones produced by hydration of but-l-yne and but-2-yne.
Answer:

Question 39.
How is acetaldehyde prepared from acetyl chloride?
Answer:
Acetyl chloride is reduced to acetaldehyde by hydrogen in presence of Pd catalyst poisoned with BaSO₄. This reaction is called Rosenmund reduction.

Question 40.
How is benzaldehyde obtained from benzoyl chloride?
OR
Write chemical equation for Rosenmund reduction.
Answer:
When benzoyl chloride is hydrogenated in the presence palladium on barium sulphate (Pd/BaSO₄), benzaldehyde is obtained. This reaction is called Rosenmund reduction.

Question 41.
How will you prepare acetophenone from benzene? (Friedel – Crafts acylation).
Answer:
When benzene is treated with acetyl chloride in the prcscncc of anhydrous aluminium chloride, acetophenonc is obtained. This reaction is known as Friedel – Crafts acylation.

Question 42.
How will you convert benzene into 1-phenylethanone?
Answer:

Question 43.
How will you obtain 4-Nitrobenzaldehyde from 4-Nitrotoluene ? (Friedel- Crafts reaction).
Answer:
When 4-nitrotoluene is treated with chromium oxide in acetic anhydride, a diacetate derivative is obtained which on acid hydrolysis produces 4-nitrobenzaldehyde.

Question 44.
How will you prepare Propanone (acetone) from Grignard reagent?
Answer:
Grignard reagent (methyl magnesium iodide) reacts with cadmium chloride to give dimethyl cadmium. When acetyl chloride reacts with dimethyl cadmium, propanone (acetone) is obtained.

Question 45.
How is acetophenone obtained from Grignard reagent ?
Answer:
Grignard reagent (methyl magnesium iodide) reacts with cadmium chloride to give dimethyl cadmium. When benzoyl chloride reacts with dimethyl cadmium, acetophenone is obtained.

Question 46.
How is benzyl methyl ketone obtained from Grignard reagent ?
OR
Convert: Acetyl chloride to benzyl methyl ketone.
Answer:
Grignard reagent (Benzyl magnesium chloride) reacts with cadmium chloride to give diphenyl cadmium. When acetyl chloride reacts with dibenzyl cadmium, benzyl methyl ketone is obtained.

Question 47.
How is an aldehyde obtained from alkyl nitrile ?
OR
What is Stephen reaction ?
OR
Write a note on Stephen reaction.
Answer:
(1) An ethereal solution of a nitrile is reduced to imine hydrochloride by SnCl₂ in the presence of HCl gas. Further, imine hydrochloride on acid hydrolysis gives aldehyde. This reaction is called Stephen reaction.


(2) Alternatively, nitriles are selectively reduced by diisobi.ityl aluminium hydride (DIBAI-H) lo imines which on acid hydrolysis to aldehydes.

Question 48.
How are following compounds prepared using Gngnard reagent
(1) Acetone
(2) Benzophenone?
Answer:
(1) Acetone: Acetoniti-ile (ethanenitrile) reacts with methyl magnesium iodide in presence of dry ether to give imine complex which on hydrolysis gives acctonc. During reaction acetonitrile and methyl magnesium iodide should be
taken in equimolecular proportion.

(2) Benzophenone: Benzonitrile reacts with phenyl magnesium bromide in presence of dry ether to give an imine complex which on acid hydrolysis gives a benzophenone. During reaction bcnzonitrile and phenyl magnesium bromide should be aken in equimolecular proportion.

Question 49.
Write the structures and IUPAC names of ketones produced by Friedel-Crafts acylation of benzene with
(i) C₂H₅COCl
(ii) C₆H₅COCl.
Answer:

Question 50.
Predict the products of the following conversions.
Answer:

Question 51.
How are the following preparations carried out ?

(1) Benzaldehyde from toluene. (Etard oxidation)
Answer:
When toluene is treated with solution of chromyl chloride (CrO₂Cl₂) in Cs₂, brown chromium complex is obtained, which on acid hydrolysis gives benzaldehyde. This reaction is called Etard reaction.

(2) Benzaldehyde from methyl arene.
Answer:
Methylarene is converted into a benzyllidene diacetate on treatment with chromium oxide in acetic anhydride at 273-278 K. The diacetate derivative on acid hydrolysis gives benzaldehyde.

(3) Benzaldehyde from toluene (commerical method).
Answer:
Side chain chlorination of toluene gives benzal chloride which on hydrolysis gives benzaldehyde.

(4) Benzaldehyde from benzene (Gattermann-Koch synthesis).
OR
Write chemical equation for Gatter- mann-Koch formylation.
Answer:
When benzene is treated with vapours of carbon monoxide and hydrogen chloride in the presence of a catalyst mixture of A1Cl₃ and CuCl under high pressure, benzaldehyde is obtained. This reaction is called Gattermann- Koch synthesis.

Preparation of aromatic ketones from hydrocarbons

Question 52.
Explain Friedel-Craft’s acylation reaction.
Answer:
The reaction in which hydrogen atom of benzene is replaced by an acyl group I in the presence of anhydrous AlCl₃ is called Friedel-Craft’s acylation. When benzene is heated with an acetyl chloride or acetic anhydride in the presence of anhydrous AlCl₃, forms acetophenone (1-Phenyl ethanone).

Electrophile : R – C + = O acylium ion Formation of the electrophile :

Question 53.
Give the preparation of acetophenone from benzene using
(i) acetyl chloride
(ii) acetic anhydride.
Answer:

The reaction in which hydrogen atom of benzene is replaced by an acyl group I in the presence of anhydrous AlCl₃ is called Friedel-Craft’s acylation. When benzene is heated with an acetyl chloride or acetic anhydride in the presence of anhydrous AlCl₃, forms acetophenone (1-Phenyl ethanone).

Electrophile : R – C + = O acylium ion Formation of the electrophile :

Question 54.
How will you prepare propionaldehyde from ethyl propionate?
Answer:
When ethyl propionate is reduced in presence of diisobutyl aluminium hydride (DIBAI-H), propionaldehyde is obtained.

Question 55.
Explain the structure of carboxyl group.
Answer:
In carboxyl group, the carboxyl carbon is sp²-hybridised and the bonds to the carboxyl carbon lie in one plane. The C-C = O and O = C-O bond angles are 120°. The carboxylic carbon is less electrophilic than carbonyl carbon because of the resonance structures shown below :

Question 56.
How is carboxylic acid obtained by the acid hydrolysis of an alkyl cyanide ?
Answer:
Alkyl cyanides or alkyl nitriles on acid or alkaline hydrolysis give corresponding carboxylic acids.

Acid Hydrolysis of Alkyl ankle: When alkyl cyanide is boiled wiLh dilute mineral acid, it gives corresponding carboxylic acid. In this, acid amide is obtained as the intermediate product.

Question 57.
How is ethanoic add obtained from methyl cyanide by acid hydrolysis?
Answer:
When methyl cyanide is heated with dilutc hydrochloric acid or dilute sulphuric acid. ethanoic acid is obtained.

Question 58.
How Is proplonlc acid obtained from an alkyl nitrile?
Answer:
When ethyl cyanide (propionitrile) is boiled with dilute HCI or dilute H₂SO₄, propionic acid is obtained.

Question 59.
How Is benzoic acid obtained from bcnzamide?
Answer:
When benzarnide is heated with dil. HCl. benzoic acid is obtained.

Question 60.
How is carboxylic acid obtained from acyl chlorides and acid anhydrides?
Answer:
When acyl chloride is hydrolysed with water, carboxylic acid is obtained. The reaction is carried out in presence of a base pyridine or NaOH to remove HCl generated.

Acetyl chloride reacts with water almost explosively while benzoyl chloride very slowly.

Acid anhydrides react with water to give carboxylic acid.

Question 61.
How is benzoic acid obtained from
(i) ethyl benzoate
(ii) styrene?
Answer:
(i) Benzoic acid from ethyl benzoate : When an ethyl benzoate is heated with dil. H₂SO₄, undergoes hydrolysis to form benzoic acid and ethyl alcohol.

(ii) Benzoic acid from styrene:

Question 62.
How is propanoic acid obtained from phenyl propanoate?
Answer:
When phenyl propanoate is heated with dil. NaOH, sodium salt is obtained, which on hydrolysis gives propanoic- acid.

Question 63.
How is propanoic acid obtained from methyl propanoate ?
OR
When methyl propanoate is heated with dil. NaOH, sodium, salt is obtained, which on hydrolysis gives propanoic acid.
Answer:

Question 64.
How is benzoic acid obtained from phenyl ethene?
Answer:
When phenyl ethene is heated with strong oxidising agents like acidic KMnO₄ or acidic K₂Cr₂O₇, benzoic acid is obtained.

Question 65.
How is adipic acid obtained from cyclohexene?
Answer:
When cyclohexene is heated with acidified KMnO₄, adipic acid (Hexane-1, 6-dioic acid) is obtained.

Question 66.
What is carbonation of Grignard reagent ? How is acetic acid prepared by this reaction ? How is ethanoic acid pepared from dry ice?
Answer:
Addition reaction of carbon dioxide (0 = C = 0) to Grignard reagent, forming a complex and further formation of carboxylic acid is called carbonation of Grignard reagent.

Example : When methyl magnesium iodide is added to solid carbon dioxide, a complex is formed which on acid hydrolysis forms acetic acid.

Question 67.
How is benzoic acid prepared from Grignard reagent?
OR
Write the preparation of benzoic acid from dry ice.
Answer:
When phenyl magnesium bromide is treated with dry ice (solid carbon dioxide) in the presence dry ether, complex is obtained which on acidification gives benzoic acid.

Question 68.
What are soaps ? How are they prepared ?
Answer:
The sodium or potassium salts of higher fatty acids are known as soaps. Soaps contain more than twelve carbon atoms.

When fat or oil is hydrolysed using sodium or potassium hydroxide solution, soap obtained remains in colloidal form. Soap and glycerol are separated by adding sodium chloride. Soap precipitates out due to common ion effect, and glycerol remains in the solution can be recovered by fractional distillation.

Question 69.
How is benzoic acid prepared from alkyl benzenes ?
OR
How will you convert the following :
(1) n-butyl benzene to benzoic acid.
(2) Toluene to benzoic acid.
(3) Cumene to benzoic acid ?
Answer:
When an alkyl benzene is heated with strong oxidizing agents like acidic or alkaline KMnO₄ or acidified K₂Cr₂O₇ etc. gives aromatic carboxylic acid. The alkyl side chain gets oxidised to -COOH group irrespective of the size of the chain.

Question 70.
Lower aldehydes and ketones are water soluble whereas higher homologues are insoluble. Explain, why.
Answer:
(1) The oxygen atom of shows hydrogen bonding with water molecule.

(2) As a result of this, the lower aldehydes and ketones are water soluble (example – acetaldehyde, acetone). As the molecular mass increases, the proportion of hydrocarbon part of the molecule increases which cannot form hydrogen bonding with water and the solubility of higher homologues in water decreases.

Question 71.
Carboxylic acids have higher boiling points than those of alcohols, aldehydes, ketones, ethers, hydrocarbons of comparable molecular masses. Explain, why.
Answer:
(1) Carboxylic group (-COOH) in acids is highly polar. in liquid state, pair of carboxylic acid molecules is held together by two intermolecular hydrogen bonds, have higher aggregations and in the vapour.

(2) Intermolecular hydrogen bonding in carboxylic acids state most of the carboxylic acid.s exist as dimmers in which two molecules are held by two hydrogen R – R bonds. Acidic hydrogen of one molecule forms hydrogen bond with carbonyl oxygen of the other molecule. This doubles the size of the molecule resulting in increase in o – intermolecular van der Waals forces, which in turn results in high boiling points. Therefore, carboxylic acids possess higher boiling points than those of alcohols, aldehydes, ketones. ether, hydrocarbons of comparable molecular masses.

Question 72.
Lower aliphatic carboxylic acids are miscible with water while higher carboxylic acids are immiscible.
Answer:
(1) Lower aliphatic carboxylic acids are miscible with water due to the formation of hydrogen bonding with water molecules.

(2) Hydrogen bonding between acid and water. As the molecular mass increases (he solubility of carboxylic acids in water decreases. The insolubility of carboxylic acids is due to increased hydrophobic interaction of hydrocarbon parts with water.

Question 73.
Explain why carboxylic acids are much weaker acids than mineral acids.
Answer:
Carboxylic acids are the organic compounds which are acidic in nature. However, compared to mineral acids like HCI or H₂SO₄. the carboxylic acids are weaker acids.

The strength of acidity depends upon their ability to release H⁺ ions. Greater the ease with which they release H⁺ ions, stronger is the acid.

Carboxylic acids when dissolved in water, pcoduce H⁺ due to its dissociation. (it does not dissociate completely.)

The mineral acid-s like HCI, H₂SO₄ release H⁺ ion to a larger extent as they dissociatc almost complctcly in aqueous solution for e.g. HCl → H⁺ Cl^– thus carboxylic acids are weaker than mineral acids. The equilibrium exists in aqueous solution of carboxylic acid as

Since concentration of water practically remains constant

where K_a is acidity constant
Larger the value K_a greater is the extent of ionization and stronger is the acid. But strength of acids is expressed in ternis of their pK_a values. Smaller the value of pK_a. the stronger is the carboxylic acid. Here pK_a value of carboxylic acids is higher than mineral acids. Hence, carboxylic acids arc weaker than mineral acids.

Question 74.
Carboxylic acids are more acidic than phenols and alcohols. Explain. Why?
Answer:
(1) Carboxylic acid loses a proton as compared to phenol. Consider the ionization of carboxylic acid and phenol

Due to delocalization the negative charge over the ortho and para positions of aromatic ring, phenoxide anion is more stable than phenol. Thus phenol easily undergoes ionization

However, alcohol and alkoxide ion are single structures. In an alkoxide anion the negative charge is localized on a single oxygen atom. Hence, phenols are more acidic than alcohols.

(2) Carboxylic acid has two resonance hybride non equivalent structures (I & II) while carboxylate anion has two resonance hybrid equivalent structures (III & IV). The carboxylate ion is more stable than carboxylic acid and equilibrium is shifted towards the direction of increased ionization.

(3) Carboxylate ion has two equivalent resonance structures with nejative charge is delocalized over two electro negative oxygen atoms. Phenoxide anion has non-equivalent resonance structures in which negative charge is
delocalized over one oxygen atom and less electronegative carbon atom. As a result carboxylate anion is more stable than phenoxide ion. Hence carboxylic acids ionize to the greater extent than phenol furnishing higher concentration of H⁺ ions. Therefore, carboxylic acids are more acidic than phenols and alcohols.

Question 75.
Trichloro acetic acid is a stronger acid than acetic acid. ExplaIn.
Answer:
(1) The acidic nature of carboxylic acid is due to the ability to release H ions. Greater the ease with which they release H⁺ ions, stronger is the acid. Any factor that stabilizes the carboxylate ion would help the release of H⁺
ions and thus increase the strength of the acid. The electron-withdrawing group attached to -carbon atom increases the strength of the acid. In trichloroacetic acid, three chloro substituents on s-carbon atom of acetic acid makes the electrons withdrawing effect more pronounced and the negative charge of carboxylate ion formed gets dispersed.

Thus, increases the stability of carboxylate ion.

(2) The acetate ion formed gets destabilised due to the electron releasing effect of a methyl group (+ I effect). As a result, acetic acid dissociates to a lesser extent.

(3) The trichloro acetate ion formed gels stabilised due to electron-withdrawing effect of three 3Cl atoms (- I effect). As a result. trichloro acetic acid dissociates to a greater extent. Trichioro acetic acid having lower pK_a value than acetic acid. Hence. trichloro acetic acid is a stronger acid than acetic acid.

Question 76.
Which is the stronger acid in each of the following pairs ?
(1) CH₃-COOH and CH₂ = CH-COOH
Answer:
CH₂ = CH-COOH is the stronger acid than CH₃-COOH

(2) C₆H₅-COOH and C₆H₅-CH₂-COOH
Answer:
C₆H₅-COOH is the stronger acid than C₆H₅-CH₂-COOH

Answer:

(4) CH₃-CH₂-COOH and NC-CH₂-COOH
Answer:
NC-CH₂-COOH is the stronger acid than CH₃-CH₂-COOH

(5) (CH₂)₂CH-CH₂-COOH and (CH₃)₂NH-CH₂-COOH
Answer:
(CH₃)₂NH-CH₂-COOH is the stronger acid than (CH₃)₂CH-CH₂-COOH

(6) O₂N-CH₂-C00H and Cl-CH₂-COOH
Answer:
NO₂-CH₂-COOH is the stronger acid than Cl-CH₂-COOH.

Question 77.
Arrange the following acids in their increasing order of acidic strength.

(1) Acetic acid, chloroacetic acid, propionic acid, formic acid.
Answer:
Propionic acid < acetic acid < formic acid < chloroacetic acid

(2) Bromoacetic acid, chloroacetic acid, fluoroacetic acid, iodoacetic acid.
Answer:
Iodoacetic acid < bromoacetic acid < chloroacetic acid < fluoroacetic acid.

(3) Acetic acid, chloroacetic acid, dichloroacetic acid, trichloroacetic acid.
Answer:
Acetic acid < chloroacetic acid < dichloroacetic acid < trichloroacetic acid.

(4) n-butyric acid, 3-chlorobutyric acid, 2-chlorobutyric acid, 3-chlorobutyric acid.
Answer:
n-Butyric acid < 3-chlorobutyric acid < 2-chlorobutyric acid < 1-chlorobutyric acid.

(5) Acetic acid, benzoic acid, p-methoxy benzoic acid, p-nitrobenzoic acid.
Answer:
Acetic acid < benzoic acid < p-methoxy benzoic acid < p-nitrobenzoic acid.

(6) Acetic acid, phenyl acetic acid, p-nitro phenyl acetic acid.
Answer:
Acetic acid < phenyl acetic acid < p-nitro phenyl acetic acid.

(7) Benzoic acid, p-toluic acid, p-chlorobenzoic acid.
Answer:
p-toluic acid < benzoic acid < p-chlorobenzoic acid.

Question 78.
Arrange the following carboxylic acids in order of increasing acidity : m-Nitrobenzoic acid, Trichloroacetic acid, benzoic acid, a-Chlorobutyric acid.
Answer:
Acidity in the increasing order : Benzoic acid, m-nitrobenzoic acid, a-chlorobutyric acid, trichloroacetic acid.

Question 79.
Arrange the following carboxylic acids with increasing order of their acidic strength and justify your answer.

Question 80.
Explain polarity of carbonyl group.
Answer:
The polarity of a carbonyl group is duc to higher electronegativity of oxygen compared to carbon. It is explained on the basis of resonance involving neutral and dipolar structures.

The carbonyl carbon has positive polarity (see structures (A) and (D)). Therefore, it is electron deficient. As a result, this carbon atom is electrophilic (electron loving) and is susceptible to attack by a nucleophile (Nu : ^–).

Question 81.
Explain SchifPs reagent test.
OR
What is a SchifPs reagent ? How is it used to detect aldehydes ?
Answer:

• Schiff’s reagent is prepared by dissolving pink p-rosaniline hydrochloride (dye Fuchsin) in water and passing SO₂ gas till the pink solution is decolourised.
• Schiff s reagent is an oxidising agent.
• When an aldehyde is added to Schiff s reagent, the colourless solution turns pink or in magenta colour and aldehyde is oxidised to a carboxylic acid.
• This test is not given by ketones, hence, used to distinguish between aldehyde and ketone.

Question 82.
Which colour is obtained when SchifFs reagent is treated with acetaldehyde?
Answer:
When Schiff s reagent is treated with acetaldehyde, pink colour is obtained.

Question 83.
What is Tollen’s reagent?
Answer:

• Tollen’s reagent is an ammoniacal silver nitrate, [Ag(NH₃)₂]⁺ OH^–.
• It is prepared by adding NH₄OH solution to silver nitrate solution.
  
• It is a stronger oxidising agent than Fehling solution. Aldehyde when heated with Tollen’s reagent, silver mirror is deposited.

Question 84.
Explain Tollen’s reagent test.
OR
Explain silver mirror test.
Answer:

• Tollen’s reagent is an (ammoniacal silver nitrate) [Ag(NH₃)₂]⁺ OH^–.
• When an aldehyde, like acetaldehyde is heated with Tollen’s reagent, it is oxidised to acetic acid and silver ions Ag⁺ in Tollen’s reagent complex are reduced to silver Ag giving greyish black precipitate or deposition of silver on inner surface of the test tube which shines like a mirror. Hence this test is also called silver mirror test.
  
• This test is not given by ketones.
• Hence Tollen’s reagent is used to distinguish between aldehydes and ketones.

Question 85.
What is Fehling’s solution? How is it prepared?
Answer:

• Fehling’s solution is a complex of cupric ions with tartaric acid.
• It is a mild oxidising agent.
• It is prepared by mixing equal amount of Fehling’s solution ‘A’ containing CuSO₄ solution and Fehling’s solution ‘B’ containing sodium potassium tartrate (Rochelle salt) in caustic soda (NaOH) solution.
• It is used to detect aldehydes that decolourise deep blue colour of the solution and give red precipitate of Cu₂O.

Question 86.
Explain Fehling’s solution test.
Answer:

• Fehling’s solution is a mixture of Fehling’s solution ‘A’ containing CuSO₄ solution and Fehling’s solution ‘B’ containing sodium potassium tartrate (Rochella salt) in caustic soda (NaOH) solution.
• When an aldehyde is heated with Fehling’s solution, the deep blue colour of the solution disappears and Cu⁺² (cupric ion) is reduced to Cu⁺ ion a red precipitate of cuprous oxide, Cu₂O is obtained while aldehyde is oxidised to a carboxylate ion.
• For example,
  
• This test is not given by ketones, since they cannot be oxidised by Fehling solution.
• Aromatic aldehydes are not oxidised by Fehling solution.
• Hence this test is used to distinguish between aldehydes and ketones.

Question 87.
What is the action of the following reagents on ethanal :
(1) Fehling’s solution,
(2) Tollen’s reagent or Ammonical silver nitrate ?
Answer:

1. When ethanal is heated with Fehling’s solution, the deep blue colour of the solution disappears and a red precipitate of Cu₂O is obtained.
   
2. When ethanal is heated with Tollen’s reagent a greyish black precipitate or deposition of silver is obtained.
   

Question 88.
Why is benzaldehyde NOT oxidized by Fehling solution ?
Answer:
When benzaldehyde is treated with Fehling solution, it does not reduce cupric ion (Cu⁺²). Fehling solution does not oxidise benzaldehyde. Thus, Fehling test cannot be used for aromatic aldehyde.

Question 89.
Explain laboratory test for ketonic group or sodium nitroprusside test.
Answer:
Laboratory test for ketonic group : Sodium nitroprusside test : When a freshly prepared sodium nitroprusside solution is added to a ketone, mixture is shaken well and basified by adding sodium hydroxide solution drop by drop, red colour appears in the solution, which indicates the presence of ketonic (> C = O) group.

The anion of ketone formed by alkali reacts with nitroprusside ion to form a red coloured complex which indicates the presence of the ketonic group.

Question 90.
What is the action of hydrogen cyanide on the following :
(1) Acetaldehyde
(2) Acetone
(3) Benzaldehyde?
Answer:
(1) Action of HCN on acetaldehyde : When acetaldehyde is treated with hydrogen cyanide, acetaldehyde cyanohydrin is formed.

(2) Action of HCN on acetone : When acetone is treated with hydrogen cyanide, acetone cyanohydrin is formed.

(3) Action of HCN on benzaldehyde : When benzaldehyde is treated with hydrogen cyanide, benzaldehyde cyanohydrin is formed.

Question 91.
What is the action of hydrogen cyanide in basic medium on (1) butanone (2) 2,4-dichlorobenzaldehyde?
Answer:
(1) Action of hydrogen cyanide on butanone : When butanone is treated with hydrogen cyanide, butanone cyanohydrin is obtained.

(2) Action of hydrogen cyanide on 2,4-dichlorobenzaldehyde : When 2, 4-dichloro benzaldehyde is treated with hydrogen cyanide, cyanohydrin of 2,4-dichloro benzaldehyde is obtained.

Question 92.
What is the action of sodium bisulphite on :
(1) Acetaldehyde
(2) Acetone (propanone)?
Answer:
(1) Acetaldehyde reacts with saturated aqueous solution of sodium bisulphite (NaHSO₃) and forms crystalline acetaldehyde sodium bisulphite. It is water soluble crystalline solid.

(2) Acetone reacts with saturated aqueous solution of sodium bisuiphite (NaHSO₃) and forms crystalline acetone sodium bisuiphite. It is water soluble crystalline solid.

Question 93.
A carbonyl compound ‘A’ having molecular formula C₅H₁₀O forms crystalline precipitate with sodium bisulphite and gives positive iodoform test but does not reduce Fehling solution. Write the structure of carbonyl compound.
Answer:
A carbonyl compound C₅H₁₀O has two structures.

Pentan-2-one forms crystalline precipitate with sodium bisulphite and gives positive iodoform test. But does not reduce Fehling solution.

Pentan-3-one does not react with iodine and NaOH because it does not contain group.

Question 94.
How does alcohols react with aldehydes and ketones?
Answer:
Aldehyde reacts with one molecule of anhydrous monohydric alcohol in presence of dry hydrogen chloride to give alkoxyalcohol known as hemiacetal, which further reacts with one more molecule of anhydrous monohydric alcohol to give a geminaldialkoxy compound known as acetal as shown in the reaction.

Ketones react with 1, 2 – or 1, 3 – diols in presence of dry hydrogen chloride to give five or six-membered cyclic ketals.

Question 95.
What is the action of ethanol on acetaldehyde ? What is the action of ethylene glycol on acetone ?
Answer:
Acetaldehyde reacts with one equivalent of monohydric alcohol in the presence of dry hydrogen chloride to form an intermediate known as hemiacetal, which further adds another molecule of alcohol to form a gem-dialkoxy compound known as acetal. Acetone reacts with ethylene glycol under similar conditions to form cyclic products known as ethylene glycol ketals.

Question 96.
Write the structure of product in the following reactions:

Answer:

Answer:

Question 97.
How does Grignard reagent react with the carbonyl compounds (or aldehydes and ketones)?
Answer:
The carbonyl compounds like aldehydes and ketones react with Grignard reagent (R – Mg – X) in dry ether and form a complex which on further hydrolysis with acid forms the corresponding alcohol.

Question 98.
What is the action of Grignard reagent, CH₃ – Mg – I on : (1) formaldehyde (2) acetone?
Answer:
(1) Grignard reagent with formaldehyde gives a primary alcohol.
Formaldehyde on reaction with Grignard reagent, CH₃ – Mg – I in dry ether forms a complex which on hydrolysis with dilute HCl forms ethyl alcohol.

(2) Acetone on reaction with Grignard reagent, CH₃ – Mg – I in dry ether forms a complex which on hydrolysis with dilute HC1 forms tert-butyl alcohol.

Question 99.
Explain the mechanism of addition reactions of ammonia derivatives H₂N-Z with carbonyl compounds (aldehydes or ketones).
Answer:
Derivatives of ammonia H₂N-Z reacts with carbonyl compounds (aldehydes or ketones) in weakly acidic medium to give addition products, which loses a water molecule to give a final product imine derivatives. A substituted imine is called a Schiff base. Schiff bases are solids and have sharp melting points.

