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Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 10 Halogen Derivatives Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Chemistry Solutions Chapter 16 Green Chemistry and Nanochemistry

1. Choose the most correct option.

Question i.
The development that meets the needs of the present without compromising the ability of future generations to meet their own need is known as
a. Continuous development
b. Sustainable development
c. True development
d. Irrational development
Answer:
b. Sustainable development

Question ii.
Which of the following is ϒ-isomer of BHC?
a. DDT
b. lindane
c. Chloroform
d. Chlorobenzene
Answer:
b. lindane

Question iii.
The prefix ‘nano’ comes from
a. French word meaning billion
b. Greek word meaning dwarf
c. Spanish word meaning particle
d. Latin word meaning invisible
Answer:
(b) Greek word meaning dwarf

Question iv.
Which of the following information is given by FTIR technique?
a. Absorption of functional groups
b. Particle size
c. Confirmation of formation of nanoparticles
d. Crystal structure
Answer:
(a) Absorption of functional groups

Question v.
The concept of green chemistry was coined by
a. Born Haber
b. Nario Taniguchi
c. Richard Feynman
d. Paul T. Anastas
Answer:
(d) Paul T. Anastas

2. Answer the following

Question i.
Write the formula to calculate % atom economy.
Answer:

Question ii.
Name the ϒ-isomer of BHC.
Answer:
Lindane

Question iii.
Ridhima wants to detect structure of surface of materials. Name the technique she has to use.
Answer:
Scanning electron microscopy (SEM)

Question iv.
Which nanomaterial is used for tyres of car to increase the life of tyres?
Answer:
Carbon black

Question v.
Name the scientist who discovered scanning tunneling microscope (STM) in 1980.
Answer:
Gerd Binning and Heinrich Rohrer. (Nobel prize 1986)

Question vi.
1 nm = …..m?
Answer:
1 nm = 10⁹ m

3. Answer the following

Question i.
Define
(i) Green chemistry
(ii) sustainable development.
Answer:
(i) Green chemistry : Green chemistry is the use of chemistry for pollution prevention and it designs the use of chemical products and processes that reduce or eliminate the use or generation of hazardous substances.

(ii) Sustainable development : Sustainable development is the development that meets the needs of the present, without compromising the ability of future generations to meet their own needs.

Question ii.
Explain the role of green chemistry.
Answer:
When the waste and pollution that society generates exceeds the Earth’s natural capacity for dealing with it, the green chemistry approach plays an important role.

• To reduce or eliminate the use or generation of hazardous substances in the design, manufacture and use of chemical products by promoting innovative chemical technologies.
• Capital expenditure required for prevention of pollution is controlled by the use of green chemistry.
• Since green chemistry incorporates and promotes pollution prevention practices in the manufacturing process of chemicals it helps industrial ecology.
• Green chemistry helps to protect the presence of ozone in the stratosphere. Ozone layer is essential for the survival of life on the earth.
• Global warming (Greenhouse effect) is controlled by green chemistry. At present it is the beginning of the green revolution.
• It is an exciting time with the new challenges for chemist involved with the discovery, manufacturing and use of chemicals. Green chemistry helps us to save environment and save earth, which is important for our future.

Question iii.
Give the full form (long form) of the names for the following instruments.
a. XRD
b. TEM.
c. STM
d. FTIR
e. SEM
Answer:
a. XRD-X-ray diffraction
b. TEM-Tunneling Electron Microscope
c. STM – Scanning Tunneling Microscope
d. FTIR-Fourier Transform Infrared Spectroscope
e. SEM-Scanning Electron Microscope

Question iv.
Define the following terms :
a. Nanoscience
b. Nanotechnology
c. Nanomaterial
d. Nanochemistry
Answer:
a. Nanoscience : The study of phenomena and manipulation of materials at atomic, molecular and macromolecular scales where properties differ significantly from those at a larger scale is called nanoscience.

b. Nanotechnology : The design, characterization, production and application of structures, device and system by controlling shape and size at nanometer scale is called nanotechnology.

c. Nanomaterial : A material having structural components with at least one dimension in the nanometer scale that is 1 -100 nm is called the nanomaterial. Nanomaterials are larger than single atoms but smaller than bacteria and cells.

d. Nanochemistry : It is the combination of chemistry and nanoscience. It deals with designing and synthesis of materials of nanoscale with different size and shape, structure and composition and their organization into functional architectures.

Question v.
How nanotechnology plays an important role in water purification techniques?
Answer:

1. Water purification is an important issue as 1.1 billion people do not have access to improved water supply. Water contains water bom pathogens like viruses, bacteria.
2. Silver nanoparticles are highly effective bacterial disinfectant to remove E. Coli from water. Hence, filter materials coated with silver nanoparticles is used to clean water.
3. Silver nanoparticles (AgNps) is a cost effective alternative technology (for e.g. water purifier).

Question vi.
Which nanomaterial is used in sunscreen lotion? Write its use.
Answer:
Zinc oxide (ZnO) and Titanium dioxide (TiO₂) nanoparticles are used sunscreen lotions. The chemicals protect the skin against harmful u.v (ultraviolet) rays by absorbing or reflecting the light and prevent the skin from damage.

Question vii.
How will you illustrate the use of safer solvent and auxiliaries?
Answer:

• Use of safer solvents and auxiliaries – is a principle of green chemistry it states that safer solvent like water, supercritical CO₂ should be used in place of volatile halogenated organic solvents, like CH₂CI₂, CHCI₃, CCI₄ for chemical synthesis and other purposes.
• Solvents dissolve solutes and form solutions, they facilitate many reactions. Water is a safer benign solvent while solvents like dichloromethane (CH₂CI₂), chloroform (CHCI₃) etc are hazardous.
• Use of toxic solvents affect millions of workers every year and have implications for consumers and the environment. A large amount of waste is created by their use and they also have huge environmental and health impacts.
• Finding safer solvents or designing processes which are solvent free is the best way to improve the process and the product.

Question viii.
Define catalyst. Give two examples.
Answer:
A substance which speeds up the rate of a reaction without itself being changed chemically in the reaction is called a catalyst. It helps to increase selectivity, minimise waste and reduce reaction time and energy demands. For example : Hydrogenation of oil the catalyst used are platinum or palladium, Raney nickel.

4. Answer the following

Question i.
Explain any three principles of green chemistry.
Answer:

1. Environment protection is the prime concern which has lead to the need for designing chemicals that degrade and can be discarded easily. These chemicals and their degradation products should be non-toxic, non-bioaccumulative or should not be environmentally persistent.
2. This principle aims at waste product being automatically degradable to clean the environment. Thus the preference for biodegradable polymers and pesticides.
3. To make the separation and segregation easier for the consumer an international plastic recycle mark is printed on larger items.
4. There is a dire need to develop improvised analytical methods to allow for real time, in process monitoring and control prior to the formation of hazardous substances.
5. It is very much important for the chemical industries and nuclear reactors to develop or modify analytical
   methodologies so that continuous monitoring of the manufacturing and processing unit is possible.
6. It is needed to develop chemical processes that are safer and minimize the risk of accidents. It is important to select chemical substances used in a chemical reaction in such a way that they can minimize the occurrence of chemical accidents, explosions, fire and emissions.
7. For example : Chemical process that works with the gaseous substances can lead to relatively higher possibilities of accidents including explosion as compared to the system working with nonvolatile liquid and solid substances.

Question ii.
Explain atom economy with suitable example.
Answer:
(1) Atom economy is a measure of the amount of atoms from the starting material that are present in the final product at the end of a chemical process. Good atom economy means most of the atoms of the reactants are incorporated in the desired products. Only small amount of waste is produced, hence lesser problem of waste disposal.

(2) The atom economy of a process can be calculated using the following formula.

The atom economy of the above reacijon is less than 50% and waste produced is higher.

Question iii.
How will you illustrate the principle, minimization of steps?
Answer:
(1) The technique of protecting or blocking group is commonly used in organic synthesis. Finally on completion of reaction deprotection of the group is required. This leads to unnecessary increase in the number of steps and decreased atom economy.

(2) The green chemistry principle aims to develop processes to avoid necessary steps i.e. (minimization of steps). When biocatalyst is used very often there is no need for protection of selective group. For example, conversion of m-hydroxyl benzaldehyde to m-hydroxybenzoic acid.

Question iv.
What do you mean by sol and gel? Describe the sol-gel method of preparation for nanoparticles.
Answer:
(1) Sol : Sols are dispersions of colloidal particles in a liquid. Colloids are solid particles with diameter of 1-100 nm.

(2) Gel : A gel is interconnected rigid network with pores of submicrometer dimensions and polymeric chains whose average length is greater than a micrometer.

(3) Sol-gel Process : A sol-gel process is an inorganic polymerisation reaction. It is generally carried out at room temperature, it includes four steps : Hydrolysis, polycondensation, drying and thermal decomposition. This method is widely used to prepare oxide materials.

The reactions involved in the sol-gel process are as follows :
MOR + H₂O → MOH + ROH (hydrolysis)
metal alkoxide
MOH + ROM → M-O-M + ROH (condensation)

• Formation of different stable solution of the alkoxide or solvated metal precursor.
• Gelation involves the formation of an oxide or alcohol-bridged network (gel) by a polycondensation reaction.
• Aging of the gel means during that period gel transforms into a solid mass.
• Drying of the gel involves removal of water and other volatile liquids from the gel network.
• Dehydration is achieved when the material is heated at temperatures up to 800°C.

Question v.
Which flower is an example of self-cleaning?
Answer:

• Lotus is an example of self cleansing.
• Nanostructures on the lotus plant leaves are super hydrophobic, they repel water which carries dirt as it rolls off.
  Thus though lotus plant (Nelumbonucifera) grows in muddy water, its leaves always appear clean.

Activity :
Collect information about the application of nanochemistry in cosmetics and pharmaceuticals

12th Chemistry Digest Chapter 16 Green Chemistry and Nanochemistry Intext Questions and Answers

Do you know? (Textbook page 343)

Question 1.
Does plastic packaging impact the food they wrap ?
Answer:
Phthalates leach into food through packaging so you should avoid microwaving food or drinks in plastic and not use plastic cling wrap and store your food in glass container whenever possible. Try to avoid prepackaging, processed food so that you will reduce exposure to the harmful effects of plastic.

Used Catalyst (Textbook page 342)

Question 18.
Complete the chart:

Answer:

Balbharti Maharashtra State Board Class 11 Secretarial Practice Solutions Chapter 2 Joint Stock Company Textbook Exercise Questions and Answers.

Maharashtra State Board Class 11 Secretarial Practice Solutions Chapter 2 Joint Stock Company

1A. Select the correct answer from the options given below and rewrite the statements.

Question 1.
A sole proprietorship has ______________ owner/owners.
(a) one
(b) two
(c) five
Answer:
(a) one

Question 2.
The head of Joint Hindu Family Business is called as ______________
(a) Proprietor
(b) Director
(c) Karta
Answer:
(c) Karta

Question 3.
Indian Partnership Act was passed in the year ______________
(a) 1923
(b) 1932
(c) 1956
Answer:
(b) 1932

Question 4.
The members of Hindu Undivided Family Business are called ______________
(a) Karta
(b) partners
(c) co-parceners
Answer:
(c) co-parceners

Question 5.
The liability of shareholders in the public limited joint stock company is ______________
(a) Limited
(b) Unlimited
(c) Collective
Answer:
(a) Limited

Question 6.
The minimum number of members required for a co-operative society is ______________
(a) 10
(b) 20
(c) 50
Answer:
(a) 10

Question 7.
The ______________ is/are elected representative of shareholders who manage affairs of company.
(a) Secretary
(b) Directors
(c) Auditors
Answer:
(b) Directors

Question 8.
State Bank of India is the example of ______________ Company.
(a) Chartered
(b) Statutory
(c) Foreign
Answer:
(b) Statutory

1B. Match the pairs.

Question 1.

Answer:

Question 2.

Answer:

1C. Write a word or a term or a phrase that can substitute each of the following statements.

Question 1.
The owner is the sole manager and decision-maker of his business.
Answer:
Sole Trader

Question 2.
The senior-most family member of Joint Hindu Family Business.
Answer:
Karta

Question 3.
The members of Joint Hindu Family Business.
Answer:
Co-parceners

Question 4.
An artificial person created by law.
Answer:
Joint Stock company

Question 5.
The persons who have entered into an agreement of partnership.
Answer:
Partners

Question 6.
A person who purchases shares of a Joint Stock Company.
Answer:
Shareholder

Question 7.
The official signature of Joint Stock Company.
Answer:
Common seal

Question 8.
Name a company that is created by special legislation of parliament or state assembly.
Answer:
Statutory company

1D. State whether the following statements are True or False.

Question 1.
A Joint Stock company is a voluntary association of persons.
Answer:
True

Question 2.
A Joint Stock company is a formal form of business organization.
Answer:
True

Question 3.
Registration of a Joint Stock company is compulsory.
Answer:
True

Question 4.
A Joint Stock company is a natural person.
Answer:
False

Question 5.
A Joint Stock company does not enjoy independent legal status.
Answer:
False

Question 6.
The liability of shareholders of a public limited company is limited.
Answer:
True

Question 7.
A Joint Stock company has a long and stable life.
Answer:
True

Question 8.
There is no separation of ownership and management in a Joint Stock company.
Answer:
False

Question 9.
Board of Directors manages the Company.
Answer:
True

1E. Complete the sentences.

Question 1.
A company is a creation of law, hence it is called as ______________
Answer:
Legal Person or Artificial Person

Question 2.
A company which is incorporated under a Special Act is called as ______________
Answer:
Statutory Company

Question 3.
A company which has only one member is called as ______________
Answer:
One Person Company

Question 4.
A listed company must follow the provisions of Companies Act and ______________
Answer:
SEBI Guidelines

1F. Select the correct option from the bracket.

Question 1.

(Foreign Company, Minimum 7 members, Maximum 200 members, Co-operative society, Unlimited Liability Company)
Answer:

1G. Answer in one sentence.

Question 1.
How many member/s can be there in a One Person company?
Answer:
There can be only one member in a Person Company.

Question 2.
What is a Holding company?
Answer:
A company that holds more than one-half of the total share capital of another company or carries the power to appoint or remove all or majority of directors of another company is called a Holding Company.

Question 3.
What is meant by a Foreign company?
Answer:
A company incorporated outside India, but conducting business in India, called a foreign company.

1H. Correct the underlined word and rewrite the following sentences.

Question 1.
Statutory companies are registered under the Companies Act.
Answer:
Statutory companies are registered under Special Act passed by Central or State legislative.

Question 2.
A Subsidiary company holds more than half of the total share capital of another company.
Answer:
A Holding company holds more than half of the total share capital of another company.

Question 3.
A private company must have a minimum of 7 Members.
Answer:
A private company must have a minimum of 2 members.

Question 4.
A public company can have a maximum of 200 members.
Answer:
A private company can have a maximum of 200 members.

2. Explain the following terms/concepts.

Question 1.
Dormant company
Answer:

• It is registered for future projects.
• It has not made any accounting transactions in the last two years.
• It has not submitted a financial statement or annual report in the last two years.
• Section 455 of Companies Act, 2013 is applicable to a Dormant Company.

Question 2.
Holding company
Answer:

• A company holds more than half of the share capital of another company.
• Such a company may have the power to appoint a director of another company.
• It has the power to remove directors of another company.

Question 3.
Foreign company
Answer:

• A company that is incorporated/registered outside India.
• It may conduct business in India.
• Bata India Limited, Nestle India Limited, Whirlpool Corporation, etc. are examples of foreign companies.

Question 4.
Company limited by guarantee
Answer:

• Such a company is formed under Section 2(21).
• This company may or may not have share capital.
• Member promises to pay a fixed amount at the time of liquidation.
• This fixed amount is mentioned in the Memorandum of Association.
• This amount is used to pay debts and liabilities.

Question 5.
Associate company
Answer:

• The firm over which another firm exercises control, which is less than the degree of control exercised over a subsidiary company.
• A company in which another company has a significant influence.
• It means control of at least 20% of total capital or of decisions under an agreement.

Question 6.
Limited Liability
Answer:

• The liability of shareholders is limited in Joint Stock Company.
• Personal property cannot be used to pay the debts of the company.
• Liability is limited to the unpaid part of the face value of shares held by a shareholder.
• Shareholders are not liable to pay debts and liability of the company.

Question 7.
Perpetual Succession
Answer:

• It means continuous existence.
• Joint Stock Company has perpetual succession.
• The life/existence of the company is not affected by the death, insolvency, or retirement of any member or director.
• The company enjoys long and stable life.

