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Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 15 Introduction to Polymer Chemistry Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 15 Introduction to Polymer Chemistry

Question 1.
What are polymers?
Answer:
Polymers are high molecular mass macromolecules (10³ – 10⁷ u) and consist of repeating units of monomers.

Question 2.
Write the classification of polymers based on the source. Give examples.
OR
What are natural and synthetic polymers? Give two examples of each type.
Answer:
Based on the source polymers are classified as natural, semisynthetic and synthetic polymers.

1. Natural polymers : These polymers are obtained either from plants or animals, e.g., cellulose, jute, linene, rubber, silk.
2. Semisynthetic polymers : The fibres obtained by giving special chemical treatment to natural fibres (cellulose) and further regenerated are called semisynthetic polymers e.g., acetate rayon, viscose rayon, cuprammonium silk.
3. Synthetic polymers : The man made fibres prepared by polymerization of one monomer or copolymerization of
   two or more monomers are called synthetic polymers, e.g., nylon, terylene, polythene, etc.

Question 3.
Write the classification of polymers based on structure. Give examples
OR
Write the reactions involved in the preparation of (1) Polyvinyl chloride (PVC) (2) Polypropylene.
Answer:
Based on structure polymers are classified as linear chain polymers, branched chain polymers and network or cross linked polymers.

(1) Linear chain polymers : When the monomer molecules are joined together in a linear arrangement, the resulting polymer is straight-chain or long-chain polymer, e.g., polythene, PVC.

They have high melting points; high densities and high tensile strength.

(2) Branched-chain polymers : These polymers consist of long and straight chain with smaller side chains give rise to branched-chain polymers. They have low density. They have lower melting points and tensile strength. Polypropylene having methyl groups as branches.

(3) Network or cross-linked polymers : These polymers consist of cross-linking of chains by strong covalent bonds leading to a network-like structure. Cross-linking results from polyfunctional monomers, e.g., melamine, bakelite, vulcanization of rubber. These polymers are hard rigid and brittle.

Question 3.
How are polymers classified on the basis of mode of polymerization?
OR
Write the classification of polymers based on mode of polymerization.
Answer:
There are three modes of polymerization depending upon the types of reactions taking place between the monomers.

1. Addition polymerization (or chain growth polymerization)
2. Condensation polymerization (or step growth polymerization)
3. Ring opening polymerization
4. Addition polymerization or chain growth polymerization : It is a process of polymers by the repeated addition of a large number of monomers is called addition polymerization (like alkenes) without loss of any small molecules.
   Example : Formation of polyethylene from ethylene is well known example of addition polymerization. It is a chain growth polymerization.
5. Condensation polymerization or step growth polymerization : The process of formation of polymers from polyfunctional monomers with the elimination of some small molecules such as water, hydrochloric acid, methanol, ammonia is called condensation polymerization.
   
   Example : The formation of terylene, a polyester polymer, from ethylene glycol and terephthalic acid. Condensation polymerization is a step growth polymerization.
6. Ring opening polymerization : The process of formation of polymers from cyclic compounds (like lactams, cyclic ethers, etc.) by ring opening is called ring opening polymerization.
   Example : Polymerization of e-caprolactam.
   Ring opening polymerization proceeds by addition of a single monomer unit to the growing chain molecules. It is a step growth polymerization.

Question 4.
Classify the polymers given below as addition, condensation and ring opening polymers:
PVC, polythene, cyclic ethers, polyester, polyacrylonftrile. polystyrene.
Answer:

• Addition polymers: PVC, polythene. polyacrylonitrile. polystyrene.
• Condensation polymers: Polyester.
• Ring opening polymers : Cyclic ethers

Question 5.
Write the classification of polymers based on intermolecular forces. Give examples.
OR
In which dasses, are the polymers classified on the basis of Inter molecular forces?
Answer:
Molecular forces bind the polymer chains either by hydrogen bonds or Vander Waal’s forces. These forces are called intermolecular forces. On the basis of magnitude of intcrmolccular forces, polymers are further classified as ebstomers, fibres, thermoplastic polymers. thermosetting polymers.

(1) Elastomers: Weak van der Waals type of intermolecular forces of attraction between the polymer chains are observed in cbstomcrs. When polymer is stretched, the polymer chain stretches and when the strain is relieved the chain returns to its odginal position, Thus, polymer shows elasticity and is called elastomers. Elastomers, the elastic polymers, have weak van der Waals type of intermolecular forces which permit the polymer to be stretched. Lilastorners are soft and stretchy and used in making rubber bands. E.g.. neoprene, vulcanized rubber, buna.S, buna-N.

(2) Fibres : It consists of strong intermolecular forces of’ attraction due to hydrogen bonding and strong dipole-dipole forces. These polymers possess high tensile strength. Due to these strong intermolecular forces the fibres are crystalline in nature. They are used in textile industries, strung tyres. etc.. e.g., nylon, terylene.

(3) Thermoplastic polymers: These polymer possess moderately strong intermolecular forces of attraction between those of elastomers and fibres. These polymers arc called thermoplastic because they become soft on heating and hard on cooling. They are either linear or branched chain polymers. They can be remoulded and recycled. E.g. polyethenc, PVC, polystyrene.

(4) Thermosetting polymers: These polymers are cross linked or branched molecules and are rigid polymers. During their formation they have property of being shaped on heating. but they get hardened while hot. Once hardened these become infusible, cannot be softened by heating and therefore, cannot be remoulded and recycled.
This shows extensive cross linking by covalent bonds formed in the moulds during hardening/setting process while hot. E.g. Bakelite, urea formaldehyde resin.

Question 6.
What is meant by the term homopolymer?
Answer:
A polymer made from only one type of repeating units is called homopolymer. in some cases the repeating unit is formed by condensation of two distinct monomers. Examples. Polythene, PVC. Nylon-6.

Question 7.
What is meant by the term copolymer?
Answer:
A polymer made from two or more types of repeating units is called a copolymer. The different monomer units are randomly sequenced in the copolymer, e.g., Terylene, Nylon-6, 6, Buna-S, Buna-N.

Question 8.
Write the classification of polymers on the basis of biodegradability?
OR
(1) What are biodegradable polymers?
(2) What are nonbiodegradable polymers?
Answer:
Based un biodegradability, polymers are classified as biodegradable polymer and nonbiodegradable polymers.

(1) Biodegradable polymers: Polymers that are affected by microbes or disintegrate by themselves afler a certain period of time due to environmental degradation are called biodegradable polymers.

Examples: PHBV i.e., Polyhydroxy butyrate-CO-β-hydroxy valerate Dextron. Nylon-2-nylon-6.

(2) Non biodegradable polymers: Synthetic polymers do not disintegrate by themselves after a certain period or not affected by microbes, are called nonbiodegradhle polymers.

Examples: Bakelite, Nylon, Terylene.

Question 9.
Explain the following terms :
Answer:

1. Branched chain polymer : The polymer consists of long and straight chain with smaller side chains give rise to branched chain polymers, e.g. Polypropylene
2. Addition polymer : The polymer formed by the repeated addition of a large number of monomers (like alkenes) without loss of any small molecules are called addition polymers, e.g. polythene -[-CH₂ – CH₂-]_n. It is a chain growth polymerization.
3. Condensation polymer : The polymers formed by the repeated condensation reaction between polyfunctional monomers with the elimination of some molecules such as water, hydrochloric acid, methanol, ammonia are called condensation polymers, e.g. Nylon-6, 6.
4. Elastomers : Polymers in which the intermolecular forces of attraction between the polymer chains are the weakest. When polymer is stretched, it has ability to stretch and when the strain is relieved it returns to its original position. Thus, polymer shows elasticity and is called elastomers, e.g. natural rubber, neoprene, vulcanized rubber.
5. Homopolymer : A polymer made from only one type of repeating unit of one monomer is called homopolymer e.g. Polythene, PVC, etc.
6. Biodegradable polymer : Polymers which are affected by microbes or disintegrate by themselves after a certain period of time due to environmental degradation are called biodegradable polymers.
   Example : PHSV i.e. Polyhydroxy butyrate -CO-β-hydroxy valerate Dextron.

Question 10.
What is natural rubber?
Answer:
Natural rubber is a high molecular mass linear polymer of isoprene (2-methyl-1, 3-butadiene).

Question 11.
Write a note on natural rubber.
Answer:
Natural rubber is manufactured from rubber latex obtained from the rubber tree in the form of colloidal suspension. Reaction involved in the formation of natural rubber by the process of addition polymerization is as follows :

Natural rubber is -1, 4- polyisoprene. It exhibits elastic properties.

Question 12.
State properties of natural rubber.
Answer:

1. Polyisoprene molecule has cis configuration of the C = C double bond. It consists of various chains held together by weak van der Waals forces and has coiled structure.
2. It can be stretched like a spring and exhibits elastic property.
3. Its molecular mass varies from 130,000 u to 340,000 u.

Question 13.
Write a note on vulcanization of rubber. OR Discuss the main purpose of vulcanization of rubber.
Answer:
The tensile strength, toughness and elasticity of natural rubber can be increased by adding 3 to 5% sulphur and heating at 100-150°C, resulting in cross linking of cis-1, 4-polypropene chains through disulphide bonds, (-S-S-). This process is known as vulcanization of rubber. The physical properties of rubber can be changed by controlling the amount of sulphur in the vulcanization process. Rubber made with 20-30% sulphur is hard, 3 to 10% sulphur is little harder and is used in making tyres and 1 to 3% sulphur is used in making rubber bands.