General reaction :

Question 100.
What Is the action of ethylamine on :
(1) acetaldehyde
(2) acetone ?
Answer:
(1) Acetaldehyde on reaction with ethyl amine forms imine (Schiff base).

(2) Acetone on reaction with ethyl amine forms imine (Schiff base).

Question 101.
What are oximes? Which functional group do they contain?
Answer:
Oximes : These are the compounds obtained by the reactions of carbonyl compounds namely aldehydes and ketones with hydroxyl amine NH₂OH.

Question 102.
What is the action of hydroxyl amine (NH₂OH) on (1) acetaldehyde (2) acetone?
Answer:
(1) Acetaldehyde on reaction with hydroxyl amine (in weakly acidic medium) forms crystalline acetaldoxime.

(2) Acetone on reaction with hydroxyl amine (in weakly acidic medium) forms crystalline acetoxime.

Question 103.
What are hydrazones?
Answer:
Carbonyl compounds like aldehydes and ketones react with hydrazine forming compounds like hydrazones. For example, acetaldehyde on reaction with hydrazine gives acetaldehyde hydrazone.

Question 104.
Which compound can convert
Answer:
The compound which can convert.

Question 105.
What is the action of hydrazine on (1) formaldehyde (2) acetone ?
Answer:
(1) Formaldehyde on reaction with hydrazine forms formaldehyde hydrazone.

(2) Acetone with hydrazine forms acetone hydrazone.

Question 106.
What are phenylhydrazones?
Answer:
The carbonyl compounds like aldehydes and ketones on reaction with phenylhydrazine form hydrazones. For example, acetaldehyde on reaction with phenylhydrazine forms acetaldehydephenylhydrazone.

Question 107.
What is the action of phenylhydrazine on (1) formaldehyde (2) acetone (propanone) ?
Answer:
(1) Formaldehyde on reaction with phenylhydrazine forms formaldehydephenylhydrazone.

(2) Acetone on reaction with phenylhydrazine forms acetone phenylhydrazone.

Question 108.
What is the action of 2,4-Dinitrophenylhydrazine on
(1) Acetaldehyde
(2) Acetone
(3) Butanone
(4) Benzaldehyde ?
OR
Complete and rewrite the balanced chemical equation : Butanone + 2, 4-Dinitrophenylhydrazine.
Answer:
(1) Acetaldehyde on reaction with 2,4-Dinitrophenylhydrazine forms 2,4-Dinitrophenylhydrazone.

(2) Acetone on reaction with 2,4-Dinitrophenylhydrazine forms 2,4-Dinitrophenylhydrazone.

(3) Butanone on reaction with 2,4-Dinitrophenylhydrazine forms 2,4-Dinitrophenylhydrazone.

Question 109.
What is the action of semi carbazide on (1) Acetaldehyde (2) Acetone?
Ans.
(1) Acetaldehyde on reaction with semicarbazide forms scrnicarbazone derivative.

(2) Acetone on reaction with sernicarbazide forms semicarbazone derivative.

Question 120.
Write the structures of carbonyl compounds and ammonia derivatives that combine to give following imines.
Answer:

Question 121.
Write the structure of the products obtained from the following ketones by action of hydrazine in presence of (1) slightly acidic medium (2) strong base KOH.

Answer:
(1) In slightly acidic medium

(2) In the presence of a strong base KOH

Question 122.
Explain haloform reaction.
Answer:
A ketone containing -COCH₃ group is oxidised by sodium hypohalite a mixture of (sodium hydroxide and halogen) results in the formation of sodium salt of carboxylic acid having one carbon atom less than that of ketone and methyl group is converted to haloform.

Acetaldehyde is the only aldehyde which gives haloform reaction. In this reaction, R may be hydrogen, methyl group or aryl group and X may be Cl, Br or I. The reaction is given by all methyl ketones (CH₃ – CO – R) and all alcohols containing CH₃ – (CHOH) group.

When a methyl ketone is warmed with iodine and sodium hydroxide, a yellow precipitate of iodoform is obtained. The iodoform reaction is used as a qualitative test for detection of CH₃CO-group in a organic compound.

Question 128.
Identify the compounds, amongst the following, that give positive iodoform test.
(CH₃)₂CHOH, (CH₃)₃COH, CH₃COCH₂CH₂CH₃, CH₃CH₂CHO, CH₃CH₂CH(OH)CH₂CH₃, CH₃CH₂OH, C₆H₅COCH₂CH₃, CH₃CHO, C₆H₅CH₂CH₂OH and CH₃CH(OH)CH₂CH₂CH₃.
Answer:
For an iodoform test, the carbonyl compound must have  group.
The compounds that give positive iodoform test :

Question 129.
Explain cross aldol condensation.
Answer:
(1) An aldol condensation between two different carbonyl compounds (aldehydes and or ketones) takes place even though one of the two carbonyl compounds molecules does not contain a-hydrogen atom e.g. HCHO and C₆H₅CHO.

(2) If both aldehydes or ketones contain two a-hydrogen atoms each, then a mixture of four products, is formed. For example, a mixture of ethanal and propanal on reaction with dilute alkali followed by heating gives a mixture of four products.

Self aldol condensation:

Cross aldol condensation:

Question 130.
Write the structure of the major product of the following crossed aldol condensation.
Answer:
(1) Formaldehyde and propionaldehyde :

(2) Benzaldehyde with acetone:

Question 131.
Explain aldol condensation reaction of propionaldehyde.
Answer:
Since propionaldehyde has an a-hydrogen atom it undergoes aldol condensation with alkali Ba(OH)₂, forming 3-Hydroxy-2-methylpentanal.

3-Hydroxy-2-methylpentanal on heating undergoes dehydration and forms 2-Methylpent-2-enal.

Question 132.
If a mixture of formaldehyde and acetaldehyde is subjected to aldol condensation, predict the products formed and draw their structures.
Answer:
Since formaldehyde, does not have α-hydrogen atom it will not undergo self aldol condensation. Since acetaldehydehas a-hydrogen atom, it will undergo self aldol condensation.

Formaldehyde and acetaldehyde undergo cross aldol condensation.

Hence two aldol condensation products will be obtained.

Question 133.
Indicate by equations, what happens when a mixture of acetaldehyde and acetone are treated with alkali.
Answer:
When a mixture of acetaldehyde and acetone is treated with alkali, Ba(OH)₂, they undergo self aldol condensation and cross aldol condensation.
(1) Self aldol condensation acetaldehyde :

(2) Self aldol condensation of acetone:

(3) Cro5s aldol condensation:

Question 134.
Explain Cannizzaro reaction.
OR
Write a note on Cannizzaro reaction.
OR
Write a note on self oxidation-reduction reaction of aldehyde with suitable example.
Answer:

• Aldehydes which do not have a-hydrogen atom, on heating with concentrated alkali (50% aqueous or ethanolic solution of NaOH or KOH) undergo self oxidation and reduction reaction or redox reaction.
• This self redox reaction or disproportionation reaction is called Cannizzaro reaction.
• In this reaction one molecule of the aldehyde is oxidised to carboxylic acid while the second molecule of the aldehyde is reduced to alcohol (carboxylic acid formed, reacts with alkali, NaOH and forms a salt R – COONa).
• When formaldehyde (methanal) is heated with 50% NaOH solution, methanol (reduction product) and sodium formate (oxidation product) are formed.
  
• Ketones and aldehydes like acetaldehyde, propionaldehyde, etc. having a – H atom do not give Cannizzaro reaction.

Question 135.
Explain cross Cannizzaro reaction with example.
OR
Write the reaction for the action of 50 % NaOH on a mixture of formaldehyde and benzaldehyde.
Answer:
The reaction between two different aldehydes, not having a-hydrogen atoms is called cross Cannizzaro reaction. These two aldehydes undergo disproportionation in presence of concentrated alkali to give four products. However, if one of the aldehydes is formaldehyde, the reaction yields exclusively formate and alcohol to corresponding aldehyde.

Formaldehyde and benzaldehyde since do not have a-hydrogen atom, will undergo Cannizzaro (redox) reactions.

(1) Self Cannizzaro (redox) reaction :

(2) Cmss Cannizzaro (redox) reaction:

Question 136.
What is the action of cone, potassium hydroxide on benzaldehyde?
Answer:
When benzaldehyde is heated with concentrated potassium hydroxide in presence of methanol, a mixture of potassium benzoate and phenyl methanol (benzyl alcohol) is obtained.

Question 137.
Differentiate between Cannizzaro reaction and Aldol reaction.
Answer:

Cannizzaro reaction	Aldol reaction
1. It is given by aldehydes not having alpha hydrogen atom. 2.  In this reaction an aldehyde is converted to the corres­ponding acid and an alcohol. 3.  It is a disproportionate ion reaction. 4.  It requires concentrated alkali as a catalyst.	1. It is given by aldehydes and ketones possessing alpha hydrogen atom. 2.  In this reaction aldehydes and ketones are converted into aldol and ketols, respectively. 3.  It is an addition reaction. 4.  It requires dilute alkali as a catalyst.

Question 138.
Write the chemical equations for aldol condensation or Cannizzaro reaction that the following compounds undergo :
(1) Propanal
(2) 2-Methyl propanal (isobutyraldehyde)
(3) Pentanal
(4) 3-Methylbutanal
(5) Acetophenone
(6) p-Methoxybenzaldehyde
(7) 2-Methyl cyclohexanone
(8) Chloral
(9) Cyclopentanone
(10) Phenyl acetaldehyde
(11) 1-Phenyl propan-l-one.
Answer:
(1) Propanal (Aldol condensation) : Propanal contains α-H atom. Two molecules of propanal undergo self condensation in presence of dil. alkali to form 3-Hydroxy-2-methyl pentanal.

(2) 2-NIethvl propanal (Canniuaro reaction) : Two molecules of them undergo cannizzaro reaction in the presence of 50% alkali to form sodium isobutyrate and isohutyl alcohol.

(3) Pentanal (Aldol condensation) : Pentanal contains a-H atom. Two molecules of them undergo self condensation in the presence of dil. alkali to form 3-Hydroxy-2-propyl heptanal.

(4) 3-Methyl butanal (Aldol condensation) : 3-Methyl butanal contains a-H atom. Two molecules of them undergo self condensation in the presence of dil. alkali to form 3-Hydroxy-2-isopropyl-5-methyl hexanol.

(5) Acetophenone (Aldol condensation) : Acetophenone contains a-H atom. Two molecules of them undergo self condensation in the presence of base to form 3-Hydroxy-1, 3-diphenyl but-l-one.

(6) p-Methoxybenzaldehyde (Cannizzaro reaction) :

(7) 2-Methyl cyclohexanone

(8) Chloral (Cannizzaro’s reaction) : There is no α-H atom CCl₃CHO, therefore it undergoes Cannizzaro’s reaction.

(9) Cyclopentanone (Aldol condensation) : Cyclopentanone contains a-H atom. Two molecules of them undergo self-condensation in the presence of base to form 2-(l-Hydroxy-1 cyclopentyl) cyclo pentane-l-one.

(10) Phenyl acetaldehyde (Aldol condensation) : Phenyl acetaldehyde contains a-H atom. Two molecules of them undergo self-condensation in the presence of base to form 3-Hydroxy-2, 4-diphenylbutanal.

(11) 1-phenyl propan-l-one (Aldol condensation) : 1-phenyl propan-l-one contains a-H atom. Two molecules of them undergo self-condensation in the presence of base to form 3-Hydroxy-2-methyl-l, 3-diphenyl pentan-l-one
3-Hydroy-2-methyl-1, 3-diphenyl pentan-l-one

Question 139.
Write a note on Clemmensen reduction.
OR
Explain Clemmeusen’s reduction.
OR
Explain the reduction of carbonyl group into methylene group.
Answer:

• The carbonyl groupon reduciion with zinc amalgam (Zn – Hg) in concentrated hydrochloric acid is converted into methylene group ( – CH₂ -).
  
• Aldehydes and kctoncs on reaction with Zn – Hg in concentrated HCl forms corresponding alkanes. ibis reduction is called Clemmensen reduction.
  
• Acetaldehyde on reduction with Zn – Hg in concentrated HCl forms ethane.
  
• Acetone on reduction with Zn – Hg in concentrated HCl forms propane.
  

Question 140.
Explain Wolff-Kishner reduction.
Answer:
Hydrazine (NH₂-NH₂) reduces carbonyl groupof aldehydes or ketones to metylene group  When aldehyde or ketone is heated with hydrazine in the presence of base such as potassium hydroxide and ethylene glycol, an alkane is obtained due to reduction of carbonyl compound.

Question 141.
Compound A (C₅H₁₀O) form a phenyl hydrazone and gives a negative Tollen’s reagent test and iodoform test. On reduction with Zn-Hg/HCl, compound A gives n-pentane. Write the structure of ‘A’.
Answer:
Since A (C₅H₁₀O) forms a phenyl hydrozone, it is a carbonyl compound. Since it gives negative Tollen’s reagent test, it is not an aldehyde but it must be a ketone.

Since it doesn’t give iodoform test, it doesn’t have group.
Hence the structure of ‘A’ will be,

Question 142.
Identify A and B in the following reactions :

Answer:

 

Answer:

 

Answer:

 

Answer:

Question 143.
What is the action of concentrated nitric acid on (1) Benzaldehyde (2) Benzophenone?
Answer:

1. Benzaldehyde on reaction with concentrated nitric acid in presence of cone. H₂SO₄ forms m-nitrobenzaldehyde
   
2. Benzophenone on reaction with concentrated nitric acid in presence of cone. H₂SO₄ forms m-nitrobenzophenone
   

Question 144.
Explain laboratory tests for carboxyl (- COOH) group.
Answer:
The presence of – COOH group in carboxylic acids is identified by the following tests :

(1) Litmus test : (valid for water soluble substances)
Aqueous solution of Organic compound containing – COOH group turns blue litmus red which indicates the presence of acidic functional group.

(2) Sodium bicarbonate test : When sodium bicarbonate is added to an organic compound containing – COOH group, a brisk effervescence of carbon dioxide gas is evolved. Water insoluble acid goes in solution and gives precipitate an acidification with cone. HCl. This indicates the presence of -COOH group.

(3) Ester test : One drop of concentrated sulfuric acid is added to a mixture of given organic compound containing – COOH group and one mL of ethanol, the reaction mixture is heated for 5 minutes in hot water bath. After this, hot solution is poured in a beaker containing water, fruity smell of ester confirms the presence of carboxylic acid.

Question 145.
How is acid chloride obtained from carboxylic acid?
Answer:
Carboxylic acid on heating with thionyl chloride (SOCl₂), phosphorus trichloride (PCl₃) or phosphorus pentachloride (PCl₅) give corresponding acid chlorides. In this reaction – OH of carboxyl group is replaced by -Cl.

The reactions are :
(1) Action on SOCl₂ on carboxylic acid :

Example : Acetic acid reacts with thionyl chloride to give acetyl chloride.

(2) Action of PCl₃ on carboxylic acid (ethanoic acid) :

Example : Action of phosphorus trichloride on acetic acid gives acetyl chloride.

(3) Action of PCI₅ on carboxIic acid :

Question 146.
How will you convert 3,5-Dinitrobenzoic acid to 3,5-Dinitrobenzoyl chloride?
Answer:
When 3,5-Dinitrobenzoic acid is heated with phosphorus pentachloride, 3,5-Dinitrobenzoyl chloride is obtained.

Question 147.
How is acid amide obtained from carboxylic acid?
Answer:
Carboxylic acid or acid chloride with ammonia salts, which on further strong heating at high temperature decompose to give amides.

When acetic acid is treated with ammonia, ammonium acetate is obtained. Ammonium acetate on strong heating decomposes to form acetamide.

When acetyl chloride is treated with ammonia, acetamide is obtained

Question 148.
How is acid anhydride obtained from carboxylic acid?
Answer:
When carboxylic acid is heated with strong dehydrating agent like phosphorus pentoxide or concentrated sulphuric acid, an acid anhydride is obtained,

The reaction is reversible and anhydride is hydrolysed back to acid.

Alternatively, when sodium acetate is heated with acetyl chloride, acetic anhydride is obtained. This reaction is irreversible.

Question 149.
What is decarboxylation of an acid ? How is it done?
OR
What happens when sodium acetate is heated with soda lime ?
Answer:
Removal of a carboxylic group from acid is called decarboxylation. Decarboxylation of an acid is carried out by heating anhydrous sodium salts of carboxylic acids with soda lime (NaOH + CaO). The product hydrocarbons obtained contain one carbon atom less than the carboxylic acid.

When sodium acetate is heated with soda lime, methane is obtained.

Question 150.
How is alcohol obtained from carboxylic acid ?
OR
What is the action of LiAlH₄/H₃O⁺on ethanoic acid? Write balanced equation for the conversion :
Cyclopropane carboxylic acid to Cyclopropylmethanol.
Answer:
Carboxylic acids are reduced to primary alcohols using powerful reducing agent lithium aluminium hydride.

(i) When ethanoic acid is reduced in the presence of LiAlH4 in dry ether, forms ethanol.

(ii) When cyclopropane carboxylic acid is reduced in the presence of lithium aluminium hydride in dry ether, forms cyclopropyl methanol.

Question 151.
What is the action 9f following compounds on cyclohexanone in presence of dry hydrogen chloride?
(1) Ethyl alcohol
(2) Ethylene glycol
Answer:
(1) With Ethyl alcohol : Cyclohcxanonc reacts with one equivalent of monohydric ethyl alcohol lo form hemi ketal, which further adds another molecule of alcohol to form a gem-dialkox compound known as ketal.

(2) With Ethylene glycol : cyclohexanone reaCts with ethylene glycol to form cyclic ketal.

Question 152.
Answer the following in one sentence.

1. Name the compound which reacts with formaldehyde to produce ethyl alcohol.
Answer:
The compound which reacts with formaldehyde to produce ethyl alcohol is methyl magnesium iodide.

2. What are imines ?
Answer:
These are the compounds obtained by the reactions of carbonyl compounds namely aldehydes and ketones with primary amine.

3. Why does skin have burning sensation, when an ant bites ?
Answer:
When an ant bites, formic acid is released from an ant which gives burning sensation as the acid comes in contact with the skin.

4. What is the percentage of acetic acid in vinegar?
Answer:
The percentage of acetic acid in vinegar is 6 to 8%.

5. Which reagent is used to distinguish formic acid and acetic acid?
Answer:
The reagent used to distinguish formic acid and acetic acid is ammoniacal silver nitrate.

6. What happens when acetyl chloride is treated with dibenzyl cadmium ? Give reaction.
Answer:

7. Complete the following reaction :

Answer:

8. Give reason : Aromatic carboxylic acids do not undergo Friedel-Crafts reaction.
Answer:
Aromatic carboxylic acids do not undergo Friedel-Crafts reaction because the catalyst aluminium chloride (Lewis acid) gets bonded to carboxyl group.

9. Write the name of two compounds which do not contain carbonyl group but show iodoform test.
Answer:
The name of two compounds which do not contain carbonyl group but show iodoform test are ethanol

10. Give reason : In semicarbazide, – NH₂ group bonded to carbonyl group is not involved in the formation of semicarbazone.
Answer:
NH₂ group attached to – NH group in semicarbonide is more active than NH2 group attached to carbonyl group due to electron density difference.

11. Fehling solution does not oxidise benzaldehyde but Tollen’s reagent oxidises benzaldehyde. Give reason.
Answer:
When benzaldehyde is heated with Fehling solution, there is no change in colour of the solution, Cu²⁺ ion is not reduced, hence Fehling solution does not oxidise benzaldehyde. However, Tollen’s reagent oxidises benzaldehyde to give silver mirror test.

12. Give reason : Direct attachment of vinyl group to carboxylic group increases the acidity of corresponding acids.
Answer:
Direct attachment of vinyl group to carboxylic group increases the acidity of corresponding acids due to greater electronegativity of sp²-hybridised carbon to which carboxyl carbon is attached.

Multiple Choice Questions

Question 153.
Select and write the most appropriate answer from the given alternatives for each sub-question :

1. IUPAC name of

(a) l-Phenylhexan-2-one
(b) 6-Phenylhexan-5-one
(c) l-Benzylhexan-2-one
(d) Dodecan-5-one
Answer:
(a) l-Phenylhexan-2-one

2. The general formula of carbonyl compounds is
(a) C_nH_2n+1OH
(b) C_nH_2nO
(c) C_nH_2nO₂
(d) C_nH_2n+1O
Answer:
(b) C_nH_2nO

3. Aldehydes and ketones are
(a) chain isomers
(b) functional isomers
(c) geometrical isomers
(d) position isomers
Answer:
(b) functional isomers

4. Identify ‘C’ in the following reaction,
\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br} \stackrel{\mathrm{KCN}}{\longrightarrow} \mathrm{A} \stackrel{\mathrm{H}^{+} \mathrm{H}_{2} \mathrm{O}}{\longrightarrow} \mathrm{B} \stackrel{\mathrm{LiAlH}_{4}}{\longrightarrow} \mathrm{C}\)
(a) Propan – 1 – ol
(b) Propanone
(c) 2-Ethyl-3-pentanone
(d) Propanal
Answer:
(d) Propanal

5. Grignard reagent when reacted with alkyl cyanide followed by hydrolysis gives
(a) an aldehyde
(b) a ketone
(c) a primary alcohol
(d) a secondary alcohol
Answer:
(b) a ketone

6. Identify ‘ B ’ in the following reaction :
\(\mathrm{CH}_{3}-\mathrm{C}=\mathrm{N} \stackrel{\mathrm{CH}_{3} \mathrm{MgI}}{\longrightarrow} \mathrm{A} \underset{\mathrm{HCl}}{\stackrel{2 \mathrm{HOH}}{\longrightarrow}} \mathrm{B}+\mathrm{NH}_{3}+\mathrm{MgIOH}\)
(a) Magnesium intermediate
(b) Ethanol
(c) Propanal
(d) Propanone
Answer:
(d) Propanone

7. \(\mathrm{A}+\mathrm{B} \stackrel{\text { dry ether }}{\longrightarrow} \mathrm{Col}\) Complex \(\mathrm{H}_{2} \mathrm{O} / \mathrm{H}^{+}\) Ethyl Methyl Ketone. In the above reaction, A and B are
(a) Formonitrile, Propyl magnesium bromide
(b) Ethyl cyanide, Ethyl magnesium bromide
(c) Hydrogen cyanide, Ethyl magnesium bromide
(d) Acetonitrile, Ethyl magnesium bromide
Answer:
(d) Acetonitrile, Ethyl magnesium bromide

8. A dilute solution of p-rosaniline hydrochloride in water whose pink colour has been discharged by passing sulphur dioxide, does not restore its colour by
(a) HCHO
(b) CH₂CHO
(c) (CH₃)₂COCH₃
(d) CCl₃CHO
Answer:
(c) (CH₃)₂COCH₃

9. The reagent with which both acetaldehyde and acetone reacts easily is –
(a) Fehling’s solution
(b) Tollen’s reagent
(c) Grignard reagent
(d) Schiff s reagent
Answer:
(c) Grignard reagent

10. Isopropyl methyl ketone when treated with Zn-Hg and concentrated hydrochloric acid give
(a) iso-butane
(b) iso-pentane
(c) n-pentane
(d) neo-pentane
Answer:
(b) iso-pentane

11. The formation of acetone cyanohydrin from acetone is an example of
(a) Nucleophilic addition
(b) Nucleophilic substitution
(c) Electrophilic addition
(d) Electrophilic substitution
Answer:
(a) Nucleophilic addition

12. Which of the following is Fehling solution ‘A’?
(a) CuSO₄ solution
(b) CaSO₄ solution
(c) NaOH solution
(d) Sodium potassium tartarate solution
Answer:
(a) CuSO₄ solution

13. The compound ‘X’ upon alkaline hydrolysis gives a product which reacts with phenylhydrazine but does not reduce ammoniacal silver nitrate solution. A possible structure for ‘X’ is
(a) CH₃CHCl CH₂Cl
(b) CH₃CCl₂CH₃
(c) CH₃CH₂CH₂Cl
(d) CH₃CH₂CHCl₂
Answer:
(b) CH₃CCl₂CH₃

14. Which of the following is the correct statement with respect to aldehyde and ketones ?
(a) Ketones are reducing agents
(b) Aldehydes are good reducing agents
(c) Cannizzaro reaction is an addition reaction
(d) Ketones do not react with Grignard reagent
Answer:
(b) Aldehydes are good reducing agents

15. Acetaldehyde acts as
(a) a catalyst
(b) a reducing agent
(c) an oxidizing agent
(d) a mordant
Answer:
(b) a reducing agent

16. An organic compound (A) C₃H₈0 on oxidation gives (B) C₃H₆O₂. The compound A may be
(a) an ester
(b) an alcohol
(c) an aldehyde
(d) a ketone
Answer:
(b) an alcohol

17. Identify ‘B’ from the following reaction :
\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{CHO}+\mathrm{NH}_{2} \mathrm{OH} \rightarrow \mathrm{A} \stackrel{\mathrm{Na} / \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}}{\Delta} \mathrm{B}\)
(a) Propan-1-amine
(b) Propan-2-amine
(c) Isopropylamine
(d) Dimethylamine
Answer:
(a) Propan-1-amine

18. Sodium borohydride does not reduce
(a) – COOH group
(b) – NO₂ group
(c) – X atom
(d) – CHO group
Answer:
(a) – COOH group

19. An aldehyde when warmed with Zn/Hg and cone. HCl gives
(a) alcohol
(b) hydrocarbon
(c) carboxylic acid
(d) ketone
Answer:
(b) hydrocarbon

20. Acetaldol is
(a) 3-hydroxy butanol
(b) 3-hydroxy butanal
(c) 2-hydroxy propanal
(d) 3-hydroxy pentanal
Answer:
(b) 3-hydroxy butanal

21. Acetone can be reduced to propane, the reduction is called
(a) Clemmensen’s reduction
(b) catalytic reduction
(c) Rosenmund’s reduction
(d) partial reduction
Answer:
(a) Clemmensen’s reduction

22. Which of the following reagents can react with acetaldehyde to give water soluble white crystal-line solid?
(a) NaHSO₄
(b) NaHSO₃
(c) Na₂SO₃
(d) Na₂SO₄
Answer:
(b) NaHSO₃

23. Which of the following compounds does NOT undergo aldol condensation?

Answer:
(a)

24. Formalin is 40% aqueous solution of :
(a) Methanal
(b) Methanoic acid
(c) Methanol
(d) Methanamine
Answer:
(a) Methanal

25. Which of the following compounds do not produce pink colour with Schiff s reagent?
(a) Formaldehyde
(b) 2-propanone
(c) 3-pentanone
(d) Both (b) and (c)
Answer:
(d) Both (b) and (c)

26. Both aldehydes and ketones can react with
(a) Tollen’s reagent
(b) the Grignard reagent
(c) Fehling’s solution
(d) Schiffs reagent
Answer:
(b) the Grignard reagent

27. Aldol reaction is.
(a) an addition reaction
(b) an elimination reaction
(c) a self-reduction reaction
(d) a disproportionate ion reaction
Answer:
(a) an addition reaction