Question 8.
Listed company
Answer:

• It means a company that has any of its securities listed on any recognized stock exchange.
• A public company may be a listed or unlisted company.
• The listed company needs to follow the guidelines of SEBI.
• They have to follow the Companies Act.

Question 9.
One Person company
Answer:

• In this company, only one person is a member.
• It should follow all rules and formalities of a private company.
• It is conducted by one promoter with limited liability.
• Such a business organization is better as compared to a sole trader.

Question 10.
Government company
Answer:

• More than 51% of paid-up share capital is held by the Government.
• The government may be Central or State Government or partly Central Government and partly one or more State Government.
• It may be a subsidiary company of a Government company.
• It may be a Private company or a Public company.

3. Study the following case/situation and express your opinion.

1. Two promoters got ‘Super Drinks Pvt. Ltd.’ incorporated on 18th January 2018. The company has 100 members as of 31st March 2019.

Question (a).
What is the maximum number of members this company can have?
Answer:
The maximum number of members for this company is 200 because this is a private company.

Question (b).
Can this company invite the general public to subscribe for shares?
Answer:
This company cannot invite the general public to subscribe for shares.

Question (c).
Can the shareholders of the company sell its shares to outsiders?
Answer:
Being a private company, there are restrictions to shareholders to sell these to outsiders.

2. Kali VFX Ltd. was incorporated on 1st January 2019 as a public limited company.

Question (a).
How many minimum numbers of members must be there in this company?
Answer:
A minimum number of members must be 7 in this company because it is a Public company.

Question (b).
Can the members of this company sell their shares to outsiders?
Answer:
Being a Public company member can sell its shares to outsiders. There is no restriction on the transferability of shares.

Question (c).
How many maximum numbers of members can this company have?
Answer:
In a public company, there is no limit for the number of maximum members as it is a Public company.

3. Sunset Printers Pvt. Ltd. was incorporated on 5th December 2015 as per the provisions of the Companies Act, 2013. Mr. Manoj was the only subscriber to the Memorandum and Articles of Association and he was also the only member of the company.

Question (a).
Is this company a One Person company?
Answer:
Yes, this is a One Person company, because only one/single person is a member of this company.

Question (b).
Will the liability of Mr. Manoj be limited or unlimited?
Answer:
The liability of Mr. Manoj is limited.

Question (c).
Will the company close down on the death, insanity, or insolvency of Mr. Manoj?
Answer:
No, Company will not close down on the death, insanity, or insolvency of Mr. Manoj.

4. On 1st January 2018 Mr. John bought 100 shares of TIPS Paints Ltd. The face value of each share was ₹ 10. Mr. John paid the full amount of ₹ 1,000. In December 2018 the company suffered a loss of ₹ 10 crores.

Question (a).
Can the company ask Mr. John to pay any further money to the company?
Answer:
No, Company cannot ask Mr. John for further payment, because, he has already paid the full amount of face value.

Question (b).
Which feature of a Joint Stock company is referred to in this example?
Answer:
‘Limited Liability is a feature, which is referred to in this example.

Question (c).
Explain the feature briefly.
Answer:
As per ‘Limited liability’ member of the company is not liable to debts of the company. Member is liable only up to the unpaid amount of share capital. Members’ personal property will not be used for the liability of a company.

4. Distinguish between the following.

Question 1.
Sole Trading Concern and Joint Hindu Family Business.
Answer:

Question 2.
Sole Trading Concern and Partnership Firm
Answer:

Question 3.
Partnership Firm and Joint Stock Company
Answer:

Question 4.
Joint Stock Company and Co-operative Society
Answer:

Question 5.
Private Company and Public Company
Answer:

5. Answer in brief.

Question 1.
How is LLP different from a partnership firm?
Answer:
LLP and Partnership Firm both look alike but are separate and have separate features:

• Legal base: Limited Liability Partnership is based on “Limited Liability Partnership Act, 2008” and regular partnership is based on “Partnership Act, 1932”.
• A number of partners: In a Partnership firm, a minimum of two partners and a maximum of fifty partners are allowed and in LLP minimum of two partners and a maximum no limit.
• Liability: In LLP, partners have limited liability while in a partnership firm, partners have unlimited liability.
• Transfer of ownership: There is no restriction on joining and leaving the LLP, but in a partnership firm, partners cannot transfer their shares without the permission of other partners.

Question 2.
Explain the different types of companies on the basis of the liabilities of members.
Answer:
There are three types of companies on the basis of liabilities of members which are explained as under:
(i) Companies Limited by Shares: Such companies are formed as per Section 2(22) of the Companies Act, 2013. Such companies have to share capital and their members have limited liabilities up to unpaid part of the face value of shares held by them. At the time of winding up of the company, the personal property of shareholders is not used. They are liable only for the unpaid part of the number of shares purchased.

(ii) Company Limited by Guarantee: As per Section 2(21), such companies may or may not have share capital. Every Member promises to pay a specific amount for liabilities and debts of the company at the time of liquidation. Such amount is mentioned in Memorandum. Member simply gives guarantee and carries a specific amount of liability. Generally, such companies work for the promotion of sports, art, culture, charity, etc.

(iii) Unlimited Liability Company: As per section 2(92), members of these companies have unlimited liability. Members are fully liable to liabilities and debts of the company. It may be a private, public, or one-person company.

Question 3.
What are holding and subsidiary companies?
Answer:
Holding company:

• A company holds more than half of the share capital of another company.
• Such a company may have the power to appoint a director of another company.
• It has the power to remove directors of another company.

Subsidiary company:

• Such companies are controlled by holding companies.
• Holding company purchases more than half of the total share capital of the Subsidiary company.
• Holding Company has the power to appoint or remove all or a majority of its directors.
• A subsidiary company is just the opposite of a Holding company.

6. Justify the following statements.

Question 1.
Registration of Joint Stock Company is compulsory.
Answer:

• In India, the Joint Stock Companies are governed by the Companies Act, 2013.
• Every company has to be registered under the Companies Act, 2013.
• Registration gives birth to a company.
• On registration, the company gets a separate legal entity/identity.
• Without registration, no company can come into existence.
• After getting a registration certificate, it becomes a corporate body.
• So, the registration of a Joint Stock Company is compulsory and not optional.
• Thus, registration of a Joint Stock Company is compulsory.

Question 2.
A Joint Stock Company is an artificial person.
Answer:

• A Joint Stock Company is an incorporated association, which is an artificial person created by law, having a separate name, a separate legal entity, and perpetual succession.
• It is an artificial person because it is the creation of law.
• It does not have a physical existence but has legal existence.
• It enjoys certain rights and also conducts business like any human being.
• A company is an artificial person because it is not developed by the process of a natural person.
• It is the law of the land that gives birth to a company, hence, it is an artificial but legal person.
• A company has a distinct name and a common seal.
• It can make contracts, appoint staff, borrow money, open an account in the bank, acquire assets and conduct other business activities by its office bearers and staff.
• It can sue and be sued by others. So, a Joint Stock Company is an artificial person, created by law.
• Thus, a Joint Stock Company is an artificial person.

Question 3.
The liability of shareholders of the company is limited.
Answer:

• The liability of shareholders of the company is always limited.
• It is limited to the extent of the unpaid part of the face value of the shares held by them. Personal property of shareholders will not be taken or sold to pay the creditors or loan of the company.
• Shareholders are not concerned with other liabilities of the company.
• eg. If a shareholder has bought 100 shares of ₹ 10/- each, then he is liable only for ₹ 1000/- and not more.
• Thus, the liability of shareholders of a Joint Stock Company is limited.

Question 4.
The ownership and management are separated in Joint Stock Company.
Answer:

• A Joint Stock Company is a voluntary association of individuals for profit, having its capital divided into transferable shares, the ownership of which is the condition of membership.
• The members of a joint-stock company are many and they are scattered all over.
• It becomes very difficult for them to manage the business of the company.
• Lots of time may be wasted as all the shareholders are not able to come together at one place at the same time and as a result, there is a delay in decision making.
• To avoid these problems, the shareholders elect their representatives known as “Board of Directors”.
• The Board of Directors handles the business of the company on behalf of the shareholders.
• But, all important decisions pertaining to the company are not taken without the consent of all the shareholders.
• The owners of the company are the shareholders.
• The managers of the company are the Board of Directors.
• So, there is a separation of ownership and management in Joint Stock Company.
• Thus, the ownership and management are separated in Joint Stock Company.

Question 5.
The Joint Stock Company collects huge capital from the public.
Answer:

• A company requires large capital to carry out its activities.
• This capital is being collected from the public as it is not possible for an individual to contribute such a big amount.
• A company issues shares, debentures, bonds and also accepts public deposits for raising its capital.
• A company can also take loans from banks and financial institutions.
• a Large number of members bring in a large number of funds that can be profitably invested in the expansion of the company.
• So, for a large-scale business, large capital is collected by the Joint Stock Company.
• Thus, the Joint Stock Company collects huge capital from the public.

Question 6.
There is more Government control and supervision over the working of a Joint Stock company.
Answer:

• A Joint Stock Company is controlled and supervised by the Government.
• A company has to follow numerous provisions of the Companies Act and other Acts.
• The company has to follow all rules and regulations of the Government.
• If any of the legal provisions are violated, various charges are levied on the company.
• Government control protects the financial interest of a large number of investors.
• If any of the business of a company is carried out illegally, strict actions are taken by the Government authorities on the working of the company.
• So, there is more Government control and supervision over the working of Joint Stock companies.

7. Answer the following questions.

Question 1.
State the features of Sole Trading Concern.
Answer:
Features of Sole Trading Concern.

• No separate law: There is no separate law or act for sole trading concern. But, while conducting the business, it should follow routine laws which are applicable.
• Ownership: In a sole trading concern, only one person is the ail owner.
• Capital: Capital is contributed by the owner and an owner is a single person. Hence, capital is collected by a single person. So the size of capital is very small.
• Division of earnings: A sole trader is a single person, so there is no division of profit or loss. All profit is enjoyed by the owner and also bears all the losses of the business.
• Management: The business activities of Sole Trading Concern are managed by a single owner. Such an owner is a decision-maker.
• Secrecy: Sole Trader can ensure maximum business secrecy. The owner is not required to discuss the business matter with any outsider. Thus, maximum secrecy can be maintained.
• Liability: The liability of Sole Trader is unlimited. The owner’s personal property can be used for debts and liabilities of the business concern.
• Legal Status: Sole Trading Concern does not enjoy a separate legal status. There is no business registration by law. So, it does not have legal status.
• Suitable for small-scale business: Sole Trading Concern is suitable for small-scale business activity. Sole Traders can collect limited capital and thus cannot undertake large-scale business activity.
• Government control: There is no much government control over such type of business.

Question 2.
State the features of Joint Hindu Family Business.
Answer:
Features of Joint Hindu Family Business:

• Membership: Membership of Joint Hindu Family business is possible only by birth. Every child born in a family is considered a member of the Joint Hindu Family business.
• Karta: ‘Karta’is the head of the family. Generally, a senior person of the family plays the role of Karta in the Joint Hindu Family business.
• Co-parceners: ‘Co-parceners’ are the members of the family, rather than ‘Karta’. They play a supporting role in the family and have limited liability.
• The number of members: There is no limit on the minimum and maximum number of members.
• Liability: Liability of ‘Karta’ is unlimited, while the liability of ‘Co-parceners’ is limited in Joint Hindu Family business.
• Decision making: Being ahead of a Joint Hindu Family business, ‘Karta’ is the sole decision-maker. Thus, quick decision-making is possible in such type of business.
• Management: ‘Karta’ is head of Joint Hindu Family business activity, so such person is considered as manager, controller, and co-ordinator of business.
• Profit-sharing: The profit-sharing ratio keeps on changing in the Joint Hindu Family business. This ratio depends on the number of family members, which keeps on changing on birth and death in a family member.
• Legal Act: Joint Hindu Family business is governed by Hindu Succession Act, 1956. But, it does not need any registration. There is less government control on the Joint Hindu Family business.

Question 3.
State the features of the Partnership Firm.
Answer:
Features of Partnership Firm:

• Meaning: The business organization, which is owned, managed, and controlled by two or more persons under Partnership Act, 1932 is called a partnership firm.
• Registration: As per Partnership Act, 1932, registration is not compulsory. But, registration has been made compulsory in the state of Maharashtra.
• A number of members: The minimum number of members is 2 partners and a maximum of 50 partners for general business.
• Capital: The capital of a partnership firm is contributed by all the partners.
• Liability: In a partnership firm, the liability of partners is unlimited and is joint and several. The personal property of partners can be used to pay off the liabilities and debts of the partnership firm.
• Legal status: Partners of partnership firm enter into business with an agreement which is made as per Partnership Act, 1932. But, such firms do not enjoy separate legal statuses.
• Transfer of shares: In a partnership firm, partners cannot transfer their shares without the permission of other partners.
• Management: The business of a partnership firm is conducted, managed and controlled by all the partners. They are the joint owners and joint managers of a firm.
• No separation of ownership and management: As per the Partnership Act, 1932 all partners of the firm are owners, and as such it is the responsibility of all the partners to manage the business. So there is no separation of ownership and management.
• Profit-sharing: Being an owner of a firm, all the partners share profits and losses as per the ratio which is mentioned in the partnership deed.

Question 4.
State the features of Co-operative Society.
Answer:
Features of Co-operative Society:
(i) Meaning: Co-operative society is a voluntary association of persons, formed to provide services and economic welfare to its members.
(ii) Registration: The registration of a Co-operative society is compulsory in the state of Maharashtra, under Maharashtra State Co-operative Societies Act, 1960.
(iii) Membership: Membership of a Cooperative Society is open to all. Any person of sound mind can enter in Cooperative society.
(iv) Number of Membership: Minimum ten members are required and maximum there is no limit to join Co-operative Society.
(v) Liability: Members of the Cooperative society carry limited liability. The personal property of a member cannot be used for liability or debts of the society.
(vi) Aim of society: Generally, another form of business organization has its aim as profit. But, a Cooperative society has no aim of maximization of profit. They are formed with an aim of providing service and economic welfare to the members.
(vii) Legal status: Co-Operative societies are formed under a specific act, so they enjoy independent legal status different from its members.
(viii) Management: Management in a Cooperative society is based on the principle of ‘Democracy’. Shareholder/member enjoys equal voting right to decide management authority. ‘One Person One Vote’ principle is followed by the Cooperative society.
(ix) Government control: There is strict Government control and supervision on working of Co-operative society. Generally, the state Government controls the activity of Cooperative society.

Question 5.
State the features of a Limited Liability Partnership.
Answer:
Features of Limited Liability Partnership (LLP):

• Meaning: It is a combination of features of a partnership firm and a Joint Stock company. The liability of all partners in such a partnership firm is limited.
• Legal entity: Limited Liability Partnership has a separate legal entity. Partners and Limited Liability Partnership are distinct from each other i.e. it is a body corporate.
• the number of members: In a Limited Liability Partnership, a minimum of two partners are required. There is no limit on a maximum number of partners in a Limited Liability partnership.
• Capital: The capital of a business organization is collected from all partners. There is no requirement for minimum capital contribution.
• Business operation: It is simple to form and easy to operate.
• (vi) Liability: Limited Liability Partnership carry limited liability. The liability of each partner is limited to his share as written in the agreement.
• Cost of formation: As compared to a Joint Stock Company, the formation of a Limited Liability Partnership is very simple and easy. It has a low cost of formation.
• Transfer of share: In a Limited Liability Partnership, there is no restriction on the transfer of ownership/shares, except to follow rules of the partnership agreement, In short, there is less restriction on joining and leaving the Limited Liability Partnership.
• In a Limited Liability Partnership, double taxation is avoided and there is no tax on share in profit.

Question 6.
Define Joint Stock Company and explain its features.
Answer:
Definition of Joint Stock Company:

• As per Section 2(20) of the Companies Act, 2013: “Company means a company incorporated under this Act or under any previous company law”.
• According to Prof. H.L.Haney: “A Joint Stock Company is a voluntary association of individuals for profit, having its capital divided into transferable shares, the ownership of which is the condition of membership”.

Features of Joint Stock Company:
(i) Voluntary association: It is a voluntary association of individuals. Membership is open to all. Any person can join and leave the company subject to rules of the Articles of Association of the company.

(ii) Incorporated Association: Company is an association of persons formed and incorporated/registered under the Companies Act, 2013. Registration is compulsory. After incorporation, an association obtains the status of a Joint Stock Company.