Question 14.
Write the name and structure of one of the initiators used in free radical polymerisation.
Answer:
The initiator used in free radical polymerisation is acetyl peroxide.

Question 15.
What is LDP? How is it prepared? Give its properties and uses.
Answer:
LDP means low-density polyethene. LDP is obtained by polymerization of ethylene under high pressure (1000 – 2000 atm) and temperature (350 – 570 K) in presence of traces of O₂ or peroxide as initiator.

The mechanism of this reaction involves free radical addition and H-atom abstraction. The latter results in branching. As a result the chains are loosely held and the polymer has low density.

Properties of LDP :

• LDP films are extremely flexible, but tough chemically inert and moisture resistant.
• It is poor conductor of electricity with melting point 110 °C.

Uses of LDP :

• LDP is mainly used in preparation of pipes for agriculture, irrigation, domestic water line connections as well as insulation to electric cables.
• It is also used in submarine cable insulation.
• It is used in producing extruded films, sheets, mainly for packaging and household uses like in preparation of squeeze bottles, attractive containers, etc.

Question 16.
Whatis HDP ? How is it prepared ? Give its properties and uses ?
Answer:
HDP means high density polyethylene. It is a linear polymer with high density due to close packing.

HDP is obtained by polymerization of ethene in presence of Zieglar-Natta catalyst which is a combination of triethyl aluminium with titanium tetrachloride at a temperature of 333K to 343K and a pressure of 6-7 atm.

Properties of HDP :

• HDP is crystalline, melting point in the range of 144 – 150 °C.
• It is much stiffer than LDP and has high tensile strength and hardness.
• It is more resistant to chemicals than LDP.

Uses of HDP :

• HDP is used in manufacture of toys and other household articles like buckets, dustbins, bottles, pipes, etc.
• It is used to prepare laboratory wares and other objects where high tensile strength and stiffness is required.

Question 17.
How is polyacrylonitrile (PAN) prepared? Give its uses.
Answer:
Acrylonitrile (monomer) on polymerization (addition polymerization) in the presence of peroxide initiator gives polyacrylonitrile.

Uses : Polyacrylonitrile resembles wool and is used as wool substitute and for making orlon or acrilan.

Question 18.
How is nylon-6 prepared?
Write the reaction for the preparation of nylon 6.
Answer:
When epsilon (ε)-caprolactam is heated with water at high temperature it undergoes ring opening polymerization to form the polyamide polymer called nylon-6.

The name nylon-6 is given on the basis of six carbon atoms present in the monomer unit. Nylon-6 has high tensile strength and luster, nylon-6 fibres are used for manufacture of tyre cords, fabrics and ropes.

Question 19.
Draw the structures of polymers formed using the following monomers.

Answer:

Question 20.
How is Novolac prepared?
OR
Write the reaction to prepare Novolac polymer.
Answer:
The monomers phenol and formaldehyde undergo polymerisation in the presence of alkali or acid as catalyst.

Phenol reacts with formaldehyde to form ortho or p-hydroxy methyl phenols, which further reacts with phenol to form a linear polymer called Novolac. It is used in paints.

Question 21.
How is bakelite prepared?
Answer:
In the third stage, various articles are shaped from novolac by putting it in appropriate moulds and heating at high temperature (138 °C to 176 °C) and at high pressure forms rigid polymeric material called bakelite. Bakelite is insoluble and infusible and has high tensile strength.

Bakelite is used in making articles like telephone instrument, kitchenware, electric insulators frying pans, etc.

Question 22.
How is a melamine-formaldehyde polymer (melamine) prepared?
Answer:
The monomers formaldehyde and melamine undergo condensation polymerisation to form cross-linked melamine formaldehyde. It is used in making crockeries, decorative table tops like formica and plastic dinner-ware.

Question 23.
Write the preparation of the following synthetic rubbers and give their uses :
(1) Buna-S or styrene-butadiene rubber (SBR) (2) Neoprene rubber
Answer:
(1) Buna-S rubber : Its trade name is SBR (Styrene-butadiene rubber) Buna-S is a copolymer of styrene and 1, 3-butadiene. When 75 parts of butadiene and 25 parts of styrene subjected to addition polymerization by the action of sodium.

Uses : Buna-S is superior to natural rubber with regard to mechanical strength and has abrasion resistance. Hence, it is used in tyre industry.

(2) Neoprene : Neoprene, a synthetic rubber, is a condensation polymer of chloroprene (2-chloro-l, 3-butadiene), Chloroprene polymerizes rapidly in presence of oxygen.

Vulcanization of neoprene takes place in presence of magnesium oxide.

Uses : Neoprene is resistant to petroleum, vegetable oils, light as well as heat. It is used in making hose pipes for transport of gasoline and making gaskets. It is used for manufacturing insulator cable, jackets, belts for power transmission and conveying.

Question 24.
Write structure of natural rubber and neoprene rubber along with the name and structure of their monomers.
Answer:

Question 25.
Write the preparation of viscose rayon.
Answer:
Viscose rayon is a semisynthetic fibre. It is a regenerated cellulose. The molecular formula of cellulose is (C₆H₁₀O₅)_n. A modified representation of the molecular formula of cellulose Cell-OH is used in the reactions involved in viscose formation.

Cellulose (wood pulp) is treated with the concentrated NaOH to form alkali cellulose. It is then converted to xanthate by treating with CS₂. Further xanthate is mixed with dilute NaOH to form viscose solution which is extruted through spinnerates of spinning machine into acid bath, when regenerated cellulose fibres precipitate, i.e. viscose rayon.

Question 26.
How is PHBV polymer prepared?
Answer:
It is a copolymer. The monomers β-hydroxy butyric acid (3-hydroxy butanoic acid) and β-hydroxy valeric acid (3-hydroxy pentanoic acid) undergo polymerization to form PHBV polymer. It has an ester linkage. Hydroxyl group of one monomer forms ester link by reacting with carboxyl group of the other. Thus PHBV is an aliphatic polyester i.e. poly β-hydroxybutyrate-co-β-hydroxy valerate (PHBV). It is a biodegradable polymer.

Question 27.
Write the name/s of monomer/s, polymer structure and one use of each of the following polymers (trade name) :
Answer:

Question 28.
Write the names of monomers used in preparing following polymers :
(1) Dacron.
Answer:
Monomers : Ethylene glycol and Dmiethyl terephthalate (DMT)

(2) Bakelite.
Answer:
Monomers : o-hydroxy methyl phenol and formaldehyde.

(3) Nylon-6, 8.
Answer:
Monomers : Hexamethylene diamine and Hexamethylene dicarboxylic acid.
H₂N-(CH₂)₆-NH₂ HOOC-(CH₂)₆-COOH

(4) Melamine.
Answer:
Monomers : Melamine and Formaldehyde

(5) Buna-S.
Answer:

(6) Buna-N.
Answer:

(7) Butyl rubber.
Answer:

(8) Teflon.
Answer:
Monomers : F₂C = CF₂ Tetrafluoroethene

(9) Natural rubber.
Answer:
Monomers : 1,3-Butadiene
CH₂ = CH – CH = CH₂

(10) Neoprene.
Answer:
Monomers : Chloroprene or 2-Chloro-l,3-butadiene

Question 29.
Write the structures of monomers used in the preparation of following polymers.

Answer:
The monomers used in the preparation of above polymer are :

Answer:
The monomers used in the preparation of above polymer are :

Answer:
The monomer used in the preparation of above polymer are :

Answer:
The monomer used in the preparation of above polymer is

Question 30.
Following monomers are used to prepare polymers. Predict the structures of polymers:

(1) Ethylene glycol.
Answer:
Ethylene glycol is used in the preparation of polyester (terylene or dacron).

Terylene is polyester fibre formed by the polymerization of terephthalic acid and ethylene glycol.

Terylene is obtained by condensation polymerization of ethylene glycol and terephthalic acid in presence of catalyst zinc acetate and antimony trioxide at high temperature.

Properties :

• Terylene has relatively high melting point (265 °C)
• It is resistant to chemicals and water.

Uses :

• It is used for making wrinkle-free fabrics by blending with cotton (terycot) and wool (terywool), and also as glass reinforcing materials in safety helmets.
• PET is the most common thermoplastic which is another trade name of the polyester polyethylenetereph- thalate.
• It is used for making many articles like bottles, jams, packaging containers.

(2) ε-Caprolactam.
Answer:
ε-caprolactam is used in the preparation of nylon-6.

When epsilon (ε)-caprolactam is heated with water at high temperature it undergoes ring opening polymerization to form the polyamide polymer called nylon-6.

The name nylon-6 is given on the basis of six carbon atoms present in the monomer unit. Nylon-6 has high tensile strength and luster, nylon-6 fibres are used for manufacture of tyre cords, fabrics and ropes.

(3) Ethene.
Answer:
Ethene is used in the preparation of polythene
Polymer:

Linear chain polymers : When the monomer molecules are joined together in a linear arrangement, the resulting polymer is straight-chain or long-chain polymer, e.g., polythene, PVC.

They have high melting points; high densities and high tensile strength.

(4) Formaldehyde.
Answer:
Formaldehye is used in the preparation of bakelite.

The monomers phenol and formaldehyde undergo polymerisation in the presence of alkali or acid as catalyst.

Phenol reacts with formaldehyde to form ortho or p-hydroxy methyl phenols, which further reacts with phenol to form a linear polymer called Novolac. It is used in paints.