28. The reaction in which two molecules combine to form a new molecule with the elimination of a small molecule like water is called
(a) an oxidation reaction
(b) a condensation reaction
(c) a hydrolysis reaction
(d) a redox reaction
Answer:
(b) a condensation reaction

29. Benzaldehyde undergoes Cannizzaro’s reaction to give
(a) sodium benzoate and methyl alcohol
(b) sodium benzoate and benzyl alcohol
(c) benzyllic acid and benzyl alcohol
(d) phenol and benzoic acid
Answer:
(b) sodium benzoate and benzyl alcohol

30. Identify ‘X’ in the following reaction :
CH₃-CHO + X → CH₃-CH = N-NH-C₆H₅ + H₂O
(a) C₆H₅-NH₂
(b) C₆H₅-NH-NH₂
(c) C₆H₅-N = NH
(d) C₆H₅-NH-NH-CH₃
Answer:
(b) C₆H₅-NH-NH₂

31. What happens when propanal is treated with zinc amalgam and conc.HCl ?
(a) Propan-l-ol
(b) Propan-2-ol
(c) Propane
(d) Propanone
Answer:
(c) Propane

32. Identify ‘ B ’ in the following reaction :
\(2 \mathrm{CH}_{3}-\mathrm{CHO} \frac{\text { dil. base or acid }}{300 \mathrm{~K}} \mathrm{~A} \underset{\text { dehydration }}{\text { dehy }} \mathrm{B}+\mathrm{H}_{2} \mathrm{O}\)
(a) CH₃-CH(OH)-CH₂-CHO
(b) CH₃-CH₂-CH(OH)-CHO
(c) CH₃-CH = CH-CHO
(d) CH₃-CO-CH₃
Answer:
(c) CH₃-CH = CH-CHO

33. The blue colour of Fehling’s solution is due to
(a) Cu₂O
(b) CuCO₃
(c) CuO
(d) Cu⁺⁺ ions
Answer:
(d) Cu⁺⁺ ions

34. How is Schiff s reagent prepared?
(a) By passing CO₂ through p-rosaniline solution
(b) By passing NO₂ through p-rosaniline solution
(c) By passing SO₂ through p-rosaniline solution
(d) By passing NH₃ through silver nitrate solution
Answer:
(c) By passing SO₂ through p-rosaniline solution

35. Benzaldehyde when treated with cone. HNO₃ gives
(a) o-nitrobenzaldehyde
(b) p-nitrobenzaldehyde
(c) m-nitrobenzaldehyde
(d) a mixture of -o and -p-nitrobenzaldehyde
Answer:
(c) m-nitrobenzaldehyde

36. Which of the following carbonyl compounds undergoes aldol condensation ?
(a) Benzaldehyde
(b) Benzophenone
(c) Acetophenone
(d) tert-Butyl phenyl ketone
Answer:
(c) Acetophenone

37. Which of the following carbonyl compounds undergoes self redox reaction in presence of concentrated base ?
(a) 3-Methylpentanal
(b) 2-Chlorobutanal
(c) 2,2-Dimethylpropanal
(d) tert-Butyl methyl ketone
Answer:
(c) 2,2-Dimethylpropanal

38. The smell of bitter almond is given by the compound.
(a) Benzoic acid
(b) Benzaldehyde
(c) Vanillin
(d) Cinnamaldehyde
Answer:
(b) Benzaldehyde

39. Which of the following will not give yellow precipitate when treated with NaOH and H?
(a) 3-Methylbutan-2-one
(b) 2-methylpentan-3-one
(c) Propanone
(d) Hexan-2-one
Answer:
(b) 2-methylpentan-3-one

40. A β-hydroxyl carbonyl compound is obtained by the action of NaOH on
(a) HCHO
(b) C6H₅CHO
(c) CR₃CHO
(d) CH₃CHO
Answer:
(d) CH₃CHO

41. Decarboxylation of sodium propionate gives
(a) methane
(b) ethane
(c) propane
(d) ethene
Answer:
(b) ethane

42. Ester on hydrolysis with dil HCl gives
(a) RCOOH + ROH
(b) RCOR + ROH
(c) ROH + ROH
(d) RCOR + RCOOH
Answer:
(a) RCOOH + ROH

43. \(\mathrm{CH}_{3}-\mathrm{CO}-\mathrm{CH}_{3} \stackrel{\mathrm{CrO}_{3}}{\longrightarrow} \mathrm{A}\) The compound A is
(a) acetic acid
(b) propionic acid
(c) formic acid
(d) benzoic acid
Answer:
(a) acetic acid

44. The reaction of C₆H₅CH = CHCHO with LiAlH₄ gives
(a) C₆H₅CH₂CH₂CH₂OH
(b) C₆H₅CH₂CH₂CHO
(c) C₆H₅CH = CHCH₂OH
(d) C₆H₅CH₂CHOHCH₃
Answer:
(a) C₆H₅CH₂CH₂CH₂OH

45. A mixture of sodium benzoate and sodalime on heating yields
(a) methane
(b) benzene
(c) sodium benzoate
(d) calcium benzoate
Answer:
(b) benzene

46. Which is the strongest acid?
(a) CH₃COOH
(b) CH₃CH₂COOH
(c) (CH₃)₃CCOOH
(d) CICH₂COOH
Answer:
(d) CICH₂COOH

47. Benzaldehyde when treated with alkaline KMnO₄ yields
(a) Benzyl alcohol
(b) Benzoic acid
(c) CO₂ and H₂O
(d) Salicylic acid
Answer:
(a) Benzyl alcohol

48. Acetonitrile on acidic hydrolysis gives
(a) HCOOH
(b) CH₃NC
(c) CH₃COONa
(d) CH₃COOH
Answer:
(d) CH₃COOH

49. The organic compounds A and B reacts with sodium metal and liberates hydrogen gas. A and B reacts together to give ethyl acetate. The A and B are
(a) CH₃COOH & C₂H₅OH
(b) HCOOH & C₂H₅OH
(c) CH₃COOH & HCOOH
(d) CH₃COOH & CH₃OH
Answer:
(a) CH₃COOH & C₂H₅OH

50. Which one of the following undergoes reaction with 50% NaOH to give to corresponding alcohol and acid?
(a) Phenol
(b) Benzoic acid
(c) Benzaldehyde
(d) Butanal
Answer:
(c) Benzaldehyde

51. Identify the reactant in the following reaction.

Answer:
(c) CO

52. The strongest acid is

Answer:
(c)

53. Predict the product in the following reaction

The compound A is
(a) butane
(b) propane
(c) ethane
(d) propene
Answer:
(b) propane

54. Ethyl benzoate when heated with dil H₂SO₄ gives
(a) acetic acid
(b) benzoic acid
(c) ethanoic acid
(d) phenyl methanol
Answer:
(b) benzoic acid

55. Margarine contains
(a) acetaldehyde
(b) propionaldehyde
(c) butyraldehyde
(d) formaldehyde
Answer:
(c) butyraldehyde

56. Monocarboxylic acids have the general formula
(a) C_nH_2n+1O₂
(b) C_nH_2nO₂
(c) C_nH_2nO
(d) C_nH_2n-1O₂
Answer:
(b) C_nH_2nO₂

57. Formic acid is obtained from
(a) vinegar
(b) red ants
(c) butter
(d) oil
Answer:
(b) red ants

58. Butter contains
(a) lactic acid
(b) butyric acid
(c) citric acid
(d) acetic acid
Answer:
(b) butyric acid

59. Glacial acetic acid is
(a) HCOOH
(b) CH₃COOH
(c) CH₃CH₂COOH
(d) C₃H₇COOH
Answer:
(b) CH₃COOH

60. Which of the following acids is optically active?
(a) Oxalic acid
(b) Salicylic acid
(c) Acetic acid
(d) Lactic acid
Answer:
(d) Lactic acid

61. Lactic acid is
(a) propionic acid
(b) α-hydroxy propionic acid
(c) p-hydroxy benzoic acid
(d) butyric acid
Answer:
(b) α-hydroxy propionic acid

62. The carbon atom of the carboxylic group is
(a) sp³-hybridized
(b) sp²-hybridized
(c) sp-hybridized
(d) unhybridized
Answer:
(b) sp²-hybridized

63. The common name of carboxylic fatty acids is derived from
(a) the name of parent alkanes
(b) the name of corresponding aldehydes
(c) from their original sources
(d) the name of alkyl group present in them
Answer:
(c) from their original sources

64. The IUPAC name of a-methylpropionic acid is
(a) Propanoic acid
(b) Butanoic acid
(c) 2-Methylpropanoic acid
(d) 2-Methylbutanoic acid
Answer:
(c) 2-Methylpropanoic acid

65. For the nomenclature of carboxylic acids, the suffix used is
(a) -ane
(b) -oic
(c) -al
(d) -ol
Answer:
(b) -oic

66. Propionic acid can be prepared by the
(a) action of propyl magnesium chloride on dry ice
(b) alkaline hydrolysis of propyl cyanide
(c) acid hydrolysis of ethyl cyanide
(d) oxidation of Propanone
Answer:
(c) acid hydrolysis of ethyl cyanide

67. The intermediate compound formed during hy-drolysis of acetonitrile to acetic acid is
(a) acetone
(b) acetamide
(c) ammonium acetate
(d) ethyl ammonium chloride
Answer:
(b) acetamide

68. Carbonation of CH₃MgI gives organic compound. The same compound can also be obtained by
(a) oxidation of Methanol
(b) oxidation of Methanal
(c) acid hydrolysis of acetonitrile
(d) alkaline hydrolysis of ethyl cyanide
Answer:
(c) acid hydrolysis of acetonitrile

69. The acid that cannot be prepared by the action of Grignard reagent on dry ice is
(a) methanoic acid
(b) ethanoic acid
(c) propanoic acid
(d) butanoic acid
Answer:
(a) methanoic acid

70. The compound which on acid hydrolysis followed by oxidation gives acetic acid is
(a) CH₃I
(b) CH₂Cl₂
(c) ClCH₂CH₂C1
(d) CH₃CHCl₂
Answer:
(d) CH₃CHCl₂

71. The hydrolysis product of alkyl cyanide is
(a) primary amine
(b) amides
(c) aldehyde
(d) carboxylic acid
Answer:
(d) carboxylic acid

72. To prepare acetic acid,
(a) methyl alcohol is oxidized with KMnO₄
(b) calcium acetate is distilled with calcium for-mate under dry conditions
(c) acetaldehyde is oxidized in the presence of K₂Cr₂O₇ and dil. H₂SO₄
(d) glycerol is heated with H₂SO₄
Answer:
(c) acetaldehyde is oxidized in the presence of K₂Cr₂O₇ and dil. H₂SO₄

73. Solid carbon dioxide when treated with etheral solution of C₂H₅MgBr followed by acid hydrolyzis gives
(a) propanoic acid
(b) ethanoic acid
(c) propionic acid
(d) both (a) and (c)
Answer:
(d) both (a) and (c)

74. Which of the following compound does not give acetic acid on oxidation ?
(a) Ethanol
(b) Propan-l-ol
(c) Propan-2-ol
(d) 2-Methyl propan-2-ol
Answer:
(b) Propan-l-ol

75. A carboxylic acid resembles an alcohol with respect to its reaction with
(a) acidified K₂Cr₂O₇
(b) washing soda
(c) caustic soda
(d) sodium metal
Answer:
(d) sodium metal

76. Acetic acid can be converted into acetic anhydride on heating with
(a) POCl₃
(b) PCl₃
(c) PCI₅
(d) P₂O₅
Answer:
(d) P₂O₅

77. Acetyl chloride reacts with ammonia to give
(a) ammonium acetate
(b) ethylammonium chloride
(c) ethylamine
(d) acetamide
Answer:
(d) acetamide

78. The reagent that reacts with acetic acid to give sodium acetate with liberation of carbon dioxide gas is
(a) sodium metal
(b) caustic soda
(c) caustic potash
(d) baking soda
Answer:
(d) baking soda

79. An alkene on hydration gives a compound, which reacts with propionic acid to produce isopropyl propionate. The alkene is
(a) CH₂ = CH₂
(b) CH₃-CH = CH₂
(c) CH₃ – CH₂ – CH = CH₂
(d) CH₃ – CH = CH – CH₃
Answer:
(b) CH₃-CH = CH₂

80. Both the compounds ‘A’ and ‘B’ react with sodium metal to liberate hydrogen gas and react with each other to give Methylethanoate. The compounds ‘A’ and ‘B’ are
(a) C₂H₅ – COOH and CH₃ – OH
(b) C₂H₅ – COOH and C₂H₅ – OH
(c) CH₃ – COOH and C₂H₅ – OH
(d) CH₃ – COOH and CH₃ – OH
Answer:
(d) CH₃ – COOH and CH₃ – OH

81. Identify the product ‘D’ in the following series of reactions.

(a) CH₃COOCH₃
(b) CH₃COOC₂H₅
(c) C₂H₅COOCH₃
(d) C₂H₅COOC₂H₅
Answer:
(b) CH₃COOC₂H₅

82. Acetyl chloride on heating with sodium acetate gives
(a) ethyl acetate
(b) acetamide
(c) acetic anhydride
(d) acetaldehyde
Answer:
(c) acetic anhydride

83. Carboxylic acids are acidic in nature because
(a) it dissociates to give H⁺ ions
(b) it donates proton
(c) it reacts with active metal and liberates hydro-gen gas
(d) all of these
Answer:
(d) all of these

84. Carboxylic acids on heating with P₂O₅ give
(a) acid chlorides
(b) alkyl halides
(c) acid amides
(d) acid anhydrides
Answer:
(d) acid anhydrides

85.

Answer:
(c)

86. The compound having general formula is called

(a) diester
(b) acid anhydride
(c) hemiacetal
(d) acetal
Answer:
(d) acetal

87. Identify the strongest acid amongst the following :
(a) Chloroacetic acid
(b) Acetic acid
(c) Trichloroacetic acid
(d) Dichloroacetic acid
Answer:
(c) Trichloroacetic acid

88. Acetaldehyde, when treated with which among the following reagents does ‘not’ undergo addition reaction?
(a) ammonia
(b) hydroxyl amine
(c) ammoniacal silver nitrate
(d) semicarbazide
Answer:
(c) ammoniacal silver nitrate

89. Popcorn has butter flavour which contains
(a) butan-l-one
(b) butane-2, 3-dione
(c) butan-2-one
(d) butyric acid
Answer:
(b) butane-2, 3-dione

90. On acid hydrolysis, propane nitrile gives
(a) propanal
(b) acetic acid
(c) propionamide
(d) propanoic acid
Answer:
(d) propanoic acid

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

Question 1.
What are alcohols? How are they classified?
Answer:
Alcohols are the hydroxy derivatives of hydrocarbons in which one or more hydrogen atoms are replaced by hydroxyl group.

Examples : CH₃ – OH methyl alcohol, CH₃ – CH₂ – OH ethyl alcohol. Depending on the basis of hydroxyl groups present in a molecule, alcohols are classified into monohydric, dihydric, trihydric and polyhydric alcohols.

Question 2.
What are monohydric alcohols? How are they classified?
Answer:
Alcohols having only one hydroxyl group in their molecules are called monohydric alcohols. Monohydric alcohols are classified according to the type of hybridization of the carbon atom to which the hydroxyl group is attached.

(1) Alcohols containing C_sp³ – OH bond : In these alcohols -OH group is attached to a sp³ – hybridised carbon atom of alkyl group. These alcohols are represented as R-OH. They are further classified as primary, secondary and tertiary alcohols in which – OH group is attached to primary, secondary and tertiary carbon atoms respectively.

(a) Allylic alcohols : In these alcohols -OH group is attached to a sp³ -hybridised carbon atom next to the carbon-carbon double bond i.e., to allylic carbon. Allylic alcohols may be primary, secondary and tertiary alcohols.

(b) Benzylic alcohols : In these alcohols -OH group is attached to a sp³ -hybridised carbon atom next to an aromatic ring. Benzylic alcohols may be primary, secondary and tertiary alcohols.

(2) Alcohols containing C_sp³ – OH bond: In these alcohols -OH group is attached to a sp² – hybridised carbon atom,
i.e., vinylic carbon. These alcohols are also called vinylic alcohols.
e.g., CH₂ = CH – OH vinyl alcohol.

Question 3.
What are phenols (carbolic acids)?
Answer:
Hydroxy derivatives of aromatic hydrocarbons in which the hydroxyl group is directly attached to the aromatic ring are called phenols.
Examples:

Question 4.
What is phenol? OR Define carbolic acid.
Answer:
The hydroxy derivative of benzene in which the OH group is directly attached to benzene ring is called phenol.

Question 5.
How are phenols classified? Give suitable examples.
Answer:
Phenols are classified on the basis of number of hydroxyl (- OH) groups present in a molecule of phenol.
(1) Monohydric phenols : Phenols contain one hydroxyl group in their molecule.

(2) Dihydric Phenols : Phenols contain two hydroxyl groups in their molecule.

(3) Trihydric phenols : Phenols contain three hydroxyl groups in their molecule.

Question 6.
What are ethers? How are ethers classified?
Answer:
They are alkoxy derivatives of alkanes in which a hydrogen atom of alkane (R – H) is replaced by alkoxy group ( – O – R) and divalent oxygen atom is attached to two alkyl groups or two aryl groups or one alkyl and one aryl group. Ethers are organic oxides R – O – Ar, Ar – O – Ar.
E.g. R – O – R or C₂H₅ – O – C₂H₅.

Ethers are classified into two groups as follows :
(1) Simple or symmetrical ether : The ethers in which both alkyl (or aryl) groups attached to the oxygen atom are same are called simple ethers.
E.g. (R – O – R), CH₃ – O – CH₃, dimethyl ether; C₆H₅ – O -C₆H₅ diphenyl ether.

(2) Mixed or unsymmetrical ethers : The ethers in which the two alkyl (or aryl) groups attached to the oxygen atom are different are called mixed ethers.
E.g. (R – O – R), CH₃ – O – C₂H₅ ethyl methyl ether; C₂H₅ – O – C₆H₅ ethyl phenyl ether.

Question 7.
What is the general formula of ethers?
Answer:
The general formula of ethers is C_nH_{2n + 2} O. For example, dimethyl ether CH₃ – O – CH₃ has molecular formula C₂H₆O.

Common nomenclature system :
(1) In this system, monohydric alcohols (R — OH) are named as alkyl alcohols.
(2) According to the attachment of hydroxyl group to a carbon atom they are named with prefixes as n-(normal or primary) alcohol, sec-(secondary) alcohol, tert-(tertiary) alcohol.
(3) Alcohols with two hydroxyl groups are named as glycols.

Carbinol system : In this system alcohols are considered as derivatives of methyl alcohol which is called carbinol. The alkyl group attached to the carbon carrying – OH group are named in alphabetical order. Then the suffix carbinol is added.

For example :

IUPAC system of nomenclature :

• In this system, alcohols are named as alkanols.
• A longest continuous chain of carbon atoms containing – OH group is chosen as a parent hydrocarbon and alcohol is considered as a hydroxy derivative of this alkane.
• The carbon atoms are numbered from a terminal carbon atom nearest to a carbon atom attached to – OH group so that the position of – OH group is indicated by the lowest locant.
• ‘e’ of an alkane is preplaced by ‘oT, giving alkanol. The number of OH groups is indicated by prefix, di, tri, etc. before ‘oT. The positions of -OH groups are indicated by appropriate locants.
• The different substituents are arranged in the alphabetical order, and their positions are indicated by proper numbers.
• Their names are hyphened on either sides except the last substituent.
• For cyclic alcohols are named by using prefix cyclo to the parent alkane considering – OH group attached to carbon atom C – 1.

Question 8.
Give common and IUPAC names for the following :
Answer:

Question 9.
Classify the following alcohols as primary, secondary and tertiary and write their IUPAC names.
Answer:

Question 10.
Give the structures of the following :
(1) Ethanol
(2) Propan-l-ol
(3) 2-Methylbutan-l-ol
(4) Pentan-3-ol
(5) Butan-2-ol
(6) 3-Methylbutan-2-ol
(7) Hexan-l-ol
(8) 3-Methylpentan-3-ol
(9) 2-Methylpropan-2-ol
(10) Butan-l-ol (H-Butyl alcohol)
(11) 2-Methylpropan-l-ol
(12) 2,3-Dimethylbutan-l-ol
(13) 2, 3-Dimethylbutan-2-oI
(14) 2-MethyIhexan-l-ol
(15) 2,2,3-Trimethylpentan-3-ol
(16) 2,3,3-Trimethylbutan-2-ol
(17) 3-EthyI-4-methylpentan-l-ol.
Answer:

Question 11.
Give the IUPAC names of the following alcohols. Classify them as primary (1°), secondary (2°) and tertiary (3°) alcohols. Identify allylic and benzylic alcohols amongst them.
Answer:

Question 12.
Write all the possible structural isomers of alcohol having molecular formula C6H140. Give their IUPAC names. Classify them as primary, secondary and tertiary alcohols. Identify optically active alcohols amongst them.
Answer:

The optical isomers of C₆H₁₄O are (2), (3), (6), (7), (12).

Question 13.
Give the IUPAC names of the following compounds :
Answer:

Question 14.
Write IUPAC names of following alcohols :
Answer:

Question 15.
Write the structures of following alcohols : (1 mark each)
Answer:

The IUPAC system name of phenol is benzenol. The common name phenol is also accepted by IUPAC. The common names have prefixes ortho, meta and para in substituted phenols. IUPAC system uses the locant 2-, 3-, 4-, etc. to indicate the positions of substituents.

Question 16.
Write the IUPAC names of the following compounds :
Answer:

Question 17.
Write IUPAC name of the following compounds :
Answer:

Question 18.
Give the structures and IUPAC names of isomeric phenols represented by the molecular formula C₈H₁₀O.
Answer:

Common and IUPAC system of nomenclature of ethers

In the common system of nomenclature, the ethers are named by writing names of the alkyl groups attached to the oxygen atom in alphabetical order and word ether is added. If two alkyl groups are same, prefix di- is used. According too the IUPAC system of nomenclature, ethers are named as alkoxyalkanes. The larger alkyl group is considered to be parent alkane. The name of the smaller alkane is prefixed by the name of alkoxy group and its locant.

Question 19.
Give common name and JUPAC name for the I11owing ethers :
Answer:

Question 20.
Give the IUPAC name of the following ethers.
Answer:

Question 23.
Give the structures and IUPAC names of all metameric ethers represented by formula C₅H₁₂O.
Answer:

Question 24.
How many isomeric compounds can be represented by formula C₄H₁₀O?
Answer:
A compound with the molecular formula C₄H₁₀O can show two functional isomers as a monohydric alcohol and ether.

Isomers of C₄H₁₀O as alcohol :

Isomers of C₄H₁₀O as ether :

Hence, the total number of isomers of C₄H₁₀O are seven

Question 25.
Write the structural formula and IUPAC names of all possible isomers of the compound with molecular formula C₃H₈O.
Answer:
Possible isomers of C₃H₈O with structural formulae and IUPAC names :

Question 26.
Write structures of alcoholic and ether isomers of a compound having molecular formula C₇H₈O.
Answer:

Question 27.
How is alkyl halide converted into alcohol by using
(1) Aqueous NaOH (or KOH),
(2) Moist silver oxide?
Answer:
(1) When an alkyl halide (R – X), is boiled with aqueous NaOH (or KOH) an alcohol is obtained,

(2) Alkyl halide when heated with moist Ag₂O, undergoes hydrolysis and forms an alcohol.

Question 28.
How are following compounds prepared by hydrolysis of alkyl halides
(1) Ethanol
(2) Isopropyl alcohol
(3) tert-butyl alcohol
(4) methyl alcohol
(5) butan-2-ol?
Answer:
(1) Ethanol When ethyl bromide (bromoethane) is refluxed with aqueous potassium hydroxide, ethyl alcohol is formed. The reaction is called a hydrolysis reaction.

(2) Isopropyl alcohol : When isopropyl bromide (2-bromopropane) is boiled with aqueous potassium hydroxide, isopropyl alcohol is formed.

(3) Tert-butyl alcohol : When tert-butyl chloride is refluxed with aqueous potassium hydroxide, tert-butyl alcohol is formed.

(4) Methyl alcohol : When methyl bromide (bromomethane) is heated with aq KOH, it is hydrolysed to methyl alcohol (methanol).

(5) Butan-2-ol : When 2-Chlorobutane is boiled with aqueous KOH, Butan-2-ol is obtained.

Question 29.
How are following compounds prepared from alkyl halides using moist silver oxide?
(1) Ethanol
(2) Propan-2-ol.
Answer: ‘
(1) Bromoethane (C₂H₅Br) when boiled with moist Ag₂O undergoes hydrolysis and forms C₂H₅OH.

(2) When 2-chloropropane is boiled with moist Ag20, propan-2-ol is formed.

Question 30.
What is hydration of alkenes or olefins? How is it carried out? Explain with an example.
OR
How are alcohols prepared from alkenes?
Answer:
The addition of a water molecule across the double bond in an alkene is called hydration of alkenes or olefins.

Hydration does not take place directly. It is carried out by passing an alkene through cold and concentrated H₂SO₄ which forms deliquescent solid, alkyl hydrogen sulphate, which when boiled with water forms an alcohol.

Question 31.
How are the following compounds obtained by hydration of alkenes :
(1) Ethyl alcohol
(2) Isopropyl alcohol
(3) Tert-butyl alcohol?
Answer:
(1) When ethene is passed through cold 98 % H₂SO₄, ethyl hydrogen sulphate is formed, which on heating with water gives ethanol.

(2) Propene with cold 80% H₂SO₄ gives isopropyl hydrogen sulphate which further on boiling with water gives isopropyl alcohol.

(3) 2-Methylpropene (isobutylene) directly reacts with 50 % H₂SO₄ giving tert-butyl hydrogen sulphate, which when heated with water gives tert-butyl alcohol.

Question 32.
Identify C in the following reaction.

Answer:

Question 33.
Explain hydroboration-oxidation of alkene.
Answer:
When diborane is treated with alkene in the presence of tetrahydrofuran (THF) solvent, an addition product trialkyl borane is formed. Trialkyl borane is then oxidised with alkaline peroxide forms primary alcohol.

The addition of diborane to the double bond takes place in such a way that the boron gots attached to the less substituted carbon. The overall reaction gives Anti-Markovnikov’s product from unsymmetrical alkenes.

Question 34.
How is propan-l-ol prepared using diborane?
Answer:
When diborane is treated with propene, in the presence of THF an addition product tripropyl borane is formed. Tripropyl borane is then oxidised to propan-l-ol using hydrogen peroxide in the presence of dil NaOH.

The addition of diborane to the double bond takes place in such a way that the boron gets attached to the less substituted carbon. The alcohol formed by the addition of water to the alkene in a way opposite to the Markovnikov’s rule.

Question 35.
How is ethanol prepared using diborane?
Answer:
When diborane is treated with ethene in the presence of THF an addition product triethylborane is formed.Triethylborane is then oxidised with hydrogen peroxide to form ethyl alcohol.

Question 36.
Predict the major product when 2-methylbut-2-ene is converted into an alcohol in each of the following methods :
(1) acid catalysed hydration
(2) hydroboration by BH₃ – THF complex.
Answer:
(1) Acid catalysed hydration :

(2) Hydroboration by BH₃ – THF complex :

Question 37.
How are alcohols prepared from aldehydes and ketones?
OR
How are the following compounds obtained using Ni as catalyst and at high temperature
(1) Ethanol
(2) Propan-2-ol?
Answer:
Aldehyde and ketones are carbonyl compounds containing a carbonyl group  C = O. The reduction of the carbonyl group gives an alcohol.