(iii) Separate legal entity: The company enjoy a separate legal status different from its members and directors. Though the members are the owners, yet they are not liable for the actions of the company.

(iv) Artificial person: A company is a creation of law. A company does not have a physical existence, but it can conduct various activities like a human being.
E.g. enter into a contract, open a bank account, purchase or sell assets, appoint employees, etc. The company has corporate existence.

(v) Perpetual succession: A company has a perpetual succession means continuous existence. The company can enjoy a long and stable life. It is not affected by the death, insolvency, or retirement of any member.

(vi) Common seal: A company has a common seal of its own and all its activities are conducted under this seal. A company is an artificial person, its seal is the substitute for its signature. This seal is a name or any other recognition of a company.

(vii) Limited liability: The liability of members/shareholders of the company is limited. It is limited up to the unpaid part of the face value of shares held by shareholders. The personal property of a shareholder cannot be used for repayment of debts of the company.

(viii) Separation of ownership and management: As per the Companies Act, shareholders are the owners of the company, but they are unable to manage the day-to-day business activities as they are large in number, scattered and they keep on transferring shares. So, they appoint directors for management purposes. Thus, ownership and management are separate in the case of a Joint Stock Company.

(ix) Transferability of shares: The shares of a public company are transferable. They can be transferred freely whenever shareholder desires to sell. Shares of private companies are not freely transferable.

(x) Number of members: A company is owned by a large number of members. For private companies, minimum of 2 members and a maximum of 200 members are required and for the public company a minimum of 7 members and a maximum no limit.

(xi) Capital: Due to a large number of members, a huge amount of capital can be collected by the company in the form of shares, debentures, bonds, public deposits, etc. It can also borrow loans from banks and financial institutions.

(xii) Government control: There is strict control and supervision by the Government on the working of the company. The company has to follow the regulations and file Profit and Loss Account, Balance Sheet, and other financial statements with the Registrar. It should maintain all required books of accounts.

Question 7.
Define a ‘company’. Explain the types of companies on the basis of the liability of members.
Answer:
Definition of Joint Stock Company:

• As per Section 2(20) of the Companies Act, 2013: “Company means a company incorporated under this Act or under any previous company law’.
• According to Prof. H.L.Haney: “A Joint Stock company is a voluntary association of individuals for profit, having it’s capital divided into transferable shares, the ownership of which is the condition of membership”.

Types of Companies on the basis of liability of members:
(i) Companies Limited by Shares: Such companies are formed as per Section 2(22) of the Companies Act, 2013. Such companies have to share capital and their members have limited liabilities up to unpaid part of the face value of shares held by them. At the time of winding up of the company, the personal property of shareholders is not used.

(ii) Company Limited by Guarantee: As per Section 2(21) of Companies Act, 2013 such companies may or may not have share capital. Every Member promises to pay a specific amount for liabilities and debts of the company on liquidation. Such amount is mentioned in the Memorandum of Association. Members give guarantees and they carry a specific amount of liability. Generally, such companies work for the promotion of sports, art, culture, charity, etc.

(iii) Unlimited Liability Companies: As per Section 2(92) of the Companies Act, 2013 such companies have members with unlimited liability. Members are fully liable to liabilities and debts of the company. It may be a private, public, or one-person company.

Question 8.
Explain any four types of companies.
Answer:
Following are the types of companies:
(A) On the basis of Incorporation:

• Statutory Company: Statutory companies are incorporated by Special Act. Such Act is passed in Central or State legislation.
• E.g. Reserve Bank of India, State Bank of India, Unit Trust of India, Life Insurance Corporation, etc.
• Registered Company: Such companies are formed under the Companies Act, 2013 or any previous company law.

(B) On the basis of Number of Members:
(i) Private Company: It is a company having minimum paid-up capital as prescribed by its Articles. Such companies restrict the rights of their members to transfer their shares and also restrict the maximum number of its members up to 200. Such companies are also prohibited to invite the public to subscribe to their securities or deposits.

(ii) Public Company: It is a company having a minimum paid-up share capital as prescribed by its Articles. Such companies do not restrict the rights of their members to transfer their shares. It requires minimum of 7 members to form a company and there is no limit on the maximum number of members. Such companies can
invite the public to subscribe for its securities or deposits.

(iii) One Person Company: It is a company, in which one person is a member. Such a company is managed by a single person, having limited liability. It should follow the rules of a private company. It may have one or more directors.

(C) One the basis of Liability of Members:
(i) Companies Limited by Shares: Such companies are formed as per Section 2(22) of the Companies Act, 2013. Such companies have to share capital and its members have limited liabilities up to unpaid part of the face value of shares held by them. At the time of winding up of the company, the personal property of shareholders is not used.

(ii) Company Limited by Guarantee: As per Section 2(21) of Companies Act, 2013 such companies may or may not have share capital. Every Member promises to pay a specific amount for liabilities and debts of the company on liquidation. Such amount is mentioned in the Memorandum of Association. Members give guarantees and they carry a specific amount of liability. Generally, such companies work for the promotion of sports, art, culture, charity, etc.

(iii) Unlimited Liability Companies: As per Section 2(92) of the Company Act, 2013 such companies have members with unlimited liability. Members are fully liable to liabilities and debts of the company. It may be a private, public, or one-person company.

(D) On the basis of Control:
(i) Holding Company: A company holding more than half of the share capital of another company is called a Holding Company. This company has the power to appoint directors of another company and remove directors of another company.

(ii) Subsidiary Company: The company which is controlled by a holding company is called a Subsidiary Company.

Activity (Text Book Page No. 25)

Identify the type of following companies:

1. Bajaj Auto Limited
2. Coal India Limited
3. Microsoft India
4. Kirloskar Foundation

Answer:

1. Bajaj Auto Limited – Public Limited Company
2. Coal India Limited – Government Company
3. Microsoft India – Subsidiary Company of American Software Company Microsoft Corporation
4. Kirloskar Foundation – Company Not for Profit (Corporate Social Responsibility)

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 10 Halogen Derivatives Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 10 Halogen Derivatives

Question 1.
What are halogen derivatives of hydrocarbons?
Answer:
The replacement of hydrogen atom/s in aliphatic or aromatic hydrocarbons by halogen atom/s results in the formation of halogen derivatives of hydrocarbons.

Question 2.
How are halogen derivatives of hydrocarbons classified?
Answer:
Halogen derivatives of alkane are classified as :

1. Monohalogen derivative (or alkyl halide) : It is a halogen derivative of an alkane in which one hydrogen atom is replaced by one halogen atom and it is also called alkyl halide. E.g. C₂H₅Br.
2. Poh halogen derivatives : These are halogen derivatives in which more than one hydrogen atoms of an alkane are substituted by corresponding number of halogen atoms.

They are classified as follows :
(i) Dihalogen derivatives : The compounds formed by the substitution of two hydrogen atoms of an alkane by two halogen atoms are called dihalogen derivatives.

They are further classified as :

• Vicinal dihalides (Two halogen atoms on vicinal or adjacent carbon atoms)
• Geminal dihalides : CH₃ – CHCI₂ (Two halogen atoms on the same carbon atom)

(ii) Trihalogen derivatives : The compounds formed by the substitution of three hydrogen atoms of an alkane by three halogen atoms are called trihalogen derivatives.

(iii) Tetrahalogen derivatives : The compounds formed by the substitution of four hydrogen atoms of an alkane by four halogen atoms are called tetrahalogen derivatives. E.g. CCI₄.

Question 3.
What are alkyl halides? How are they classified?
Answer:
The compound formed by the replacement of one hydrogen atom in an alkane by a halogen atom is called an alkyl halide. The halogen atom is bonded to sp³ hybridised carbon. Alkyl halides are classified into the following three classes depending on the type of the carbon to which halogen atom is bonded.

(1) Primary (1°) alkyl halide : Alkyl halide in which a halogen atom is bonded to a primary carbon atom is called primary alkyl halide.

[Primary (1°) carbon atom i.e., the carbon atom which is attached to only one carbon atom.]

They are represented by the general formula R – ^1°CH₂ – X.

(2) Secondary (2°) alkyl halide : Alkyl halide in which a halogen atom is bonded to a secondary carbon atom is called secondary alkyl halide. [Secondary (2°) carbon i.e., the carbon atom which is attached to two other carbon atoms.]

They are represented by the general formula (R and R’ can be same or different)

(3) Tertiary (3°) alkyl halide : Alkyl halide in which halogen atom is bonded to a tertiary carbon atom is called tertiary alkyl halide. [Tertiary (3°) carbon i.e., the carbon atom which is attached to three other carbon atoms.]

They are represented by the general formula  (R, R’ and R” may be same or different)

Question 4.
Explain the following :
(1) Alkyl halide or haloalkanes
(2) Allylic halides
(3) Benzylic halide
(4) Vinylic halide
(5) Haloalkyne
(6) Aryl halide or haloarenes.
Answer:
(1) Alkyl halide or haloalkanes : In alkyl halides or haloalkanes the halogen atom is bonded to sp3 hybridized carbon which is a part of saturated carbon chain.
Example :

(2) Allylic halides : In allylic halides, halogen atom is bonded to a sp³ hybridized carbon atom next to a carbon-carbon double bond.

Example :

(3) Benzylic halide : In benzylic halides, halogen atom is bonded to a sp³ hybridized carbon atom which is further bonded to an aromatic ring.
Example :

(4) Vinylic halides : In vinylic halides, halogen atom is bonded to a sp² hybridized carbon atom of aliphatic chain. Vinylic halide is a haloalkene.
Example :

(5) Haloalkyne : In haloalkynes, halogen atom is bonded to a sp hybridized carbon atom.
Example :

(6) Aryl halides or haloarenes : In aryl halides, halogen atom is directly bonded to the sp² hybridized carbon atom of aromatic ring.
Example :

Question 5.
Give the IUPAC names of the following :
Answer:

Question 6.
Draw the structures of the following compounds:
Answer:

Question 7.
Write the structure of-
(a) 3-chloro-3-ethylhex-l-ene
(b) 1-Iodo-2, 3-dimethylbutane
(c) 1, 3, 5-tribromobenzene
Answer:

Question 8.
Write structures of
(a) 2-iodo-3-methyl pentane
(b) 3-chiorolleNane
(c) 1-chloro-2, 2-dimethyl propane
(d) 1-chloro-4-ethyl cyclohexane.
Answer:

Question 9.
Write the possible isomers of monochloro derivatives of 2,3-Dimethylbutane and write their IUPAC names.
Answer:
The given parent hydrocarbon has molecular formula, C₆H₁₄. The monochloro derivative of this compound has molecular formula C₆H₁₃CI.
The parent hydrocarbon is,

Hence the structures of isomers of monochioroderivative are.

Question 10.
Write structures and IUPAC names of all possible isomers of C₅H₁₁Br and classify them as l°/2°/3°.
Answer:
C₅H₁₁Br is a monohalogen derivative.

Question 11.
How are following compounds obtained from alcohols :
(1) ethyl chloride C₂H₅CI
(2) isopropyl chloride (CH₃ CHCI – CH₃)
(3) tert-butyl chloride (CH₃)₃ – CI?
Answer:
Alcohols in the presence of Lucas reagent which is a solution of concentrated HCI and ZnCI₂ form alkyl halides. Hydrogen chloride is used with zinc chloride (Grooves’ process) for primary and secondary alcohols.

(3) Tertiary alcohols don’t need ZnCI₂ to react with HCI.

The order of reactivity of alcohols with a given halo acid is 3° >2°> 1°.

Question 12.
How are following compounds prepared from alcohols:
(1) ethyl bromide (C₂H₅Br)
(2) isopropyl bromide (CH₃ – CHBr – CH₃)
(3) tert-butyl bromide (CH₃)₃ C – Br?
Answer:
(1) Ethyl alcohol on heating with conc. hydrobromic acid (48%) forms ethyl bromide.
OR
When ethyl alcohol is treated with a mixture of NaBr and H₂SO₄, ethyl bromide is formed. Here HBr is generated in situ.

(2) Isopropyl alcohol, on reaction with NaBr and dil. H₂SO₄ forms isopropyl bromide.

(3) Tertiary alcohol on reaction with sodium bromide and dil. H₂SO₄ forms tert-butyl bromide.

Question 13.
How is ethyl iodide obtained from ethyl alcohol?
Answer:
When ethyl alcohol is treated with sodium or potassium iodide in 95 % phosphoric acid, ethyl iodide is formed. Here HI is generated in situ.

Question 14.
How will you prepare the following :

(1) Ethyl chloride (chloroethane) from ethyl alcohol using
(i) PCI₃
(ii) PCI₅ and
(iii) SOCI₂.
Answer:
(i) When ethyl alcohol is refluxed with phosphorus trichloride, ethyl chloride is formed.

(ii) When ethyl alcohol is refluxed with phosphorus pentachloride, ethyl chloride is formed.

(iii) When ethyl alcohol is refluxed with thionyl chloride, in the presence of pyridine, ethyl chloride is formed. The by-products obtained are gases. Therefore, this method is preferred for preparation of alkyl chloride.

(2) Isopropyl chloride (2-chloropropane) from isopropyl alcohol using
(i) PCI₃
(ii) PCI₅
(iii) SOCI₂.
Answer:
When isopropyl alcohol is refluxed with phosphorus trichloride, isopropyl chloride is formed.

When isopropyl alcohol is refluxed with phosphorus pentachloride, isopropyl chloride is formed.

When isopropyl alcohol is refluxed with thionyl chloride, in the presence of pyridine, isopropyl chloride is formed. The by-products obtained are gases. Therefore, this method is preferred for the preparation of alkyl chloride.

(3) Ethyl bromide (bromoethane) from ethyl alcohol.
Answer:
When ethyl alcohol is treated with a mixture of red phosphorus and bromine or hydrobromic acid (phosphorus tribromide is generated in situ), ethyl bromide is formed.

(4) Ethyl iodide (do ethane) from ethyl alcohol.
Answer:
When ethyl alcohol is heated with a mixture of red phosphorus and iodine, (phosphorus triiodide is generated in situ), ethyl iodide is formed.

Question 15.
Explain halogenation of methane.
Answer:
Halogenation : A reaction of alkanes with halogens (CI₂, Br₂, I₂) in the presence of appropriate conditions forming a mixture of alkyl halides.
(1) Chlorination:

(2) When excess of chlorine is used, tetrachioro methane, a major product is obtained. When excess of methane is used, chioromethane, a major product is obtained. The order of reactivity of halogens towards alkane is

(3) lodination:

However, iodination reaction is a reversible reaction. HI being a strongest reducing agent reduces methyl iodide back to methane.

(4) Fluorination: A reaction of alkane with fluorine is explosive and also hydrofluoric acid is poisonous and corrosive. Hence, alkyl fluorides are not prepared by halogenation of alkane.

Question 16.
Predict the possible products of the following reaction :
(1) Bromination of propane
(2) Bromination of n-butane
(3) Bromination of 2-Methyl propane.
Answer:
(1) Bromination of propane

(2) Bromination of n-butane

(3) Bromination of 2-Methyl propane

Question 17.
How are following compounds prepared by halogenation of ethane :
(1) Chloroethane
(2) Bromoethane
(3) Iodoethane?
Answer:
(1) Chlorination of ethane : When ethane (excess) is reacted with a limited quantity of chlorine in the presence of diffused sunlight or U.V. light or at high temperature, chloroethane is obtained.

(2) Bromination of ethane : When ethane is heated with Br₂ in the presence of anhydrous AlBr₃, bromoethane is obtained.

(3) Iodination : When ethane is reacted with I₂ in the presence of suitable oxidising agents like-HgO or HIO₃ or dilute HNO₃ iodoethane is obtained.

Question 18.
Direct iodination of alkanes is not possible.
Answer:
(1) Direct iodination of alkanes using iodine is highly reversible.
\(\mathrm{RH}+\mathrm{I}_{2} \rightleftharpoons \mathrm{RI}+\mathrm{HI}\)
(2) Hydroiodic acid HI being strong reducing agent, it reduces RI to alkane RH.
(3) The reaction takes place only in the presence of a suitable oxidizing agent like HgO, HIO₃ or dilute HNO₃ which decomposes HI. Hence, direct iodination of alkanes is not possible.

Question 20.
How are following compounds obtained from alkenes :

Answer:
(1) Ethene on reaction with hydrogen chloride forms

(2) Ethene on reaction with hydrogen bromide forms

(3) Propene on reaction with hydrogen iodide forms

(4) but-2-ene on reaction with hydrogen iodide forms 2-iodobutane.