In the third stage, various articles are shaped from novolac by putting it in appropriate moulds and heating at high temperature (138 °C to 176 °C) and at high pressure forms rigid polymeric material called bakelite. Bakelite is insoluble and infusible and has high tensile strength.

Bakelite is used in making articles like telephone instrument, kitchenware, electric insulators frying pans, etc.

Question 31.
Classify the following polymers as step growth or chain growth polymers :
(1) Nylon-6,6
(2) Terylene
(3) Polyethene (4) PVC.
Answer:
Step growth polymers : Nylon-6,6, terylene
Chain growth polymers : Polythene, PVC.

Question 32.
Classify the following as linear, branched or cross linked polymers :
(1) Bakelite
(2) Starch
(3) Polythene
(4) Nylon.
Answer:
Linear polymers : Polythene, nylon.
Cross-linked polymers : Bakelite, starch.

Question 33.
Classify the following as addition and condensation polymers :
(1) Bakelite
(2) polyvinyl chloride
(3) polythene
(4) terylene.
Answer:
Addition polymers : Polyvinyl chloride, polythene
Condensation polymers : Bakelite, terylene.

Question 34.
Arrange the polymers in increasing order of their intermolecular forces :
Nylon-6,6, Polythene, Buna-S.
Answer:
The increasing order of their intermolecular forces of attraction follows the order :
Buna-S, Polythene, Nylon-6,6.

Question 35.
Classify the following as natural, synthetic and semisynthetic polymers :
Terylene, cuprammonium silk, jute, melamine
Answer:
Natural polymers : Jute
Synthetic polymers : Terylene, melamine
Semisynthetic polymers : Cuprammonium silk

Question 36.
Complete the following statements :
(1) Chemically teflon is …………………………. .
(2) …………………………. is the catalyst used to obtain HDP by polymerisation of ethene.
(3) Viscose rayon is a …………………………. .
Answer:
(1) polytetrafluoroethylene
(2) Zieglar-Natta
(3) semisynthetic fibre (regenerated fibre).

Question 37.
Answer the following in one sentence each.

(1) Name a polymer used for making LCD screen?
Answer:
The polymer used for making LCD screen is Perspex.

(2) Which of the two is a condensation polymer? Bakelite or Polythene?
Answer:
The condensation polymer is bakelite.

(3) Which of the two is a linear polymer? Nylon or Starch.
Answer:
The linear polymer is nylon.

(4) Which of the two is a step growth polymer? Nylon-6,6 or PVC.
Answer:
The step growth polymer is Nylon-6,6.

(5) Write the use of polyacrylamide gel.
Answer:
Polyacrylamide gel is used in electrophoresis.

(6) Write the use of urea formaldehyde resin.
Answer:
Urea formaldehyde resin is used in making unbreakable dinnerware and decorative laminates.

(7) Give an example of semisynthetic polymer.
Answer:
Semisynthetic polymer : Viscose rayon, cuprammonium silk.

(8) Write the monomer unit of teflon.
Answer:
Monomer unit of teflon : Tetrafluoroethene (F₂C = CF₂).

(9) Write the equation for the preparation of polypropylene.
Answer:

(10) Name a synthetic polymer which contains amide linkage.
Answer:
Polymer that contains amide linkage : Nylon-6; Nylon 6,6.

(11) Name a synthetic polymer which contains ester linkage.
Answer:
Polymer that contains ester linkage : Terylene or Dacron.

(12) Name one thermosetting and one thermoplastic polymer.
Answer:
Thermosetting polymer : Bakelite.
Thermoplastic polymer : Polythene, polystyrene.

(13) State the uses of biodegradable polymers.
Answer:
Biodegradable polymers are used as orthopaedic devices, implants, sutures and drug release matrices.

(14) Name a copolymer which is used for making nonbreakable crockeries.
Answer:
The polymer used in making nonbreakable crockeries : Melamine formaldehyde polymer.

(15) Write the structure of monomer used in the preparation of
Answer:
The structure of monomer :

(16) Write the structure of melamine.
Answer:
The monomers formaldehyde and melamine undergo condensation polymerisation to form cross-linked melamine formaldehyde. It is used in making crockeries, decorative table tops like formica and plastic dinner-ware.

(17) What does SBR stand for?
Answer:
SBR stand for styrene(S)butadiene (B) rubber (R).

(18) Draw the structure of repeating unit in nylon-6,10.
Answer:
The structure of repeating unit in nylon-6,10 is

(19) What are the monomers used to prepare nylon given below?

Answer:
Monomers used in the preparation of nylon are

(20) Write the monomers used to prepare nylon given below :

Answer:
Monomers used in the preparation of nylon are

(21) Name the catalyst which is formed from titanium chloride and triethyl aluminium.
Answer:
The catalyst Zieglar Natta is formed from titanium chloride and triethyl aluminium.

(22) Define molecular mass of polymer.
Answer:
Molecular mass of a polymer is an average of the molecular masses of the constituent molecules.

(23) Which factor determines the molecular mass of polymer.
Answer:
Molecular mass of polymer depends upon the degree of polymerization (DP). DP is the number of monomer units in a polymer molecule.

Multiple Choice Questions

Question 38.
Select and write the most appropriate answer from the given alternatives for each subquestion :

1. Which one is the natural polyamide polymer?
(a) Cuprammonium silk
(b) Wool
(c) Perlon-L
(d) Jute
Answer:
(b) Wool

2. The synthetic fibres are prepared from
(a) cellulose
(b) starch
(c) chemical compounds
(d) polymers
Answer:
(c) chemical compounds

3. Which of the following is NOT a vegetable fibre?
(a) Hemp
(b) Jute
(c) Cotton
(d) Keratin
Answer:
(d) Keratin

4. Which of the following fibres are made up of polyamides?
(a) Dacron
(b) Rayon
(c) Nylon
(d) Terylene
Answer:
(c) Nylon

5. An example of natural cellulose fibre is
(a) cotton
(b) wool
(c) silk
(d) rayon
Answer:
(a) cotton

6. Cellulose is the main constituent of
(a) nylon-6
(b) cotton
(c) terylene
(d) wool
Answer:
(b) cotton

7. Of the following, which group contains two cellulosic fibres and one protein fibre?
(a) Cotton, keratin, wool
(b) Linen, keratin, wool
(c) Cotton, linen, rayon
(d) Cotton, keratin, linen
Answer:
(d) Cotton, keratin, linen

8. Which of the following is not a vegetable fibres?
(a) Hemp
(b) Jute
(c) Cotton
(d) Keratin
Answer:
(d) Keratin

9. Which of the following is polyamide?
(a) Teflon
(b) Nylon 6, 6
(c) Terylene
(d) Bakelite
Answer:
(b) Nylon 6, 6

10. The monomer of e-caprolactam is
(a) styrene
(b) amino acid
(c) aminocaproic acid
(d) adipic acid O
Answer:
(c) aminocaproic acid

11. is the formula of _ Jn
(a) Nylon-66 salt
(b) Nylon-66
(c) DMT
(d) Nylon-6
Answer:
(d) Nylon-6

12. Nylon-6 is a
(a) polyester fibre
(b) protein fibre
(c) poly caprolactum fibre
(d) poly amine fibre
Answer:
(c) poly caprolactum fibre

13. The condensation product of e-caprolactum is
(a) teflon
(b) nylon-6
(c) nylon-66
(d) bakelite
Answer:
(b) nylon-6

14. \(\left[\overline{\mathrm{O}} \mathrm{OC}-\left(\mathrm{CH}_{2}\right)_{4}-\mathrm{COO}-\mathrm{N} \mathrm{H}_{3}-\left(\mathrm{CH}_{2}\right)_{6}-\mathrm{N} \mathrm{H}_{3}\right]\) is the formula of
(a) nylon-6
(b) nylon-6 salt
(c) nylon-66 salt
(d) nylon-66
Answer:
(b) nylon-6 salt

15. The starting material required for the preparation of Nylon-66 is
(a) glycol
(b) α-amino acid
(c) adipic acid and hexamethylene diamine
(d) dimethyl terephthalate and ethylene glycol
Answer:
(c) adipic acid and hexamethylene diamine

16. Terylene is also known as
(a) styrene
(b) butadiene
(c) dacron
(d) teflon
Answer:
(c) dacron

17. Terylene is
(a) vegetable fibre
(b) protein fibre
(c) polyester fibre
(d) polyamide fibre
Answer:
(c) polyester fibre

18. Terylene polymer is formed in
(a) the presence of nitrogen
(b) the presence of hydrogen
(c) the presence of oxygen
(d) the vacuum
Answer:
(d) the vacuum

19. The by-product formed during the preparation of terylene fibre is
(a) glycerol
(b) propylene glycol
(c) ethylene glycol
(d) ethyl alcohol
Answer:
(c) ethylene glycol

20. Nylon polymer cannot be used for making
(a) tyre cords
(b) films
(c) dress materials
(d) fishing nets
Answer:
(b) films

21. Glycol is an important constituent of
(a) nylon-6
(b) nylon-66
(c) terylene
(d) hexamethylene diammonium adipate
Answer:
(c) terylene

22. Terylene is prepared by the process of
(a) halogenation
(b) condensation
(c) esterification
(d) hydrogenation
Answer:
(b) condensation

23. What are the steps during polymerisation to form terylene?
(a) Terephthalic acid is condensed with ethylene glycol at 420 K-460 K.
(b) Ethylene glycol displaces methanol to form dihydroxy diethyl terephthalic acid
(c) Zinc acetate – antimony trioxide is used as catalyst
(d) All of these
Answer:
(d) All of these