(1) Primary alcohols are prepared by the reduction of aldehydes.

Example : When acetaldehyde is reduced with hydrogen in the presence of nickel as catalyst and at high temperature, ethyl alcohol (ethanol) is obtained.

(2) Secondary alcohols are prepared by the reduction of ketones.

Example : When acetone is reduced with hydrogen in the presence of nickel as catalyst and at high temperature, isopropyl alcohol (propan-2-ol) is obtained.

Question 38.
How will you convert carboxylic acids and esters to primary alcohols? Explain with suitable examples.
Answer:
Carboxylic acids and esters are not easily reduced by catalytic hydrogenation or by NaBH₄. However, relatively more reactive and an expensive LiAlH₄ is used to convert carboxylic acids and esters to primary alcohols. When acetic acid is reduced in the presence of LiAlH₄ and followed by their acid hydrolysis, ethyl alcohol is obtained.

When ethyl acetate is reduced in the presence of LiA1H₄ and followed by their acid hydrolysis. n-propyl alcohol and ethyl alcohol is obtained.

[Since LiA1H₄ is an expensive reagent. commercially acids are reduced to alcohol by converting them to esters, followed by their reduction. (catalytic hydrogenation)

Question 39.
How is Crotoflyl alcohol obtained from crotonaldehyde?
Answer:
When crotonaldehyde is reduced in the presence of lithium aluminium hydride, the produc obtained is hydrolysed to give crotonyl alcohol. Here. LiA1H₄ does not reduce carbon-carbon double bond.

Question 40.
WrIte the structure of aldehyde that yields

Answer:
The structure of aldehyde:

Question 41.
How are the following compounds prepared using Grignard reagent
(1) Ethanol
(2) Propan-l-ol
(3) Propan-2-ol
(4) 2-Methyl propan-2-ol?|
Answer:
(1) Ethanol : Formaldehyde on reaction with Grignard reagent, CH₃ – Mg – I in dry ether forms a complex which on further hydrolysis with dilute HCl forms ethanol.

(2) Propanol-l-ol : Formaldehyde on reaction with Grignard reagent, C₂H₅ – Mg – I in dry ether forms a complex which on further hydrolysis with dilute HC1 forms Propan-l-ol.

(3) Propan-2-ol : An acetaldehyde on reaction with Grignard reagent in dry ether forms a complex which on further hydrolysis with dilute acid HC1, forms propan-2-ol.

(4) 2-Methyl propan-2-ol : Acetone on reaction with Grignard reagent in dry ether forms a complex which on further hydrolysis with dilute acid HCl, forms a tertiary butyl alcohol.

Question 42.
Give a mechanism of following reaction.

Answer:
Grignand reagent reacts with aldehyde or ketone to form an adduct which on hydrolysis with dil. acid gives the corresponding alcohol.

In the first step, the nucleophilic addition of Grigard reagent to the carbonyl group resulting in the formation of an adduct, which on hydrolysis yields an alcohol.

Question 43.
Write the structure of carbonyl compounds that can be converted by reduction methods into following alcohols :


Answer:

Answer:

Question 44.
Using Grignard reagent, suggest synthesis of following alcohols from aldehydes or ketones. Wherever possible, suggest more than one combination.

Answer:

(b) Synthesis of propan- 1-01:

(c) Synthesis of butan-2-ol:

(d) Synthesis of 2-methylhexan-2-oI:

Question 45.
How will you obtain butan-2-ol from
(1) Propanal
(2) butan-2-one
(3) but-2-ene?
Answer:
(1) Propanal : When propanal is treated with methyl magnesium iodide in the presence of dry ether, a complex is formed, which on acid hydrolysis butan 2-ol is obtained.

(2) Butan-2-one: When butan-2-one is hydrogenated at 413 K in the presence of catalyst finely divided nickel butan-2-ol is obtained.

(3) But-2-ene : When but-2-ene is passed through cold concentrated sulphuric acid, isobutyl hydrogen sulphate is formed. Isobutyl hydrogen sulphate on heating with water gives butan-2-ol.

Question 46.
Write the structure of aldehyde, carboxylic acid and ester that yield the following alcohol.

Answer:

Question 47.
How will you prepare?
(1) 2-Ntethylbutan-1-oI from an alkene.
Answer:

(2) CycIoheyImethanoI from a Grignard reagent.
Answer:

(3) 1-Phenyl ethanol from acelaldehyde.
Answer:

Question 48.
How are following conversions brought about?
(1) Benzyl chloride to benzyl alcohol. (NCERT)
OR
How is benzyl alcohol prepared from benzyl chloride?
Answer:

(2) Benzyl alcohol to Benzoic acid.
Answer:

(3) 1-Ethyl cyclohexanol from cyclohexanone.
Answer:

Question 49.
How is phenol (carbolic acid) prepared from chlorobenzene (Dow’s process)? OR
Write chemical reaction for the preparation of phenol from chlorobenzene.
Answer:
Preparation of phenol from chlorobenzene (Dow’s process) : Chlorobenzene is fused with NaOH at about 623 K under a pressure of about 150 atmospheres (1.5 x 10⁷ Nm^-2), when sodium phenoxide is formed. Sodium phenoxide is acidified with dil.HCl to obtain phenol.

Question 51.
How is phenol (carbolic acid) prepared from benzene sulphonic acid?
Answer:
Preparation of phenol from benzene sulphonic acid : Benzene sulphonic acid is neutralized with the requisite quantity of soda ash (Na₂CO₃) or NaOH and the solution is evaporated to obtain sodium benzene sulphonate salt. Dry sodium benzene sulphonate is fused with an excess of caustic soda (NaOH) at about 573 K when sodium phenoxide is formed. The fused mass of sodium phenoxide on treatment with dilute HC1 gives phenol.

Question 52.
How is phenol (carbolic acid) prepared from aniline (diazotization)?
OR
How is carbolic acid prepared from amino benzene?
Answer:
Preparation of phenol from aniline (diazotization) : When aniline is treated with sodium nitrite and hydrochloric acid (NaNO₂ + HC1) at low temperature (0°C – 5°C), benzene diazonium chloride is formed. This reaction is called diazotization. An aqueous solution of benzene diazonium chloride on warming with water or dil. H2S04 gives phenol.

Question 54.
Describe the physical properties of alcohols and phenols.
Answer:
The properties of alcohols and phenols are mainly due to the hydroxyl group.

(1) Nature of intermolecular forces : Due to presence of – OH groups, alcohols and phenols are polar molecules. The polar – OH groups are held together by the strong intermolecular forces i.e. hydrogen bonding.

(2) Physical State : Lower alcohols are colourless, toxic liquids having characteristic alcoholic odour. Pure phenol is colourless, toxic, low melting solid having characteristic carbolic or phenolic odour.

(3) Boiling Points : The boiling points of alcohols and phenols increase with increase in their molecular mass.
Methyl alcohol – 65 °C
Phenol -182° C
n-Butyl alcohol-118 °C
o-nitrophenol-217 °C

(4) Solubility : Solubility of alcohols and phenols in water due to their ability to form intermolecular hydrogen bonding.

Question 55.
Arrange the following compounds in order of their increasing boiling points.
Butan-2-ol, ethanol, pentan-l-ol, butan -l-ol, propan-l-ol, methanol.
Answer:
Methanol, ethanol, propan:l-ol, butan-2-ol, butan-1 -ol pentan-l-ol.

Question 56.
Explain the following :
(1) Ethanol has higher boiling point than ethane.
Answer:
(1) The hydroxyl group in alcohols is highly polar. The H-atom has partial positive charge and the oxygen atom has partial -ve charge. The hydroxyl group in ethanol is extensively hydrogen bonded.

(2) Large number of ethanol molecules associated together by intermolecular hydrogen bonding. The energy required to separate the molecules by breaking hydrogen bond into vapour state is higher. This results in increasing the boiling point of ethanol.

(3) On the other hand, in ethane (alkane) there is no hydrogen bonding between the molecules. The molecules of ethane are held together by weak van der Waals forces of attraction. Hence, these molecules can be easily separated. Thus, ethanol has higher boiling point than that of ethane.

(2) Methanol is more soluble in water than propan-l-ol.
Answer:
(1) Methanol being lower members of alcohols is more soluble in water, but as the size of an alkyl group or molecular weight of alcohol increases the solubility decreases.

(2) The solubility of methanol in water is due to polar characters of alcohols (R – O^-δ -H^+δ ) and water (H^+δ – O^-δ – H^+δ).

(3) The solubility of methanol is due to the formation of intermolecular hydrogen bonding between polar molecules of methyl alcohol and water. Hence, methyl alcohol is a associated liquid.

(4) In methyl alcohol, size of methyl group being very small, -OH group constitutes major part of the molecule giving more solubility. As size of alkyl group increases, the non-polar character increases the solubility decreases. Hence, methanol is more soluble in water than propan-l-ol.

Question 57.
Which of the following pair is more acidic and why?

Answer:
Due to high electronegativity of sp²-hybridized carbon, electron density on oxygen in phenol (I) decreases. This increases the polarity of O – H bond and results in more ionization of phenol than that of cyclohexyl alcohol (II). Therefore, phenol is more acidic than cyclohexyl alcohol.

Answer:
In p-nitrophenol, nitro group (N02) is an electron-withdrawing group present at ortho position which enhances the acidic strength (- I effect). The O – H bond is under strain and release of proton (H⁺) becomes easily. Hence, o-nitrophenol is more acidic than phenol.

Question 58.
Draw intramolecular hydrogen bonding structures in the following compounds :
(a) o-nitrophenol
(b) o-hydroxy benzoic acid.
Answer:
(a) o-nitrophenol :

(b) o-hydroxy benzoic acid :

Question 59.
Explain laboratory test of alcohols and phenols.
Answer:
Laboratory test : Aqueous solution of alcohols and phenols can be tested with litmus paper. Aqueous solution of alcohols is neutral to litmus (neither blue nor red litmus change colour). Aqueous solutions of phenols turn blue litmus red. Thus, phenols have acidic character.

Question 62.
Write the action of aq NaOH on phenol and the product obtained is acidified.
Answer:
Phenols dissolve in aqueous NaOH by forming water soluble sodium phenoxide and are reprecipitated/ regenerated as phenols on acidification with HC1.

Question 63.
Explain the action of sodium on ethanol.
Answer:
When cthanol is treated with sodium metal, sodium ethoxide is formed and hydrogen gas is liberated.

Liberation of H₂ gas is used to detect the presence of alcoholic -OH group of a molecule.

Question 64.
How are the following compounds obtained from alcohols using HCI:
(1) C₂H₅CI
(2) Isopropyl chloride
(3) tert-butyl chloride?
Answer:
(1) Ethyl alcohol, C2H5OH in the presence of Lucas reagent (ZnCl₂ + HCI conc.) forms ethyl chloride.

(2) Isopropyl alcohol reacts with Lucas reagent forms isopropyl chloride.

(3) Tert-butyl alcohol reacts with Lucas reagent forms tert-butyl chloride

Question 65.
What is Lucas reagent? What are its uses?
Answer:
Lucas reagent is composed of a mixture of concentrated HCl and Lewis acid. anhydrous ZnCI₂.

It is used to prepare alkyl chlorides and distinguish between 10, 2° and 3° alcohols.

Question 66.
How can alcohols be distinguished with the help of Lucas reagent?
Answer:
Lucas reagent is a mixture of concentrated HCI and anhydrous ZnCl₂. It is used to distinguish between primary, secondary and tertiary alcohols.
Alcohol with Lucas reagent forms an alkyl chloride, R -Cl which is insoluble and gives cloudiness and forms separate layer.

The time required for cloudiness to appear is based on the type of the alcohol and its reactivity :

(1) Primary alcohol : It reacts with the Lucas reagent very slowly and on heating forms alkyl chloride. The cloudiness and separation of a layer takes place after a long time on heating.

(2) Secondary alcohol : It reacts with the Lucas reagent much faster to form alkyl chloride. The cloudiness and separation of layer takes place slowly.

(3) Tertiary alcohol : It reacts immediately with the Lucas reagent at room temperature to form alkyl chloride. The cloudiness and separation of layer takes place instantaneously.

Question 67.
What is esterificatlon? how Is an ester obtained from alcohol or phenol?
Answer:
(1) When an alcohol or phenol is heated with a carboxylic acid in the presence of conc.sulphuric acid an ester is obtained. The reaction is called esterification. This is reversible reaction and the formation of an ester is favoured
using excess of alcohol in the presence of conc. H₂SO₄.

(2) Alcohols and phenols react with acid anhydrides in presence of acid catalyst to form eser.

(3) The reaction of alcohol and phenols with acid chloride is carried out in the presence of pyridine (base), which neutralizes HCl.

Question 68.
Explain the action of the following on ethanol :
(1) Acetic acid
(2) Acetic anhydride
(3) Acetyl chloride.
Answer:
(1) Acetic acid : When ethanol is treated with acetic acid in the presence of cone, sulphuric acid, ethyl acetate (ester) is formed.

(2) Acetic anhydride : When ethanol is treated with acetic anhydride in the presence of cone, sulphuric acid, ethyl acetate (ester) is formed.

(3) Acetyl chloride : When ethanol is treated with acetyl chloride in the presence of pyridine, ethyl acetate (ester) is formed.

(Pyridine neutralises HC1 formed during reaction)

Question 69.
Explain the action of the following phenol :
(1) Acetic acid
(2) Acetic anhydride
(3) Acetyl chloride
Answer:
(1) Acetic acid : When phenol is treated with acetic acid in the presence of cone, sulphuric acid, (ester) is formed.

(2) Acetic anhydride : When phenol is treated with acetic anhydride in the presence of cone, sulphuric acid, (ester) is formed.

(3) Acetyl chloride : When phenol is treated with acetyl chloride in the presence of cone, sulphuric acid, (ester) is formed.

(Pyridine neutralizes HC1 formed during reaction.)

Question 70.
What is the action of acetic anhydride on salicyclic acid?
Answer:
When acetic anhydride is treated with salicyclic acid in presence of glacial acetic acid, acetyl salicyclic acid (aspirin) is obtained.

Aspirin is a common analgesic, antipyretic drug. Reactions involving breaking of C – O bond of alcohol :

Question 71.
How are the following compounds prepared by using HBr from corresponding alcohols :
(1) Ethyl bromide
(2) Isopropyl bromide
(3) Tert-butyl bromide (2-Methyl-propan-2-ol)?
Answer:
(1) When ethyl alcohol is heated with HBr, ethyl bromide is formed. (HBr is prepared in situ by adding NaBr to HCl or H₂SO₄)

(2) Isopropyl alcohol on heating with HBr forms isopropyl bromide.

(3) Tert-butyl alcohol on heating with HBr forms tert-butyl bromide.

Question 72.
Explain the action of hydroiodic acid on the following :
(1) Propan-2-ol
(2) 3-Methyl butan-2-ol.
Answer:
(1) Propan-2-ol : When propan-2-ol is heated with hydroiodic acid, 2-iodopropane is formed.

(2) 3-Methyl butan-2-ol : When 3-Methyl butan-2-ol is heated with hydroiodic acid, 2-Iodo-2-methyl butane is obtained. Here, secondary alcohol is converted into a tertiary alkyl halide.

Question 73.
Describe the action of PCl₃ on
(1) ethanol
(2) Propan-l-ol
(3) Propan-2-ol.
Answer:
(1) Ethanol : When ethanol is treated with PCl₃, ethyl chloride is obtained.

(2) Propan-l-ol : When propan-l-ol is treated with PCl₃, n-propyl chloride is obtained.

(3) Propan-2-ol : When Propan-2-ol is treated with PCl₃, 2-Chloropropane is obtained.

Question 74.
Describe the action of PCl₅ on
(1) ethanol
(2) propan-2-ol.
Answer:
(1) Ethanol : When ethanol is treated with PCl₅, ethyl chloride is obtained.

(2) Propan-2-ol : When propan-2-ol is treated with PCl₅, 2-chloropropane is obtained.

Question 75.
Describe the action of SOCl₂ on
(1) ethanol
(2) propan-l-ol.
Answer:
(1) Ethanol : When ethanol is treated with SOCl₂ in the presence of pyridine, ethyl chloride is obtained.

(2) Propan-l-ol : When propan-l-ol is treated with SOCl₂ in the presence of pyridine, w-propyl chloride is obtained.

Question 76.
Explain dehydration of alcohols.
OR
What is dehydration of alcohols? Give the chemical reactions showing dehydration of primary (1°), secondary (2°) and tertiary (3°) alcohols.
Answer:
Removal of water from an alcohol is called dehydration of alcohol. Alcohols having a /i-hydrogen is heated with dehydrating agents like concentrated H₂SO₄ (or H₃PO₄ or P₂O₅ or Al₂O₃). The ease of dehydration of alcohols is in the following order : tert-alcohol (3°) > secondary (2°) > primary (1°)
(1) Primary (1°) alcohol is dehydrated by heating it with 95% H₂SO₄ at 453 K.

(2) Secondary alcohol (2°) is dehydrated by heating with 60% H₂SO₄ at 373 K.

(3) A ternary alcohol can be easily dehydrated by heating with 20% H₂SO₄ at 363 K.

An alcohol can be dehydrated by passing vapours alcohols over heated alumina (Al₂O₃).

Question 77.
Explain oxidation of primary and secondary alcohols.
Answer:
(1) Primary alcohol on oxidation with CrO₃ forms aldehyde. However, a better reagent to bring about this oxidation is PCC (pyridinium chlorochromate).

(2) Secondary alcohol on oxidation with chromic anhydride (CrO₃) forms ketone.

Question 78.
What is the action of acidified K₂Cr₂O₇ on the following :

How are following alcohols distinguished :
(1) ethyl alcohol
(2) isopropyl alcohol
(3) tert-butyl alcohol?
OR
How will you distinguish primary, secondary and tertiary alcohols by oxidation process?
Answer:
Primary, secondary and tertiary alcohols are distinguished on the basis of their oxidation products, when their oxidation is carried out using K₂Cr₂O₇ and dil. H₂SO₄. Acidified potassium dichromate, K₂Cr₂O₇ is an oxidising agent.

(1) Ethyl alcohol is a primary alcohol and on oxidation, it first forms acetaldehyde, which on further oxidation forms acetic acid. In this both, aldehyde and acid have same number of carbon atoms.

(2) Isopropyl alcohol is a secondary alcohol and on oxidation it gives a ketone, acetone with the same number of carbon atoms. Acetone resists further oxidation as it involves breaking of C-C bond.

(3) The oxidation of tert-butyl alcohol is difficult, since it does not have a-hydrogen atom. It is oxidised by using acidic and stronger oxidising agents like KMn04, CrO₃ at high temperature, which dehydrate tertiary alcohol to alkene and then oxidise it to a ketone, acetone with less number of carbon atoms.

Question 79.
What happens when vapours of primary, secondary and tertiary alcohols are passed over heated copper at 573 K?
Answer:
When vapours of primary, secondary and tertiary alcohols are passed over heated copper at 573 K, dehydrogena¬tion of primary and secondary alcohol takes place while tertiary alcohols undergo dehydrogenation to given an alkene.

Primary alcohol :

Secondary alcohol :

Reactions of phenols :

Question 80.
(1) Explain the action of bromine in carbon disulphide (CS₂) on phenol (carbolic acid). OR Give equation of the reaction of bromine in CS₂ with phenol.
Answer:
When phenol is stirred at a low temperature with bromine dissolved in a polar solvent such as carbon disulphide or CCl₄ at (273 K), a mixture of o-bromophenol and p-bromophenol is formed, p-bromophenol is the major product.

Question 81.
Explain the action of bromine water on phenol (carbolic acid).
OR
Name the reagent used in the bromiriation of phenol to 2, 4, 6 tribromophenol.
Answer:
When phenol is treated with bromine water, a yellowish white precipitate of 2, 4, 6 – tribromophenol is formed.

Question 82.
Explain the action of dilute nitric acid on phenol (carbolic acid).
OR
Give equation of the reaction of dilute HNO₃ with phenol.
Answer:
When phenol is treated with dilute nitric acid, a mixture of o-nitrophenol and p-nitrophenol is formed. In this reaction, p-nitrophenol is formed as the major product.

Question 83.
Explain the action of cone, nitric acid (nitrating mixture) on phenol (carbolic acid).
OR
How is phenol converted into picric acid?
Answer:
When phenol is warmed with a mixture of cone, nitric acid and cone, sulphuric acid (a nitrating mixture or the mixed acid), 2, 4, 6 – trinitrophenol, commonly called picric acid, is formed.

Question 84.
Explain the action of concentrated sulphuric acid on phenol at different temperatures.
Answer:
(a) At room temperature : When phenol is treated with cone. H₂SO₄ at room temperature (about 300 K), o-phenol sulphonic acid is formed.

(b) At 373 K : When phenol is treated with cone. H₂SO₄ at about 373 K, p-phenol sulphonic acid is formed.

Question 85.
Explain (1) Kolbe’s reaction (2) Reimer-Tiemann reaction.
Answer:
(1) Kolbe’s reaction : When phenol reacts with sodium hydroxide, sodium phenoxide is obtained. Phenoxide ion being more reactive than phenol towards electrophilic substitution. Phenoxide undergoes electrophilic substitution with carbon dioxide at 398 K under pressure of 6 atm (a weak electrophile) forms salicylic acid as major product.

(2) Reimer-Tiemann reaction : Phenol is heated with chloroform along with aqueous NaOH, this is followed by acidification with dil. HC1 when salicyladehyde (2-hydroxy benzaldehyde) is formed as the major product, which can be separated from p-isomer by steam distillation. The stability of o-isomer is due to intramolecular hydrogen bonding.

Question 86.
Explain the action of zinc dust on phenol.
OR
How is phenol converted into benzene?
Answer:
When phenol is heated with zinc dust, benzene is obtained.

Question 87.
Explain the action of chromic acid on phenol.
OR
How is phenol converted into benzoquinone?
Answer:
When phenol is oxidised by chromic acid, a diketone, p-benzoquinone is formed. It is a conjugated diketone.

Question 88.
Explain catalytic hydrogenation of phenol.
Answer:
When a mixture of vapours of phenol and hydrogen is passed over nickel catalyst at 433 K., cyclohexanol is obtained.

Question 89.
How is diethyl ether (ethoxyethane) obtained from alcohol?
Answer:
When excess of ethyl alcohol is distilled with concentrated sulphuric acid (H₂SO₄) at 413 K, diethyl ether is formed.

Question 91.
Write a note on Williamson’s synthesis.
OR
How are ethers prepared from Alkyl halides?
OR
How are simple ethers and mixed ethers prepared by Williamson’s synthesis?
Answer:
Williamson’s synthesis : When an alkyl halide (R – X) is heated with sodium alkoxide (R – O – Na), an ether is obtained, this reaction is known as Williamson’s synthesis. This method is used to prepare simple (or symmetrical) ethers and mixed (unsymmetrical) ethers.

Sodium alkoxide is obtained by a reaction of sodium with an alcohol.

(A) Simple (Symmetrical) ether : When an alkyl halide and sodium alkoxide having similar alkyl groups are heated, symmetrical ether is obtained.

Sodium ethoxide on heating with ethyl bromide gives diethyl ether.

(B) Mixed (Unsymmetrical) ether : When an alkyl halide and sodium alkoxide or sodium phenoxide having different alkyl groups are heated, unsymmetrical ether (dialkyl ethers or alkyl aryl ether) is obtained.

Sodium ethoxide on heating with methyl bromide gives ethyl methyl ether.

Sodium phenoxide on heating with ethyl bromide gives ethyl phenyl ether.

Question 92.
How is anisole obtained from phenol?
OR
How is methoxy benzene prepared from carbolic acid?
Answer:
Phenol reacts with sodium hydroxide, sodium phenoxide is formed. When sodium phenoxide is heated with methyl iodide, anisole is obtained.

Question 93.
Explain the limitations for the preparation of unsymmetrical ethers.
OR
What care is to be taken in the preparation of unsymmetrical ethers by Williamson’s synthesis? Explain.
OR
Illustrate with examples the limitations of Williamson’s synthesis for the preparation of certain types of ethers.
Answer:
(1) In the preparation of unsymmetrical ethers by Williamson’s synthesis, the proper choice of the reactants namely alkyl halide and sodium alkoxide is necessary.

(2) The best yield of unsymmetrical ether is obtained when primary alkyl halide and tertiary alkoxide are heated, since primary alkyl halides are more susceptible to SN² reaction.

(3) Secondary or tertiary alkyl halide undergo a and fi (halogen and hydrogen) elimination reaction giving an alkene instead of an ether since a carbon atom is sterically hindered by bulky alkyl groups.

(4) For example : t-butyl methyl ether can be synthesised by reaction of methyl bromide with sodium t-butoxide.

Question 94.
State physical properties of ethers.
Answer:
(1) Dimethyl ether and ethyl methyl ether are gases. Other ethers are colourless liquids with pleasant odour.
(2) Lower ethers are highly volatile and highly inflammable substances.
(3) Boiling points of ethers show gradual increase with the increase in molecular mass.

(4) The solubility/miscibility of ethers in water is similar to that of alcohols of comparable molecular mass.

Question 95.
Explain, ethers posses a small net dipole moment.
Answer:
In ethers, bond angle is 110° and not 180°, bond dipole moments of the two C – O bonds do not cancel each other, therefore, ethers possess a smal net dipole moment, (for example, dipole moment of diethyl ether is 1.18 D)

Question 96.
Explain, the solubility/miscibility of ethers in water is similar to that of alcohols of comparable molecular mass.
Answer:
The solubility/miscibility of ethers in water is similar to that of alcohols of comparable molecular mass. This is because ethers can form hydrogen bonds with water through ethereal oxygen.

For example, diethyl ether and n-butyl alcohol have respective miscibilities of 7.5 and 9 g per 100 g of water.

Question 97.
Explain laboratory test for ethers.
Answer:
Ethers are neutral compounds in aqueous medium. Ethers do not react with bases, cold dilute acids, reducing agents, oxidizing agents and active metals. However, ethers dissolve in cold concentrated H₂SO₄ due to formation of oxonium salts.

This property distinguishes ethers from hydrocarbons.

Question 98.
What is the action of atmospheric oxygen on diethyl ether?
Answer:
When atmospheric oxygen combines with diethyl ether, peroxide of diethyl ether is obtained.

Question 99.
What is the action of dilute sulphuric acid on
(1) Dimethyl ether
(2) Diethyl ether
(3) Ethyl methyl ether?
(4) Anisole?
OR
Explain hydrolysis of ethers.
Answer:
Simple ethers on heating with dilute sulphuric acid under pressure undergoes hydrolysis to give alcohol.

(1) Dimethyl hydrolysis on hydrolysis give methanol.

(2) Diethyl ether on hydrolysis give ethanol.

(3) A mixed ether on heating with dii. H₂SO₄ under pressure undergoes hydrolysis to give mixture of two different alcohols.

Ethyl methyl ether on hydrolysis give a mixture of ethanol and methanol.

(4) Anisole on hydrolysis give a mixture of phenol and methanol.