Question 19.
State and explain Markovnikov’s rule.
Answer:
Markovnikov’s rule : When an unsymmetrical reagent is added to an unsymmetrical alkene, the negative part of the reagent gets attached to that carbon atom of the double bond which carries less number of hydrogen atoms.

Example : Addition of HBr’to unsymmetrical alkene like propene gives two products.

Isopropyl bromide is the major product, since the negative part (Br^–) of HBr is attached to carbon atom of a double bond with less number of hydrogen atoms.

Question 20.
Explain peroxide effect.
OR
Write a note on the Kharasch-Mayo effect.
OR
Explain the addition of HBr to (unsymmetrical alkene) propane in the presence of benzoyl peroxide.
Answer:
The addition of HBr to an unsymmetrical alkene (propane) in the presence of benzoyl peroxide takes place in the opposite orientation to that of Markovnikov’s rule and this is known as Kharasch-Mayo effect or peroxide effect or Anti-Markovnikov addition.

Question 21.
Write the structure of alkyl halide obtained by the action hydrogen bromide on 2-Methyiprop-1-ene in the presence of peroxide.
Answer:
In the presence of peroxide. HBr to 2-Methyl prop-I-cne forms l-Bromo-2-methylpropane.

Question 22.
How are alkyl iodides prepared from alkyl chlorides/bromides?
Answer:
Alkyl iodide is prepared by treating alkyl chloride or alkyl bromide with sodium iodide, in the presence of dry acetone, sodium chloride or sodium bromide precipitates from the solution and can be separated by filtration. This reaction is known as Finkelstein reaction.

Question 23.
How are alkyl fluorides prepared with alkyl chlorides/alkyl bromides?
Answer:
When alkyl chloride or alkyl bromide is heated with metallic fluorides like AgF, CaF₂, CoF₂ or Hg₂F₂, alkyl fluoride is formed.

This reaction is known as Swarts reaction.

Question 24.
Explain the preparation of haloarenes using electrophilic substitution.
Answer:
When arene is treated with chlorine or bromine in dark at ordinary temperature in the presence of lewis acid as a catalyst like Fe, FeCI₃ or anhydrous AlCI₃, aryl chloride or aryl bromide is formed.

When toluene is brominated in dark at ordinary temperature in the presence of iron, a mixture of ortho and para bromo tolerene is obtained.

Ortho and para isomers can be easily separated as there is large difference in melting points of ortho and para isomers.

Question 25.
Write a note on Sandmeyer’s reaction.
Answer:
Aryl halides are most commonly prepared by replacement of nitrogen of diazonium salt. The replacement of diazonium group by -Cl or -Br using cuprous salt is called Sandmeyer’s reaction. When a primary aromatic amine (like aniline) suspended in cold F1C1, is treated with sodium nitrite, a diazonium salt (benzene diazonium chloride) is formed. When diazonium salt is treated with cuprous chloride or cuprous bromide, aryl halide (chlorobenzene or bromobenzene) is formed.

When benzene diazonium salt is mixed with potassium iodide, iodobenzene is formed.

Question 26.
Define the following :
Answer:
(1) Monochromatic light : It consists of rays of single wavelength vibrating in different planes perpendicular to the direction of propagation of the light.

(2) Plane polarized light : A light having oscillations only in one plane perpendicular to direction of propagation of light is known as plane polarized light.

(3) Optical isomerism : The steroisomerism in which the isomers have different spatial arrangements of groups/atoms around a chiral atom is called optical isomerism.

(4) Optical activity : The property of a substance by which it rotates plane of polarization of incident plane polarized light is known as optical activity.

(5) Optically active compound : The compound which rotate the plane of plane polarized light is called optically active compound.

(6) Enantiomers : The optical isomers which are non-superimposable mirror images of each other are called enantiomers or enantiomorphs or optical antipodes.
Example : 2-chlorobutane, lactic acid

(7) Chiral carbon atom : Carbon atom in a molecule which carries four different groups/atoms is called chiral carbon atom.

Chiral atom in a molecule is marked with asterisk (*)

For example : C in lactic acid

(8) Chiral molecule : When a molecule contains one chiral atom, it acquires a unique property i.e. it is non- superimposable with its mirror image is said to be chiral molecule.

(9) Chirality : The relationship between a chiral molecule and its mirror image is similar to the relationship between left and right hands. Therefore it is called handedness or chirality.

(10) Dextrorotatory substance or r/-Isomer : An optically active substance (or isomer) which rotates the plane of a plane polarized light to the right hand side (RHS) is called dextrorotatory substance (or isomer) and denoted by d or (+) sign.

(11) Laevorotatory substance or /-Isomer : An optically active substance (or isomer) which rotates the plane of a plane polarized light to the left hand side (LHS) is called laevorotatory substance (or isomer) and denoted by / or (-) sign.

(12) Racemic mixture or Racemate : A mixture containing equimolar quantities of dextro (d) and laevo (/) optical
isomers which is optically inactive due to molecules of one enantiomer is cancelled by equal and opposite optical rotation due to molecules of the other enantiomer is called a racemic mixture or racemate. It is represented as (dl) or (+).

Question 27.
Calculate the number of isomers for 2-chlorobutane.
Answer:
The number of optical isomers possible for a compound is 2ⁿ where n = number of asymetric carbon atoms.
As n = 1 for 2-ehlorobutane, 2ⁿ = 2¹ = 2. Hence, it has two optical isomers.

Question 28.
How many optical isomers are possible for C₅H₁₁ CI?
Answer:
The number of optical isomers : 3.

Question 29.
How many optical isomers are possible for glucose?
Answer:
The number of optical isomers : 16.

Question 30.
Draw the structures and indicate the chiral carbon atoms in
(1) Lactic acid
(2) 2-Chlorobutane.
Answer:
(1) In lactic acid structure, the starred carbon atom is chiral carbon atom as it is attached to four different substituents, COOH, OH, CH₃ and H.
(2) In 2-chlorobutane structure, the starred carbon atom is chiral carbon atom as it is attached to four different substituents, -CH₂ – CH₃ (ethyl), CH₃ (methyl), Cl and H.

Question 31.
Identify chira! and achiral molecules.

Question 32.
Complete the following reactions and explain optical activity of the products formed:
(i) Pent-1-ene with HBr
Answer:

(ii) Pent-2-ene with HBr
Answer:

Question 33.
C₆H₁₂ (A) on treatment with HCI produced a compound Y. Which is optically active, what is structure A?
Answer:

Question 34.
A racemic mixture is optically inactive. Explain.
Answer:

• A racemic mixture contains equimolar (or equimolecular) quantities of the dextrorotatory (d-) and laevorotatory (l-) isomers (enantiomers) of a compound.
• The d-enantiomer rotates the plane of plane-polarized light to the right, while the l-enantiomer rotates the same to the left to the same extent.
• The quantities of the d- and l-enantiomers being the same, both the rotations are of the same magnitude, but of opposite directions. Hence, they cancel each other. Hence, a racemic mixture is optically inactive.
• It is represented as dl or ( + ). Example : ( ± ) lactic acid

Question 35.
Explain Fischer projection formula with illustration.
OR
Write a note on Fischer projection formula.
Answer:
Fischer projection formula or cross formula : The three dimensional (3-D) view of a molecule is presented on plane of paper. A Fischer projection formula can be drawn by visualizing the main carbon chain verical in the molecule. Each carbon on the vertical chain is represented by a cross.

Conventionally the horizontal lines of the cross represent bonds projecting up from the carbon and the vertical lines represent the bonds going below the carbon.

Question 36.
Explain Wedge formula with illustration
OR
Write a note on Wedge formula.
Answer:
Wedge formula : When a tetrahedral carbon is imagined to be present in the plane of paper all the four bonds at this carbon cannot lie in the same plane. The bonds in the plane of paper are represented by normal lines, the bonds projecting above the plane of paper are represented by solid wedges (or simply by bold lines) while bonds going below the plane of paper are represented by broken wedges (or simply by broken lines).

Question 37.
Give a laboratory test to confirm the presence of halogen in the original organic compound.
Answer:
Haloalkanes are of neutral type in aqueous medium. On warming with aqueous sodium or potassium hydroxide the covalently bonded halogen in haloalkane is converted to halide ion.

When this reaction mixture is acidified by adding dilute nitric acid and silver nitrate solution is added a precipitate of silver halide is formed which confirms presence of halogen in the original organic compound.

Question 38.
Define the following :
Answer:
(1) Mechanism of a reaction : It is a step by step description of exactly how the reactants are transformed into products in as much details as possible.
(2) Substitution reaction : When a group bonded to a carbon in a substrate is replaced by another group to get a product with no change in state of hybridization of that carbon, the reaction is called substitution reaction.

Question 39.
Describe the action of aqueous KOH (or NaOH) on :
(1) ethyl bromide
(2) isopropyl bromide
(3) tert-butyl chloride
(4) methyl bromide
(5) 2-chlorobutane.
Answer:
(1) Ethyl bromide : When ethyl bromide (bromoethane) is refluxed with aqueous potassium hydroxide, ethyl alcohol is formed. The reaction is called a hydrolysis reaction.

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

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

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

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

Question 40.
Describe the action of sodium ethoxide on
(1) ethyl bromide
(2) methyl bromide :
OR
Write a note on Williamson’s synthesis.
OR
How are ethers prepared from alkyl halides?
Answer:
Williamson’s synthesis : When an alkyl halide (R – X) is heated with sodium alkoxide (R – O – Na), an ether is obtained. In this reaction halide (-X) of alkyl halide is replaced by an alkoxy group (-OR). This reaction is known as Williamson’s synthesis. This method is used to prepare simple (or symmetrical) ethers and mixed (or unsymmetrical) ethers.

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

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

e.g., When ethyl bromide is heated with sodium ethoxide, diethyl ether is formed.

(2) Mixed (unsymmetrical) ether : When an alkyl halide and sodium alkoxide having different alkyl groups are heated, unsymmetrical ether is obtained.

When methyl bromide is heated sodium ethoxide, ethyl methyl ether is formed.
When ethyl bromide is heated with sodium meihoxide, ethyl methyl ether is formed.

Question 41.
What is the action of silver salt of carboxylic acid on alkyl halide?
Answer:
When an alkyl halide (R – X) is heated with silver salt of carboxylic acid (R -COOAg). an ester is obtained.

Question 42.
Describe the action of alcoholic silver acetate on
(1) methyl bromide
(2) ethyl bromide.
Answer:
(1) Methyl bromide : When methyl bromide is heated with an alcoholic silver acetate, methyl acetate is formed.

(2) Ethyl bromide : When ethyl bromide is heated with an alcoholic silver acetate, ethyl acetate is formed.

Question 43.
What is the action of alcoholic silver propionate on ethyl bromide?
Answer:
When ethyl bromide is heated with an alcoholic silver propionate. ethyl propionate is formed.

Question 44.
Describe the action of excess of ammonia on (I) ethyl bromide (2) n.propyl bromide.
Answer:
(1) Ethyl bromide : When ethyl bromide is boiled under pressure with an excess of alcoholic ammonia, ethylamine (ethanamine) is formed. This is known as ‘ammonolysis of ethyl bromide.

(2) n-propyl bromide : When n-propy1 bromide is boiled under pressure with an excess of ammonia, n-propyl amine (propanamine) is formed.

Question 45.
What is ammonolysis? Give a suitable example for the reaction.
Answer:
When an alkyl halide is boiled under pressure with an excess of alcoholic solution of ammonia (NH₃), corresponding (primary amine) alkyl amine is formed. This reaction is known as ammonolysis of alkyl halide.

(1) Ethyl bromide : When ethyl bromide (bromoethane) is refluxed with aqueous potassium hydroxide, ethyl alcohol is formed. The reaction is called a hydrolysis reaction.

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

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

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

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

Question 46.
Describe the action of aqueous alcoholic potassium cyanide on
(1) ethyl bromide
(2) methyl iodide.
Answer:
Ethyl bromide : When ethyl bromide (bromoethane) is boiled with alcoholic solution of potassium cyanide in aqueous ethanol, ethyl cyanide (ethyl nitrile) is formed.

(2) Methyl iodide : When methyl iodide is boiled with alcoholic solution of potassium cyanide, methyl cyanide is formed.

Question 47.
Describe the action of alcoholic silver cyanide on
(1) ethyl bromide
(2) methyl chloride.
OR
Explain isocyanide reaction of
(1) ethyl bromide
(2) methyl chloride.
Answer:
(1) Ethyl bromide : When ethyl bromide is heated with alcoholic silver cyanide, ethyl isocyanide is formed.

(2) Methyl chloride : When .methyl chloride is heated with alcoholic silver cyanide, methyl isocyanide is formed.

The above reactions (1) and (2) are called isocyanide reaction.

Question 48.
Describe the action of potassium nitrite on
(i) ethyl bromide,
(ii) methyl chloride.
Answer:
(1) Ethyl bromide : When ethyl bromide is treated with potassium nitrite, ethyl nitrite is formed.

(2) Methyl chloride : When methyl chloride is treated with potassium nitrite, methyl nitrite is formed.

Question 49.
Describe the action of silver nitrite on (1) ethyl chloride (2) n-propyl bromide.
Answer:
(1) Ethy chloride : When ethyl chloride is treated with silver nitrite, nitroethane is obtained.

(2) n-Propyt bromide: When n-propyl bromide is treated with silver nitrate, nitropropane is obtained.

Question 50.
How will you convert (the following:

(1) Ethyl bromide to ethanol.
Answer:

(2) Ethyl bromide to propane nitrile.
Answer:

(3) Ethyl bromide to ethyl amine.
Answer:

(4) Ethyl bromide to ethyl acetate.
Answer:

(5) Ethyl bromide to ethyl isocyanide.
Answer:

(6) Ethyl bromide to ethyl methyl ether.
Answer:

(7) Ethyl bromide to n-butane.
Answer:

(8) Ethyl bromide to Ethyl magnesium bromide.
Answer:

Question 51.
Define the following :
Answer:
(1) Nucleophilic bimolecular reaction (SN²) : The substitution reaction in which a nucleophile reacts with the substrate and the rate of the reaction depends on the concentration of the substrate and the nucleophile is called a nucleophilic bimolecular reaction.

Example :

(2) SN¹ reaction : The substitution reaction in which a nucleophile reacts with the substrate and the rate of the reaction depends only on the concentration of the substrate is called nucleophilic unimolecular or first order reaction or SN¹ reaction.

Question 52.
Explain, the mechanism of alkaline hydrolysis (reaction with aqueous KOH) of tert-butyl bromide (2-Bromo-2-methylpropane) with energy profile diagram.
OR
Explain only reaction mechanism for alkaline hydrolysis of tert-butyl bromide.
Answer:
(i) Consider alkaline hydrolysis of tert-butyl bromide (2-Bromo-2 methylpropane) with aqueous NaOH or KOH.

(ii) Kinetics of the reaction : Due to steric hindrance of voluminous three methyl groups around carbon, nucleophile OH- cannot attack carbon atom directly. Hence, the reaction takes place in two steps.

Step I : This involves heterolytic fission of C – Br covalent bond in the substrate forming carbocation and Br^–. This is a slow process.

Step II : This step involves attack of nucleophile OH- or carbocation forming C – OH bond and product tert-butyl alcohol. Since it involves ionic charge neutralisation, it is a fast step.

Rate Determining Step (R.D.S.) : Since the first step is a slow step, it is R.D.S., and therefore the rate of the reaction depends on the concentration of only one reactant, (CH₃) C – Br.
Rate = R = k [(CH₃)₃ C – Br] where k is a rate constant of the reaction.

SN¹ reaction : The reaction between tert.butyl bromide and hydroxide ion to form tert.butyl alcohol follows a first-order kinetics. The rate of this reaction depends only on the concentration of one substance (tert-butyl bromide) and is independent of the concentration of alkali added. It is an unimolecular first (1st) order Nucleophilic Substitution reaction denoted as SN reaction.

Stereochemistry and mechanism of the reaction : The reaction takes place in two steps and both the steps involve formation of transition states (T.S.).

T.S. -1 for first step :

In this transition state, C – Br bond is partially broken, so that carbon atom carries partial positive charge (+δ) and Br carries partial negative charge (-δ) which further breaks forming carbocation and Br . Tert-butyl cation (carbocation) has a planar structure and the CH₃ – C – CH₃ bond angle is 120°. It is the intermediate of the reaction. It is unstable. In this step, hybridisation of carbon atom changes from sp³ (tetrahedral geometry) to sp (planar geometry).