24. During polymerisation of nylon salt to nylon-66, the conditions are
(a) room temperature and pressure
(b) temperature 503 K
(c) temperature 553 K in presence of Nitrogen
(d) heating in an autoclave at 373 K
Answer:
(c) temperature 553 K in presence of Nitrogen

25. Which one of the following is a condensation polymer?
(a) Nylon
(b) Polythene
(c) PVC
(d) Teflon
Answer:
(a) Nylon

26. Which one of the following is an addition polymer?
(a) Bakelite
(b) Nylon-6,6
(c) Polystyrene
(d) Terylene
Answer:
(c) Polystyrene

27. A polymer of butadiene and Acrylonitrile is called
(a) Buna-S
(b) Buna-N
(c) Buna-B
(d) Buna-A
Answer:
(b) Buna-N

28. Natural rubber is a polymer of
(a) Styrene
(b) Butadiene
(c) Vinyl chloride
(d) Isoprene
Answer:
(d) Isoprene

29. In which of the following pairs both are copolymers?
(a) PHBV, bakelite
(b) Polythene, terylene
(c) Polyacrylonitrile, nylon-6,6
(d) Polystyrene, melamine
Answer:
(a) PHBV, bakelite

30. The polymer used in paints is
(a) Nylon
(b) Glyptal
(c) Neoprene
(d) Terylene
Answer:
(b) Glyptal

31. Which of the following contains biodegradable polymers only?
(a) Cellulose, dextron, PHBV
(b) Starch, PHBV, PVC
(c) Bakelite, nylon-2-nylon-6, nylon-6,6
(d) Cellulose, starch, terylene
Answer:
(a) Cellulose, dextron, PHBV

32. Thermosetting polymer is
(a) Nylon-6
(b) Nylon-6,6
(c) Bakelite
(d) SBR
Answer:
(c) Bakelite

33. Nylon thread contains the polymer
(a) Polyamide
(b) Polyvinyl
(c) Polyester
(d) Polyethylene
Answer:
(a) Polyamide

34. Polythene, PVC, teflon and neoprene are all
(a) Monomers
(b) Homopolymers
(c) Copolymers
(d) Condensation polymers
Answer:
(b) Homopolymers

35. Which one of the following is NOT a biodegrad¬able polymer?
(a) Starch
(b) Cellulose
(c) Dextron
(d) Decron
Answer:
(d) Decron

36. The polymer used in making blankets (artificial wool) is
(a) Polyester
(b) Polyacrylonitrile
(c) Polythene
(d) Polystyrene
Answer:
(b) Polyacrylonitrile

37. Which one of the following is a linear polymer?
(a) Bakelite
(b) LDP
(c) Nylon
(d) Formaldehyde melamine polymer
Answer:
(c) Nylon

38. Which one of the following is a branched polymer?
(a) PVC
(b) Polyester
(c) Nylon
(d) Polypropylene
Answer:
(d) Polypropylene

39. The polymer used to make non-stick cookware is
(a) Polyethene
(b) Polystyrene
(c) Polytetrafluoroethylene
(d) Polyvinyl chloride
Answer:
(c) Polytetrafluoroethylene

40. The monomer used to prepare orlon is
(a) CH₂ = CH-CN
(b) CH₂ = CHCl
(c) CH₂ = CH-F
(d) CH₂ = CF₂
Answer:
(a) CH₂ = CH-CN

41. Buna-N rubber is a copolymer of

Answer:
(b)

42. Bakelite is a polymer of
(a) Formaldehyde and phenol
(b) Benzaldehyde and phenol
(c) Formaldehyde and benzyl alcohol
(d) Acetaldehyde and phenol
Answer:
(a) Formaldehyde and phenol

43. The process involving heating of natural rubber with sulphur is known as
(a) Sulphonation
(b) Vulcanisation
(c) Galvanisation
(d) Calcination
Answer:
(b) Vulcanisation

44. A polymer of bisphenol and phosgene is called
(a) Polyamide
(b) Glyptal
(c) Polycarbonate
(d) Polystyrene
Answer:
(c) Polycarbonate

45. Thermocol is a homopolymer of
(a) terephthalic acid
(b) acrylonitrile
(c) methyl a-cyanoacrylate
(d) styrene
Answer:
(d) styrene

46. The polymer is used to prepare shatter resistant glass is called
(a) Perspex/acrylic glass
(b) Soda glass
(c) Buna N
(d) Polyacrylamide
Answer:
(a) Perspex/acrylic glass

47. A polymer used in making shoe soles is
(a) Glyptal
(b) Buna-N
(c) Buna-S
(d) Poly carbonate
Answer:
(b) Buna-N

48. The Zieglar-Natta catalyst is used in the preparation of
(a) LDPE
(b) PHBV
(c) PAN
(d) HDPE
Answer:
(d) HDPE

49. Which of the following is natural rubber?
(a) cis-1, 4-polyisoprene
(b) neoprene
(c) Trans-1. 4-polyisoprene
(d) Butyl rubber
Answer:
(a) cis-1, 4-polyisoprene

50. Which one from the following is the Terylene polymer?

Answer:
(b)

51. Equivalent amount of Hexamethylene diamine and adipic acid on complete neutralization produces :

Answer:
(a)

52. Polyhydroxy butyrate-CO-β-hydroxy valerate represents
(a) Dextron
(b) Nylon-6, 6
(c) Nylon-2-nylon-6
(d) PHBV
Answer:
(d) PHBV

53. Among the following, the biodegradable polymer is
(a) PVC
(b) Nylon-6, 6
(c) Nylon-2-nylon-6
(d) Neoprene
Answer:
(c) Nylon-2-nylon-6

54. The monomers of Nylon-2-nylon-6 are
(a) glycine and ω-amino caproic acid
(b) lactic acid and glycolic acid
(c) glycolic acid and co-amino caproic acid
(d) isobutylene and isoprene
Answer:
(a) glycine and ω-amino caproic acid

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 14 Biomolecules Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 14 Biomolecules

Question 1.
What are biomolecules? Give examples of biomolecules.
Answer:
Biomolecules: The lifeless, complex organic molecules which combine in a specific manner to produce life or control biological reactions are called biomolecules.

Examples: Carbohydrates, lipids (fats and oils), nucleic acids, enzymes.

Question 2.
What is the importance of biomolecules?
Answer:
Biomolecules are organic molecules which combine in a particular fashion to give complex substances which help to sustain life and produce identical daughter cells and play an important role in the actions of an organism.

• Carbohydrates are the major constituents of food and source of energy.
• Proteins help in proper functioning of living beings. They are important constituents of skin, hair, muscles. Enzymes which catalyse chemical reactions that take place in cells are proteins.
• Lipids (fats and oils) function as the storehouses of energy.
• Nucleic acids, the ribonucleic acid (RNA), and deoxyribonucleic acid (DNA) are responsible for genetic characteristics and synthesis of proteins.

Question 3.
What are carbohydrates?
OR
Define the term : Carbohydrates.
Answer:
Carbohydrates : Carbohydrates are optically active polyhydroxy aldehydes or polyhydroxy ketones, or the compounds which on hydrolysis produce polyhydroxy aldehydes or polyhydroxy ketones.

Examples : Glucose, sucrose, fructose.

Question 4.
What is monosaccharide?
Answer:
The basic unit of all carbohydrates which is a simple carbohydrate and cannot be hydrolysed further is known as monosaccharide. The monosaccharide is crystalline and soluble in water. E.g. Glucose, fructose, ribose.

Question 5.
Mention the names of monosaccharides or simple carbohydrates.
Answer:
Monosaccharides are (1) glucose (2) fructose (3) ribose.

Question 6.
State the basic unit of all carbohydrates.
Answer:
The basic unit of all carbohydrates which is a simple carbohydrate and cannot be hydrolysed further is known as monosaccharide.

Question 7.
How are carbohydrates classified?
OR
Classification of carbohydrates with examples.
Answer:
Carbohydrates are classified as monosaccharides oligosaccharides and polysaccharides.
(1) Monosaccharides : These carbohydrates cannot be further hydrolysed into smaller units. They are basic units of all carbohydrates, and are called monosaccharides.

Examples : Glucose, fructose, ribose

(2) Oligosaccharides : An oligosaccharide is a carbohydrate (sugar) which on hydrolysis gives two to ten monosaccharide units.
Depending on the number of monosaccharides produced on hydrolysis, oligosaccharides are further classified as :

Oligosaccharide is homogeneous. In this, each molecule of oligosaccharide contains the same number of monosaccharide units joined together in the same order as every other molecule of the same oligosaccharide.

(3) Polysaccharides : These are carbohydrates which on hydrolysis give a large number of monosaccharides.

Polysaccharides are tasteless, amorphous, insoluble in water. They are long chain, naturally αcurring polymers of carbohydrates.

Example : Cellulose, starch, glycogen.

Question 8.
Classify the following carbohydrates into Monosaccharide, Disaccharide, Oligosaccharide, Polysaccharide.
(1) Glucose
(2) Starch
(3) Sucrose
(4) Maltose
(5) Galactose
(6) Lactose
(7) Ribose.
Answer:

Question 9.
Classify the following carbohydrates.
(1) Cellulose,
(2) Maltose,
(3) Raffinose,
(4) Fructose.
Answer:

Question 10.
Classify the following into monosaccharides, oligosaccharides and polysaccharides.
(1) Starch
(2) Glucose
(3) Stachyose
(4) Maltose
(5) Raffinose
(6) Cellulose
(7) Sucrose
(8) Lactose.
Answer:

Question 11.
Classify the following into monosaccharides and disaccharides.
Ribose, maltose, galactose, fructose and lactose (~2 mark each)
Answer:

Question 12.
Give the preparation of glucose from sucrose or cane sugar.
OR
Describe the laboratory method of preparation of glucose.
Answer:
Preparation of glucose from sucrose (cane sugar) : Laboratory method.