Question 100.
State the combustion products of diethyl ether.
Answer:
The combustion products of diethyl ether are CO₂ and H₂O.

Question 101.
What is the action of phosphorus pentachloride on
(1) Diethyl ether
(2) Ethyl methyl ether
(3) Methyl phenyl ether (anisole)?
Answer:
(1) Diethyl ether : When diethyl ether is heated with ethyl methyl ether, ethyl chloride is formed.

(2) Ethyl methyl ether : When ethyl methyl ether is heated with phosphorus pentachloride, a mixture of ethyl chloride and methyl chloride is formed.

(3) Methyl phenyl ether (Anisole) : When methyl phenyl ether is heated with phosphorus pentachloride, a mixture of methyl chloride, chlorobenzene is formed.

Question 102.
Describe the action of hot concentrated HI on
(i) Dialkyl ether
(ii) Alkyl aryl ether.
Answer:
Ether reacts with excess of hot concentrated hydrogen halide to give two alkyl halide molecules.
R – O – R + HX → RX + R – OH
R – OH + HX → RX + H₂0
Alkyl aryl ether reacts with hot concentrated hydrogen halide to give phenol and alkyl halide.

Ethers with two different alkyl groups react with hot.conc. HI to give alkyl halides.
R – O – R’+ 2H – X → R – X + R’ – X + H₂O
The order of reactivity of HX is HI > HBr > HC1

Question 103.
What is the action of hot HI on isopropyl methyl ether?
Answer:
When isopropyl methyl ether is treated with excess of hot hydroiodic acid, a mixture of isopropyl iodide and methyl iodide is formed.

Question 104.
Describe the action of hydroiodic acid on the following :
(1) Diethyl ether.
Answer:
When diethyl ether (ethoxy ethane) is treated with hydroiodic acid, a mixture of ethanol and ethyl iodide is formed.

If excess of hydroiodic acid is available, then ethyl alcohol further reacts with hydroiodic acid at higher temperature to form ethyl iodide and water.

(2) Ethyl methyl ether.
Answer:
When ethyl methyl (methoxy ethane) is treated with hydroiodic acid, a mixture of ethyl alcohol and methyl iodide is formed.

If excess of hydroiodic acid is available, then ethyl alcohol further reacts with hydroiodic acid at higher temperature to form ethyl iodide and water.

(3) Methyl n-propyl ether.
Answer:
When methyl n-propyl ether (1-Methoxy propane) is treated with hydroiodic acid, a mixture of n-propyl alcohol and methyl iodide is formed.

If excess of hydroiodic acid is available then n-propyl alcohol further reacts with hydroiodic acid at higher temperature to form n-propyl iodide and water.

(4) Methyl phenyl ether (anisole).
Answer:
When methyl phenyl ether (anisole) is treated with hydroiodic acid, phenol and methyl iodide is formed. Here, phenol does not react further with HI because – OH group is attached to sp²-hybridised carbon atom and it cannot be replaced by iodide (nucleophile).

Question 105.
Draw the resonance structures of aromatic ethers.
Answer:
The alkoxy group in aromatic ether is a ring activating and ortho-, paradirecting group toward electrophilic aromatic substitution.

Resonance structures :

+ R Effect of – OR group results in increased electron density at the para- and two ortho-posotions (see resonance structures II, III and IV).

Question 106.
Describe the action of bromine in acetic acid on anisole.
OR
Write the equation of the reaction of bromination of anisole in ethanoic acid medium.
Answer:
When anisole is treated with bromine in acetic acid, /i-bromoanisole (major product) is obtained.

Question 107.
Describe the action of methyl chloride on anisole (Friedel-Crafts reaction).
OR
Write the equation of Friedel-Crafts reaction-alkylation of anisole.
Answer:
When anisole is treated with alkyl halide in the presence of anhydrous aluminium chloride (a Lewis acid) as catalyst, 4-Methoxy toluene is formed as major product. The alkyl groups are introduced at -ortho and -para positions in anisole, the reaction is known as Friedel-Crafts alkylation reaction.

Question 108.
Describe the action of Acetyl chloride on anisole (Friedel-Crafts acylation).
OR
Write the equation of the reaction Friedel-Crafts acylation of anisole.
OR
Write a note on Friedel-Crafts acylation.
Answer:
When anisole is treated with acetyl chloride in the presence of anhydrous aluminium chloride (a Lewis acid), 4-Methoxy acetophenone (major product) is obtained. The acetyl groups are introduced at -ortho and -para positions in anisole, the reaction is known as Friedel Craft’s acylation reaction.

Question 109.
Describe the action of cone. HNO₃ on anisole.
OR
Write the equation of nitration of anisole.
Answer:
When anisole is reacted with nitrating mixture (cone. HNO₃ + cone. Fl₂SO₄), a mixture of p-nitroanisole and o-nitroanisole is obtained.

Question 110.
An organic compound with the formula C₄H₁₀O₃ shows properties of ether and alcohol. When treated with an excess of HBr yields only one compound 1,2 dibromomethane. Write structural formula of ether and that of alcohol.
Answer:
When C₄H₁₀O₃ is treated with excess of HBr, a single compound 1,2-dibromoethane is formed.

Question 111.
An organic compound ‘A’ having molecular formula C₄H₁₀O does not react with sodium metal. On hydrolysis with dilute H₂SO₄ it gives only one organic compound ‘B’ The compound B on heating with red phosphorus and iodine gives compound ‘C’. The compound C can also be obtained from compound ‘A’ on heating with excess HI. Identify the compounds A, B and C.
Answer:
(1) The organic compound ‘A’ with molecular formula C₄H₁₀O may be an alcohol or ether.

(2) Since ‘A’ does not react with sodium (Na), it is not an alcohol. Hence ‘A’ may be an ether.
(3) Since an ether ‘A’ on hydrolysis with dilute H₂SO₄ gives only one compound ‘B’, the compound ‘A’ must be a symmetrical (simple) ether. Hence compound ‘A’ may be, C₂H₅ – O – C₂H₅

Question 112.
How will you affect the following two-step conversions? Diethyl ether into n-butane :
Answer:

Question 114.
Answer in one sentence/word.

(1) Name the alcohol that is used to make propan-2-one.
Answer:
The alcohol used to make propan-2-one is iso-propyl alcohol.

(2) Which is the first oxidation product of secondary alcohol?
Answer:
The first oxidation product of secondary alcohol is ketone.

(3) Name the alcohol that is used to make acetic acid.
Answer:
The alcohol that is used to make acetic acid is ethyl alcohol.

(4) Write the structure of cyclohexane-1, 4-diol.
Answer:

(9) Which of the following isomers is more volatile : o-nitrophenol or p-nitrophenol?
Answer:
The isomer o-nitrophenol with lower boiling point is more volatile.

(10) Identify the product of the following reaction.

Answer:

(11) Identify the product obtained by industrial synthesis of carboxylation of phenoxide ion followed by acidification.
Answer:
The product is phenol

(12) What is the product A obtained in the following reaction?

Answer:
The product is phenol

(13) Which positions are occupied by – NO₂ group during nitration of carbolic acid?
Answer:
– o – (ortho) and – p – (para) positions are occupied by – NO₂ group during nitration of carbolic acid.

(14) Write the name of reactants used for the preparation of ethyl-tert-butyl ether.
Answer:

are used for the preparation of ethyl-tert-butyl ether.

(15) The product formed in the reaction of reverse of dehydration of alcohol is
Answer:
The product formed in the reaction of reverse of dehydration of alcohol is alkene.

(16) Which rule is obeyed by hydroboration oxidation process?
Answer:
The rule obeyed by hydroboration oxidation process is opposite to the Markovnikov’s rule.

(17) Name the reagents for the complete hydroboration-oxidation reaction in step 1 and step 2.
Answer:
Step 1 : Diborane
Step 2 : Hydrogen peroxide and dil. NaOH.

(18) Write the name of the test by which methanol can be distinguished from ethanol.
Answer:
Iodoform test by which methanol can be distinguished from ethanol.

(19) Write the name of reactant used for preparation of phenol, which gives byproduct used as solvent.
Answer:
Reactant used in the preparation of phenol :

(20) Ether is a good solvent for Grignard reagent. Which property makes it a good solvent?
Answer:
Ether has a low polarity, this property makes it a good solvent.

(21) The C – O – C bond angle in dimethyl ether is
Answer:
The C – O – C bond angle in dimethyl ether is 110°.

Question 115.
State the uses of methyl alcohol.
Answer:
(1) Methyl alcohol is used as an industrial solvent for dissolving oils, fats, gums, etc.
(2) It is used for dry cleaning and preparation of perfumes and varnishes.
(3) It is used as antifreeze agent for automobile radiators at low temperature.
(4) It is used in the preparation of methyl chloride, dimethyl sulphate and formaldehyde.
(5) It is used to denature ethyl alcohol.

Question 116.
State the uses of ethyl alcohol. OR Write two uses of ethyl alcohol.
Answer:
(1) Ethyl alcohol is used as solvent for dyes, oils, perfumes, cosmetics and drugs.
(2) A mixture of 10-20% ethyl alcohol with petrol is used as motor fuel.
(3) A mixture of ethyl alcohol and calcium acetate in gel form is used as solid fuel.
(4) It is widely used in beverages.
(5) Since ethyl alcohol has low freezing point, it is used in thermometer.
(6) It is an effective tropical antiseptic therefore it is used in many mouth washes.
(7) It kills micro-organisms on wound surface and in the mouth but its low toxicity does not kill the cells of the skin or mouth tissues.
(8) It is used in the preparation of chloroform, iodoform, acetic acid and ethers.
(9) It is used as fuel.

Question 117.
Give the important uses of phenol.
OR
Write two uses of phenol.
Answer:
(1) Phenol is used in the preparation of phenol-formaldehyde polymer which is used in a plastic bakelite.
(2) It is used in the preparation of phenol-phthalein-an indicator and in certain dyes.
(3) It is used in the preparation of drugs such as salol, aspirin, etc.
(4) It is used in the preparation of dettol, which is an antiseptic.
(5) It is used in the preparation of 2,4-dichlorophenoxy acetic acid which is used as selective weed killer.
(6) It is used to prepare picric acid which is used as explosive.

Multiple Choice Questions

Question 118.
Select and write the most appropriate answer from the given alternatives for each sub-question:

1. Which one of the following is a tertiary alcohol?
(a) Pentan-l-ol
(b) Pentan-2-ol
(c) 2-Methylpentan-2-ol
(d) 3-Methylpentan-2-ol
Answer:
(c) 2-Methylpentan-2-ol

2. Which of the following is a primary alcohol?
(a) 3-ethyl-3-hexanol
(b) 2-butanol
(c) 3-methyl-l-butanol
(d) 1-hexanol
Answer:
(d) 1-hexanol

3. The molecular formula C₄H₁₀O represents
(a) aldehydes
(b) alcohols
(c) ethers
(d) both (b) and (c)
Answer:
(d) both (b) and (c)

4. The general formula of primary, secondary and tertiary alcohols is
(a) C_nH_2nOH
(b) C_nH_2n-1OH
(c) C_nH_2n+1OH
(d) C_nH_n+1OH
Answer:
(c) C_nH_2n+1OH

5. Which of the following alcohols cannot be prepared by hydration of the corresponding alkene?
(a) Ethanol
(b) Propan- l-ol
(c) Propan-2-ol
(d) 2-Methylpropan-2-ol
Answer:
(b) Propan- l-ol

6. Which of the following compounds when treated with CH₃MgI in dry ether followed by the hydroly¬sis, will give Propan-2-ol?
(a) HCHO
(b) CH₃CHO
(c) CH₃CH₂OH
(d) CH₃COCH₃
Answer:
(b) CH₃CHO

7. To prepare 3-Ethylpentan-3-ol, the reagents needed are
(a) CH₃CH₂MgBr + CH₃COCH₂CH₃
(b) CH₃MgBr + CH₃CH₂CH₂COCH₂CH₃
(c) CH₃CH₂MgBr + CH₃CH₂COCH₂CH₃
(d) CH₃CH₂CH₂MgBr + CH₃COCH₂CH₃
Answer:
(c) CH₃CH₂MgBr + CH₃CH₂COCH₂CH₃

8. How is 1-propanol obtained?
(a) Using propanal
(b) Using propanone
(c) Using propene
(d) All of these
Answer:
(a) Using propanal

9.

Answer:
(a)

10. Ketone on reduction gives
(a) 1° alcohol
(b) 2° alcohol
(c) 3° alcohol
(d) all of these
Answer:
(b) 2° alcohol

11. Primary alcohols are prepared by catalytic hydro-genation of aldehydes in presence of ……………………………… as a catalyst
(a) aluminium bromide
(b) fluoroboric acid
(c) dry ether
(d) palladium
Answer:
(d) palladium

12. Lower member of alcohols are
(a) insoluble in water
(b) soluble in water
(c) insoluble in acetaldehyde
(d) insoluble in petrol
Answer:
(b) soluble in water

13. Which of the following compounds contain hydro¬gen bonds?
(a) Ethane
(b) Ethanol
(c) Methoxymethane
(d) Ethylene
Answer:
(b) Ethanol

14. Which of the following is a trihydric alcohol?
(a) n-propyl alcohol
(b) Glycerol
(c) Glycol
(d) Glycine
Answer:
(b) Glycerol

15. Aldehydes are first oxidation products of
(a) primary alcohols
(b) secondary alcohols
(c) tertiary alcohols
(d) carboxylic acids
Answer:
(a) primary alcohols

16. The oxidation product of alcohol depends on
(a) – OH group of an alcohol
(b) number of carbon atoms in alcohol
(c) number of hydrogen atoms attached to hydroxyl bearing carbon
(d) all of these
Answer:
(c) number of hydrogen atoms attached to hydroxyl bearing carbon

17. The order of ease of oxidation is
(a) primary > secondary > tertiary
(b) primary < secondary < tertiary
(c) primary > tertiary > secondary
(d) secondary > tertiary > primary
Answer:
(b) primary < secondary < tertiary

18. Ethyl alcohol \(\frac{\text { conc. } \mathrm{H}_{2} \mathrm{SO}_{4}}{443 \mathrm{~K}} \mathrm{~A} \stackrel{\mathrm{HBr}}{\longrightarrow}\) Ethyl bromide. Identify A.
(a) Ethyl hydrogen sulphate
(b) Ethylene
(c) Isopropyl hydrogen sulphate
(d) Acetic acid
Answer:
(b) Ethylene

19. One mole of PCl₅ reacts with one mole of ethyl alcohol to give
(a) 1 mole Cl₂
(b) 1/2 mole Cl₂
(c) 1 mole HCl
(d) 1/2 mole HCl
Answer:
(c) 1 mole HCl

20. When ethyl alcohol is reacted with sodium metal and the compound so formed is treated with ethyl bromide, the product formed is
(a) an acid
(b) an alkane
(c) an ether
(d) an ester
Answer:
(c) an ether

21. Dehydration occurs at the lowest temperature and concentration for
(a) methyl alcohol
(b) n-propyl alcohol
(c) iso-propyl alcohol
(d) tert-butyl alcohol
Answer:
(d) tert-butyl alcohol

22. One mole of sodium when reacts with one mole of methyl alcohol, gives
(a) one mole of oxygen
(b) one mole of hydrogen
(c) half mole of hydrogen
(d) half mole of oxygen
Answer:
(c) half mole of hydrogen

23. A compound ‘X’(C₃H₈O) on oxidation gives com-pound ‘ Y’(C₃H₆O₂), hence compound ‘X’ must be
(a) a ketone
(b) an aldehyde
(c) a primary alcohol
(d) a secondary alcohol
Answer:
(c) a primary alcohol

24. The toxicity of alcohols
(a) increases with an increase in their molecular weight
(b) decreases with an increase in their molecular weight
(c) increases with a decrease in their molecular weight
(d) does not depend on their molecular weight
Answer:
(a) increases with an increase in their molecular weight

25. The structural formula of 2-phenyl ethanol is

Answer:
(c)

26. Compound that fails to give effervescence with NaHCO₃ is
(a) C₆H₅COO
(b) CH₃COOH
(c) C₆H₅OH
(d) Picric acid
Answer:
(c)

27. Phenol is a bifunctional compound because
(a) it gives reactions of hydroxyl group (-OH) as well as aromatic ring
(b) it is a strong acid
(c) it is insoluble in NaOH
(d) it is readily soluble in water
Answer:
(a) it gives reactions of hydroxyl group (-OH) as well as aromatic ring

28. Phenol gives characteristic colour with
(a) iodine solution
(b) bromine water
(c) ammonium hydroxide
(d) aqueous ferric chloride solution
Answer:
(d) aqueous ferric chloride solution

29. Ethanol and phenol are distinguished from each other by the action of
(a) neutral ferrous chloride
(b) neutral ferric chloride
(c) ferric hydroxide
(d) ferrous hydroxide
Answer:
(b) neutral ferric chloride

30. Which of the following compounds is used to prepare bakelite resin?
(a) Acetaldehyde
(b) Ethanol
(c) Phenol
(d) Methyl amine
Answer:
(d) Methyl amine

31. Increasing order of acid strength among p-methoxy-phenol, p-methyl phenol and p-nitro- phenol is as
(a) p-nitrophenol > p-methoxyphenol > p-methylphenol
(b) p-methylphenol > p-methoxyphenol > p-nitrophenol
(c) p-nitrophenol > p-methylphenol > p-methoxyphenol
(d) p-methoxyphenol > p-methylphenol > p-nitrophenol
Answer:
(d) p-methoxyphenol > p-methylphenol > p-nitrophenol

32. The IUPAC name of the compound is
(a) 3,3-dimethyl-l-hydroxy cyclohexane
(b) l,l-dimethyl-3-cyclohexanol
(c) 3,3-dimethyl-1-cyclohexanol
(d) l,l-dimethyl-3-hydroxy cyclohexane
Answer:
(c) 3,3-dimethyl-1-cyclohexanol

33. Which of the following is the most reactive towards electrophilic attack?

Answer:
(a)

34. In the following reaction, Ethanol \(\stackrel{\mathrm{PBr}_{3}}{\longrightarrow} \mathrm{A}\)\(\stackrel{\text { Alc-KOH }}{\longrightarrow} \text { B } \frac{\text { (i) } \mathrm{H}_{2} \mathrm{SO}_{4} \text {, room temp. }}{\text { (ii) } \mathrm{H}_{2} \mathrm{O}, \text { heat }} \mathrm{C} \text {. }\). The product C is
(a) CH₃ – CH₂ – O – CH₂ – CH₃
(b) CH₃ – CH₂ – OSO₃H
(c) CH₃CH₂OH
(d) CH₂ = CH₂
Answer:
(c) CH₃CH₂OH

35. What is the general formula of ethers?
(a) C_nH_2nO
(b) C_nH_2n+2O
(c) C_nH_2n-1O
(d) C_nH_2n-2O
Answer:
(b) C_nH_2n+2O

36. The linkage is present in
(a) proteins
(b) ketones
(c) ethers
(d) aldehydes
Answer:
(c) ethers

37. Methoxy ethane is the functional isomer of
(a) CH₃CHOHCH₃
(b) CH₃CH₂CH₂OH
(c) CH₃ – O – CH₃
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

38. An oxygen atom in ether is
(a) sp²-hybridized
(b) sp³-hybridized
(c) sp-hybridized
(d) sp²-d-hybridized
Answer:
(b) sp³-hybridized

39. C₅H₁₂O represents
(a) only an alcohol
(b) only an aldehyde
(c) only an ether
(d) an alcohol and an ether
Answer:
(d) an alcohol and an ether

40. Ether molecules are
(a) tetrahedral
(b) angular
(c) pyramidal
(d) diagonal
Answer:
(b) angular

41. Which one of the following compounds is not isomeric with Ethoxypropane?
(a) 1-Methoxypropane
(b) 2-Methoxypropane
(c) 2-Methylpropane-2-ol
(d) 2-Methylbutan-2-ol
Answer:
(d) 2-Methylbutan-2-ol

42. The IUPAC name of C₂H₅ – O – CH₂ – CH(CH₃)₂ is
(a) 1-Ethoxy-1-butane
(b) 2-Ethoxy-2-butane
(c) l-Ethoxy-2-methylpropane
(d) 3-Ethoxy-2-methylpropane
Answer:
(c) l-Ethoxy-2-methylpropane

43. The IUPAC name of CH₃OC₆H₅ is
(a) methoxy phenyl ether
(b) phenoxy methane
(c) methoxy benzene
(d) methyl phenyl ether
Answer:
(c) methoxy benzene

44. In CH₃(CH₂)₃ – O -CH₃, the parent hydrocarbon of large alkyl group is
(a) n-butane
(b) butane
(c) pentane
(d) n-pentane
Answer:
(a) n-butane

45. Which one of the following alkyl halide gives best yield in Williamson’s synthesis?
(a) CH₃ – Br
(b) CH₃ – CH(Br) – CH₃
(c) CH₂ – CH(Br) – CH₃ – CH₃
(d) (CH₃)₃ C – Br
Answer:
(a) CH₃ – Br

46. Which one of the following ethers cannot be pre¬pared by using diazomethane?
(a) Dimethyl ether
(b) Diethyl ether
(c) Ethyl methyl ether
(d) t-Butyl methyl ether
Answer:
(b) Diethyl ether

47. The continuous etherification is carried out at
(a) 473 K
(b) 413 K
(c) 498 K
(d) 403 K
Answer:
(b) 413 K

48. Williamson’s synthesis is used for the preparation of
(a) only unsymmetrical ethers
(b) only symmetrical ethers
(c) both symmetrical and unsymmetrical ethers
(d) only methyl ethers
Answer:
(c) both symmetrical and unsymmetrical ethers

49. Which of the following ethers on hydrolysis gives two different products that are successive members of a homologous series?
(a) Methoxymethane
(b) Ethoxythane
(c) Methoxyethane
(d) 2-Methoxypropane
Answer:
(c) Methoxyethane

50. Which one of the following compounds dissolves in hot dilute sulphuric acid but does not react with sodium metal?
(a) Ethyl bromide
(b) Acetic acid
(c) Ethyl alcohol
(d) Diethyl ether
Answer:
(d) Diethyl ether

51. Identify ‘A’ in the following reaction :

Answer:
(c)

52. The C-O-C bond angle in an ether is
(a) 180
(b) 90°
(c) 104.5°
(d) 109.5°
Answer:
(d) 109.5°

53. Ether free from moisture and alcohol is known as ,
(a) dry ether
(b) absolute ether
(c) pure ether
(d) spirit ether
Answer:
(b) absolute ether

54. The geometry of ether is similar to
(a) ammonia
(b) methane
(c) water
(d) ethyne
Answer:
(c) water

55. Sodium metal does not react with
Answer:
C₂H₅ – O – C₂H₃

56. Ethers have boiling points
(a) lower than those of alkanes of comparable molecular masses
(b) higher than those of isomeric alcohols
(c) lower than those of isomeric alcohols
(d) higher than those of alkanes of comparable molecular masses
Answer:

57. Ether on hydrolysis gives
(a) aldehyde
(b) alcohol
(c) acid
(d) ester
Answer:
(c) acid

58. Which of the following reactions represent Williamson’s reaction?
(a) R – O – R’ + HI →
\(\text { (b) } \mathrm{R}^{\prime}-\mathrm{OH}+\stackrel{\mathrm{H}_{2} \mathrm{SO}_{4}}{\longrightarrow}\)

(d) R – O – Na + R’X →
Answer:
(c)

59. Which of the following reactions represent the continuous etherification process?

Answer:
(b)

60. Ethers on complete combustion produce
(a) an alcohol and water
(b) an alkene and water
(c) an alkane and an alkene
(d) carbon dioxide and water
Answer:
(d) carbon dioxide and water

61. Diethyl ether is used as a solvent in many organic reactions because it
(a) is liquid at room temperature
(b) has lower boiling point
(c) contains divalent oxygen atom
(d) is inert in nature
Answer:
(d) is inert in nature

62. Which of the following reagents can be prepared using ether as a solvent?
(a) Tollen’s reagent
(b) Grignard’s reagent
(c) Schiff’s reagent
(d) Millon’s reagent
Answer:
(b) Grignard’s reagent

63. Diethyl ether is used as
(a) a hypnotic
(b) an antiseptic
(c) an anaesthetic
(d) an antipyretic
Answer:
(c) an anaesthetic

64. To obtain fuel, diethyl ether is mixed with
(a) ester
(b) ethanal
(c) ethanol
(d) 2-propanone
Answer:
(c) ethanol

65. Which of the following alcohols is prepared by acid catalyzed hydration of alkenes?
(a) Butan-l-ol
(b) Propan-l-ol
(c) Ethanol
(d) Methanol
Answer:
(c) Ethanol

66. Which of the following alcohols can be prepared by direct hydration of corresponding alkene in presence of 50 % sulphuric acid?
(a) Butan-l-ol
(b) Butan-2-ol
(c) 2-Methylpropan-1 -ol
(d) 2-Methylpropan-2-ol
Answer:
(d) 2-Methylpropan-2-ol

67. Which of the following alcohols cannot be prepared by reduction of carbonyl compounds?
(a) Pentan-l-ol
(b) Pentan-2-ol
(c) 2-Methylpentan-2-ol
(d) 3-Methylpentan-2-ol
Answer:
(c) 2-Methylpentan-2-ol

68. Which of the following conversions explains the acidic nature of alcohols?

Answer:
(b)

69. Which of the following compounds gives 3-ethyl- pentan-3-ol by the action of ethyl magnesium iodide followed by acid hydrolysis?
(a) Propanone
(b) Butanone
(c) Pentan-2-one
(d) Pentan-3-one
Answer:
(d) Pentan-3-one

70. Benzyl phenyl ether reacts with hydrogen bromide to give
(a) benzyl bromide and phenol
(b) benzyl alcohol and bromobenzene
(c) benzyl bromide and bromobenzene
(d) benzyl alcohol and phenol
Answer:
(a) benzyl bromide and phenol

71. Ethers are considered as
(a) monoalkyl derivatives of water
(b) divalent oxygen atom is attached to two alkyl groups
(c) alkyl derivatives of fatty acids
(d) condensation products of acid and alcohol
Answer:
(b) divalent oxygen atom is attached to two alkyl groups

72. Which of the following compounds is not isomeric with ethoxyethane?
(a) 1-methoxypropane
(b) 2-methoxypropane
(c) 2-methylpropan-2-ol
(d) 2-methylbutan-2-ol
Answer:
(d) 2-methylbutan-2-ol

73. Which of the following compounds dissolves in hot dilute sulphuric acid but does not reacts with sodium metal?
(a) Ethyl bromide
(b) Acetic acid
(c) Ethyl alcohol
(d) Diethyl ether
Answer:
(d) Diethyl ether

74. Which of the following alcohol will have the fastest rate of dehydration?

Answer:
(c)

75. The phenol having lowest acidity is

Answer:
(b)

76. Which of the following reagents is best for the following conversion?

Answer:
(a)

77. 3-Methyl butane-2-ol on heating with HI gives
(a) 2 -iodo-3-methyl butane
(b) 2-iodo-2-methyl butane
(c) l-iodo-3-methyl butane
(d) l-iodo-2-methyl butane
Answer:
(b) 2-iodo-2-methyl butane

78. In phenol carbon atom attached to -OH group undergoes –
(a) sp³-hybridisation
(b) sp-hybridisation
(c) sp²-hybridisation
(d) No hybridisation
Answer:
(c) sp²-hybridisation

79. Which among the following reducing agents is ‘not’ used to reduce acetaldehyde to ethyl alcohol?
(a) Na-Hg and water
(b) Zn-Hg and cone. HCl
(c) H₂-Raney Ni
(d) Li-A1H₄/H⁺
Answer:
(b) Zn-Hg and cone. HCl

80. Identify the weakest acidic compound amongst the following :
(a) p-nitrophenol
(b) p-chlorophenol
(c) p-cresol
(d) p-aminophenol
Answer:
(d) p-aminophenol

81. Natalite is a mixture of
(a) diethyl ether and methanol
(b) diethyl ether and ethanol
(c) dimethyl ether and methanol
(d) dimethyl ether and ethanol
Answer:
(b) diethyl ether and ethanol

82. The alcohol used in thermometers is
(a) methanol
(b) ethanol
(c) propanol
(d) butanol
Answer:
(b) ethanol

83. Which of the following is the first oxidation product of secondary alcohol?
(a) Alkene
(b) Aldehyde
(c) Ketone
(d) Carboxylic acid
Answer:
(c) Ketone

84. When phenol is heated with cone. HNO₃ in presence of cone. H₂SO₄ it yields
(a) o-nitrophenol
(b) p-nitrophenol
(c) 2, 4, 6-trinitrophenol
(d) m-nitrophenol
Answer:
(c) 2, 4, 6-trinitrophenol

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 8 Transition and Inner Transition Elements Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 8 Transition and Inner Transition Elements

Question 1.
What are d-block elements? Give their general electronic configuration.
Answer:
Definition: d-block elements are defined as the elements in which the differentiating electron enters d-orbital of the penultimate shell i.e. (n – 1) d-orbital where ‘n is the last shell.