T.S. – II for second step :

In this transition state, C – OH bond is partially fonned so that carbon atom carries partial positive charge (+ δ) and OH carries partial negative charge ( -δ) which further forms tert-butyl alcohol.

Formation of a racemic mixture : Since OH^– has equal probability of the attack on carbocation from frontside and from backside, the products obtained are equal. In case of optical active alkyl halide, a racemic mixture is obtained.

Question 53.
Discuss SN² mechanism of methyl bromide using aqueous KOH. Draw energy profile diagram.
OR
Discuss the mechanism of alkaline hydrolysis of methyl bromide or Bromomethane.
Answer:
(1) Consider alkaline hydrolysis of methyl bromide (Bromomethane). CH₃Br with aqueous NaOH or KOH.

(2) Stereochemistry and Kinetics of the reaction iR.D.S.) : This hydrolysis reaction takes place only in one step which is a rate determining step i.e. R.D.S. The rate of hydrolysis reaction depends on the Concentration of CH3Br
and 0H which are present in the R.D.S. of the reaction.
Rate = R = k [CH₃Br] (OH]
where k is rate constant of the reaction.

SN² reaction : The reaction between methyl bromide and hydroxide ion to form methanol follows a second order kinetics, since the rate of the reaction depends on the concentrations of two reacting species, namely methyl bromide and hydroxide ion it is bimolecular second order (2nd) Nucleophilic Substitution reaction denoted by SN².

(3) Mechanism of the reaction :
(i) It is a single step mechanism. The reaction takes place in the following steps :

(ii) Backside attack of the nucleophile : Nucleophile, OH^– attacks carbon atom of CH₃Br from back side i. e. from opposite side to that of the leaving group i.e. Br^– to experience minimum steric repulsion and electrostatic repulsion between the incoming nucleophile (OH^–) and leaving Br^–.

(iii) Transition state : When a nucleophile, OH^– approaches carbon atom of CH₃Br, the potential energy of the system increases until a transition state (T.S.) of maximum potential energy is formed in which C – Br bond is partially broken and C – OH bond is partially formed. The negative charge is equally shared by both incoming nucleophile- OH^– and outgoing, leaving group-Br^–. (Thus, the total negative charge is diffused.)

(iv) In CH₃Br, carbon atom is sp³ -hybridized and CH₃Br molecule is tetrahedral. The hybridisation of carbon atom changes to sp² hybridisation. The transition state contains pentacoordinate carbon having three δ (sigma) bonds in one plane making bond angles of 120° with each other i.e., H₁; H₂ and H₃ atoms lie in one plane while two partial covalent bonds containing Br and OH lie collinear and on opposite sides perpendicular to the plane.

(v) Inversion of configuration : The transition state decomposes fast by the complete breaking of the C-Br bond and the new C-OH bond is formed on the other side. The breaking of C-Br bond and the formation of C-OH bond take place simultaneously. The energy required to break the C-Br bond is partly obtained from the energy released when C-OH bond is formed. The formation of product CH₃OH is accompanied by complete or 100% inversion of configuration forming again sp³-hybridized carbon atom giving tetrahedral CH₃OH molecule. But in this structure the positions of H₂ and H₃ atoms in the reactant (CH₃Br) and in product are on the opposite side. This inversion of configuration is called Walden inversion.

Question 54.
Discuss the factors influencing S_N1 and S_N2 mechanism.
Answer:
(1) Nature of substrate : S_N2 : The transition state (T.S.) of S_N2 mechanism + is pentacoordinate, it is crowded. As a result S_N2 mechanism is favoured in primary halides and least favoured in tertiary halides.

S_N1 : A planar carbocation intermediate is formed in S_N1 reaction. Bulky alkyl groups can be easily accommodated in planar carbocation and it has no steric crowding. As a result SN1 mechanism is favoured in tertiary halides and least favoured in primary halides.

The carbocation intermediate is stabilized by + effect of alkyl substituents and also by hyperconjugation y effect of alkyl substituents containing a-hydrogens. As a result, S_N1 mechanism is favoured in tertiary halides and least favoured in primary halides.

Thus, tertiary alkyl halides undergo nucleophilic substitution by S_N1 mechanism while primary halides follow S_N2 mechanism.

(2) Nucleophilicity of the reagent : A strong nucleophile attacks the substrate faster and favours S_N2 mechanism. The rate of S_N1 mechanism is independent of the nature of nucleophile. Nucleophile does not react in the 1st step (slow step) of S_N1. Nucleophile reacts fast after the carbocation intermediate is formed.

(3) Solvent polarity : (1) S_N1 reaction proceeds more rapidly in polar protic solvents than in aprotic solvent. Polar protic solvent decreases the rate of S_N2 reaction. (2) In S_N2 mechanism, rate depends on substrate as well as nucleophile. A polar solvent stabilizes nucleophile by solvation. Thus solvent deactivates the nucleophile by stabilizing it. Hence, aprotic solvents or solvent of low polarity will favour S_N2 mechanism.

Question 55.
How does relative reactivity for alkaline hydrolysis with respect to S_N2 and S_N1 vary in the following alkyl halides :
(1) Bromomethane
(2) Bromoethane
(3) 2-Bromopropane
(4) 2-Bromo-2-methylpropane ?
Answer:
(A) Relative reactivity for S_N2 mechanism decreases in the order of :

(B) Relative reactivity for S_N1 mechanism decreases in the order of :
2-Bromo-2-methylpropane > 2-Bromopropane > Bromoethane > Bromomethane

Question 56.
Explain with reason the relative order of reactivity of l°/2°/3° alkyl halides by S_N1 mechanism.
Answer:
In alkaline hydrolysis of an alkyl halide by S_N1 mechanism, the formation of carbocation as an intermediate product is involved.

The increasing order of a stability of carbocation is,

The stability order for carbocation is 3° > 2° > 1°.
Therefore the increasing order of reactivity by S_N1 mechanism of alkyl halides is
(1°) primary < (2°) secondary < (3°) tertiary

Question 57.
Which one of the following is more easily hydrolysed in S_N1 and S_N2 reaction by aqueous KOH, C₆H₅ CHCIC₆H₅ and C₆H₅CH₂CI?
Answer:
In S_N1 reaction C₆H₅CHCI C₆H₅ will be more easily hydrolysed than C₆H₅CH₂CI
In S_N2 reaction C₆H₅ CH₂CI will be more easily hydrolysed than C₆H₅CHCIC₆H₅.

Question 58.
Choose the member that will react faster than the following pairs by S_N1 mechanism.
(1) l-bromo-2, 2-dimethyl propane or 2-bromopropane.

Answer:
The reactivity of S_N1 reaction depends on the steric hindrance, in 2-bromopropane, a-carbon atom is attached to two methyl groups suffers greater steric hindrance to nucleophilic attack than l-bromo-2, 2-dimethyl propane. Hence, 2-bromopropane react faster by S_N1 mechanism.

(2) 2-Iodo-2-methyl butane or 2-iodio-3-methyl butane.

Answer:
Since, 2-Iodo-2-methyl butane is a tertiary alkyl halide, it undergoes SN^-1 reaction faster than 2-iodo-3-methyl butane.

(3) 1-Chloro propane or 2-chloropropane.

Answer:
Since, 2-chloropropane is a secondary alkyl halide, it undergoes SN^-1 reaction faster than 1-chloropropane.

(4) 2-Iodo-2-methyl butane or tert-butyl chloride.

Answer:
Since, iodine is a better leaving group than chloride 2-iodo-2-methyl butane undergo SN^-1 reaction faster than tert-butyl chloride.

Question 59.
Write a note on elimination reaction.
OR
Explain dehydrohalogenation reaction.
Answer:
When alkyl halide having at least one β-hydrogen is boiled with alcoholic solution of potassium hydroxide (KOH), an alkene is formed due to elimination of hydrogen atom from β-carbon and halogen atom from α-carbon, is called dehydrohalogenation.

Tertiary butyl bromide when heated with alcoholic solution of potassium hydroxide forms isobutylene.

This reaction is called β-elimination (or 1,2-elimination) reaction as it involves elimination of halogen and a β-hydrogen atom.

As hydrogen and halogen is removed in this reaction it is also known as dehydrohalogenation reaction.

Question 60.
Describe the action of alcoholic potassium hydroxide (aic. KOH) on
(1) ethyl bromide
(2) n-propyl bromide
(3) isopropyl bromide
(4) tert-butyl chlorIde.
Answer:
(1) Ethyl bromide : When ethyl bromide (bromoethane) is heated with alcoholic potassium hydroxide (alcoholic alkali). ethene (gas) is formed by the dehydrobrominaion reaction.

(2) n-PropI bromide : When n-propyl bromide is heated with alcoholic potassium hydroxide, propene is formed.

(3) Isopropvl bromide : When isopropyl bromide (2-bromopropane) is boiled with alcoholic potassium hydroxide, propcne is formed.

(4) Tert-hutyl chloride: When ten-butyl chloride (2-chloro-2-methyl propanc) is hcatcd with alcoholic KOH.

Question 61.
Describe the action of alc.KOH on 2-bromobutane.
When 2-bromobutane is boiled with alc.KOH on 2-bromobutane, a mixture of but-l-ene and but-2-ene is formed.
Answer:

Question 62.
Explain Saytzelf’s rule with suitable example.
Answer:
Saytzcff’s rule : In dehydrohalogenation reaction the preferred product is that alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms.

Hence the number of alkyl substituents on doubly bonded Carbon atoms increases, the stability of the alkene giving its major products.

Hence the increasing stability of alkenes is.

There are two types of fi hydrogens (β₁ and β₂) therefore two alkenes are expected.

Question 63.
What is a Grignard reagent ?
Answer:
Grignard reagent : An organometallic compound in which the divalent magnesium is directly linked to an alkyl group (R -) and a halogen atom (X), and has general formula R – Mg – X is called Grignard reagent. OR When alkyl halide is treated with magnesium in dry ether as solvent, it gives alkyl magnesium halide. It is known as Grignard reagent.

The carbon-magnesium bond is highly polar and magnesium-halogen bond is in ionic in nature. Grignard reagent is highly reactive. It is an important reagent and used in the preparation of a large number of organic compounds.

Question 64.
How is Grignard reagent prepared ?
Answer:
Grignard reagent is an alkyl magnesium halide, R – Mg – X obtained by the reaction of alkyl halide R – X with magnesium (Mg) in dry ether.

When an alkyl halide like CH₃I is added from a dropping funnel to a flask containing pieces of pure Mg in pure and dry ether (diethyl ether) and a trace of iodine, Grignard reagent, CH₃ – Mg – I is formed.

Ethyl iodide when treated with magnesium in presence of dry ether forms ethyl magnesium iodide.

Question 65.
Write a note on Grignard reagent.
Answer:
(1) Ethyl bromide : When ethyl bromide (bromoethane) is heated with alcoholic potassium hydroxide (alcoholic alkali). ethene (gas) is formed by the dehydrobrominaion reaction.

(2) n-PropI bromide : When n-propyl bromide is heated with alcoholic potassium hydroxide, propene is formed.

(3) Isopropvl bromide : When isopropyl bromide (2-bromopropane) is boiled with alcoholic potassium hydroxide, propcne is formed.

(4) Tert-hutyl chloride: When ten-butyl chloride (2-chloro-2-methyl propanc) is hcatcd with alcoholic KOH.

Question 66.
Describe the action of water on
(1) methyl magnesium iodide
(2) ethyl magnesium iodide.
Answer:
(1) Methyl magnesium iodide : When methyl magnesium iodide is treated with water, methane is obtained

(2) Ethyl magnesium iodide : When ethyl magnesium iodide is treated with water, ethane is obtained.

Question 67.
Describe the action of ammonia on
(1) ethyl magnesium bromide
(2) n-propyl magnesium chloride.
Answer:
(1) Ethyl magnesium bromide : When ethyl magnesium bromide is treated with ammonia, ethane is formed.

(2) n-Propyl magnesium chloride : When n-propyl magnesium chloride is treated with ammonia, propane is formed.

Question 68.
Explain Wurtz reaction. OR Explain the action of sodium with alkyl halides.
Answer:
(1) When an alkyl halide is treated with metallic sodium in dry ether, the corresponding higher alkane is formed. This is called Wurtz reaction or Wurtz coupling reaction.

(2) In this reaction the alkyl radicals from two molecules of the reacting alkyl halide combine or couple to form the higher alkane.

(3) Thus, methyl bromide reacts with sodium in ether to form ethane (C₂H₆), while ethyl bromide under the same conditions forms n-butane (C₄H₁₀).

(4) If a mixture of two different alkyl halides is treated with Na in dry ether, then a mixture of alkanes is obtained called self coupling products. For example, a mixture of CH₃Br and C₂H₅Br gives propane along with C₂H₆ and C₄H₁₀.

Question 69.
Explain the reaction of haloarene with alkyl halide and sodium metal.
Write a note on Wurtz-Fittig reaction.
Answer:
When an alkyl halide and an aryl halide is treated with sodium metal in dry ether the corresponding alkylarene (alkyl benzene) is formed. The reaction is known as Wurtz-Fittig reaction. This reaction allows alkylation of alkyl halides.

Question 70.
Describe the action of aryl halide on sodium metal.
Answer:
Aryl halide reacts with sodium metal in dry ether, biphenyl is formed. This reaction is known as Fittig reaction.

Question 71.
Identify the product A of following reaction.

Answer:

Question 72.
Explain the following substitution reactions of chlorobenzene :
(1) Halogenation
(2) Nitration
(3) Sulphonation.
Answer:
(1) Halogenation : When chlorobenzene is reacted with chlorine in presence of anhydrous ferric chloride, a mixture of ortho and para-dichlorobenzene (major product) is formed.

(2) Nitration : When chlorobenzene is heated with nitrating mixture (cone, nitric acid -I- cone, sulphuric acid) a mixture of l-chloro-4-nitro benzene (major product) and l-chloro-2-nitrobenzene is formed.

(3) Sulphonation : When chlorobenzene is heated with concentrated sulphuric acid, a mixture of 4-chlorobenzene sulphonic acid (major product) and 2-chlorobenzene sulphonic acid is formed.

Question 73.
Describe the action of the following on chlorobenzene :
(1) Methyl chloride in the presence of anhydrous AICI₃
(2) Acetyl chloride in the presence of anhydrous AICI₃.
Answer:
(1) Methyl chloride in the presence of anhydrous AICI₃ : When chlorobenzene is treated with methyl chloride in the presence of anhydrous AICI₃, a mixture of l-chloro-4-methyl benzene (major product) and l-chloro-2-methyl benzene is formed. Since, the alkyl group is introduced in the benzene ring, the reaction is termed as Friedel Craft’s alkylation.

(2) Acetyl chloride in the presence of anhydrous AICI₃ : When chlorobenzene is reacted with acetyl chloride in the presence of anhydrous AICI₃, a mixture of 2-chloro acetophenone and 4-chloro acetophenone (major product) is formed. Since, the acetyl group is introduced in the benzene ring, the reaction is termed as Friedel Craft’s acylation.

Question 74.
Write a note on Friedel Craft’s reaction.
Answer:

(1) Methyl chloride in the presence of anhydrous AICI₃ : When chlorobenzene is treated with methyl chloride in the presence of anhydrous AICI₃, a mixture of l-chloro-4-methyl benzene (major product) and l-chloro-2-methyl benzene is formed. Since, the alkyl group is introduced in the benzene ring, the reaction is termed as Friedel Craft’s alkylation.

(2) Acetyl chloride in the presence of anhydrous AICI₃ : When chlorobenzene is reacted with acetyl chloride in the presence of anhydrous AICI₃, a mixture of 2-chloro acetophenone and 4-chloro acetophenone (major product) is formed. Since, the acetyl group is introduced in the benzene ring, the reaction is termed as Friedel Craft’s acylation.

Question 75.
Convert 1-chlorobutane into the following compounds :

Answer:
(1) 1-Chlorobutane to butan-l-ol :

(2) 1-Chlorobutane to 1-iodobutane :

(3) 1-Chlorobutane to n-butyl cyanide (CH₃ – CH₂ – CH₂ – CH₂ – CN) :

(4) l-Chlorobutane to

Question 76.
Predict the expected product of substitution reactions :
(1) Isobutyl chloride + sodium ethoxide.
Answer:

(2) n-butyl chloride + sodium.
Answer:

(3) 1-chloropropane + aq. potassium hydroxide.
Answer:

(4) Aniline + NaNO₂/HCl.
Answer:

Question 77.
Write the products:

Question 78.
Identify A and B in the following :

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Question 79.
State the uses of the following compounds :
(1) Dichloromethane (CH₂CI₂)
(2) Trichloromethane or Chloroform (CHCI₃)
(3) Tetrachloromethane or carbon tetrachloride (CCI₄)
(4) Iodoform (CHI₃)
(5) Freons
(6) DDT (p, p’-Dichlorodiphenyl trichloroethane).
Answer:
(1) Dichloromethane (CH₂CI₂) :

• Dichloromethane dissolves wide range of organic compounds, hence it is used as solvent for many chemical reactions.
• It is used as a solvent as a paint remover and degreaser.
• It is used as propellant in aerosols and as a fumigant pesticide for grains and strawberries.
• It is used to decaffinate tea or coffee.