Glucose is prepared in the laboratory by hydrolysis of sucrose by boiling it with dilute hydrαhloric acid or dilute sulphuric acid for about two hours. On hydrolysis, sucrose gives one molecule of glucose and one molecule of fructose.

Alcohol is added during cooling to separate glucose and fructose since, glucose is almost insoluble in alcohol, hence it crystallizes out first. Fructose remains in the solution as it is more soluble than glucose.

Crystals of glucose are separated out by filtration and purified by recrystallization.

Question 13.
Give the preparation of glucose from starch.
OR
How is glucose prepared on commercial scale?
Answer:
Commercially, on a large scale, glucose is prepared by hydrolysis of starch with dilute sulphuric acid. Starchy material is mixed with water and dilute sulphuric acid and heated at 393 K under 2 to 3-atm pressure. Starch is hydrolysed to give glucose.

Question 14.
Explain the structure of glucose.
Answer:
Molecular formula of glucose is C₆H₁₂O₆.

Glucose has an aldohexose structure. In other words, glucose molecule contains one aldehydic, that is, formyl group and the remaining five carbons carry one hydroxyl group (-OH) each. The six carbons in glucose form one straight chain.

Question 15.
Describe the action of following reagents on glucose :
(1) HI
(2) Hydroxyl amine (NH₂OH)
(3) Hydrogen cyanide
(4) Bromine water
(5) dil. Nitric acid
(6) Acetic anhydride.
Answer:
(1) Action of HI : Glucose on prolonged heating with HI gives n-hexane, indicates that all the six carbon atoms are linked in straight chain.

(2) Action of hydroxyl amine : Glucose reacts with hydroxyl amine in an aqueous solution to form glucose oxime. This indicates the presence of CHO group in glucose.

(3) Action of hydrogen cyanide : Glucose reacts with hydrogen cyanide to form glucose cyanohydrin.

(4) Action of bromine water : Glucose on oxidation with mild oxidising agent like bromine water gives gluconic acid, which shows that the carbonyl group in glucose is aldehyde group.

(5) Action of dll. nitric acid : Glucose on oxidation with dilute nitric acid forms dicarboxylic acid, saccharic acid. This indicates the presence of a primary alcoholic group (-CH₂OH) in glucose.

(6) Action of acetic anhydride : When glucose is heated with acetic anhydride in the presence of catalyst pyridine, glucose penta acetate is formed. It indicates that glucose is a stable compound and contains five hydroxyl groups.

Question 16.
Write Fischer projection formulae for
(1) Glucose
(2) Gluconic acid
(3) Saccharic acid.
Answer:
Fischer projection formulae of glucose, gluconic acid and saccharic acid :

Question 17.
Explain D and L configuration in sugars.
Answer:
The simplest carbohydrates glyceraldehyde is chosen as the standard, to assign D and L configuration to monosaccharides. Glyceraldehyde contains one asymmetric carbon atom and exist in two enantiomeric forms

The dextro entantiomer is represented as (+) glyceraldehyde and it is referred as D-configuration i.e., D-glyceraldehyde. The laevo enantiomer of glyceraldehyde is represented as ( -) glyceraldehyde and it corelated as L-configuration i.e., L-glyceraldehyde.

In Fischer projection formula, a monosaccharide is assigned D-configuration if the (- OH) hydroxyl group at the last chiral carbon and lies towards right hand side. On the other hand it is assigned L-configuration if the – OH group on the last chiral carbon atom and lies on the left hand side. In monosaccharides, the most oxidised carbon (i.e., -CHO) is at the top.

Examples :

Question 18.
Write Fischer projection formulae for (a) L-( + )-erythrose (b) L-( +) ribulose.
Answer:

Question 19.
Is the following sugar, D-sugar or L-sugar?

Answer:
The compound is L-sugar.
The compound is L-sugar.

Question 20.
Assign D/L configuration to the following monosaccharides.


Answer:

Question 21.
Explain ring structure of glucose.
Answer:

Glucose has two cyclic structures (II and III) which are in equilibrium with each other through the open chain structure (I) in aqueous solution.

The ring structure of glucose is formed by reaction between the formyl ( – CHO) group and the alcoholic (- OH) group at C – 5. Thus, the ring structure is called a hemiacetal. The two hemiacetal structures (II and III) differ only in the configuration of C – I (Fig.), the additional chiral centre resulting from ring closure. The two ring structures are called α- and β- anomers of glucose and C-l is called the anomeric carbon. The ring of the cyclic structure of glucose contains five carbons and one oxygen. Thus, it is a six membered ring. It is called pyranose structure, in analogy with the six membered heterαyclic compound pyran (IV). Hence glucose is also called glucopyranose.

Question 22.
Write the structures of α-D-( + )-glucopyranose and β-D-( +) glucopyranose.
Answer:

Question 23.
Explain Haworth formula of glycopyranose.
Answer:

In the Haworth formula the pyranose ring is considered to be in a perpendicular plane with respect to the plane of paper. The carbons and oxygen in the ring are in the places as they appear in figure. The lower side of the ring is called α-side and the upper side is the β-side. The α-anomer has its anomeric hydroxyl (- OH) group (at C-l) on the α-side, whereas the β-anomer has its anomeric hydroxyl (- OH) group (at C-l) on the β-side. The groups which appear on right side in the Fischer projection formula appear on α-side in the Haworth formula, and the groups which appear on left side in the fischer projection formula appear on a β-side in the Haworth formula.

Question 24.
Explain the structure of fructose.
Answer:
Fructose has molecular formula C₆H₁₂O₆. It contains ketonic functional group at carbon number 2 and six carbon atoms in straight chain. It belongs to D-series and is a laevo rotatory compound. It is written as D-( – )-fructose. Being an α-hydroxy keto compound fructose is a reducing sugar.

Question 25.
Draw mirror images of glucose and fructose.
Answer:
(1) Glucose

(2) Fructose

Question 26.
Write the two cyclic structures of α-D-( – )-fructofuranose and β-D-( – )-fructofuranose exist in equilibrium with open chain structure.
Answer:

Question 27.
Write the Haworth projection formulae for α -D-( -) – Fructofuranose and β – D – ( -) – Fructo- furanose.
Answer:

Question 28.
Explain the structure of sucrose.
Answer:
Sucrose is a hexasaccharide and has molecular formula C₁₂H₂₂O₁₁. The structure of shcrose contains glycosidic linkage between C – 1 of α-glucose and C – 2 of β-fructose. Since aldehyde and ketone groups of both monosaccharide units are involved in the formation of glycosidic bond, sucrose is a nonreducing sugar.

Sucrose is dextrorotatory, on hydrolysis with dilute acid or an enzyme invertase gives equimolar mixture of dextrorotatory glucose and laevorotatory fructose.

The solution is laevorotatory because laevo rotation of fructose (- 92.4°) is more than dextrorotation of glucose ( + 52.50), hence the sign of rotation is changed from (+) to (-) after hydrolysis, the product is called invert sugar.

Question 29.
Explain the structure of maltose.
Answer:
Maltose is another disaccharide obtained by partial hydrolysis of starch or made of two units of D-glucose. In maltose, C-l of one α-D-glucose is linked to C-4 of another α-D-glucose molecule by glycosidic linkage. The glucose ring which uses its hydroxyl group at C-1 is α – 1 → 4 glycosidic linkage. It is a reducing sugar because a free aldehyde group can be produced at C₁ of second glucose molecule. Maltose on hydrolysis with dilute acids gives glucose.

Question 30.
Draw a neat diagram for Haworth formula of maltose.

Question 31.
Explain the structure of lactose.
Answer:

Lactose (C₁₂H₂₂O₁₁) is a disaccharide. It is found in milk, therefore, it is also known as milk sugar. It is formed from two monosaccharide units, namely D – galactose and D – glucose. The glycosidic linkage is formed between C-l of β-D-galactose and C -4 of glucose. Therefore the linkage in lactose is called β – 1,4 – glycosidic linkage. The hemiacetal group at C-l of the glucose unit is not involved in glycosidic linkage but is free. Hence lactose is a reducing sugar. The above figure shows Haworth formula of lactose.

Question 32.
What are the hydrolysis products of (1) lactose (2) sucrose?
Answer:
(1) Lactose on hydrolysis in presence of an acid or enzyme lactase gives one molecule each of glucose and galactose

(2) Sucrose on hydrolysis in the presence of dii. acid or the enzyme invertase gives one molecule each of glucose and fructose.

Question 33.
Explain the structure of starch.
Answer:
Starch is found in cereal grains, roots, tubers, potatoes, etc. It is a polymer of α-D-glucose and consists of two components, amylose and amylopectin.

Amylose is water soluble component forms blue coloured complex with iodine. It constitutes about 20 % of starch. Amylose contains 200 to 1000 α-D-glucose units linked together by glycosidic linkage between C-l of one unit and C-4 of another unit. i.e. α-1, 4 glycosidic linkages.

Amylopectin is insoluble in water and constitutes about 80 % starch which forms blue-violet coloured complex with iodine. It is a branched chain polymer. In amylopectin, α-D-glucose molecules are linked together by glycosidic linkage between C₁ – of one unit and C-4 of another unit to form long chain and branching αcurs by glycosidic linkage between C-l and C6 glycosidic linkage.