The general electronic configuration can be represented as, (n – n) d^{n – 10}, ns^{n – 2}

Question 2.
What is the position of the transition elements in the periodic table?
Answer:
The transition elements are placed in periods 4 to 7 and groups 3 to 12 of the periodic table.

Question 3.
In which block of the modern periodic table are the transition elements placed?
Answer:
Transition elements are placed in d-block of the modern periodic table.

Question 4.
Why are most of the d-block elements called transition elements?
Answer:

• d-block elements have electronic configuration,(n – n) d^{n – 10}, ns^{l – 2}. They are all metals.
• In the periodic table, they are placed between the ,s-block and p-block elements, i.e., in the groups between 2 and 13.
• They show characteristic properties which are intermediate between those of the elements of s-block and p-block.
• Hence, they show a transition in the properties from those of the most electropositive .v-block elements and less
• electropositive (or electronegative) p-block elements.
• Therefore, most of the d-block elements are called transition elements.

Question 5.
How many series of d-block elements are present in the long-form periodic table? Give their general electronic configuration.
Answer:
There are four series of d-block elements which are placed between 5 and p-block elements in the long-form periodic table as follows :

d-series	Period	Electronic configuration
(1) 3d-series	fourth	[Ar] 3d1 – 10, 4s1 – 2
(2) 4d-series	fifth	[Kr] 4d1 – 10, 5s1 – 2
(3) 5d-series	sixth	[Xe] 4f14 5d1 – 10 6s1 – 2
(4) 6d- series	seventh	[Rn] 5f14 6d1 – 10 7s2

Modern periodic table :

Question 6.
Represent the elements in the four series of transition elements.
Answer:

Question 7.
In which period of the periodic table, will an element, be found whose differentiating electron is a 4d-electron?
Answer:
An element whose differentiating electron is a 4d-electron will be present in fifth period of the periodic table.

Question 8.
Write the condensed electronic configuration of each series of transition elements.
Answer:
Condensed Electronic Configuration of Transition Elements

Question 9.
Write expected and observed electronic configuration of 3d-series block elements.
Answer:
Electronic configuration of 3d-series of d-block elements

Question 10.
Explain why transition elements with electronic configuration 3d44s2 and 3d94s2 do not exist.
Answer:
(1) d-orbitals are degenerate orbitals and they acquire extra stability when half-filled (3d⁵) or completely filled (3d¹⁰). Hence 3d⁴ and 3d⁹ electronic configurations are less stable.
(2) The energy difference between 3d and 4.s’ subshells is very low, hence there arises a transfer of one electron from 45 orbital to 3d orbital.
The electronic configuration changes as,
3d⁴, 4s² → 3d⁵ 4s¹
3d⁹, 4s² → 3d¹⁰ 45¹
Therefore transition elements, with electronic configurations 3d⁴, 4s² and 3d⁹, 4s² do not exist.

Question 11.
Write observed electronic configuration of elements from first transition series having half-filled d-orbitals.
Answer:
There are two elements namely Cr and Mn which have half-filled d-orbitals.
₂₄Crls²2s²2p⁶3s²3p⁶3d⁵4s¹
₂₅Mnls²2s²2p⁶3s²3p⁶3d⁵4s²

Question 12.
Explain the variable oxidation states of metals of first transition series.
Answer:

• The transition metals (or, elements) exhibit variable oxidation states due to their electronic configuration, (n – 1) d^{1 – 10} ns^{1 – 2} for the first row.
• They show only positive oxidation states due to loss of electrons from outer 45-orbital and the penultimate 3rf-orbital.
• Loss of one 45 electron forms M+ ion. Loss of two 45 electrons form M²⁺ ion.
• +2 is the common oxidation state of these elements.
• Higher oxidation states are due to loss of 3 d-electrons along with 45 electrons.
• As the number of unpaired electrons increases, the number of oxidation states shown by the element also increases.
• Sc has only one unpaired electron and it shows two oxidation states ( + 2 and + 3)
• Mn with 5 unpaired d electrons show six different oxidation states. They are +2, +3, +4, +5, +6 and + 7. Thus Mn has the highest oxidation state.
• From Fe onwards variable oxidation states decreases as the number of unpaired electron decreases.
• The last element in the series, Zn shows only one oxidation state ( + 2).

Question 13.
Show different oxidation states of 3d-series of transition elements.
Answer:
The following table shows, different oxidation states of 3d-series of transition elements.

Question 14.
Write oxidation states and outer electronic configuration of first transition series elements.
Answer:
Oxidation states of first transition series elements

Question 15.
Zinc shows only one oxidation slate. Explain.
Answer:

• The electronic conliguration of zinc is, ₃₀Zn Is² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² or [Ar] 3d¹⁰ 4s².
• Due lo loss of two electrons from 4s suhshell Zn shows oxidation state +2. with elcctronic configuration. [Ar] ¹⁸3d¹⁰.
• Since Zn⁺² acquires an extra stability of completely fIlled 3d¹⁰ orbital. it shows only one oxidation state + 2.

Question 16.
Why is manganese more stable in the + 2 state than the + 3 state and the reverse is true for iron?
Answer:

• The electronic configuration of Mn is ₂₅Mn [Ar] 3d⁵ 4s²
• In + 2 and + 3 oxidation states, the electronic configuration of Mn is, Mn²⁺ [Ar] 3d⁵ and Mn³⁺ [Ar] 3d⁴
• Since half-filled d-orbital (3d⁵) has more stability and lower energy than 3d⁴, Mn²⁺ is more stable than Mn³⁺.
• The electronic configuration of Fe is ₂₆Fe [Ar] 3d⁶ 4s² In + 2 and + 3 oxidation states of Fe, the electronic configuration is, Fe²⁺ [Ar] 3d⁶ and Fe³⁺ [Ar] 3d⁵ Since half-filled orbital is more stable, + 3 state of Fe is more stable than + 2 state.

Question 17.
What are the electronic configurations of various ions of 3d-elements?
Answer:
Electronic configuration of various ions of 3d elements

Question 18.
Scandium shows only two oxidation states. Explain.
Answer:
Scandium has electronic configuration, ₂₁Sc : Is², 2s², 2p⁶, 3s², 3p⁶, 3d¹, 4s² Sc shows only two oxidation states namely + 2 and + 3.

• Due to the loss of two electrons from the 4s-orbital, Sc acquires + 2 oxidation state Sc^{2 +} : Is² 2s² 2p⁶ 3s² 3p⁶ 3d¹.
• Due to the loss of one more electron from the 3d-orbital, it acquires + 3 oxidation state with the extra stability of an inert element ₁₈Ar. Sc⁺³ : Is² 2s² 2p⁶ 3s² 3p⁶.
• Since Sc³⁺ acquires extra stability of inert element [Ar]¹⁸, it does not form higher oxidation state.

Question 19.
Write different oxidation states of iron.
OR
Write the electronic configurations of
(i) Fe
(ii) Fe²⁺ and
(iii) Fe³⁺.
Answer:
Oxidation states of iron are +2, +3, +4, +5, +6.
(i) ₂₆Fe : ls²2s²2p⁶3s²3p⁶3d⁶4s²
(ii) Fe²⁺ : Is² 2s² 2p⁶ 3s² 3p⁶ 3d⁶
(iii) Fe³⁺ : Is² 2s² 2p⁶ 3s² 3p⁶ 3d⁵.

Question 20.
Explain different oxidation states of chromium.
Answer:

• The observed electronic configuration of chromium is, ₂₄Cr [Ar] 3d⁵ 4s¹.
• Different possible oxidation states of Cr are 4-1 (3d⁵), + 2 (3d⁴), + 3 (3d³), + 4 (3d²), + 5 (3d¹) and + 6 (3d°).
• Although in + 1 state, Cr gets extra stability of half-filled 3d⁵-orbital, it does not exhibit + 1 state in common except with pyridine.
• Cr⁺² has few stable salts like CrCl₂, CrSO₄ while Cr⁺³ forms very stable salts like CrCl₃.
• Cr⁺⁴ and Cr ^{+ 5} are unstable oxidation states.
• Cr⁺⁶ is the most stable state due to inert gas [Ar] electronic configuration and form the salts like K₂Cr₂O₇.

Question 21.
Manganese shows variable oxidation states. Give reasons.
Answer:

• Manganese (₂₅Mn) has electronic configuration. ₂₅Mn [Ar]¹⁸ 3d⁵ 4s².
• Mn has stable half-filled d-subshell.
• Due to a small difference in energy between 3d and 4s-orbitals, Mn can lose or share electrons from both the orbitals, hence shows variable oxidation states.
• Mn shows oxidation states ranging from + 2 to + 7.

Question 22.
Write the different oxidation states of manganese. Why is + 2 oxidation state of manganese more stable than Mn³⁺?
Answer:

• The different oxidation states of Mn are +2, +3, +4, +5, + 6 and +7.
• The electronic configuration of Mn is Is² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s²
• + 2 oxidation state is very stable due to higher stability of half-filled 3d orbital.
• Mn³⁺ has electronic configuration, ls²2s² 2p⁶3s²3p⁶3d^A which is less stable.

Question 23.
Write the physical properties of first transition series.
Answer:
Physical properties of first transition series :

• All transition elements of the first series are metals.
• Except Zn, they are very hard and have low volatility.
• They show characteristic properties of metals. They are lustrous, malleable and ductile.
• They are good conductors of heat and electricity.
• They have high melting points and boiling points.
• Except Zn and Mn, they have one or more typical metallic structures at normal temperatures.

Question 24.
Which elements in the transition elements, 3d-series has
(i) the lowest density
(ii) the highest density?
Answer:
In 3d transition elements,
(i) Scandium (Sc) has lowest density and
(ii) Zinc (Zn) has the highest density.

Question 25.
Explain the variation in density of d-block elements.
Answer:
The densities of d-block elements are higher than 5-block elements due to higher nuclear charge which results in reduction in atomic size.

Question 26.
Explain the variation in melting points of the transition elements.
Answer:

1. All transition elements are metals and the strength of metallic bonding increases as the number of unpaired electrons increases.
2. In transition elements as atomic number increases, the number of unpaired electrons increases from (n – 1)d¹ to (n – 1 )d⁵.
   For example in 3d-series, melting points increase from ₂₁Sc to ₂₄Cr in 4d-series from ₃₉Y to ₄₂Mo, and in 5d-series from ₇₂Hf to ₇₄W.
3. After (n – l)d⁵ electronic configuration, the electrons start pairing, hence the number of unpaired electrons decrease, hence metallic character, melting points decrease from (n-1 )d6 to (n – 1)d¹⁰.
4. In all transition series the melting point increases steadily up to d⁵ configuration and after this melting point decreases regularly.
   

Question 27.
The first ionisation enthalpies of third transition series elements are much higher than those of the elements of first and second transition series. Explain.
Answer:

1. Third transition series elements have electronic configuration, 4f¹⁴ 5d^{1 – 10} 6s².
2. Thus, atoms of third series elements possess filled 4f-orbitals.
3. 4f-orbitals due to their diffused shape, exhibit poor shielding effect and give rise to lanthanide contraction. Hence the valence electrons experience greater nuclear attraction and greater amount of energy is required to ionise the elements of third transition series namely (H_f to Au).
4. Therefore the ionisation enthalpies of third transition series elements are much higher than those of the first and second transition series.

Question 28.
Explain the metalic character of transition metals.
Answer:

• All the transition elements are metals.
• They are hard, lustrous, malleable, ductile and they have high tensile strength.
• They have high melting points and boiling points.
• Their metallic character is due to vacant or partially filled (n – 1) d-orbitals, and they involve both metallic and covalent bonding.
• Since the strength of metallic bonds depends upon the number of unpaired electrons, it increases up to middle i.e., up to (n – 1 )d⁵, hence accordingly melting points and boiling points also increase.
• After (n – l)d⁵ configuration, the electrons start pairing, hence the metallic strength, melting points and boiling points decrease with the increase in atomic number.

Question 29.
How does metallic character vary in 3d transition elements?
Answer:

1. In 3d-series elements as atomic number increases from scandium (Sc [Ar]¹⁸ 3d¹ 4s²) the number of unpaired electrons increases up to 3d⁵ in chromium.
2. As the number of unpaired electrons increases, the metallic character increases, hence the melting points and boiling points increase from ₂₁Sc(3d¹) to ₂₄Cr (3d⁵).
3. After chromium the number of unpaired electrons goes on decreasing due to the pairing of electrons, hence metallic character, melting points and boiling points decrease from ₂₅Mn to ₂₉Cu.
4. Zinc has all electrons paired, hence it is soft, has a low melting and boiling points.

Question 30.
Which are the common arrangement of the atoms in the structure of transition metals?
Answer:
Most of the transition metals have simple hexagonal closed packed (hep), cubic closed packed (ccp) or body centred cubic (bcc) lattices.

Question 31.
Why do the compounds of transition metals exhibit magnetic properties?
Answer:
The compounds of transition metals exhibit magnetic properties due to the presence of unpaired electrons in their atoms or ions.

Question 32.
What is the cause of paramagnetism and ferromagnetism?
Answer:
Paramagnetism and ferromagnetism is due to the presence of unpaired electrons in species.

Question 33.
When does species become diamagnetic?
Answer:
When there is no unpaired electron, i.e. all electron spins are paired, the species become diamagnetic.

Question 34.
How do metals Fe, Co, Ni acquire permanent magnetic moment?
Answer:
The transition metals Fe, Co and Ni are ferromagnetic. When the magnetic field is applied, all the unpaired electrons in these metals (and their compounds) align in the direction of the applied magnetic field. Due to this the magnetic susceptibility is enhanced and these metals can be magnetised, that is, they acquire permanent magnetic properties.

Question 35.
In which oxidation state, is vanadium diamagnetic?
Answer:

• The electronic configuration of vanadium is, ₂₃V [Ar] 3d³ 4s².
• In +5 oxidation state, the electronic configuration is, V⁵⁺ [Ar].
• Since in V⁵⁺ state, vanadium has all electrons paired, it is diamagnetic.

Question 36.
How is a magnetic moment expressed?
Answer:
The magnetic moment is expressed in Bohr magneton (B.M.). It is denoted by μ.

Question 37.
What is Bohr magneton (B.M.)?
Answer:
Bohr magneton (B.M.) is a unit of magnetic moment :
\(1 \mathrm{~B} . \mathrm{M} .=\frac{e h}{4 \pi m_{\mathrm{e}} c}\)
where, h : Planck’s constant (h = 6.626 x 10^-34 Js)
e : electronic charge (1.60218 x 10^-19 C)
me : mass of an electron (9.109 x 10^-31 kg)
c : velocity of light. (2.998 x 10⁸ ms^-1)

Question 38.
Explain the magnetic properties of transition (or d-block) elements.
Answer:

• Most of the transition metal ions and their compounds are paramagnetic in nature due to the presence of one or more unpaired electrons in their (n – 1)d-orbitals. Hence they are attracted in the magnetic field.
• As the number of unpaired electrons increases from 1 to 5 in J-orbitals, the paramagnetic character and magnetic moment increase.
• The transition elements or their ions having all electrons paired are diamagnetic and they are repelled in the magnetic field.
• Metals like Fe, Co and Ni possess very high paramagnetism and acquire permanent magnetic moment hence they are ferromagnetic.

Question 39.
Explain the effective magnetic moment of the species.
Answer:

• The magnetic moment in the species arises due to the presence of unpaired electrons.
• The magnetic moment depends upon the sum of orbitals and spin contribution for each unpaired electron present in the species.
• In transition metal ions, the contribution of orbital magnetic moment is suppressed by the electrostatic field of other atoms, molecules or ions surrounding the metal ion in the compound.
• Hence the net or effective magnetic moment arises mainly due to spin of electrons. The effective magnetic moment μ_eff, of a paramagnetic substance is given by 1 spin only’ formula represented as, \(\mu=\sqrt{n(n+2)}\) B.M. where n is the number of unpaired electrons.

Question 40.
What is the importance of magnetic moment (μ)?
Answer:

• From the measurements of the magnetic moment (μ) of the species or metal complexes of the first row of transition elements, the number of unpaired electrons can be calculated with the spin-only formula.
• As magnetic moment is directly related to the number of unpaired electrons, value of μ will vary directly with the number of unpaired electrons.
• In 2nd and 3rd transition series, orbital angular moment is significant. Hence spin-only formula for the complexes of 2nd and 3rd transition series is not useful.

Question 41.
Calculate the magnetic moment of the following species :
(1) Cr³⁺
(2) Co
(3) Co³⁺
(4) Cu^{2 +}.
Answer:

Question 42.
Explain : A slight difference in the calculated and observed values of magnetic moments.
Answer:
Magnetic moments are determined experimentally in solution or in solid state where the central atom or ion is hydrated or bound to ligands. Hence a slight difference is observed in calculated and experimentally obtained values of magnetic moment (μ).

Question 43.
Calculate the magnetic moment of a divalent ion in an aqueous solution, if its atomic number is 24.
Answer:
(1) The electronic configuration of divalent inri M²⁺ having atomic number 24 is.

The ion has number of unpaired electrons. n = 4.
By spin only’ formula, the magnetic μ is given by, \(\mu=\sqrt{n(n+2)}=\sqrt{4(4+2)}=4.90 \mathrm{~B} . \mathrm{M}\)
(This M²⁺ ion is Cr²⁺ ion)

Question 44.
When does a substance appear coloured?
Answer:
A substance appears coloured when it absorbs a portion of visible light. The colour depends upon the wavelength of absorption in the visible region of electromagnetic radiation.

Question 45.
Why do the d-block elements form coloured compounds?
Answer:

• Compounds (or ions) of many d-block elements or transition metals are coloured.
• This is due to the presence of one or more unpaired electrons in (n – 1) d-orbital. The transition metals have incompletely filled (n – 1) cf-orbitals.
• The energy required to promote one or more electrons within the d-orbitals involving d-d transitions is very low.
• The energy changes for d-d transitions lie in visible region of electromagnetic radiation.
• Therefore transition metal ions absorb the radiation in the visible region and appear coloured.
• Colour of ions of d-block elements depends on the number of unpaired electrons in (n – 1) d-orbital. The ions having equal number of unpaired electrons have similar colour.
• The colour of metal ions is complementary to the colour of the radiation absorbed.

Question 46.
How is complementary colour of a compound identified?
Answer:

1. The transition metal ions absorb the radiation in the visible region and appeared coloured.
2. Metal ion absorbs radiation of certain wavelength from the visible region. Remaining light is transmitted and the observed colour corresponds to the complementary colour of the light observed.
3. The complementary colour can be identified (with the diagram given).

For example if red colour is absorbed then transmitted complementary colour is green.

Question 47.
Write outer electronic configuration (d-orbital) and colour of 3d-series of transition metal ions.
Answer:
Colour of 3d-transition metal ions

Question 48.
Mention the factors on which the colour of a transition metal ion depends.
Answer:
The factors on which the colour of transition metal ion depends are as follows :

• The presence of incompletely filled d-orbitals in metal ions. (The compounds with the configuration d° and d’0 are colourless.)
• The presence of unpaired electrons in d-orbitals.
• d → d transitions of electrons due to absorption of radiation in the visible region.
• Nature of groups (anions) (or ligands) linked to the metal ion in the compound or a complex.
• Type of hybridisation in metal ion in the complex.
• Geometry of the complex of the metal ion.

Question 49.
Give reasons : Zinc salts are colourless.
Answer:

• Colour of the ions of d-block elements depends on the number of unpaired electrons in (n – 1) d-orbitals.
• Zinc forms salts of Zn²⁺ ions.
• The electronic configuration of Zn⁺² is [Ar] 3d¹⁰.
• Since Zn⁺² does not have unpaired electrons in 3d-orbital, d→d transition cannot take place, hence, Zn⁺² ions form colourless salts.

Question 50.
Explain : The compounds of Cu(II) are coloured.
Answer:

• The electronic configuration of ₂₉Cu [Ar] 3d¹⁰ 4s¹ and Cu²⁺ [Ar] 3d⁹.
• In copper compounds Cu²⁺ ions have incompletely filled 3d-orbital (3d⁹).
• Due to the presence of one unpaired electron in 3 d-orbital, Cu²⁺ ions absorb red light from visible spectrum and emit blue radiation due to d → d transition. Therefore, copper compounds are coloured.

Question 51.
Explain why the solution of Ti³⁺ salt is purple in colour.
OR
Why is Ti³⁺ coloured? (atomic number Ti = 22)
Answer:

• Ti²⁺ ions in the aqueous solution exist in the hydrated complex form as [Ti(H₂O)₆]²⁺.
• The electronic configuration of Ti is, ₂₂Ti [Ar]¹⁸ 3d² 4s² and Ti³⁺ [Ar]¹⁸ 3d¹. Hence in complex, Ti³⁺ has one unpaired electron in 3d subshell.
• Initially, the 3d electron occupies lower energy d-orbital (in t_2g-orbitals).
• On the absorption of radiations of about 500 nm in yellow green region by a complex, 3d¹ electron is excited to the higher energy d-orbital (e_g-orbitals).
• When the electron returns back to the lower energy d-orbital (t_2g), it transmits radiation of complementary colour i.e. red blue or purple colour. Hence, the solution of hydrated Ti³⁺ is purple.

Question 52.
What will be the colour of Cd²⁺ salts? Explain.
Answer:

• The electronic configuration of, ₄₈Cd [Kr]³⁶ 3d¹⁰ 5s² and Cd²⁺ [Kr]³⁶ 3d¹⁰.
• Cd²⁺ ions have completely filled 3d subshell and there are no unpaired electrons in 3d-orbital.
• Hence d → d transition is not possible.
• Therefore, Cd²⁺ ions do not absorb radiations in the visible region and the salts of Cd²⁺ ions are colourless (or white).

Question 53.
Indicate which of the ions may be coloured- V³⁺, Sc³⁺, Cr³¹, Cu²⁺, Ti³⁺, Cu⁺
Answer:

• V³⁺ [Ar]¹⁸ 3d²-((green)
  Since there are two unpaired electrons available, for d → d transition, it will show a Green colour.
• Sc³⁺ [Ar]¹⁸ 3d° (colourless/white).
  Since there are no unpaired electrons in the 3d subshell, it will not show colour.
• Cr³⁺ [Ar]¹⁸ 3d³ – (violet)
  There are three unpaired electrons in the 3d subshell, hence due to d → d transition, it will show violet colour.
• Cu²⁺ [Ar]¹⁸ 3d⁹ (blue)
  It has one unpaired electron that can undergo a d → d transition, hence it will show the colour blue.
• Ti³⁺ [Ar]¹⁸ 3d¹ (purple)
  It has one unpaired electron that can undergo a d → d transition, hence it will show the colour purple.
• Cu¹⁺ [Ar]¹⁸ 3d¹⁰ (colourless)
  There are no unpaired electrons in the 3d subshell, hence it will not show colour.

Question 54.
Explain why is cobalt chloride pink in colour when dissolved in water but turns deep blue when treated with concentrated hydrochloric acid.
Answer:

• The electronic configuration of ₂₇Co : [Ar] 3d¹4s² and Co²⁺ [Ar] 3d¹.
• When dissolved in water cobalt chloride, Co²⁺ forms pink complex, [Co(H₂O)₆]²⁺.
• The complex has octahedral geometry.
• Due to absorption of radiation in the visible region and d – d transition, it forms pink coloured solution.
• When CoCl₂ solution is treated with concentrated HCl solution it turns deep blue.
• This change is due to the formation of another complex, [CoC1₄]²⁺ which has a tetrahedral geometry.
• Thus due to a change in geometry of the complex formed the colour of the solution changes from pink to deep blue.

Question 55.
Explain the catalytic properties of the rf-block or transition metals.
Answer:

• d-block elements or transition metals and their compounds or complexes influence the rate of a chemical reaction and hence act as catalysts.
• In homogeneous catalysis a catalyst forms an unstable intermediate compound which decomposes into products and regenerates the catalyst. But transition metals involve heterogeneous catalysis.
• The transition metals have incompletely filled d-subshells which adsorb reactants on the surface and provide a large surface area for the reactants to react.
• Since transition metals have variable oxidation states they are very good catalysts.
• Hence, compounds of Fe, Co, Ni, Pt, Pd, Cr etc are used as catalysts in many reactions.

Question 56.
Explain the use of different transition metals as catalysts.
Answer:
The transition metals are very good catalysts.

• MnO₂ is used as a catalyst in the decomposition of KClO₃.
  
• In the manufacture of ammonia by Haber’s process, Mo/Fe is used as a catalyst.
  
• In the synthesis of gasoline by Fischer Tropsch process, Co-Th alloy is used as a catalyst.
• Finely divided Ni (formed by reduction of heated oxide in hydrogen) is very efficient catalyst in hydrogenation of ethene to ethane at 140 °C.
  
• Commercially, hydrogenation with Ninkel as catalyst is used to convert inedible oils into solid fat for the production of margarine.
• In the contact process of industrial production of sulphuric acid, sulphur dioxide and oxygen (from air) react reversibly over a solid catalyst of platinised asbestos.
  
• Carbon dioxide and hydrogen are formed by the reaction of carbon monoxide and steam at 500 °C with Fe-Cr catalyst.
  

Question 57.
What are interstitial compounds of transition metals?
Answer:

• The interstitial compounds of the transition metals are those which are formed when small atoms like H, C or N are trapped inside the interstitial vacant spaces in the crystal lattices of the metals.
• Sometimes, sulphides and oxides are also trapped in the crystal lattices of transition elements.
• Presence of these elements in the crystal lattices of metals provide new properties to the metals.