(2) Trichloromethane or Chloroform (CHCI₃) :

• Chloroform in the production of chlorofluoromethane, freon refrigerant R-22.
• It is used as solvent in pharmaceuticals, pesticides, gums, fats, resins and dye industry.
• It is a good source of dichlorocarbene species.

(3) Tetrachloromethane or carbon tetrachloride (CCI₄) :

• Carbon tetrachloride is used in the manufacture of refrigerants.
• It is used as a dry cleaning agent and as a pesticide for stored grains.
• It is very useful solvent for oils, fats and resins. It serves as a source of chlorine.

(4) Iodoform (CHI₃) :

• Iodoform is used as antiseptic, dressing of wounds and sores.
• On small scale it is used as disinfectant.

(5) Freons :

• Freons are widely used as propellants in aerosol, products of food, cosmetics and pharmaceutical industries.
• Freons containing bromine in their molecules are used as fire extinguishers.
• They are used in aerosol insecticides, solvent for cleaning clothes and metallic surfaces.
• It is used as foaming agents in the preparation of foamed plastics and in production of certain fluorocarbons.
• It is used as refrigerants and air conditioning purposes.

Question 80.
State the environmental effects of the following compounds :
(1) Dichloromethane (CH₂CI₂)
(2) Trichloromethane or chloroform (CHCI₃)
(3) Tetrachloromethane or carbon tetrachloride (CCI₄)
(4) Iodoform (CHI₃)
(5) Freon.s (CCI₂F₂, CCI₃F, CHCIF₂)

Answer:
(1) Dichloromethane (CH₂CI₂) :

• Higher levels of dichloromethane in air causes nausea, numbness in fingers and toes, dizziness.
• Lower levels of dichloromethane causes impaired vision and hearing.
• Direct contact with eyes can damage cornea.

(2) Trichloromethane or chloroform (CHCI₃) :

• When chloroform is exposed to air in the presence of sunlight, it slowly oxidised to phosgene, a poisonous compound, therefore it is stored in dark, amber coloured bottles.
• Chloroform vapour when inhaled for a short time causes dizziness, headache and fatigue and if inhaled for a long time affects central nervous system.

(3) Tetrachloromethane or carbon tetrachloride (CCI₄) :

• Exposure to carbon tetrachloride causes eye irritation, damages nerve cells, vomiting sensation, dizziness, unconciousness or death. Long exposure to chloroform may affect liver.
• When mixed with air it causes depletion of the ozone layer, which affects human skin leading to cancer.

(4) Iodoform (CHI₃) : Iodoform has a strong smell. It causes irritation to skin and eyes. It may cause respiratory irritation or breathing difficulty, dizziness, nausea, depression of central nervous system, visual disturbance.

(5) Freons (CCI₂F₂, CCI₃F, CHCIF₂) :

• Freon as refrigerant causes ozone depletion.
• Freons have low toxicity and low biological activity.
• Freons from propane group are more toxic in nature.
• Regular large inhalation of freon results in breathing problems, organ damage, loss of consciousness.

(6) DDT :

• DDT is not readily metabolised by animals.
• It is deposited and stored in fatty tissues.
• Exposure to high doses of DDT may cause vomiting, tremors or shakiness.
• Laboratory animal studies showed adverse effect of DDT on liver and reproduction.
• DDT is a pressistent organic pollutant, readily absorbed in soils and tends to accumulate in the ecosystem.
• When dissolved in oil or other lipid, it is readily absorbed by the skin. It is resistant to metabolism.
• There is a ban on use of DDT due to all these adverse effects.

Question 81.
What is the chemical name of freon?
Answer:
The chemical name of freon is Dichlorodifluoromethane.

Question 82.
What is the chemical name of DDT ?
Answer: The chemical name of DDT is p, p’-Dichlorodiphenyltrichloroethane.

Activity :
(1) Collect detailed information about Freons and their uses.
(2) Collect information about DDT as a persistent pesticide.
Reference books :
(1) Organic chemistry by Morrison, Boyd, Bhattacharjee, 7th edition, Pearson.
(2) Organic chemistry by Finar, Vol 1, 6th edition, Pearson

Multiple Choise Questions

Question 83.
Select and write the most appropriate answer from the given alternatives for each sub-question :

Question 1.

Answer:
(b) CH₃ – CH₂ – CH₂ – I

Question 2.
The rate of S_N2 reaction depends on the concentra¬tion of
(a) only the substrate
(b) only the reagent
(c) both the substrate and the reagent
(d) neither the substrate nor the reagent
Answer:
(c) both the substrate and the reagent

Question 3.
In S_N2 reaction, the hydrolysis of alkyl halide shows
(a) the retention of configuration
(b) the inversion of configuration
(c) both retention and inversion of configuration
(d) no change in the configuration
Answer:
(b) the inversion of configuration

Question 4.
The one step exothermic reaction is
(a) S_N1
(b) S_N2
(C) S_N
(d) S2_N
Answer:
(b) S_N2

Question 5.
Which of the following is correct about S_N2 mechanism?
(a) Two step reaction
(b) Complete inversion of configuration
(c) Formation of carbonium ion
(d) Favoured by polar solvent
Answer:
(b) Complete inversion of configuration

Question 6.
Which of the following is not a nucleophile?
(a) Ammonia
(b) Ammonium ion
(c) Primary amine
(d) Secondary amine
Answer:
(d) Secondary amine

Question 7.
Which of the following undergoes nucleophilic substitution exclusively by S_N2 mechanism ?
(a) ethyl chloride
(b) isopropyl chloride
(c) chlorobenzene
(d) benzyl chloride
Answer:
(d) benzyl chloride

Question 8.
Which of the following is most reactive towards nucleophilic substitution reaction ?
(a) CH₂ = CH – CI
(b) CH₃CH = CHCI
(c) C₆H₅CI
(d) CICH₂ – CH = CH₂
Answer:
(d) CICH₂ – CH = CH₂

Question 9.
The stability order of carbocation is
(a) 2° > 3° > 1°
(b) 3° > 2° > 1°
(c) 3° > 1° > 2°
(d) 1° > 3° > 2°
Answer:
(b) 3° > 2° > 1°

Question 10.

(a) ethane
(b) propane
(c) n-butane
(d) n-pentane
Answer:
(c) n-butane

Question 11.
Which of the following characteristic properties of the enantiomers is correct?
(a) The enantiomers possess same physical and chemical properties
(b) The enantiomers are optically active compounds
(c) The enantiomers have different optical rotations
(d) All of these
Answer:
(d) All of these

Question 12.
The optically inactive compound is
(a) glucose
(b) lactic acid
(c) isopropyl alcohol
(d) 2-bromo butane
Answer:
(c) isopropyl alcohol

Question 13.
A compound with the molecular formula CH₂OH(CHOH)₃CH₂OH has optically active forms
(a) 3
(b) 4
(c) 6
(d) 8
Answer:
(d) 8

Question 14.
A racemic mixture consists of
(a) equal amount of d and l isomers
(b) unequal amounts of d and / isomers
(c) unknown amounts of d and / isomers
(d) only d isomers
Answer:
(a) equal amount of d and l isomers

Question 15.
Which of the following compounds is not optically active ?
(a) Lactic acid
(b) Secondary butyl chloride
(c) n-propyl iodide
(d) Glucose
Answer:
(c) n-propyl iodide

Question 16.
Which of the following compounds shows optical activity ?
(a) n-butyl chloride
(b) isobutyl chloride
(c) sec-butyl chloride
(d) t-butyl chloride
Answer:
(c) sec-butyl chloride

Question 17.
The major product of the following reaction is

Answer:
(c)

Question 18.
The above reaction is known as

(a) Wurtz-Fittig reaction
(b) Friedel Craft’s reaction
(c) Sandmeyer’s reaction
(d) Swarts reaction
Answer:
(b) Friedel Craft’s reaction

Question 19.
Iodoform is used as
(a) an anaesthetic
(b) an antiseptic
(c) an analgesic
(d) an antibiotic
Answer:
(b) an antiseptic

Question 20.
p, p’-dichlorodiphenyl trichloroethane is used as
(a) insecticide
(b) anaesthetic
(c) antiseptic
(d) refrigerant
Answer:
(a) insecticide

Question 21.
The order of reactivity in nucleophilic substitution reaction is
(a) CH₃F < CH₃C1 < CH₃I < CH₃Br
(b) CH₃F < CH₃C1 < CH₃Br < CH₃I
(c) CH₃F < CH₃Br < CH₃C1 < CH₃I
(d) CH₃I < CH₃Br < CH₃C1 < CH₃F
Answer:
(b) CH₃F < CH₃C1 < CH₃Br < CH₃I

Question 22.
Racemate is
(a) optically active
(b) optically dextro rotatory
(c) optically inactive
(d) optically laevorotatory
Answer:
(c) optically inactive

Question 23.
The number of asymmetric carbon atoms in glucose are
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(c) 4

Question 24.
The geometry of carbon lum ion is
(a) Tetrahedral
(b) planar
(c) linear
(d) pyramidal
Answer:
(b) planar

Question 25.
In its nucleophilic substitution reaction, aryl halide resembles
(a) Vinyl chloride
(b) allyl chloride
(e) Benzyl chloride
(d) ethyl chloride
Answer:
(a) Vinyl chloride

Question 26.
The weakest C-Cl bond is present in

Answer:
(d)

Question 27.
Which alkyl halide among the following com¬pounds has the highest boiling point ?
(a) (CH₃)₃CCI
(b) CH₃CH₂CH₂CH₂CI
(c) CH₃CH₂CH₂C1
(d) CH₃CH(CH₃)CH₂CI
Answer:
(b) CH₃CH₂CH₂CH₂CI

Question 28.
It is difficult to break C-Cl bond in CH₂ = CH – CI due to
(a) Hyper conjugation
(b) Resonance
(c) Electromeric effect
(d) Inductive effect
Answer:
(b) Resonance

Question 29.
Which one of the following when heated with metallic sodium will not give the corresponding alkane ?

Answer:
(c)

Question 30.
The most reactive alkyl halide towards SN2 reac¬tion is
(a) CH₃X
(b) R₃CX
(C) R₂CHX
(d) RCH₂X
Answer:
(a) CH₃X

Question 31.
The number of electrons surrounding the carbon- ium ion is
(a) 6
(b) 8
(c) 10
(d) 7
Answer:
(a) 6

Question 32.
The lowest stability of carbocation among the compounds

Answer:
(a)

Question 33.
Carbon atom in methyl carbocation contains how many pairs of electrons?
(a) 8
(b) 4
(c) 3
(d) 5
Answer:
(b) 4

Question 34.
The optically inactive compound is
(a) Glucose
(b) Lactic acid
(c) 2-Chlorobutane
(d) 2-Chloropropane
Answer:
(d) 2-Chloropropane

Question 35.
The hydrogen halide which does not obey Markownikv rule in presence of peroxide is
(a) HC1
(b) HBr
(c) HF
(d) HI
Answer:
(b) HBr

Question 36.
Which one of the following is NOT used to prepare alkyl halide from an alcohol ?
(a) SOCl₂
(b) PC1₃
(c) HC1 + ZnCl₂
(d) NaCl
Answer:
(d) NaCl

Question 37.
The total number of electrons present in the central carbon atom of a free radical is
(a) 7
(b) 8
(c) 9
(d) 6
Answer:
(a) 7

Question 38.
In which of the following pairs both are nucleophiles ?
(a) BF₃, AICI₃
(b) NO⁺₂, Cl^–
(c) CN^–, NH₃
(d) Br⁺, BC1₃
Answer:
(c) CN^–, NH₃

Question 39.
Which one of the following alkane is NOT formed in Wurtz reaction ?
(a) Methane
(b) Ethane
(c) Propane
(d) Butane
Answer:
(a) Methane

Question 40.
Which of the following groups has highest priority according to R, S convention?
(a) CH₂OH
(b) COOH
(c) COCH₃
(d) COOCH₃
Answer:
(d) COOCH₃

Question 41.
The halogen atom in aryl halides is
(a) o- and p-di reefing
(b) m-directing
(c) o, m and p-di reefing
(d) only m-directing
Answer:
(a) o- and p-di reefing

Question 42.
Chlorobenzene can be obtained by benzene diazonium chloride by
(a) Friedel Craft’s reaction
(b) Wurtz reaction
(c) Gatterman’s reaction
(d) Fittig reaction
Answer:
(c) Gatterman’s reaction

Question 43.
Which of the following carbocations is least stable ?


Answer:
(c)

Question 44.
But-l-ene on reaction with HCI in the presence of sodium peroxide yields
(a) n-butyl chloride
(b) isobutyl chloride
(c) secondary butyl chloride
(d) tertiary butyl chloride
Answer:
(c) secondary butyl chloride

Question 45.
Carbon tetrachloride is used as
(a) anaesthetic
(b) antiseptic
(c) dry cleaning agent
(d) fire extinguisher
Answer:
(c) dry cleaning agent

Question 46.
Identify the product D in the following sequence of reactions :

(a) 2, 2-dimethyl butane
(b) 2, 3-dimethyl butane
(C) hexane
(d) 2, 4-dimethylpentane
Answer:
(b) 2, 3-dimethyl butane

Question 47.
The preparation of alkyl fluoride from alkyl chlor ide, in presence of metallic fluorides is known as
(a) Williamson’s reaction
(b) Finkeistein reaction
(c) Swarts reaction
(d) Wurlz reaction
Answer:
(c) Swarts reaction

Question 48.
UPAC name of the following compound is

(a) 3-Bromo-3, 4-dimetbyiheptane
(b) 3,4-dimethyl-3-bromoheptane
(c) 5-Bromo-4,5-dimethylheptane
(d) 4,5-dimethyl-5-bromoheptane
Answer:
(a) 3-Bromo-3, 4-dimetbyiheptane

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 2 Solutions Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Chemistry Solutions Chapter 2 Solutions

1. Choose the most correct answer.

Question i.
The vapour pressure of a solution containing 2 moles of a solute in 2 moles of water (vapour pressure of pure water = 24 mm Hg) is
(a) 24 mm Hg
(b) 32 mm Hg
(c) 48 mm Hg
(d) 12 mm Hg
Answer:
(d) 12 mm Hg

Question ii.
The colligative property of a solution is
(a) vapour pressure
(b) boiling point
(c) osmotic pressure
(d) freezing point
Answer:
(c) osmotic pressure

Question iii.
In calculating osmotic pressure the concentration of solute is expressed in
(a) molarity
(b) molality
(c) mole fraction
(d) mass per cent
Answer:
(a) molarity

Question iv.
Ebullioscopic constant is the boiling point elevation when the concentration of solution is
(a) 1 m
(b) 1 M
(c) 1 mass%
(d) 1 mole fraction of solute
Answer:
(a) 1 m

Question v.
Cryoscopic constant depends on
(a) nature of solvent
(b) nature of solute
(c) nature of solution
(d) number of solvent molecules
Answer:
(a) nature of solvent

Question vi.
Identify the correct statement
(a) vapour pressure of solution is higher than that of pure solvent.
(b) boiling point of solvent is lower than that of solution
(c) osmotic pressure of solution is lower than that of solvent
(d) osmosis is a colligative property.
Answer:
(b) boiling point of solvent is lower than that of solution

Question vii.
A living cell contains a solution which is isotonic with 0.3 M sugar solution. What osmotic pressure develops when the cell is placed in 0.1 M KCl solution at body temperature ?
(a) 5.08 atm
(b) 2.54 atm
(c) 4.92 atm
(d) 2.46 atm
Answer:
(c) 4.92 atm

Question viii.
The osmotic pressure of blood is 7.65 atm at 310 K. An aqueous solution of glucose isotonic with blood has the percentage (by volume)
(a) 5.41%
(b) 3.54%
(c) 4.53%
(d) 53.4%
Answer:
(a) 5.41%

Question ix.
Vapour pressure of a solution is
(a) directly proportional to the mole fraction of the solute
(b) inversely proportional to the mole fraction of the solute
(c) inversely proportional to the mole fraction of the solvent
(d) directly proportional to the mole fraction of the solvent
Answer:
(d) directly proportional to the mole fraction of the solvent

Question x.
Pressure cooker reduces cooking time for food because
(a) boiling point of water involved in cooking is increased
(b) heat is more evenly distributed in the cooking space
(c) the higher pressure inside the cooker crushes the food material
(d) cooking involves chemical changes helped by a rise in temperature
Answer:
(a) boiling point of water involved in cooking is increased

Question xi.
Henry’s law constant for a gas CH₃Br is 0.159 mol dm^-3 atm at 250°C. What is the solubility of CH₃Br in water at 25 °C and a partial pressure of 0.164 atm?
(a) 0.0159 mol L^-1
(b) 0.164 mol L^-1
(c) 0.026 M
(d) 0.042 M
Answer:
(c) 0.026 M

Question xii.
Which of the following statement is NOT correct for 0.1 M urea solution and 0.05 M sucrose solution ?
(a) osmotic pressure exhibited by urea solution is higher than that exhibited by sucrose solution
(b) urea solution is hypertonic to sucrose solution
(c) they are isotonic solutions
(d) sucrose solution is hypotonic to urea solution
Answer:
(c) they are isotonic solutions

2. Answer the following in one or two sentences

Question i.
What is osmotic pressure ?
Answer:
(1) Definition : The osmotic pressure is defined as the excess mechanical pressure required to be applied to a solution separated by a semipermeable membrane from pure solvent or a dilute solution to prevent the osmosis or free passage of the solvent molecules at a given temperature. The osmotic pressure is a colligative property.