Question 34.
What are polysaccharides?
Answer:
A large number of same or different monosaccharides are joined together by glycosidic linkages are called polysaccharides. They have general formula (C₆H₁₀O₅)_n.

Question 35.
Explain the structure of cellulose.
Answer:

Cellulose mainly αcurs in plants. Cell wall of plant cells is made up of cellulose. It is a long chain polymer. In cellulose, β-D-glucose units are linked by glycosidic linkage between C₁-of one unit of glucose and C₄ of another glucose unit. Thus cellulose contains 1 → 4β glycosidic linkages like those in cellobiose.

Question 36.
Explain the structure of glycogen.
Answer:
The glucose is stored in animal body in the form of glycogen. It is also known as animal starch because its structure is similar to amylopectin. Glycogen is highly branched. Whenever the body is required glucose, enzymes breaks the glycogen to glucose.

Question 37.
How is glycogen different from starch?
Answer:
Starch is the main storage molecules of plants whereas glycogen is the main storage molecule of animals. Starch is found in cereals, roots, tubers, etc. Glycogen is present in liver, muscles and brain.

Question 38.
What do you understand by the term glycosidic linkage?
Answer:
The linkage between two monosaccharide units through oxygen atom is called glycosidic linkage.

Question 39.
What is the basic structural difference between starch and cellulose?
Answer:
Starch is a polymer of a-glucose and consists of two components-amylose and amylopectin. In amylose α-D-D-( + )-glucose units held by C,-C4 glycosidic linkage and in amylopectin, α-D-glucose units held by C1-C4 glycosidic linkage whereas branching αcurs by C₁-C₆ glycosidic linkage. [Refer Question 35 (i) (ii) Fig.] Cellulose is a straight chain polysaccharide composed only of β-D-glucose units held by C₁-C₄ glycosidic linkage. (Refer Question 37 Fig.)

Question 40.
Define the term : Protein OR What are proteins?
Answer:
Chemically proteins are polyamides which are high molecular weight polymers of the monomer units i.e. α-amino acids. OR It can also be defined as Proteins are the biopolymers of a large number of a-amino acids and they are naturally occurring polymeric nitrogenous organic compounds containing 16% nitrogen and peptide linkages (-CO-NH-).

Question 41.
Write the common sources of protein.
Answer:
Common sources of proteins are milk, pulses, peanuts, eggs, fishes, cheese, cereals, etc. They are also the principal materials of muscle, nerves, tendons, skin, blood, enzymes, many hormones and antibiotics.

Question 42.
What are the products of hydrolysis of proteins?
Answer:
On hydrolysis, proteins give a mixture of α-anlino acids.

The α-carbon in α-amino acids ohtained by hydrolysis of proteins has ‘L’ configuration.

Question 43.
What are the a-amino acids?
Answer:
α-Amino acids are carboxylic acids having an amino (- NH₂) group bonded to the α-carbon, i.e. the carbon next to the carboxyl (- COOH) group.

α-amino acids are derivatives of carboxylic acids, obtained by replacing – H atom by amino group. They are bifunctional compounds containing acidic and basic – NH₂ groups.
Example : (where R is an alkyl group or aryl group).

The amino acids are colourless, crystalline, water soluble, high melting solids. These acids in their aqueous solutions behave like salts due to presence of both acidic, and basic. (- NH₂) groups in the same molecule.

Such a doubly charged ion is known as zwitter ion. Example :

Question 44.
What are the final products of hydrolysis of proteins?
Answer:
Proteins on hydrolysis with dilute solution of acids, alkalies or enzymes give a mixture of large number of a-amino acids as final products.

For example :

Question 45.
Write the classification of amino acids, giving examples.
Answer:
The amino acids are of three types : acidic, basic and neutral. The symbol ‘R’ in the structure of a-amino acids represents side chain and may contain additional functional groups.

(1) Acidic amino acids : If ‘R’ contains a carboxyl (- COOH) group the amino acid is acidic amino acid, i.e. If carboxyl groups are more in number than amino groups, then amino acids are acidic in nature.

Examples : Glutamic acid HOOC-CH₂-CH₂-; Aspartic acid HOO-CH₂–

(2) Basic amino acids : If ‘R’ contains an amino (1°, 2°, or 3°) group, it is called basic amino acid i.e. If amino groups are more in number than carboxyl groups then amino acids are basic in nature.

Examples : Arginine

(3) Neutral amino acids : The other amino acids having neutral or no functional group in ‘R’ are called neutral amino acids, i.e. The amino acids having equal number of amino and carboxyl groups are neutral amino acids.

Examples : Alanine CH₃-; Valine (CH₃)₂-CH

Question 46.
What are essential and non-essential amino acids? Give two examples of each.
Answer:
The amino acids, which cannot be synthesised in the body and are supplied through diet are called essential amino acids. Examples : Lysine H₂N-(CH₂)₄-; Valine (CH₃)₂CH- The amino acids which are synthesized in the body are called non-essential amino acids.

Examples : Glutamic acid HOO-CH₂-CH₂-; Serine HO-CH₂–

Question 47.
What is meant by Zwitter ion?
Answer:
An a-amino acid molecule contains both acidic carboxyl ( – COOH) group as well as basic amino (- NH₂) group. Proton transfer from acidic group to basic group of amino acid forms a salt, which is a dipolar ion called a zwitterion.

Question 48.
Draw zwitter ion of alanine and other two forms.
Answer:

Question 49.
What is a peptide bond (peptide linkage)?
OR
Define peptide bond.
Answer:
Proteins are the polymers of a-amino acids and they are connected to each other. The bond that connects a-amino acids to each other is called peptide bond (peptide linkage, – CONH -).

Question 50.
How is peptide linkage (dipeptide linkage) formed in proteins? How is tripeptide formed?
Answer:
Peptide linkage is formed by condensation of acidic group of one molecule of a-amino acid and basic -NH₂ group of other molecule of α-amino acid with elimination of water.

When one more molecule of amino acid combines with dipeptide, it forms tripeptide.

Thus, it forms tetra, penta and finally a polypeptide chain i.e. proteins. Hence, proteins are basically polypeptides.

Question 51.
Write the structures of all possible dipeptides which can be obtained from glycine and alanine.
Answer:
(1) Dipeptide from glycine :
Carboxylic group of glycine reacting with amino group another molecule of glycine to form dipeptide

(2) Dipeptide from alanine :
Carboxylic goup of alanine reaction with amino goup of another molecule of alamine to form dipeptide

(3) Dipeptide from glycine and alanine :
Carboxylic group of glycine reacting with amino group another molecule of alanine to form dipeptide

Question 52.
How are proteins classified on the basis of molecular shapes?
Answer:
On the basis of their molecular shapes proteins are classified as :
(1) Fibrous proteins : The proteins in which the polypeptide chains lie parallel (side by side) to form fibre-like structure, are called fibrous proteins. The polypeptide chains held together by hydrogen bonds. These proteins are insoluble in water.

The fibrous proteins are tough and insoluble in water, and dilute acids or bases.

Example : myαin (in muscles), keratin (in hair, nails, skin), fibroin (in silk), collagen (in tendons), etc.

(2) Globular proteins : The proteins have spherical shape. This shape results from coiling around of the polypeptide chain of protein, and have intramolecular hydrogen bonding are called globular proteins.

They are soluble in water and dilute acids or bases.

Example : Haemoglobin (in blood), albumin (in eggs), insulin (in pancreas), etc.

Question 53.
Distinguish between globular and fibrous proteins.
Answer:

Question 54.
Draw a neat labelled diagram for the secondary structure of protein.
Answer:
Secondary structure of proteins : The three-dimensional arrangement of lαalized regions of a long polypeptide chain is called the secondary structure of protein. Hydrogen bonding between N-H proton of one amide linkage and C = O oxygen of another gives rise to the secondary structure. There are two different types of secondary structures i.e. α-helix and β-pleated sheet.

α-Helix : In a-helix structure, a polypeptide chain gets coiled by twisting into a right handed or clαkwise spiral known as a-helixn. The characteristic features of α-helical structure of protein are :

(1) Each turn of the helix has 3.6 amino acids.
(2) A C = O group of one amino acid is hydrogen bonded to N – H group of the fourth amino acid along the chain.
(3) Hydrogen bonds are parallel to the axis of helix while R groups extend outward from the helix core.
Myosin in muscle and a-keratin in hair are proteins with almost entire a-helical secondary structure.

β-Pleated sheet : In secondary structure, when two or more polypeptide chains (strands) line up side-by-side is called β-pleated sheets. The β-picate sheet structure of protein consists of extended strands of polypeptide chains held together by intermolecular hydrogen bonding. The characteristics of β-pleated sheet structure are :

• The C = O and N – H bonds lie in the planes of the sheet.
• Hydrogen bonding occurs between the N – H and C = O groups of nearby amino acid residues in the neighbouring chains.
• The R groups are oriented above and below the plane of the sheet.

The β-pleated sheet arrangement is favoured by amino acids with small R groups.

Question 55.
What is denaturation of proteins? How is denaturation brought about?
OR
What is the effect of denaturation on the structure of proteins?
Answer:
The prαess by which the molecular shape of protein changes without breaking the amide / peptide bonds that form the primary structure is called denaturation. OR Proteins gets easily precipitated. It is an irreversible change and the prαess is called denaturation of proteins.