Question 53.
Give one example of an interstitial compound.
Answer:
Steel and cast iron are examples of interstitial compounds of carbon and iron.

Question 54.
Give examples of interstitial compounds where the property of the transition metal is changed.
Answer:
Steel and cast iron are interstitial compounds of carbon and iron (carbides of iron). Due to the presence of carbon, the malleability and ductility of iron is reduced while its tenacity increases.

Question 55.
What are the properties of the interstitial compounds of transition metals?
Answer:

• The chemical properties of the interstitial compounds are the same as that of parent transition metals.
• They are hard and show the metallic properties like electrical and thermal conductivity, lustre, etc.
• Since metal-non-metal bonds in the interstitial compounds are stronger than metal-metal bonds in pure metals, the compounds have very high melting points, higher than the pure metals.
• They have lower densities than the parent metal.
• The interstitial compounds containing hydrogen (i.e., hydrides of metals) are powerful reducing agents.
• The compounds containing carbon, hence behaving as carbides, are chemically inert and extremely hard like diamond.
• In these compounds, malleability and ductility are changed. For example steel and cast iron.

Question 56.
What are interstitial compounds? Why do these compounds have higher melting points than corresponding pure metals?
Answer:

1. The interstitial compounds of the transition metals are those which are formed when small atoms like H, C or N are trapped inside the interstitial vacant spaces in the crystal lattice of the metals.
2. Since metal-nonmetal bonds in the interstitial compounds are stronger than metal-metal bonds in pure metals, the compounds have very high melting points, higher than the pure metals.

Question 57.
Explain the formation of alloys of transition metals.
Answer:

• The transition metals form a large number of alloys among themselves, which are hard with high melting points.
• During alloy formation atoms of one metal are distributed randomly in the lattice of another metal.
• The metals with similar atomic radii and similar properties readily form alloys.
• These alloys have industrial importance.
• The alloys can be ferrous alloys or nonferrous alloys.

Question 58.
How are the transition metal alloys classIfied?
Answer:
The transition metal alloys are classified into

• Ferrous alloys
• Nonferrous alloys.

Question 59.
Explain what are
(1) ferrous alloys and
(2) nonferrous alloys.
Answer:

1. Ferrous alloys: In ferrous alloys, atoms of other elemems are distributed randomly in atoms of iron in the mixture. As the percentage of iron is more in these alloys, they are termed as ferrous alloys. For expamle : nickel steel, chromium steel, stainless steel, (All steels have abot 2% carbon)
2. onferrous alloys : These are formed by mixing atoms of transition metal other than iron with a non transition elemeni. For example, brass is an alloy of Cu and Zn. Bronze is an alloy of Cu and Sn.

Question 60.
What are the uses of alloys?
Answer:

Name of alloy	Important use in industry
(1) Bronze (Cu + Sn)	In making statues, medals and trophies (as it is tough, strong and corrosion-resistant)
(2) Cupra-nickel (Cu + Ni)	In making machinery parts of marine ships, boats, marine conden­ser tubes.
(3) Stainless steel	In the construction of the outer fuselage of ultra-high-speed aircraft.
(4) Nichrome : (Ni+ Cr in the ration 80 : 20)	For gas turbine engines.
(5) Titanium alloys	For ultra-high-speed flight, fireproof bulkheads and exhaust shrouds (as they withstand high temperatures).

Question 61.
Write the preparation of potassium permanganate.
Answer:
Potassium permanganate (KMnO₄) is prepared in the following steps,

(1) Chemical Oxidation : When finely divided manganese dioxide (Mn0₂) is heated strongly with fused caustic potash (KOH) and an oxidising agent potassium chlorate (KCIO₃), dark green potassium manganate (K₂MnO₄) is obtained. (In neutral or acidic medium K₂MnO₄ disproportionates.)

The liquid is filtered through glass wool or sintered glass and evaporated. Potassium manganate crystallises as small, blackish crystals.

(2) Oxidation of K₂MnO₄ by
(i) Electrolytic oxidation : An alkaline solution of manganate ion is electrolysed between iron electrodes separated by a diaphragm. Manganate ion \(\left(\mathrm{MnO}_{4}^{2-}\right)\) undergoes oxidation at anode forming permanganate ion \(\left(\mathrm{MnO}_{4}^{-}\right)\). Oxygen evolved at anode converts \(\left(\mathrm{MnO}_{4}^{2-}\right)\) to \(\left(\mathrm{MnO}_{4}^{-}\right)\).

The overall reaction is as follows :
2K₂MnO₄ + H₂O + [O] → 2KMnO₄ + 2KOH

The electrolytic solution is filtered and evaporated to obtain deep purple black crystals of KMn0₄.

(ii) By passing CO₂ through the solution of K₂MnO₄ :
3K₂MnO₄ + 4CO₂ + 2H₂O → 2KMnO₄ + MnO₂ + 4 KHCO₃

Question 62.
What is meant by the disproportionation of an oxidation state? Explain giving example of manganese.
Answer:

1. Disproportionation reaction is a chemical reaction in which atom or an ion of an element forms two or more species having different oxidation states, one lower and one higher.
2. Manganese (Mn) shows different oxidation states + 2 to +7.
3. When one oxidation state, lower or higher oxidation state becomes unstable as compared to another oxidation state, it undergoes disproportionation reaction.
4. For example, + 6 oxidation state of Mn is less stable than + 7 and + 4.
   • Hence, in acidic medium \(\mathrm{Mn}^{6+} \text { in } \mathrm{MnO}_{4}^{2-}\) undergoes disproportionation reaction.
     
   • In neutral medium green K₂MnO₄ disproportionates to KMn04 and MnO₂.
     

Question 63.
Give examples of oxidising reactions of KMnO₄.
Answer:
(1) KMnO₄ in acidic medium :

(2) KMnO₄ in neutral or alkaline medium in neutral or weakly alkaline medium :
(i) Iodide is oxidised to iodate ion.

(ii) Thiosulphate ion is oxidised to sulphate ion.

(iii) Manganous salt is oxidised to MnO₂.

Question 64.
Balance the following equations :
KI + KMnO₄ + H₂SO₄ → K₂SO₄ + MnSO₄ + 8H₂O + I₂
H₂S + KMnO₄ + H₂SO₄ → K₂SO₄ + MnSO₄ + H₂O + S.
Answer:
10 KI + 2KMnO₄ + 8H₂SO_{4 →} 6K₂SO₄ + 2MnSO₄ + 8H₂O + 5I₂
5H₂S + 2KMnO₄ + 3H₂SO₄ → K₂SO₄ + 2MnSO₄ + 8H₂O + 5S.

Question 65.
Give the uses of potassium permanganate.
Answer:
Uses of potassium permanganate :

• as an antiseptic.
• as a powerful oxidising agent in laboratory and industry.
• in the detection of unsaturation in organic compounds in the laboratory. (Baeyer’s reagent, alkaline KMnO₄).
• for detecting halides in qualitative analysis.
• in volumetric analysis for the estimation of H₂O₂, FeSO₄ etc.)

Question 66.
Write the formula of chromite ore.
Answer:
FeOCr₂O₃.

Question 67.
How is potassium dichromate manufactured from chromite ore (FeOCr₂O₃)?
Answer:
Manufacture of potassium dichromate (K₂Cr₂O₂) from chrome iron ore (FeOCr₂O₃) involves following steps :
(1) Concentration of ore : The chromite ore (FeOCr₂O₃) is powdered and washed with current of water.
(2) Conversion of chromite ore into sodium chromate : The concentrated ore is mixed with anhydrous sodium carbonate (Na₂CO₃) and a flux of lime in excess air and heated in a reverberatory furnace.

Sodium chromate (Na₂CrO₄) formed in the reaction is then extracted with water so that Na₂CrO₄ dissolves into solution and insoluble substances separate out.
(3) Conversion of Na2CrO₄ into sodium dichromate (Na₂Cr₄O₇) : Na₂CrO₄ solution is acidified with concentrated H₂SO₂, so that sodium chromate is converted into sodium dichromate.

Less soluble sodium sulphate crystallises out as Na₂SO₄.10H₂O. which is filtered off.
(4) Conversion of Na₂Cr₂O₇ into K₂Cr₂O₇ : Concentrated solution of Na₂Cr₂O₇ is treated with KCl on by double decomposition, K₂Cr₂O₇ is obtained.
Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl
On concentrating and cooling the solution, less soluble orange coloured K₂Cr₂O₇ crystallises out which is filtered and purified by recrystallisation.

Question 68.
What happens when hydrogen sulphide gas (H₂S) is passed through acidified K₂Cr₂O₇ solution?
Answer:
When hydrogen sulphide (H₂S) gas is passed into solution of K₂Cr₂O₇, H₂S is oxidised to a pale yellow solid (precipitate) of sulphur. Orange coloured solution becomes green due to formation of chromic sulphate (green coloured).

In the reaction, H₂S is oxidised to S and K₂Cr₂O₇ is reduced to Cr₂(SO₄)₃.

Question 69.
What are the common physical properties of d-block elements?
Answer:
The common physical properties of d-block elements are :

• All d-block elements are lustrous and shining.
• They are hard and have high density
• They have high melting and boiling points.
• They are good electrical and thermal conductors.
• They have high tensile strength and malleability.
• They can form alloys with transition and nontransition elements
• Many metals and their compounds are paramagnetic.
• Most of the metals are efficient catalysts.

Question 70.
What are the common chemical properties of d-block elements?
Answer:
The common chemical properties of the d-block elements are :

• All d-block elements are electropositive metals.
• They exhibit variable oxidation states and form coloured salts and complexes.
• They are good reducing agents.
• They form insoluble oxides and hydroxides.
• Iron, cobalt, copper, molybdenum and zinc are biologically important metals.
• They catalyse biological reactions.

Question 71.
Give examples to show that elements of first row of d-block elements differ from second and third row with respect to the stabilisation of higher oxidation states.
Answer:

• Highest oxidation state for the first row element is + 7 as in Mn.
  For the second row, the highest oxidation state is + 8 as in Ru (RuO₄).
  For the third row, the highest oxidation state is + 8 as in Os (OsO₄).
• Compounds of Mo(V) of 2nd row and W(VI) of 3rd row of transitional elements are more stable than Cr(VI) and Mn (VIII) of first row elements.

Question 72.
How do metals occur in nature?
Answer:
In nature, few metals occur in earth’s crust in free state or native state while other metals occur in the combined form.
(1) Elements in free state or native state : The metals which are non-reactive with air, water, CO₂ and non-metals occur in free state or native state. For example, gold, platinum, palladium occur in free state. Metals like Cu, Ag and Hg occur partly in the free state.

(2) Combined form : The metals which are reactive occur in the combined state with other elements forming compounds like oxides, sulphides, sulphates, carbonates, silicates, etc.

Question 73.
What are minerals?
Answer:
Minerals : They are naturally occurring chemical substances in the earth’s crust containing metal in free state or in combined form and obtainable from mining are called minerals. For example, haematite Fe203, galena PbS, etc.

Question 74.
What are ores?
Answer:
Ores : The minerals containing a high percentage of metals from which metals can be profitably extracted are called ores.
[Note : Every ore is a mineral but every mineral is not an ore.]

Question 75.
Write names of minerals and ores of Iron, Copper and Zinc.
Answer:

Metals	Mineral	Ore
Iron	Haematite Fe2O3 Magnetite Fe3O4 Limonite 2Fe2O3, 3H2O Iron pyrites FeS2 Siderite FeCO3	Haematite
Copper	Chalcopyrite CuFeS2 Chalcocite Cuprite Cu2O	Chalcopyrite Chalcocite
Zinc	Zinc blende ZnS Zincite ZnO Calamine ZnCO3	Zinc blende

Question 76.
What is metallurgy?
Answer:
Metallurgy : The process of extraction of metal in a pure state from its ore is called metallurgy.

Question 77.
Define the following:
(1) Pyrometallurgy
(2) Hydrometallurgy
(3) Electrometallurgy.
Answer:

1. Pyrometallurgy : It is a process of extraction of metal from metal oxide from concentrated ore by reduction with a suitable reducing agent like carbon, hydrogen, aluminium, etc. at high temperature.
2. Hydrometallurgy : It is a process of extraction of metals by converting their ores into aqueous solutions of metal compounds and reducing them by suitable reducing agents.
3. Electrometallurgy : It is a process of extraction of highly electropositive metals like Na, K, Al, etc. by electrolysis of fused compounds of the metals where metal ions are reduced at cathode forming metals.

Question 78.
What is gangue?
Answer:
Gangue : The earthly and undesired impurities of various substances like sand (SiO₂), metal oxides, etc. present in the ore are called gangue or matrix.

Question 79.
Define concentration of an ore.
Answer:
Concentration : A process of removal of gangue or unwanted impurities from the ore is called concentration of an ore. It is also called benefaction or dressing of an ore.

Question 80.
What are common methods of concentration of an ore?
Answer:
The concentration of an ore involves different methods depending upon the differences in physical properties of compounds or the metal present and the nature of the gangue.

The common methods of concentration of ore are as follows :

1. Gravity separation or hydraulic washing :
   This can be carried out by two processes as follows :
   • Hydraulic washing by using Wilfley’s table method
   • Hydraulic classifier methods.
2. Magnetic separation
3. Froth floatation process.
4. Leaching.

The method depends upon the nature of ore.

Question 81.
What is leaching?
Answer:
Leaching : ft is a (chemical) process used in the concentration of an ore by extracting soluble material from an insoluble solid by dissolving in a suitable solvent. This method is used in the concentration process of ores of Al, Ag, Au, etc.

Question 82.
What is roasting of an ore?
Answer:
Roasting : It is a process of strongly heating a concentrated ore in the excess of air below melting point of metal, to convert it into oxide form. It is used for a sulphide ore. For example, ZnS ore on roasting forms ZnO.

Question 83.
Write an equation to show how zinc blende (ZnS) is converted to ZnO.
Answer:
When zinc blende is roasted, it is converted to ZnO.
\(\mathrm{ZnS}+\mathrm{O}_{2} \stackrel{\Delta}{\longrightarrow} \mathrm{ZnO}+\mathrm{SO}_{2}\)

Question 84.
Explain the term : Smelting
Answer:
Smelting : The process of extraction of a metal from its ore by heating and melting at high temperature is called smelting. Reduction of ore is carried out during smelting.

Question 85.
What is calcination?
Answer:
Calcination is a process in which the ore is heated to a high temperature below the melting point of the metal in the absence of air or limited supply of air in a reverberatory furnace.

It is generally used for carbonate and hydrated oxides to convert them into anhydrous oxides.

Question 86.
Define the terms :
(1) Flux
(2) Slag
Answer:
(1) Flux : A flux is a chemical substance which is added to the concentrated ore during smelting in order to remove the gangue or impurities by chemical reaction forming a fusible mass called slag.
(2) Slag : It is a waste product formed by combination of a flux and gangue (or impurities) during the extraction of metals by smelting process.

Iron is the fourth most abundant element in the earth’s crust.

Question 87.
What is the composition of haematite ore?
Answer:
Composition of Haematite ore is Fe2O₃ + SiO₂ + Al₂O₃ + phosphates

Question 88.
Which impurities (gangue) are present in haematite ore?
Answer:
SiO₂ and Al₂O₃ are the impurities present in the haematite ore.

Question 89.
Which reducing agents are used to reduce haematite ore into metallic iron?
Answer:
Haematite ore is reduced using coke and CO. Carbon in the coke is converted to carbon monoxide. Carbon and carbon monoxide together reduce Fe203 to metallic iron.

Fe₂O₃ + 3C → 2Fe + 3CO.
Fe₂O₃ + 3CO → 2Fe + 3CO₂.

Question 90.
Why is limestone used in the extraction of iron?
Answer:

• The ore of iron contains acidic gangue or impurity of silica, SiO₂.
• To remove silica gangue, basic flux like calcium oxide CaO, is required, which is obtained from the decomposition of limestone, CaCO₃. \(\mathrm{CaCO}_{3} \stackrel{\Delta}{\longrightarrow} \mathrm{CaO}+\mathrm{CO}_{2}\)
• Silica reacts with CaO and forms a fusible slag of CaSiO₃.
  \(\mathrm{SiO}_{2}+\mathrm{CaO} \stackrel{\Delta}{\longrightarrow} \mathrm{CaSiO}_{3}\)

Therefore in the extraction of iron, lime is used.

Question 91.
Name the furnace in which iron is extracted from Haematite ore.
Answer:
Extraction of iron is carried out in Blast furnace.

Question 92.
Explain the extraction of iron from haematite.
Answer:
Iron is mainly extracted from haematite, Fe₂O₃ by reduction process.
Haematite ore contains silica (SiO₂), alumina (Al₂O₃) and phosphates as impurity or gangue.

Coke is used for the reduction of ore.

To remove acidic gangue SiO₂, a basic flux CaO is used which is obtained from lime stone CaCO₃.

The extraction process involves following steps :
(1) Concentration of an ore : The powdered ore is concentrated by gravity separation process by washing it in a current of water. The lighter impurities (gangue) are carried away leaving behind the ore.
(2) Roasting : The concentrated ore is heated strongly in a limited current of air. During this, moisture is removed and the impurities like S, As and phosphorus are oxidised to gaseous oxides which escape.

After roasting, the ore is sintered to form small lumps.
(3) Reduction (or smelting) : The roasted or calcined ore is then reduced by heating in a blast furnace.

The blast furnace is a tall cylindrical steel tower about 25 m in height and has a diameter about 5-10m lined with fire bricks inside.

Blast furnace has three parts :

• the hearth,
• the bosh and
• the stack.

At the top, there is a cup and cone arrangement to introduce the ore and at the bottom, tapping hole for withdrawing molten iron and an outlet to remove a slag.

The roasted ore is mixed with coke and limestone in the approximate ratio of 12 : 5 : 3.

A blast of hot air at about 1000 K is blown from downwards to upwards by layers arrangement. The temperature range is from bottom 2000 K to 500 K at the top. The charge of ore from top and the air blast from bottom are sent simultaneously. There are three zones of temperature in which three main chemical reactions take place.

(i) Zone of combustion : The hot air oxidises coke to CO which is an exothermic reaction, due to which the temperature of furnace rises.
C + ^1/2 O₂ → CO ΔH= – 220kJ
Some part of CO dissociates to give finely divided carbon and O₂.
2CO → 2C + O₂
The hot gases with CO rise up in the furnace and heats the charge coming down. CO acts as a fuel as well as a reducing agent.

(ii) Zone of reduction : At about 900 °C, CO reduces Fe₂O₃ to spongy (or porous) iron.
Fe₂O₃ + 3CO → 2Fe + 3CO₂
Carbon also reduces partially Fe₂0₃ to Fe.
Fe₂O₃ + 3C → 2Fe + 3CO

(iii) Zone of slag formation : At 1200 K limestone, CaCO₃ in the charge, decomposes and forms a basic flux CaO which further reacts at 1500 K with gangue (SiO₂, Al₂O₃) and forms a slag of CaSiO₃ and Ca₃AlO₃.
CaCO₃ + CaO + CO₂.
CaO + SiO₂ → CaSiO₃
12CaO + 2Al₂O₃ → 4Ca₃AlO₃ + 3O₂

The slag is removed from the bottom of the furnace through an outlet.

(iv) Zone of fusion : The impurities in ore like MnO₂ and Ca₃(PO₄)₂ are reduced to Mn and P while SiO₂ is reduced in Si. The spongy iron moving down in the furnace melts in the fusion zone and dissolves the impurities like C, Si, Mn, phosphorus and sulphure. The molten iron collects at the bottom of furnace. The lighter slag floats on the molten iron and prevents its oxidation.

The molten iron is removed and cooled in moulds. It is called pig iron or cast iron. It contains about 4% carbon.

Question 93.
Write the reaction involved in the zone of reduction in blast furnace during extraction of iron.
Answer:
Zone of reduction : At about 900 °C, CO reduces Fe₂O₃ to spongy (or porous) iron.
Fe₂O₃ + 3CO → 2Fe + 3CO₂
Carbon also reduces Fe₂O₃ to Fe.
Fe₂O₃ + 3C → 2Fe + 3CO

Question 94
Write reactions involved at different temperatures in the blast furnace.
Answer:

Temperature K	Change taking place in the blast furnace	Reactions
1. 500 K	Haematite ore loses moisture	ore xH2O → ore
2. 900 K	Reduction of ore by CO	Fe2O3 + 3CO → 2Fe + 3CO
3. 1200K	Limestone decomposes	CaCO3 → CaO + CO2
4. 1500K	Reduction of ore by C	Fe2O3 + 3C → 2Fe + 3CO
5. 1600 K	(i) Reduction of FeO by C (ii) Fusion of iron and slag formation	FeO + C → Fe + CO CaO + SiO2 → CaSiO3
6. 2000 K	Combustion of coke	2C + O2 → CO

Question 95.
What is the action of carbon on Fe203 in blast furnace?
Answer:
Fe₂O₃ + 3C → 2Fe + 3CO

Question 96.
What is refining of metals?
Answer:
Refining of metals : The purification of impure or crude metals by removing metallic and nonmetallic impurities is known as refining of metals. H

Question 97.
How is pure iron obtained from crude iron?
Answer:
Pure iron can be obtained by electrolytic refining.

Question 98.
Name the methods of refining of metals.
Answer:
Methods of refining of metals :

• Electrorefining
• Liquefaction
• Distillation
• Oxidation m

Question 99.
What are the factors that govern the choice of extraction technique of metals?
Answer:
The choice of extraction technique is governed by the following factors.

• Nature of ore
• Availability and cost of reducing agent. (Generally, cheap coke is used).
• Availability of hydraulic power.
• Purity of metal required.
• Value of by-products. For example. SO₂ obtained during the roasting of sulphide ores is important for the manufacture of H₂SO₄.

Question 100.
Which are the commercial forms of iron?
Answer:
Commercial forms of iron are :

• Cast iron
• wrought iron
• steel. H

Question 101.
(A) What are f-block elements?
(B) What are inner transition elements?
Answer:
(A)

• Elements in which differentiating electron enters into the pre-penultimate shell the (n – 2) f-orbital are known as f.block elements.
• They include 28 elements with atomic numbers ranging from 58-71 and atomic numbers 90 to 103 collectively.
• There are two f-series or two f-block elements, namely 4f and 5f series.
• The f-block includes two inner transition series namely the lanthanoid series. Cerium (58) to LuteUum (71) or the 4 f-block elements and the actinoid series. Thorium (90) to I.awrencium (103) or the 5f block elements.

(B) f-block elements are called inner transition elements since f-orbital lies much inside the f-orbital in relation to the transition metals, These elements have 1 to 14 electrons in their f-orbital.

Question 102.
What are fIrst inner transition elements?
Answer:

1. 4f-hlock elements are called (first) inner transition elements and have partly filled inner orbitaIs or (4f) orbitais.
2. They have general outer electronic configuration \((n-2) f^{1-14},(n-1) d^{0-1}, n s^{2}\).
3. There are two f-series, namely 4f and 5f series, called lanthanoids and acùnoids respectively.
4. They shos intermediate properties as compared to electropositive s-block elements and electronegative p-block elements. Hence they are called (first) inner transition elements.

Question 103.
What are lanthanoids (or lanthanides)?
OR
What is the lanthanoid series?
Answer:

• Lanthanoids or Lanthanoid series or Lanthanones : The series of fourteen elements from ₅₈Ce to ₇₁Lu in which a differentiating electron enters 4f sub-shell and follows lanthanum is called lanthanoid series and the elements are called lanthanoids.
• They have general electronic configuration, [Xe] 4f^1-14 ,5d^0-1, 6s².
• They follow Lanthanum (Z = 57) in 3d-series.

Question 104.
What are rare earths?
Answer:

• Lanthanoids or 4f-block elements are called rare earths.
• Lanthanoids are never found in free state, and their minerals are not pure.
• They exhibit similar chemical properties hence cannot be extracted and separated by normal metallurgical processes.
• Lanthanoid metals are available on small scale. Therefore they are called rare earths.

Question 105.
Explain the position of lanthanoids in the periodic table.
OR
How is the position of lanthanoids justified?
Answer:

1. Position of Lanthanoids in the periodic table : Group – 3; Period – 6.
2. They interrupt the third transition series of t/-block elements (i.e. 5 d series) in the sixth period.
3. They are 14 elements from ₅₈Ce to ₇₁Lu and their position is in between La and Hf. Since they follow lanthanum, they are called lanthanoids.
4. They are called 4f-series elements and for the convenience, they are placed separately below the main periodic table.
5. The actual position of lanthanoids is in between Lanthanum (Z = 57) and Hafnium (Z = 72).
6. Their position is justified due to following reasons :
   • All these elements have the same electronic configuration in ultimate and penultimate shells, one electron in 5d-orbital and two electrons in 6s-orbital.
   • Group valence of all lanthanoids is 3.
   • All lanthanoids from ₅₈Ce to ₇₁Lu have similar physical and chemical properties.

Question 106.
Explain the meaning of inner-transition series.
Answer:
A series of f-block elements having electronic configuration (n – 2)f^1-14 (n – I) d^0-1 ns² placed separately in the periodic table represents inner transition series. The f-orbitals lie much inside the e/ orbitals.

Since the last electron enters pre-penultimate shell, these elements are inner transition elements.

There are two inner transition series as follows :
4f-series ₅₈Ce → ₇₁Lu
5f-series ₉₀Th → ₁₀₃Lr

Question 107.
Draw a skeletal diagram of the periodic table to show the position of d and/- block elements.
Answer:

Question 108.
What are the properties of lanthanoids?
Answer:

• Lanthanoids are soft metak with silvery white colour, Colour and brightness reduces on exposure to air.
• They are good conductors of heat and electricity.
• Except promethium (P_m), all are non-radioactive in nature.
• The atomic and ionic radii decrease from La to Lu. (Lanthanoid contraction).
• Coordination numbers arc greater than 6.
• They are paramagnetic.
• They become ferromagnetic at lower temperature.
• Their magnetic and optical properties are independent of environment.
• They are called rare earths as their exiractioli was difficult.
• They are abundant in earth’s crust
• All lanthanoids fonn hydroxides which are ionic and basic. l3asicity decreases with atomic number,
• They react with nitrogen to give nitrides and with halogen to give halides.
  
• When heated with carbon at very high temperature give carbides
  

Question 109.
Explain the variations in ionisation enthalpy of lanthanoids.
Answer:

• The first ionisation enthalpy of lanthanoids is nearly same. It is very high for Gd and Yb.
• The ionisation enthalpy increases from first (IE₁] to third (IE₃).