Osmosis and osmotic pressure

(2) Explanation : Consider an inverted thistle funnel on the mouth of which a semipermeable membrane is firmly fastened. It is filled with the experimental solution and immersed in a solvent like water. As a result, solvent molecules pass through the membrane into the solution in the funnel causing rising of level in the arm of thistle funnel. This increases the hydrostatic pressure. At a certain stage this rising level stops indicating an equilibrium between the rates of flow of solvent molecules from solvent to solution and from solution to solvent. The hydrostatic pressure at this stage represents osmotic pressure of the solution in the thistle funnel.

Question ii.
A solution concentration is expressed in molarity and not in molality while considering osmotic pressure. Why ?
Answer:

1. While calculating osmotic pressure by equation, π = CRT, the concentration is expressed in molarity but not in molality.
2. This is because the measurements of osmotic pressure are made at a certain constant temperature.
3. Molarity depends upon temperature but molality is independent of temperature.
4. Hence in osmotic pressure measurements, concentration is expressed in molarity.

Question iii.
Write the equation relating boiling point elevation to the concentration of solution.
Answer:
The elevation in the boiling point of a solution is directly proportional to the molal concentration (expressed in mol kg^-1) of the solution.
Hence, if ΔT_b is the elevation in the boiling point of a solution of molal concentration m then,
ΔT_b ∝ m
∴ ΔT_b = K_b m
where K_b is a proportionality constant.
If m = 1 molal,
ΔT_b = K_b
K_b is called the ebullioscopic constant or molal elevation constant. K_b is characteristic of the solvent.

Question iv.
A 0.1 m solution of K₂SO₄ in water has freezing point of -0.43 °C. What is the value of van’t Hoff factor if K_f for water is 1.86 K kg mol^-1?
Answer:
Given : m = 0.1 m, ΔT_f = 0 – (-0.43) = 0.43 °C
K_f = 1.86 K kg mol^-1, i = ?
ΔT_f = i × K_f × m
∴ i = \(\frac{\Delta T_{\mathrm{f}}}{K_{\mathrm{f}} \times m}=\frac{0.43}{1.86 \times 0.1}\) = 2.312
van’t Hoff factor = i = 2.312

Question v.
What is van’t Hoff factor?
Answer:
Definition of the van’t Hoff factor, i : It is defined as a ratio of the observed colligative property of the solution to the theoretically calculated colligative property of the solution without considering molecular change.

The van’t Hoff factor can be represented as,

This colligative property may be the lowering of vapour pressure of a solution, the osmotic pressure, the elevation in the boiling point or the depression in the freezing point of the solution. Hence,

• When the solute neither undergoes dissociation or association in the solution, then, i = 1
• When the solute undergoes dissociation in the solution, then, i > 1
• When the solute undergoes association in the solution, then i < 1

From the value of the van’t Hoff factor, the degree of dissociation of electrolytes, degree of association of nonelectrolytes can be obtained.

van’t Hoff factor gives the important information about the solute molecules in the solution and chemical bonding in them.

Question vi.
How is van’t Hoff factor related to degree of ionization?
Answer:
Consider 1 dm³ of a solution containing m moles of an electrolyte AxBy. The electrolyte on dissociation gives x number of A^y+ ions and y number of B^x- ions. Let α be the degree of dissociation.

At equilibrium,
AxBy ⇌ xA^y+ + yB^x-
For 1 mole of electrolyte : 1 – α,  xα,  yα
and For ‘m’ moles of an electrolyte : m(1 – α), mxα, myα are the number of particles.
Total number of moles at equilibrium, will be,
Total moles = m(1 – α) + mxα + myα
= m[(1 – α) + xα + yα]
= m[1 + xα + yα – α]
= m[1 + α(x + y – 1)]

The van’t Hoff factor i will be,

If total number of ions from one mole of electrolyte is denoted by n, then (x + y) = n
∴ i = 1 + α(n – 1)
∴ α(n – 1) = i – 1
∴ α = \(\frac{i-1}{n-1}\) ……(1)
This is a relation between van’t Hoff factor i and degree of dissociation of an electrolyte.

Question vii.
Which of the following solutions will have higher freezing point depression and why ?
a. 0.1 m NaCl b. 0.05 m Al₂(SO₄)₃
Answer:
(1) Freezing point depression is a colligative property, hence depends on the number of particles in the solution.
(2) More the number of particles in the solution, higher is the depression in freezing point.
(3) The number of particles (ions) from electrolytes are,

(4) Therefore Al₂(SO₄)₃ solution will have higher freezing point depression.

Question viii.
State Raoult’s law for a solution containing a nonvolatile solute.
Answer:
Statement of Raoult’s law : The law states that the vapour pressure of a solvent over the solution of a nonvolatile solute is equal to the vapour pressure of the pure solvent multiplied by mole fraction of the solvent at constant temperature.

Question ix.
What is the effect on the boiling point of water if 1 mole of methyl alcohol is added to 1 dm³ of water? Why?
Answer:

• The boiling point of water (or any liquid) depends on its vapour pressure.
• Higher the vapour pressure, lower is the boiling point.
• When 1 mole of volatile methyl alcohol is added to 1 dm³ of water, its vapour pressure is increased decreasing the boiling point of water.

Question x.
Which of the four colligative properties is most often used for molecular mass determination? Why?
Answer:

1. Since osmotic pressure has large values, it can be measured more precisely.
2. The osmotic pressure can be measured at a suitable constant temperature.
3. The molecular masses can be measured more accurately.
4. Therefore, it is more useful to determine molecular masses of expensive substances by osmotic pressure.

3. Answer the following.

Question i.
How vapour pressure lowering is related to a rise in boiling point of solution?
Answer:
(1) The boiling point of a liquid is the temperature at which the vapour pressure of the liquid becomes equal to the external pressure, generally 1 atm (101.3 × 10³ Nm^-2).

(2) When a liquid is heated, its vapour pressure rises till it becomes equal to the external pressure.
If the liquid has a low vapour pressure, it has a higher boiling point.

Vapour pressure curve showing elevation in boiling point

(3) When a nonvolatile solute is added to a solvent, its vapour pressure decreases, hence the boiling point increases.
This is explained by graphical representation of the vapour pressure and the boiling point of the pure solvent and the solution.

If T₀ and T are the boiling points of a pure solvent and a solution, then the elevation in the boiling point is given by,
ΔT_b = T – T₀
The curve AB, represents the variation in the vapour pressure of a pure solvent with temperature while curve CD represents the variation in the vapour pressure of the solution.

(4) This elevation in the boiling point is proportional to the lowering of the vapour pressure, i.e., P₀ – P, where P₀ and P are the vapour pressures of the pure solvent and the solution.
[ΔT_b ∝ (P₀ – P) or ΔT_b ∝ ΔP]

Question ii.
What are isotonic and hypertonic solutions?
Answer:
(1) Isotonic solutions : The solutions having the same osmotic pressure at a given temperature are called isotonic solutions.

Explanation : If two solutions of substances A and B contain n_A and n_B moles dissolved in volume V (in dm³) of the solutions, then their concentrations are,
C_A = \(\frac{n_{\mathrm{A}}}{V}\) (in mol dm^-3) and
C_B = \(\frac{n_{\mathrm{B}}}{V}\) (in mol dm^-3)

If the absolute temperature of both the solutions is T, then by the van’t Hoff equation,
π_A = C_ART and π_B = C_BRT, where π_A and π_B are their osmotic pressures.
For the isotonic solutions,
π_A = π_B
∴ C_A = C_B
∴ \(\frac{n_{\mathrm{A}}}{V}=\frac{n_{\mathrm{B}}}{V}\)
∴ n_A = n_B
Hence, equal volumes of the isotonic solutions at the same temperature will contain equal number of moles (hence, equal number of molecules) of the substances.

(2) Hypertonic solutions : When two solutions have different osmotic pressures, then the solution having higher osmotic pressure is said to be a hypertonic solution with respect to the other solution.

Explanation : Consider two solutions of substances A and B having osmotic pressures π_A and π_B. If π_B is greater than π_A, then the solution B is a hypertonic solution with respect to the solution A.
Hence, if C_A and C_B are their concentrations, then C_B > C_A. Hence, for equal volume of the solutions, n_B > n_A.

Question iii.
A solvent and its solution containing a nonvolatile solute are separated by a semipermable membrane. Does the flow of solvent occur in both directions? Comment giving reason.
Answer:

1. When a solvent and a solution containing a non-volatile solute are separated by a semipermeable membrane, there arises a flow of solvent molecules from solvent to solution as well as from solution to solvent.
2. Due to higher vapour pressure of solvent than solution, the rate of flow of solvent molecules from solvent to solution is higher.
3. As more and more solvent passes into solution due to osmosis, the solvent content increases, and the rate of backward flow increases.
4. At a certain stage an equilibrium is reached where both the opposing rates become equal attaining an equilibrium.

Question iv.
The osmotic pressure of CaCl₂ and urea solutions of the same concentration at the same temperature are respectively 0.605 atm and 0.245 atm. Calculate van’t Hoff factor for CaCl₂.
Answer:
Given : π_Cacl2 = 0.605 atm;
π_Urea = 0.245 atm
For urea solution, van’t Hoff factor, i = 1
π_Cacl2 = i × (CRT)_Cacl2
π_Urea = (CRT)_Urea

van’t Hoff factor = i = 2.47

Question v.
Explain reverse osmosis.
Answer:
Reverse osmosis : The phenomenon of the passage of solvent like water under high pressure from the concentrated aqueous solution like seawater into pure water through a semipermeable membrane is called reverse osmosis.

The osmotic pressure of seawater is about 30 atmospheres. Hence when pressure more than 30 atmospheres is applied on the solution side, regular osmosis stops and reverse osmosis starts. Hence pure water from seawater enters the other side of pure water.

Purification of seawater by reverse osmosis

For this purpose of suitable semipermeable membrane is required which can withstand high pressure conditions over a long period.
This method is used successfully in Florida since 1981 producing more than 10 million litres of pure water per day.

Question vi.
How molar mass of a solute is determined by osmotic pressure measurement?
Answer:
Consider V dm³ (litres) of a solution containing W₂ mass of a solute of molar mass M₂ at a temperature T.
Number of moles of solute, n₂ = \(\frac{W_{2}}{M_{2}}\)
The osmotic pressure π is given by,
π = \(\frac{W_{2} R T}{M_{2} V}\)
∴ M₂ = \(\frac{W_{2} R T}{\pi V}\)
By measuring osmotic pressure of a solution, the molar mass of a solute can be calculated.
Since osmotic pressure can be measured more precisely, it is widely used to measure molar masses of the substances.

Question vii.
Why vapour pressure of a solvent is lowered by dissolving a nonvolatile solute into it?
Answer:
Lowering of vapour pressure of a solution :
When a nonvolatile solute is added to a pure solvent, the surface area is covered by the solute molecule decreasing the rate of evaporation, hence its vapour pressure decreases. This decrease in vapour pressure is called lowering of vapour pressure.

If P₀ is the vapour pressure of a pure solvent (liquid) and P is the vapour pressure of the solution, where P < P₀, then, (P₀ – P) is the lowering of the vapour pressure.

Question viii.
Using Raoult’s law, how will you show that ∆P = \(\boldsymbol{P}_{1}^{0}\)x₂ ? Where x₂ is the mole fraction of solute in the solution and \(\boldsymbol{P}_{1}^{0}\) vapour pressure of pure solvent.
Answer:
If x₁ and x₂ are the mole fractions of solvent and solute respectively, then
x₁ + x₂
By Raoult’s law,
P = x₁ × P₀
where P⁰ is the vapour pressure of a pure solvent and P is the vapour pressure of the solution at given temperature.

Question ix.
While considering boiling point elevation and freezing point depression a solution concentration is expressed in molality and not in molarity. Why?
Answer:

• Boiling point elevation and freezing point depression involve temperature changes, (ΔT_b and ΔT_f).
• Since molarity depends on temperature but molality is independent of temperature we use molality and not molarity in considering boiling point elevation and freezing point depression.

Question 4.
Derive the relationship between degree of dissociation of an electrolyte and van’t Hoff factor.
Answer:
Consider 1 dm³ of a solution containing m moles of an electrolyte AxBy. The electrolyte on dissociation gives number of A^y+ ions and y number of B^x- ions. Let α be the degree of dissociation.

At equilibrium,
AxBy ⇌ xA^y+ + yB^x-
For 1 mole of electrolyte : 1 – α, xα, yα and
For ‘m’ moles of an electrolyte : m(1 – α), mxα, myα are the number of particles.
Total number of moles at equilibrium, will be,
Total moles = m(1 – α) + mxα + myα
= m[(1 – α) + xα + yα]
= m[1 – xα + yα – α]
= m[1 + α(x + y – 1)]
The van’t Hoff factor i will be,

If total number of ions from one mole of electrolyte is denoted by n, then (x + y) = n
∴ i = 1 + α(n – 1)
∴ α(n – 1) = i – 1
∴ α = \(\frac{i-1}{n-1}\) ……..(1)
This is a relation between van’t Hoff factor i and degree of dissociation of an electrolyte.

Question 5.
What is effect of temperature on solubility of solids in water? Give examples.
Answer:
The solubility of a solid solute depends upon temperature.

Variation of solubilities of some ionic solids with temperature

• Generally rise in temperature increases the solubility. This is due to expansion of holes or empty spaces in the liquid solvent. Generally 10 °C rise in temperature, increases the solubility of solids two fold.
• Dissolution process may be endothermic or exothermic.
• The solubility of the substances like NaBr, NaCl, KCl, etc. changes slightly with the increase in temperature.
• The solubility of the salts like NaNO₃, KNO₃, KBr, etc. increases appreciably with the increase in temperature.
• The solubility of Na₂SO₄ first increases and after 30 °C decreases with the increase in temperature.

This variation in solubility with temperature can be used to separate the salts from the mixture by fractional crystallisation.

Question 6.
Obtain the relationship between freezing point depression of a solution containing nonvolatile nonelectrolyte and its molar mass.
Answer:
The freezing point depression, ΔT_f of a solution is directly proportional to molality (m) of the solution.
∴ ΔT_f ∝ m
∴ ΔT_f = K_f m
where Kf is a molal depression constant.
The molality of a solution is given by,

If W₁ grams of a solvent contain W₂ grams of a solute of the molar mass M₂, then the molality m of the solution is given by,

If the weights are expressed in kg then,
ΔT_f = K_f × \(\frac{W_{2}}{W_{1} M_{2}}\)
The unit of K_f is K kg mol^-1
Hence, from the measurement of the depression in the freezing point of the solution, the molar mass of the substance can be determined.

Question 7.
Explain with diagram the boiling point elevation in terms of vapour pressure lowering.
Answer:
(1) The boiling point of a liquid is the temperature at which the vapour pressure of the liquid becomes equal to the external pressure, generally 1 atm (101.3 × 10³ Nm^-2).