Denaturation uncoils the protein and destroys the shape and thus loses their characteristic biological activity. Denaturation is brought about by heating the protein with alcohol, concentrated inorganic acids or by salts of heavy metals. During denaturation secondary and tertiary and quternary structures are destroyed but primary structure remains intact.

Example : Boiling of egg to coagulate egg white, conversion of milk to curd.

Question 56.
Define : Enzymes
Answer:
All biological reactions are catalysed by bio-catalyst in living organisms called enzymes.

Question 57.
What are enzymes? Explain with suitable example.
Answer:
All biological or bio-catalysts which catalyse the reactions in living organisms are called enzymes. Chemically all enzymes are proteins. They are required in very small quantities as they are catalyst also they reduce the activation energy for a particular reaction.

Example : Enzyme maltase converts maltose to glucose.

Question 58.
Explain the catalytic action of enzymes.

Answer:
Mechanism of enzyme catalysis : Action of an enzyme on a substrate is known as lock-and-key mechanism.

Accordingly, the enzyme has active site on its surface. A substrate molecule can attach to this active site only if it has the right size and shape. Once in the active site, the substrate is held in the correct orientation, enzymes provide functional group which will attack the substrate and forms the products of reaction. The products leave the active site and the enzyme is ready to act as catalyst again.

Question 59.
Give examples of industrial application of enzyme catalysis.
Answer:

• Glucose Isomerase (enzyme) is used in conversion of glucose to sweet-tasting fructose.
• New antibiotics are manufactured using penicillin acylase (enzyme).
• Laundry detergentts are manufactured using proteases (enzyme).
• Esters used in cosmetics are manufactured using genetically engineered enzyme.

Question 60.
Draw a neat diagram for enzyme catalysis.
Answer:

Question 61.
State the main functions of enzymes.
Answer:
Enzymes are biological catalyst and they are highly specific in nature. The two main functions are as follows :
(1) They lower the requirement of activation energy.
(2) They speed up the rate of reaction.
E.g. Enzyme maltase catalyses maltose to glucose.
\(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+\mathrm{H}_{2} \mathrm{O} \stackrel{\text { Maltase }}{\longrightarrow} 2 \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\)

Question 62.
What are nucleic acids?
Answer:
Nucleic acids are unbranched polymers of repeating monomers i.e. nucleotides. In other words, nucleic acids have a polynucleotide structure which in turn consists of a base, a pentose sugar and phosphate moiety.

Nucleic acids are biomolecules which are found in the nuclei of all living cells in the form of nucleoproteins or chromosomes.

(Nucleoproteins = Proteins + Nucleic acid)
(prosthetic group)

Question 63.
State the types of nucleic acids.
Answer:
The types of nucleic acids are : Ribonucleic acid (RNA) and deoxy ribonucleic acid (DNA). DNA molecules contain several million nucleotides while RNA molecules contain a few thousand nucleotides.

Question 64.
Explain chemical composition of nucleic acids.
Answer:
Nucleic acids have a polynucleotide structure. Nucleic acids (RNA and DNA) consists of three components :
(1) monosaccharide (sugar)
(2) nitrogen containing base and
(3) phosphate group.

(1) Monosaccharides : Nucleotides of both RNA consist of five membered monosaccharide ring (furanose), called as simply sugar component.

In RNA, the sugar component of nucleotide units is D-ribose and in DNA 2-deoxy-D-ribose.
2 – deoxy means no – OH group at C₂ position.

(2) Nitrogen containing base : Total five nitrogen – containing bases are present in nucleic acids. Three bases with one ring (cytosine, uracil and thymine) are derived from the parent compound pyrimidine. Two bases with two rings (adenine and guanine) are derived from the parent compound purine. Each base in designated by a one-letter symbol. Uracil (U) αcurs only in RNA while thymine (T) ocurs only in DNA.

(3) Phosphate group : The sugar units are joined to phosphate through C₃ and C₅ hydroxyl groups.

Question 65.
What is meant by nucleosides?
OR
Write the structure of nucleoside. Give examples.
Answer:

A nucleoside contains two basic components of nucleic acids i.e. a pentose sugar and a nitrogenous base.

A nucleoside is formed when 1 -position of a pyrimidine (cytosine, thymine or uracil) or 9-position of guanine or adenine base is attached to C- l of sugar by β-linkage.

Examples: Formation of nucleoside:

Question 66.
What is meant by nucleotide?
OR
Write the structure of nucleotide. Give example.
Answer:

A nucleotide contains all three basic components of nucleic acids i.e., a pentose sugar, a phosphoric acid and a nitrogenous base. These are obtained by esterification of \(\mathrm{C}_{5}^{1}-\mathrm{OH}\) group of the pentose sugar by phosphoric acid. Nucleotides are joined together through phosphate ester linkage. Thus, nucleotides are monophosphates of nucleosides. Abridged names of some nucleotides are AMP, dAMP, UMP, dTMP and so on. Here, the first capital letter is derived from the corresponding base. MP stands for monophosphate. Small letter ‘d’ in the beginning indicates deoxyribose in the nucleotide.

Example :

Question 67.
Write the structure of nucleic acids.

Answer:
Nucleic acids, both DNA and RNA, are polymers of nucleotides, formed by joining the 3′ – OH group of one nucleotide with 5′ – phosphate of another nucleotide. Two ends of polynucleotide chain are distinct from each other. One end having free phosphate group of 5′ position is called 5′ end. The other end is 3′ end and has free OH – group at 3′ position.

Question 68.
Draw a schematic representation of polynucleotide structure of nucleic acids.
Answer:

Question 69.
Explain double helix.
OR
State the salient features of the Watson and Crick mode of DNA.
Answer:

The Salient features are :

1. DNA is made of two polynucleotide strands that wind into a right-handed double helix.
2. The two strands run in opposite directions: one from the Y end to the 3’ end, while the other from the 3’ end to the Y end.
3. Pcrpcndicular to the axis of the helix, the sugar – phosphate backbone lies on the outside of the helix and the bases lic on the inside.
4. The hydrogen bonding between the hases of the two DNA strands stabilizes the double helix. This gives rise to a ladder-like structure of DNA double helix.
5. Adenine always forms two hydrogen bonds with thymine and guanine forms three hydrogen bonds with cytosinc. Thus A – T arid C – G arc complementary hase pairs and the Two strands of the double helix arc complementary to each other.

Question 70.
Give scientific reasons :
1. In the preparation of glucose from sucrose, ethyl alcohol is added at the time of cooling.
Answer:
Hydrolysis of sucrose with dilute hydrαhloric acid gives glucose along with fructose.

Ethyl alcohol is added at the time of cooling in the preparation of glucose, to separate glucose from fructose. Glucose being insoluble in alcohol, crystallizes out first, while fructose being more soluble in alcohol, remains in the solution.

Question 71.
Answer in one sentence :

(1) How is glucose stored in the animal body?
Answer:
Glucose is stored in the form of glycogen in the animal body.

(2) Write other term used for carbohydrates.
Answer:
Carbohydrates are often termed as saccharides or sugars.

(3) How many moles of acetic anhydride will be required to form glucose penta acetate from 1 mole of glucose?
Answer:
10 moles of acetic anhydride.

(4) What are reducing sugars?
Answer:
Reducing sugars : Carbohydrates which reduce Fehling solution to red ppt of Cu₂0 or Tollen’s reagent to shining metallic silver are called reducing sugars. All monosaccharides and oligosaccharides except sucrose are reducing sugars.

(5) What are non-reducing sugars?
Answer:
Non-reducing sugars : Carbohydrates which do not reduce Fehling solution and Tollen’s reagent are called non-reducing sugars. E.g. sucrose.

(6) Give an example each of reducing and non-reducing sugars.
Answer:
Reducing sugars : Maltose or lactose
Non-reducing sugars : Sucrose.

(7) Name the linkage which joins two monosaccharide units through oxygen atom.
Answer:
The linkage which joins two monosaccharide units through oxygen atom is called glycosidic linkage.

(8) Name the sugar present in DNA.
Answer:
The sugar present in DNA is deoxyribose.

(9) A nucleotide from DNA containing thymine is hydrolysed. What are the products formed?
Answer:
When nucleotide from DNA containing thymine is hydrolysed, 2-deoxy-D-ribose, thymine and phosphoric acid is obtained.

(10) How is zwitterion formed?
Answer:
In aqueous solution, the carboxyl group loses a proton while the amino group accepts it, as a result, a dipolar or zwitter ion is formed.

(11) Name the amino acids which are synthesized in the body.
Answer:
The amino acids which are synthesized in the body are called non-essential amino acids. Examples : Glutamic acid, serine.

(12) Name the four bases present in DNA which of these is not present in RNA.
Answer:
Purines-adenine (A) and guanine (G); Pyrimidines-thymine (T) and cytosine (C), these four bases are present in DNA. Out of these, thymine (T) is not present in RNA.

(13) What are different types of RNA which are found in the cell?
Answer:
There are three different types of RNA found in the cell. (1) The messenger RNA which carries the message to the ribosome (2) Ribosomal RNA where synthesis of protein takes place (3) The transport RNA.

(14) State the functions of RNA and DNA.
Answer:
RNA and DNA are responsible for generic characteristics : DNA preserves the information and uses it by producing duplicate identical DNA molecules. RNA carries messages and transports them.