First, second and third ionization enthalpies of lanthanoids in kj/mol

Lanthanoid	IE1	IE2	IE3
La	538.1	1067	1850.3
Ce	528.0	1047	1949
Pr	523.0	1018	2086
Nd	530.0	1034	2130
Pm	536.0	1052	2150
Sm	543.0	1068	2260
Eu	547.0	1085	2400
Gd	592.0	1170	1990
Tb	564.0	1112	2110
Dy	572.0	1126	2200
Ho	581.0	1139	2200
Er	589.0	1151	2190
Tm	596.7	1163	2284
Yb	603.4	1175	2415
Lu	523.5	1340	2022

Question 110.
Give the general electronic configuration of 4f-series elements (OR lanthanoids).
Answer:

• The general electronic configuration of 4f-series elements is, Ln[Xe]⁵⁴ 4f^1-14 5d^0-1 6s² where Ln is a lanthanoid.
• Xenon has electronic configuration, [Xe] : Is² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶.
• In lanthanoids, the differentiating electron enters prepenultimate shell, 4f m

Question 111.
What are the important features of the electronic configuration of lanthanoids?
Answer:

1. Lanthanoids show two types of electronic configurations
   (a) an expected or idealized
   (b) an observed electronic configuration.
   In the idealized electronic configuration, the filling of the 4/-orbitals is regular but in the observed configuration, there is the shift of a single electron from 5d to 4/ sub-shell.
2. Lanthanum (57) has an electronic configuration [Xe] 4f° 5d¹6s². It does not have any f-electron.
3. The next incoming electron does not enter the 5d sub-shell but goes to the 4f sub-shell.
4. 14 electrons are progressively filled in the 4f sub-shell as the atomic number increases by one unit from La to Lu.
5. La, Gd and Lu are the only elements which possess one electron in a 5d orbital, while in all other lanthanoids the 5d sub-shell is empty.
6. La-(4f°), Gd-(4f⁷) and Lu-(4f¹⁴) posses extra stability due to their empty, half-filled and completely filled 4f-orbitals respectively.
7. The 4f-electrons in the prepenultimate shell are shielded by the outermost higher orbitals, 5s², 5p⁶, 5d¹, 6s², i.e. by eleven electrons, hence they are less effective in chemical bonding.

Electronic configuration (Idealised and observed)

[Xe]⁵⁴ ls²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶

Question 112.
Write the expected electronic configuration of (a) Nd (Z = 60) (b) Tm (Z = 69).
Answer:
Expected electronic configuration :
(a) Nd = [Xe] 4f³ 5d¹ 6s²
(b) Tm= [Xe] 4f¹⁴5d¹6s²

Question 113.
Write electronic configurations of
(i) Gd
(ii) Yb.
Answer:
(i) ₆₄Gd [Xe] 4f⁷5d¹6s² (Observed)
(ii) ₇₀Yb [Xe] 4f¹⁴5d°6s² (Observed)

Question 114.
Write expected and observed electronic configurations of
(i) Ce
(ii) Tb.
Answer:

Element	Expected (Idealised)	Observed
(i) 58Ce	[Xe] 4f15d16s2	[Xe] 4f25d°6s2
(ii) 65Tb	[Xe] 4f85d16s2	[Xe] 4f95d°6s2

Question 115.
Why are the expected and observed ground state electronic configurations of gadolinium and lawrencium same?
Answer:

• The degenerate orbitals like 4f and 5f acquire extra stability when they are half filled (4f⁷) or completely filled (5f¹⁴).
• The expected and observed electronic configuration of gadolinium is, ₆₄Gd [Xe] 4f⁷ 5d¹ 6s².
• The expected and observed electronic configuration of lawrencium is ₁₀₃Lr [Rn] 5f¹⁴ 6d¹ 7s².

Question 116.
Explain oxidation states of lanthanoids.
Answer:

• The common oxidation state of the Lanthanoids is 3 + due to the loss of 2 electrons from outermost 6s orbital and one electron from the penultimate 5d sub-shell.
• Gd³⁺ and Lu³⁺ show extra stability due to their half-filled and completely filled f-orbitals, Gd³⁺ = [Xe]4f⁷, Lu³⁺ = [Xe]4f¹⁴
• Ce and Tb attain the 4f° and 4f⁷configurations in the 4 + oxidation states. Eu and Yb attain the 4f⁷ and 4f¹⁴ configurations in the 2 + oxidation states. Sm and Tm also show the 2+ oxidation state although their stability can be explained based on thermodynamic factors.
• Some lanthanoids show 2 + and 4 + oxidation states even though they do not have stable electronic configuration of 4f°, 4f⁷ or 4f¹⁴. E.g. Pr⁴⁺ (4f¹), Nd²⁺ (4f⁴), Sm²⁺ (4f⁶), Dy⁴⁺ (4f⁸) etc

Question 117.
Write the. electronic configuration of the following ions :
(1) La^{3 +} ;
(2) Gd³⁺;
(3) Eu³⁺;
(4) Ce³⁺.
Answer:
(1) La^{3 +} = [Xe]
(2) Gd³⁺ = [Xe] 4f⁷
(3) Eu³⁺ = [Xe] 4f⁶
(4) Ce³⁺ = |Xe] 4f¹

Question 118.
Write the electronic configuration of
(1) Nd²⁺
(2) Nd³⁺
(3) Nd⁴⁺.
Answer:
(1) Nd²⁺ [Xe] 4f⁴
(2) Nd³⁺ [Xe] 4f³
(3) Nd⁴⁺ [Xe] 4f²

Question 119.
Among the following lathanoids, which elements show only one oxidation state 3 +? Why? Dy, Gd, Yb, Lu.
Answer:
Gd and Lu show only one oxidation state 3 +, since they acquire electronic configurations with extra stability namely 4f⁷ and 4f¹⁴ respectively.

Question 120.
Write the expected electronic configurations of :
(1) europium (Z = 63),
(2) erbium (Z = 68).
Answer:
(1) Europium (₆₃Eu) [Xe]⁵⁴4f⁶ 5d¹ 6s²
(2) Erbium (₆₈Er) [Xe]⁵⁴4f¹¹ 5d¹ 6s²

Question 121.
Why does lanthanum form La³⁺ ion, while cerium forms Ce⁴⁺ ion? (Atomic number La = 57 and Ce = 58).
Answer:

1. Electronic configuration Lanthanum is La [Xe] 4f° 5d¹ 6s². By losing three electrons, La acquires stable electronic configuration of Xe and forms La³⁺.
2. Electronic configuration of Cerium is Ce [Xe] 4f¹ 5d¹ 6s². By losing four electrons, Ce acquires stable electronic configuration of Xe and forms Ce⁴⁺.

Question 122.
₆₃EU and ₇₀Yb show 2 + oxidation state. Explain.
Answer:
₆₃EU has electronic configuration, [Xe] 4f⁷ 5d°6s². By losing 2 electrons from 6s orbital, it acquires stable configuration and 4f-orbital is half-filled.
₇₀Yb has electronic configuration, [Xe] 4f¹⁴ 5d° 6s². By losing 2 electrons from 6 s orbital, it acquires stable configuration and 4/-orbital is completely filled.
Hence Eu and Yb show 2 + oxidation states.

Question 123.
Display electronic configuration, atomic and ionic radii of lanthanoids.
Answer:
Answers are given in bold.

Electronic configuration and atomic ionic radii of lanthanoids

Question 124.
Explain the trend in atomic and ionic sizes of lanthanoids.
Answer:

• From ₅₇La (187 pm) to first element of 4f-series ₅₈Ce (183 pm), the contraction in atomic radius is very large, 4 pm.
• But from Ce onwards as atomic number increases atomic radius decreases very steadily so that total decrease in atomic radius from Ce to Lu is only 10 pm.
• In case of tripositive ions due to large pull by nucleus, the decrease in ionic radii is slightly more, i.e. 18 pm. For example, Ce³⁺ (103 pm) to Lu³⁺ (85 pm ).
• Hence all lanthanoids have similar properties. Therefore they cannot be separated from each other easily by normal metallurgical methods but require special methods.
  

Question 125.
What is meant by lanthanoid contraction?
Answer:
Lanthanoid contraction : The gradual decrease in atomic and ionic radii of lanthanoids with the increase in atomic number is called lanthanoid contraction.

Question 153.
Explain the causes of the lanthanoid contraction.
Answer:
The causes of the lanthanoid contraction are as follows :

• As the atomic number of lanthanoids or 4f-block elements increases the positive nuclear charge increases and correspondingly electrons are added to the prepenultimate 4f sub-shell.
• The attraction of nucleus on 4 f-electrons increases with the increase in atomic number.
• The outer eleven electrons namely, 5s², 5p⁶, 5d³ and 6s² do not shield inner 4 f-electrons from the nucleus.
• There is imperfect shielding of each 4f-electron from other 4 f-electrons.
• As compared to d sub-shell, the extent of shielding for 4 f-electrons is less.
• Due to these cumulative effects, 4 f-electrons experience greater nuclear attraction and hence valence shell is pulled towards the nucleus to the greater extent decreasing atomic and ionic radii appreciably.
• From 57La to 58Ce, there is a sudden contraction in atomic radius from 187 pm to 183 pm but the further decrease up to the last 4f-element, ₇₁Lu is comparatively low (about 10 pm).

Question 126.
Explain lanthanoid contraction effect with respect to (1) decrease in basicity, (2) ionic radii of post-lanthanoids.
Answer:
The lanthanoid contraction has a definite effect on the properties of lanthanoids as well as on the properties of post-lanthanoid elements.
(1) Decrease in basicity :

• In lanthanoids due to lanthanoid contraction, as the atomic number increases, the size of the lanthanoid atoms and their try positive ions decreases, i.e. from La³⁺ to Lu³⁺.
• As size of the cation decreases, according to Fajan’s rule, the polarizability increases and thus the covalent character of the M-OH bond increases, and ionic character decreases.
• Therefore the basic nature of the hydroxides decreases.
• Basicity and ionic character decrease in the order La(OH)₃ > Ce(OH)₃ > … Lu(OH)₃.

(2) Ionic radii of post-lanthanoids :

• Elements following the lanthanoids in the 6th period (third transition series, i.e. 5d-series) are known as post-lanthanoids.
• Due to lanthanoid contraction the atomic radii (size) of elements which follow lanthanum in the 6th period (3rd transition series – Hf, Ta, W, Re)-are similar to the elements of the 5th period (4d-series Zr, Nb Mo, Tc).
• Due to similarity in their size, post-lanthanoid elements (5d-series) have closely similar properties to the elements of the 2nd transition series (4d-series) which lie immediately above them.
• Pairs of elements namely Zr-Hf(Gr-4), Nb-Ta (Gr-5), Mo-W(Gr-6), Tc-Re (Gr-7) are called chemical twins since they possess almost identical sizes and similar properties.

Question 127.
Why do lanthanoids form coloured compounds?
Answer:

• The colour in lanthanoid ions is due to the presence of unpaired electrons in partially filled 4f sub-shells.
• Due to the absorption of radiations in the visible region there arises the excitations of the unpaired electrons from f-orbital of lower energy to the f-orbital of higher energy-giving f → f transitions.
• The observed colour is complementary to the colour of the light absorbed.
• The colour of try positive ions (M³⁺) depends upon the number of unpaired electrons in f-orbitals. Hence the lanthanoid ions having equal number of unpaired electrons have similar colour.
• The colours of M³⁺ ions of the first seven lanthanoids, La³⁺ to Eu³⁺ are similar to those of seven elements Lu³⁺ to Tb³⁺ in the reverse order.

Question 128.
Explain, why Ce³⁺ ion is colourless.
Answer:

• The electronic configuration of Ce³⁺ is, [Xe] 4f⁷
• Even though there is one unpaired electron in 4f sub-shell, the f → f transition involves very low energy. Hence, Ce³⁺ ion does not absorb radiation in the visible region.

Therefore Ce³⁺ ion is colourless.

Question 129.
Explain why Gd³⁺ is colourless.
Answer:

• Gd³⁺ has electronic configuration, [Xe] 4f⁷
• Due to extra stability of half filled orbital, it does not allow f → f transition, and hence does not absorb radiations in the visible region.

Hence Gd³⁺ is colourless.

Question 130.
The salts of (1) La³⁺ and (2) Lu³⁺ are colourless. Explain.
Answer:
(1) (i) La³⁺ has electronic configuration, [Xe] 4f°
(ii) Since there are no unpaired electrons in 4 f-orbital, f → f transition is not possible. Hence La³⁺ ions do not absorb radiations in visible region, and they are colourless.

(2) (i) LU³⁺ has electronic configuration [Xe] 4 f¹⁴
(ii) Since there are no unpaired electrons in 4f-orbital, f → f transition is not possible. Hence Lu³⁺ ions do not absorb radiations in visible region and they are colourless.

Question 131.
Explain giving examples, the colour of nf electrons is about the same as those having (14-n) electrons.
Answer:
(1) Consider Pr³⁺ and Tm³⁺ ions.
Tm³⁺ (4f¹²) has nf electron 12 electrons.
Pr²⁺ (4f²) has (14 – n) = (14 – 2) = 12 electrons. Both, Tm³⁺ and Pr³⁺ are green.

(2) Consider Nd³⁺ and Er³⁺ ions. Er³⁺ (4f¹¹) has nf electrons 11.
Nd³⁺ (4f³) has (14 – n) is (14 – 3) = 11 electrons. These both ions Er³⁺, Na³⁺ are pink in colour.

Question 132.
Lu³⁺ has observed magnetic moment zero. How many unpaired electrons are present?
Answer:
Since magnetic moment is zero, it has no unpaired electrons.

Question 133.
What are the application of lanthanoids?
Answer:

1. Lanthanoid compounds are used inside the colour television tubes and computer monitor. For example mixed oxide (Eu, Y)₂ O₃ releases an intense red colour when bombarded with high energy electrons.
2. Lanthanoid ions are used as active ions in luminescent materials. (Optoelectronic application)
3. Nd : YAG laser is the most notable application. (Nd : YAG = neodymium doped ytterium aluminium garnet)
4. Erbium doped fibre amplifiers are used in optical fibre communication systems.
5. Lanthanoids are used in cars, superconductors and permanent magnets.

Question 134.
What are actinoids? Give their general electronic configuration.
Answer:

• Actinoids : The series of fourteen elements from ₉₀Th to ₁₀₃Lr which follow actinium (₈₉Ac) and in which differentiating electrons are progressively filled in 5f-orbitals in prepenultimate shell are called actinoids.
• Their general electronic configuration is, [Rn]⁸⁶ 5f^1-14 6d^0-1 7s².

Question 135.
Why are actinoids called inner transition elements?
Answer:

• Actinoids are 5f-series elements in which electrons progressively enter into 5f-orbitals, which are inner orbitals.
• They have electronic configuration [Rn]⁸⁶ 5f^1-14 6d^0-1 7s².
• They show intermediate properties as compared to electropositive 5-block elements and electronegative p-block elements. Hence they are called second inner transition elements.

Question 136.
Explain the position of actinoids in the periodic table.
OR
What is the position of actinoids in the periodic table?
Answer:

• Position of actinoids in the periodic table : Group-3; Period-7.
• They interrupt the fourth transition series (6d series) in the seventh period in the periodic table.
• After Actinium, 89Ac which has electronic configuration [Rn] 6d¹7s², the electrons enter progressively 5f orbital and they have general electronic configuration, [Rn] 5f^{1 – 14} 6d^{0 – 1} 7s².
• They are fourteen elements from ₉₀Th to ₁₀₃Lr and since they follow actinium, they are called actinoids.
• They are called 5f series or second inner transition series elements and for the convenience they are placed separately below the periodic table.

Question 137.
Write idealised and observed electronic configuration of actinoids.
Answer:

Question 138.
Explain the oxidation states of actinoids.
Answer:

• Due to availability of electrons in 5f, 6d and 7s sublevels, lanthanoids show varied oxidation states.
• The most common oxidation state is + 3 due to loss of one electron from 6d and two electrons from 6s-orbitals.
• Ac, Th and Am show + 2 oxidation state.
• Th, Pa, U, Np, Pu, Am and Cm show + 4 oxidation state.
• Np and Pu show the highest oxidation state + 7.
• U, Np, Bk, Cm and Am show stable oxidation state + 4.
• In + 6 oxidation state, due to high charge density the actinoid ions form oxygenated ions, e.g. \(\mathrm{UO}_{2}^{+}, \mathrm{NpO}_{2}^{+},\) etc.

Question 139.
Why do actinoids show variable oxidation states?
Answer:

• The large number of variable oxidation states of actinoids is due to very small energy difference between 5f, 6d and 7s subshells.
• The electronic configuration of actinoids is, [Rn] 5f^1-14 6d^0-1, 7s²
• Due to the loss of three electrons from 6d¹ and 7s², the common oxidation state is + 3, but due to further loss of electrons from 5f subshell, actinoids show higher oxidation states.
• The variable oxidation states are + 2 to + 7.

Electronic configuration of actinoids and their ionic radii in + 3 oxidation state

Question 140.
What is meant by actinoid contraction?
Answer:
Actinoid contraction: The gradual decrease in atomic and ionic radii of actinoids with the increase in atomic number is called actinoid contraction.

Question 141.
The extent of actinoid contraction is greater than lanthanoid contraction. Explain Why?
Answer:

• The electronic configurations of :
  Lanthanoids [Xe] 4f^{1 – 14} 5d^{0 – 1} 6s²
  Actinoids [Rn] 5f^{1 – 14}, 6d^{0 – 1} 7s²
• The mutual screening offered in case of 5f-electrons is less than that in the 4f-electrons.
• Hence, the outer orbitals are pulled to the greater extent by nuclei in actinoids (5f-series) than in lanthanoids (4f-series).
• Therefore, actinoid contraction is greater than lanthanoid contraction.

Question 142.
Describe the important properties of actinoids.
Answer:
Properties of actinoids :

• Actinoids are silvery white ( similar to lanthanoids).
• They are highly reactive radioactive elements.
• Most of these elements are not found in nature. They are radioactive and man made.
• They experience decrease in the atomic and ionic radii from Ac to Lw, known as actinoid contraction.
• The common oxidation state is +3. Elements of the first half of the series exhibit higher oxidation states.

Question 143.
What are the applications of actinoids?
Answer:

• Thorium oxide (ThO₂) with 1% CeO₂ is used as a major source of indoor lighting, as well as for outdoor camping.
• Uranium is used in the nuclear reactors.
• The isotopes of Thorium and Uranium have very long half-life, so that we get very negligible radiation from them: Hence they can be used safely.

Question 144.
What are transuranic elements?
Answer:

• The man-made elements heavier titan Uranium (Z = 92) in the Actinoid señes are called transuranic elements.
• These are synthetically or artificially prepared (man-made) elements starting from Neptunium (Z= 93).
• Transuranic elements arc generally considered to be from Neptunium (Z = 93) to Lawrencium (Z = 103).
• Recently elements from atomic number 104 (Rf) to atomic number 118 (Og) or (Uuo) in 6 d series have also been identified as transuranic elements.
• All transuranic elements are radioactive.

Question 145.
What are post actinoid elements?
Answer:

• Elements from atomic number 104 to 118 are called postactinoid elements.
• The post actinoid elements known so far are transition metals.
• They can be synthesised in the nuclear reactions.
• As they have very short half life period, it is difficult to study their chemistry.
• Ruiherfordium forms a chloride (RfCl₄) similar to zirconium and hafnium in + 4 oxidation state.
• Dubniurn resembles niobium and protactinium.

Question 146.
Name the transuranic elements.
Answer:
Names of transuranic elements

Name	Symbol	Atomic number
Neptunium	Np	93
Plutonium	Pu	94
Americium	Am	95
Curium	Cm	96
Berkelium	Bk	97
Californium	Cf	98
Einsteinium	Es	99
Ferminum	Fm	100
Mendelevium	Md	101
Nobelium	No	102
Lawrencium	Lr	103
Rutherfordium	Rf	104
Dubnium	Db	105
Seaborgium	Sg	106
Bohrium	Bh	107
Hassium	Hs	108
Meitnerium	Mt	109
Darmstadtium	Uun/Ds	110
Roentgenium	Uuu/Rg	111
Copernicium	Uub/Cn	112
Ununtrium	Uut	113
Ununquadium	Uuq	114
Ununpentium	Uup	115
Ununhexium	Uuh	116
Ununseptium	Uus	117
Ununoctium	Uuo	118

In the transuranic elements, elements from atomic number 93 to 103 are actinoids and from atomic number 104 to 118 are called postactinoid elements.

Question 147.
What are the similarities between lanthanides and actinides.
Answer:
Lanthanides and actinides show similarities as follows :

• Both, lanthanides and actinides show+ 3 oxidation state.
• In both the series, the f-orbitals are filled gradually.
• Ionic radius of the elements in both the series decreases with increase in atomic number.
• Electronegativity in both the series is low for all the elements.
• They all are highly reactive.
• The nitrates, perchlorates and sulphates of all elements are soluble while their hydroxides, theorides and carbonates
  are insoluble.

Question 148.
Differentiate between lanthanoids and actinoids.
Answer:

Lanthanoids	Actinoids
Electronic configuration [Xe] 4f1-14 5d0-1, 6s2	Electronic configuration [Rn] 5f1-14 6d0-1, 7s2
The differentiating electron enters the 4f subshell.	The differentiating electron enters the 5f subshell.
Except for Promethium all other elements occur in nature.	Except for Uranium and Thorium, all others are synthesized in the laboratory.
The binding energy of 4f electrons is higher.	5f-orbitals have lower binding energy.
Only Promethium is radioactive.	All elements are radioactive.
Besides 3 + oxidation state they show 2 + and 4 + oxidation states.	Besides 3 + oxidation state they show 2 + , 4 + , 5 + , 6 + , 7 + oxidation states.
They have a less tendency to form complexes.	They have greater tendency to form complexes.
Many lanthanoid ions are colourless. Their colour is not as deep and sharp as actinoids.	Actinoids are coloured ions. Their colour is deep, e.g. U3+ is red and U4+ is green.
Lanthanoids cannot form oxo-cations.	Actinoids form oxo-cations such as – UO2+, PuO2+, UO22+, PuO22+.
Lanthanoid hydroxides are less basic.	Actinoid hydroxides are more basic.
Lanthanoid contraction is relatively less.	Actinoid contraction from element to element is comparatively more.
Mutual shielding of 4f electrons is more.	Mutual shielding effect of 5f electrons is less.

Question 149.
Compare Pre-transition metals, Lanthanoid and transition metals.
Answer:

Multiple Choice Questions

Question 150.
Select and write the most appropriate answer from the given alternatives for each sub-question :

1. In transition elements, the different electron enters into
(a) ns subshell
(b) np subshell
(c) (n – 1) d subshell
(d) (n – 2)f subshell
Answer:
(c) (n – 1) d subshell

2. Chromium (Z = 24) has electronic configuration
(a) [Ar]4d^A 4s²
(b) [Ar] 4d⁵ 45¹
(c) [Ar] 3d⁵ 3s¹
(d) [Ar] 3d⁵ 4s¹
Answer:
(d) [Ar] 3d⁵ 4s¹

3. Manganese achieves the highest oxidation state in its compounds
(a) Mn₃O₄
(b) KMnO₄
(c) K₂MnO₄
(d) MnO₂
Answer:
(b) KMnO₄

4. The group which belongs to transition series is
(a) 2
(b) 7
(c) 13
(d) 15
Answer:
(b) 7

5. The last electron of transition element is called
(a) s-electron
(b) p-electron
(c) d-electron
(d) f-electron
Answer:
(c) d-electron

6. Which one of the following elements does NOT belong to first transition series?
(a) Fe
(b) V
(c) Ag
(d) Cu
Answer:
(c) Ag

7. The incomplete d-series is
(a) 3d
(b) 4d
(c) 5d
(d) 6d
Answer:
(d) 6d

8. The electronic configuration of Sc is
(a) [Ar] 3d² 4s²
(b) [Ar] 3d¹ 4s²
(c) [Kr] 3d¹ 4s²
(d) [Kr] 3d² 4s¹
Answer:
(b) [Ar] 3d¹ 4s²

9. The observed electronic configuration of copper is
(a) [Ar]¹⁸ 3d⁹ 4s²
(b) [Kr] 3d¹⁰ 45¹
(c) [Kr] 3d⁹ 4s²
(d) [Ar] 3d¹⁰ 45¹
Answer:
(d) [Ar] 3d¹⁰ 45¹

10. Fe belongs to the
(a) 3d-transition series elements
(b) 4d-transition series elements
(c) 5d-transition series elements
(d) 6d-transition series elements
Answer:
(a) 3d-transition series elements

11. Which one of the following elements does not exhibit variable oxidation states?
(a) Iron
(b) Copper
(c) Zinc
(d) Manganese
Answer:
(c) Zinc

12. In KMnO₄, oxidation number of Mn is
(a) 2+
(b) 4 +
(c) 6 +
(d) 7+
Answer:
(d) 7+

13. Which one of the following transition elements shows the highest oxidation state?
(a) Sc
(b) Ti
(c) Mn
(d) Zn
Answer:
(c) Mn

14. The colour of transition metal ions is due to
(a) s → s transition
(b) d → d transition
(c) p → p transition
(d) f → f transition
Answer:
(b) d → d transition

15. Which one of the following compounds is expected to be coloured?
(a) AgNO₃
(b) CuSO₄
(c) ZnCl₂
(d) CuCl
Answer:
(b) CuSO₄

16. The metal ion which is NOT coloured, is
(a) Fe³⁺
(b) V²⁺
(c) Zn²⁺
(d) Ti³⁺
Answer:
(c) Zn²⁺

17. A pair of coloured ion is
(a)Cu²⁺, Zn²⁺
(b)Cr³⁺ , Cu⁺
(c) Cd²⁺, Mn⁵⁺
(d) Fe²⁺, Fe³⁺
Answer:
(d) Fe²⁺, Fe³⁺

18. The highest oxidation state is shown by
(a) Fe
(b) Mn
(c) Os
(d) Cr
Answer:
(c) Os

19. Transition elements are good catalysts since
(a) they show variable oxidation states
(b) they have partially filled d-orbitals
(c) they have low I.P
(d) they have small atomic radii
Answer:
(a) they show variable oxidation states

20. Highest magnetic moment is shown by the ion
(a) V³⁺
(b) Co³⁺
(c) Fe³⁺
(d) Cr³⁺
Answer:
(c) Fe³⁺

21. The most common oxidation state of lanthanoids is
(a) +4
(b) +3
(c) +6
(d) +2
Answer:
(b) +3

22. Which one of the following elements belong to the actinoid series?
(a) Cerium
(b) Lutetium
(c) Thorium
(d) Lanthanum
Answer:
(c) Thorium

23. The total number of elements in each of f-series is
(a) 10
(b) 12
(c) 14
(d) 15
Answer:
(c) 14

24. The general electronic configuration of Lanthanoids is
(a) [Xe] 4f^{1 – 14} 5d^{0 – 1} 6s²
(b) [Xe] 4f^{2 – 14} 5d^{0 – 1} 6s²
(c) [Xe] 4f^{1 – 13} 5d^{0 – 1} 6s²
(d) [Xe] 4f^{0 – 14} 5d^{0 – 1} 6s¹
Answer:
(a) [Xe] 4f^{1 – 14} 5d^{0 – 1} 6s²

25. f-block elements are called ………………….
(a) transition elements
(b) representative elements
(c) inner transition elements
(d) alkalin earth metals
Answer:
(c) inner transition elements

26. Actinoids form coloured salts due to the transition of electrons in
(a) d – d
(b) f – f
(c) f – d
(d) s – f
Answer:
(b) f – f

27. In the periodic table, Gadolinium belongs to
(a) 4th Group 6th period
(b) 4th group 4th period
(c) 3rd group 5th period
(d) 3rd group 7th period.
Answer:
(d) 3rd group 7th period.

28. The transuranic elements are prepared by
(a) addition reaction
(b) substitution reactions
(c) decomposition reaction
(d) nuclear reactions
Answer:
(d) nuclear reactions