Multiple Choice Questions

Question 72.
Select and write the most appropriate answer from the given alternatives for each subquestion :

1. Which of the following is not sugar?
(a) Sucrose
(b) Starch
(c) Fructose
(d) Glucose
Answer:
(b) Starch

2. Which of the following is the example of disaccharide?
(a) Glucose
(b) Raffinose
(c) Cellulose
(d) Sucrose
Answer:
(d) Sucrose

3. Fructose is
(a) aldopentose
(b) aldohexose
(c) ketopentose
(d) ketohexose
Answer:
(d) ketohexose

4. Oxidation product of glucose with bromine water is
(a) sorbitol
(b) gluconic acid
(c) glutamic acid
(d) saccharic acid
Answer:
(b) gluconic acid

5. The general formula of carbohydrates is
(a) C(H₂O)
(b) C_x(H₂O)_y
(c) C_x(H₂O)
(d) C_x(H₂O)_x
Answer:
(b) C_x(H₂O)_y

6. Monosaccharides containing aldehyde group are called
(a) aldoses
(b) ketoses
(c) polysaccharides
(d) disaccharides
Answer:
(a) aldoses

7. Which of the following sugars can be used to prepare glucose on a large scale?
(a) Cellulose
(b) Cane sugar
(c) Galactose
(d) Starch
Answer:
(d) Starch

8. Which of the following carbohydrates cannot undergo hydrolysis?
(a) Glucose
(b) Sucrose
(c) Cellulose
(d) Maltose
Answer:
(a) Glucose

9. Glucose differs from fructose in
(a) the functional group
(b) the number of chiral carbon atoms
(c) the number of carbon atoms
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

10. The example of aldopentose is
(a) arabinose
(b) glucose
(c) fructose
(d) sucrose
Answer:
(a) arabinose

11. Dextrose, grape sugar and blood sugar αcurs in
(a) fructose
(b) glucose
(c) sucrose
(d) starch
Answer:
(b) glucose

12. The example of ketopentose is
(a) galactose
(b) ribose
(c) raffinose
(d) maltose
Answer:
(b) ribose

13. Cane sugar on hydrolysis gives
(a) glucose and maltose
(b) glucose and lactose
(c) glucose and fructose
(d) only glucose
Answer:
(c) glucose and fructose

14. On commerical scale, glucose is prepared from
(a) starch
(b) potato pulp
(c) sucrose
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

15. The number of monosaccharide units formed on hydrolysis of glucose are
(a) zero
(b) one
(c) two
(d) three
Answer:
(a) zero

16. Which of the following is NOT TRUE about glucose?
(a) It is monosaccharide
(b) It is a polyhydroxy aldehyde
(c) It is polyhydroxy ketone
(d) It contains six carbon atoms
Answer:
(c) It is polyhydroxy ketone

17. Final hydrolysis product of simple protein is
(a) carboxylic acid
(b) α-amino acid
(c) mineral acid
(d) acetic acid
Answer:
(b) α-amino acid

18. Haemoglobin is the example of-
(a) simple protein
(b) derived protein
(c) fibrous protein
(d) conjugated protein
Answer:
(d) conjugated protein

19. Protein are also called
(a) polysaccharides
(b) polypeptides
(c) polyglycerides
(d) polyster
Answer:
(b) polypeptides

20. The simplest amino acid is
(a) glycine
(b) oxalic acid
(c) adipic acid
(d) caprolactam
Answer:
(a) glycine

21. Amino acids usually exist in the form of zwitter ion which consist of
(a) the basic group-NH₂ and the acidic group -COOH
(b) the acidic group -N⁺H₃ and the basic group COO^–
(c) the acidic group -COO⁺ and the acidic group NH^3-
(d) acidic or basic group
Answer:
(b) the acidic group -N⁺H₃ and the basic group COO-

22. The water insoluble protein is
(a) casein of milk
(b) albumin
(c) serum albumin
(d) keratin of hair
Answer:
(d) keratin of hair

23. The main structural feature of a protein molecule is the presence of
(a) an ester linkage
(b) an ether linkage
(c) a peptide linkage
(d) all of these
Answer:
(c) a peptide linkage

24. Milk sugar is
(a) sucrose
(b) lactose
(c) maltose
(d) glucose
Answer:
(b) lactose

25. The carbohydrates used for silvering of mirror is
(a) fructose
(b) starch
(c) glucose
(d) cellulose
Answer:
(c) glucose

26. Which one of the following is NOT produced by human body?
(a) DNA
(b) Hormones
(c) Enzymes
(d) Vitamins
Answer:
(c) Enzymes

27. A biological catalyst is essentially
(a) an amino acid
(b) an enzyme
(c) a nitrogen molecule
(d) a carbohydrate
Answer:
(d) a carbohydrate

28. Which one of the following is not a constituent of RNA?
(a) Ribose
(b) Uracil
(c) Thymine
(d) Phosphate
Answer:
(b) Uracil

29. DNA is a polymer of units of
(a) sugars
(b) ribose
(c) amino acids
(d) nucleotides
Answer:
(c) amino acids

30. Which one of the following molecules will form zwitter ion?
(a) CH₃COOH
(b) CH₃CH₂NH₂
(c) CCl₃NO₂
(d) NH₂CH₂COOH
Answer:
(d) NH₂CH₂COOH

31. In metabolic prαess the maximum energy is given by
(a) carbohydrates
(b) proteins
(c) vitamins
(d) fats
Answer:
(d) fats

32. DNA has a structure of helix was reported by
(a) Herman Fischer
(b) Fedrick Sauger
(c) Andreas Marggraf
(d) Watson and Crick
Answer:
(d) Watson and Crick

33. The secondary structure of a protein is determined by
(a) co-ordinate bond
(b) covalent bond
(c) ionic bond
(d) hydrogen bond
Answer:
(d) hydrogen bond

34. In maltose, glycosidic linkage is present between the two glucose units at positions
(a) 1, 2
(b) 1, 1
(c) 1, 3
(d) 1, 4
Answer:
(d) 1, 4

35. Which of the following amino acids is basic in nature?
(a) Valine
(b) Tyrosine
(c) Arginine
(d) Luecine
Answer:
(c) Arginine

36. Sucrose molecules consists of
(a) a glucofuranose and a fructopyranose
(b) a glucofuranose and a fructofuranose
(c) a glucopyranose and a fructopyranose
(d) a glucopyranose and a’ fructofuranose
Answer:
(d) a glucopyranose and a’ fructofuranose

37. Which one of the following statements is not correct about DNA molecule?
(a) It has double helix structure
(b) It serves as hereditary material
(c) The two DNA strands are exactly similar
(d) Its replication is called semi-conservative mode of replication
Answer:
(c) The two DNA strands are exactly similar

38. Glycine on heating forms


Answer:
(a)

39. Acidic amino acid is
(a) Glutamine
(b) Glutamic acid
(c) Tyrosine
(d) Lysine
Answer:
(b) Glutamic acid

40. Basic amino acid is
(a) Lysine
(b) Glycine
(c) Cystine
(d) Alanine
Answer:
(a) Lysine

41. Precipitation of protein is referred to as
(a) destruction of proteins
(b) separation of proteins
(c) denaturation of proteins
(d) fragmentation of proteins
Answer:
(c) denaturation of proteins

42. An amino acid containing sulphur is
(a) serine
(b) cysteine
(c) valine
(d) asparagine
Answer:
(b) cysteine

43. Rhamnose is a
(a) carbohydrate
(b) protein
(c) lipid
(d) vitamin
Answer:
(a) carbohydrate

44. Lactose on hydrolysis gives
(a) glucose + glucose
(b) glucose + fructose
(c) glucose + galactose
(d) fructose + galactose
Answer:
(c) glucose + galactose

45. Raffinose on hydrolysis gives
(a) glucose + glucose + galactose
(b) glucose + fructose + galactose
(c) glucose + galactose + galactose
(d) fructose + galactose + galactose
Answer:
(b) glucose + fructose + galactose

46. Naturally αcurring glucose is
(a) dextro rotatory
(b) laevo rotatory
(c) racemic mixture
(d) all of these
Answer:
(a) dextro rotatory

47. Amylopectin is
(a) soluble in water and constitutes about 80% of starch
(b) insoluble in water and constitutes about 80% of starch
(c) Soluble in alcohol and constitutes about 60% of starch
(d) in soluble in alcohol and constitutes about 60% of starch
Answer:
(b) insoluble in water and constitutes about 80% of starch

48. Insulin contains
(a) 51 amino acids
(b) 151 amino acids
(c) 15 amino acids
(d) 115 amino acids
Answer:
(a) 51 amino acids

49. Pyranose structure of glucose is
(a) an open chain structure of glucose
(b) a structure of reduction product of glucose
(c) a cyclic six-membered structure of glucose
(d) a four-membered cyclic form of glucose
Answer:
(c) a cyclic six-membered structure of glucose

50. The number of – OH groups present in ribulose is
(a) 3
(b) 4
(c) 6
(d) 5
Answer:
(b) 4

51. Peptide linkage is

Answer:
(d)

52. Stachyose is an example of
(a) monosaccharides
(b) disaccharides
(c) trisaccharides
(d) tetrasaccharides
Answer:
(d) tetrasaccharides

53. How many moles of (CH₃CO)₂O will be required to form glucose pentaacetate form 2 moles of glucose?
(a) 2
(b) 5
(c) 10
(d) 2.5
Answer:
(c) 10

54. Which of the following NOT present in DNA?
(a) Adenine
(b) Guanine
(c) Thymine
(d) Uracil
Answer:
(d) Uracil

55. Maltose is a
(a) polysaccharide
(b) disaccharide
(c) trisaccharide
(d) monosaccharide
Answer:
(b) disaccharide

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.

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:

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