(2) When a liquid is heated, its vapour pressure rises till it becomes equal to the external pressure.
If the liquid has a low vapour pressure, it has a higher boiling point.

Vapour pressure curve showing elevation in boiling point

(3) When a nonvolatile solute is added to a solvent, its vapour pressure decreases, hence the boiling point increases.
This is explained by graphical representation of the vapour pressure and the boiling point of the pure solvent and the solution.

If T₀ and T are the boiling points of a pure solvent and a solution, then the elevation in the boiling point is given by,
ΔT_b = T – T₀
The curve AB, represents the variation in the vapour pressure of a pure solvent with temperature while curve CD represents the variation in the vapour pressure of the solution.

(4) This elevation in the boiling point is proportional to the lowering of the vapour pressure, i.e., P₀ – P, where P₀ and P are the vapour pressures of the pure solvent and the solution.
[ΔT_b ∝ (P₀ – P) or ΔT_b ∝ ΔP]

Question 8.
Fish generally needs O₂ concentration in water at least 3.8 mg/L for survival. What partial pressure of O₂ above the water is needed for the survival of fish? Given the solubility of O₂ in water at 0 °C and 1 atm partial pressure is 2.2 × 10^-3 mol/L (0.054 atm)
Answer:
Given : Required concentration of O₂
= 3.8 mg/L
= \(\frac{3.8 \times 10^{-3}}{32} \mathrm{~mol} \mathrm{~L}^{-1}\)
Solubility of O₂ = 2.2 × 10^-3 mol L^-1
P = 1 atm
Partial pressure of O₂ needed = Po₂ = ?

Pressure needed = Po₂ = 0.05397 atm.

Question 9.
The vapour pressure of water at 20 °C is 17 mm Hg. What is the vapour pressure of solution containing 2.8 g urea in 50 g of water? (16.17 mm Hg)
Answer:
Given : Vapour pressure of pure solvent (water) = P0
= 17 mm Hg
Weight of solvent = W₁ = 50 g
Weight of solute (urea) = 2.8 g
Molecular weight of a solvent = M₁ = 18
Molecular weight of a solute (urea) = M₂
= 60 g mol^-1
\(\frac{P_{0}-P}{P_{0}}=\frac{W_{2} \times M_{1}}{W_{1} \times M_{2}}\)
∴ \(\frac{17-P}{17}=\frac{2.8 \times 18}{50 \times 60}\) = 0.0168
∴ 17 – P = 17 × 0.0168
17 – P = 0.2856
∴ P= 17 – 0.2856
= 16.7144 mm Hg
Vapour pressure of solution = 16.7144 mm Hg

Question 10.
A 5% aqueous solution (by mass) of cane sugar (molar mass 342 g/mol) has freezing point of 271K. Calculate the freezing point of 5% aqueous glucose solution.
Answer:
Given : W₂ = 5 g cane sugar; W₁ = 100 – 5 = 95 g
M₂ = 342 g mol^-1; T_f1 = 271 K;
ΔT_f1 = 273 – 271 = 2 K; T_f = ?
W₂ = 5 g glucose, W’₁ = 100 – 5 = 95 g,
M’₂ = 180 g mol-1, ΔT_f2 = ?

= 12.996 K kg mol^-1
≅ 13 K kg mol^-1

∴ Freezing point of solution = T_f
= 273 – 3.801 = 269.2 K
Freezing point of solution = 269.2 K.

Question 11.
A solution of citric acid C₆H₈O₇ in 50 g of acetic acid has a boiling point elevation of 1.76 K. If K_b for acetic acid is 3.07 K kg mol^-1, what is the molality of solution?
Answer:
Given : W₁ = 50 g acetic acid
ΔT_b = 1.76 K
K_b = 3.07 K kg mol^-1
m = ?
ΔT_b = K_b × m
∴ m = \(\frac{\Delta T_{\mathrm{b}}}{K_{\mathrm{b}}}\)
= \(\frac{1.76}{3.07}\)
= 0.5733 m
Molality of solution = 0.5733 m

Question 12.
An aqueous solution of a certain organic compound has a density of 1.063 gmL^-1, an osmotic pressure of 12.16 atm at 25°C and a freezing point of -1.03°C. What is the molar mass of the compound? (334 g/mol)

Question 13.
A mixture of benzene and toluene contains 30% by mass of toluene. At 30°C, vapour pressure of pure toluene is 36.7 mm Hg and that of pure benzene is 118.2 mm Hg. Assuming that the two liquids form ideal solutions, calculate the total pressure and partial pressure of each constituent above the solution at 30°C.
Answer:
Given : 30% by mass of toluene (T) and 70% by mass of benzene (B).
W_T = 30 g; W_B = 70g
\(P_{\mathrm{T}}^{0}\) = 36.7 mm Hg; \(P_{\mathrm{B}}^{0}\) = 118.2 mm Hg
M_T = 92 g mol^-1; MB = 78 g mol^-1
P_T = ?, P_B = ?, P_soln = ?

Total number of moles = n_Total = n_T + n_B
= 0.326 + 0.8974
= 1.2234 mol

Mole fractions :
x_T = \(\frac{n_{\mathrm{T}}}{n_{\text {Total }}}=\frac{0.326}{1.2234}\) = 0.2665
x_B = 1 – 0.2665 = 0.7335
Psoln = x_T + \(P_{\mathrm{T}}^{0}\) + x_B × \(P_{\mathrm{B}}^{0}\)
= 0.2665 × 36.7 + 0.7335 × 118.2
= 9.780 + 86.7
= 96.48 mm Hg

Partial pressures :
P_T = x_T × P_soln
= 0.2665 × 96.48
= 25.71 mm Hg
P_B = x_B × P_soln
= 0.7335 × 96.48
= 70.77 mm Hg
Total pressure P_soln = 96.48 mm Hg
Partial pressures : P_Toluene = 25.71 mm Hg
P_Benzene = 70.77 mm Hg

Question 14.
At 25 °C a 0.1 molal solution of CH₃COOH is 1.35% dissociated in an aqueous solution. Calculate freezing point and osmotic pressure of the solution assuming molality and molarity to be identical.
Answer:
Given : T = 273 + 25 = 298 K
C = 0.1 m ≅ 0.1 M; K_f = 1.86 K kg mol^-1
Per cent dissociation = 1.35
Freezing point = t_f = ?
π = ?
α = \(\frac{1.35}{100}\) = 0.0135

i = 1 – α + α + α = 1 + α = 1 + 0.0135 = 1.0135
(i) ΔT_f = i × K_f × m
= 1.0135 × 1.86 × 0.1
= 0.1885 °C
∴ Freezing point of solution = 0 – 0.1885
= – 0.1885 °C

(ii) n = iCRT
= 1.035 × 0.1 × 0.08206 × 298
= 2.53 atm

(i) Freezing point of solution = – 0.1885 °C
(ii) Osmotic pressure = π = 2.53 atm

Question 15.
A 0.15 m aqueous solution of KCl freezes at -0.510 °C. Calculate i and osmotic pressure at 0 °C. Assume volume of solution equal to that of water.
Answer:
Given : c = 0.15 m KCl ≅ 0.15 M KCl
ΔT_f = 0 – T_f = 0 – (-0.510) = 0.510 °C
T = 273 K; K_f = 1.86 K kg mol^-1
i = ?; π = ?
ΔT_f = i × K_f × m
∴ i = \(\frac{\Delta T_{\mathrm{f}}}{K_{\mathrm{f}} \times m}\)
= \(\frac{0.510}{1.86 \times 0.15}\)
= 1.828
π = iCRT
= 1.828 × 0.15 × 0.08206 × 273
= 6.143 atm
i = 1.828, Osmotic pressure = π = 6.143 atm

12th Chemistry Digest Chapter 2 Solutions Intext Questions and Answers

Can you tell ? (Textbook Page No. 29)

Question 1.
Why naphthalene dissolves in benzene but not in water ?
Answer:
Since naphthalene is a covalent nonpolar substance, it is soluble in a nonpolar solvent like benzene but insoluble in polar solvent like water.

Question 2.
Anhydrous sodium sulphate dissolves in water with the evolution of heat. What is the effect of temperature on its solubility ?
Answer:
Since the dissolution of anhydrous sodium sulphate in water is an exothermic process due to evolution of heat, according to Le Chatelier’s principle its solubility decreases with the increase in temperature.

(Textbook Page No. 42)

Question 1.
If 1.25 m sucrose solution has ΔT_f of 2.32 °C, what will be the expected value of ΔT_f for 1.25 m CaCl₂ solution?
Answer:
Sucrose being nonelectrolyte, it has i = 1 but for CaCl₂,
(CaCl₂ → Ca²⁺ + 2Cl^–) the value of i = 3.
Hence
ΔT_f = i × 2.32
= 3 × 2.32
= 6.92 °C
∴ ΔT_f = 6.92 °C.

(Textbook Page No. 44)

Question 1.
Which of the following solutions will have maximum boiling point elevation and which have minimum freezing point depression assuming the complete dissociation? (a) 0.1m KCl (b) 0.05 m NaCl (c) 1 m AlPO₄ (d) 0.1 m MgSO₄.
Solution :
Boiling point elevation and freezing point depression are colligative properties that depend on number of particles in solution. The solution having more number of particles will have large boiling point elevation and that having less number of particles would show minimum freezing point depression.

AlPO₄ solution contains highest moles and hence highest number particles and in turn, the maximum ΔT_b. NaCl solution has minimum moles and particles, it has minimum ΔT_f.

Question 2.
Arrange the following solutions in order of increasing osmotic pressure. Assume complete ionization. (a) 0.5 m Li₂SO₄ (b) 0.5 m KCl (c) 0.5 m Al₂(SO₄)₃ (d) 0.1 m BaCl₂.
Answer:
Consider 1 dm³ of each solution.

Osmotic pressure being a colligative property, it depends on number of particles in the solution.
Therefore, increasing order of osmotic pressure is,

Balbharti Maharashtra State Board Class 11 Secretarial Practice Solutions Chapter 1 Secretary Textbook Exercise Questions and Answers.

Maharashtra State Board Class 11 Secretarial Practice Solutions Chapter 1 Secretary

1A. Select the correct answer from the options given below and rewrite the statements.

Question 1.
The word Secretary is derived from the Latin word ____________ which means a confidential writer.
(a) Amatya
(b) Scribae
(c) Secretarius
Answer:
(c) Secretarius

Question 2.
Only ____________ can be appointed as a Secretary.
(a) an Individual
(b) a Partnership firm
(c) a Joint-stock company
Answer:
(a) an individual

Question 3.
A Secretary is a custodian of ____________ information.
(a) confidential
(b) individual
(c) investigative
Answer:
(a) confidential

Question 4.
The Secretary of a Joint-stock company should be a member of ____________
(a) IPS
(b) ICSI
(c) IAS
Answer:
(b) ICSI

Question 5.
____________ is the oldest type of Secretary.
(a) Company Secretary
(b) Personal Secretary
(c) Secretary of Government Department
Answer:
(b) Personal Secretary

Question 6.
The main objective of a co-operative society is to ____________ the interest of its members.
(a) neglect
(b) takeover
(c) protect
Answer:
(c) protect

Question 7.
Secretary of Government department must be a member of the ____________
(a) ICSI
(b) IAS
(c) ICWA
Answer:
(b) IAS

Question 8.
____________ means correctness in doing a work.
(a) Accuracy
(b) Loyalty
(c) Courteous
Answer:
(a) Accuracy

1B. Match the pairs.

Question 1.

Answer:

1C. Write a word or a term or a phrase that can substitute each of the following statements.

Question 1.
The term was used in the Roman empire was for a person who was a professional letter writer.
Answer:
Scribae

Question 2.
The sports club is an example of this type of organization.
Answer:
Non-profit association

Question 3.
A Secretary appointed by an individual.
Answer:
Personal Secretary

Question 4.
The quality that a Secretary should possess is to be faithful to his organization.
Answer:
Loyalty

Question 5.
The Latin word for a Secretary.
Answer:
Secretarius

Question 6.
The Company Secretary must be a member of this organization.
Answer:
ICSI

Question 7.
The Secretary of a Co-operative organization in Maharashtra must have knowledge of this Act.
Answer:
Maharashtra State Co-operative Societies Act, 1960

Question 8.
An association that aims to protect the interest of its members.
Answer:
Co-operative society

1D. State whether the following statements are True or False.

Question 1.
A Secretary is responsible only for typing letters.
Answer:
False

Question 2.
The Secretary of a Government department is the oldest type of Secretary.
Answer:
False

Question 3.
The Secretary should remain confidential about his organization.
Answer:
True

Question 4.
A Secretary also looks after routine and administrative duties.
Answer:
True

Question 5.
A Secretary is a paid employee.
Answer:
True

Question 6.
Chitnis is a term used in the ancient Roman period.
Answer:
False

Question 7.
A Secretary need not have knowledge of computers.
Answer:
False

Question 8.
Cultural Association is an example of a Government department.
Answer:
False

1E. Find the odd one.

Question 1.
Secretarius, Scribae, ICSI, Chitnis
Answer:
ICSI

Question 2.
Leadership, IAS, Tactfulness
Answer:
IAS

1F. Complete the sentences.

Question 1.
In Latin language, a confidential writer was called as ____________
Answer:
Secretarius

Question 2.
The oldest type of Secretary is ____________ secretary.
Answer:
Personal Secretary/Personal Assistant

Question 3.
A company secretary must be a member of ____________
Answer:
ICSI (Institute of Company Secretaries of India)

1G. Select the correct option from the bracket.

Question 1.

(IAS, Appointed by busy individuals, Secretary, Roman Empire, Company Secretary)
Answer:

1H. Answer in one sentence.

Question 1.
Who can appoint a personal secretary?
Answer:
Personal Secretary can be appointed by busy individuals, eg. Doctors, Lawyers, Engineers, Chartered Accountants, Actors, Politicians, Leaders, Ministers, etc.

Question 2.
Which company should appoint a full-time company secretary?
Answer:
Listed companies and all other companies with a paid-up share capital of five crores or more should appoint a full-time company secretary.

Question 3.
A Cultural Club is an example of which type of organization?
Answer:
A Cultural Club is an example of a Non-profit association.

Question 4.
The knowledge of which Act is a must for a secretary of Co-operative Society?
Answer:
A secretary of a co-operative society should have a good knowledge of the Co-operative Societies Act.

1I. Correct the underlined word and rewrite the following sentences.

Question 1.
Personal Secretary is appointed by a Joint Stock Company.
Answer:
Personal Secretary is appointed by a busy individual.

Question 2.
Busy individuals appoint Company Secretary.
Answer:
Busy individuals appoint Personal Secretary.

Question 3.
The Secretary of a Government Department must be a member of ICSI.
Answer:
The Secretary of a Government Department must be a member of IAS.

Question 4.
The term Chitnis was used in Roman Empire.
Answer:
The term Chitnis was used in Ancient Indian history.

2. Explain the following terms/concepts.

Question 1.
Secretary
Answer:
An individual employed to handle general administration, office management, correspondence, and confidential information of an organization.

Question 2.
Confidential officer
Answer:
Secretary is entrusted with the secret and confidential information of his employer and of the business itself.

Question 3.
Compliance officer
Answer:
A secretary must stick to or comply with various sets and its provisions that are applicable to the business like maintaining books, filing returns, etc.

Question 4.
Tactfulness
Answer:
An ability of a person to handle a situation in the right manner. Secretary must handle the different situations and people insensible way.

Question 5.
Loyalty
Answer:
Loyalty means faithfulness. As a confidential officer, a secretary should not share secret matters with anyone.

3. Study the following case/situation and express your opinion.

1. Rajiv and Sanjiv have applied for the post of Company Secretary at Pharma India Pvt. Ltd. Rajiv has experience of 10 years as a Personal Secretary. He later worked as a Secretary at Venus Co-operative Bank for 2 years. Sanjiv has passed his CS examination and is a member of ICSI. He has no working experience. Rajiv is only a commerce graduate.

Question (a).
Who is better suited for the above post? Justify your answer.
Answer:
Sanjiv is suited for the post as he has passed his CS exams and is a member of ICSI which is the basic qualification for a company secretary according to the Companies Act, 2013.

Question (b).
Can Rajiv be selected as a Personal Assistant for one of the Directors?
Answer:
Yes, Rajiv has experience of 10 years as a personal secretary, hence he can be able to assist the director.

Question (c).
Can Rajiv be appointed as Secretary to Pharma India Pvt Ltd.?
Answer:
No, Pharma India Pvt. Ltd. is a private company that can appoint a Company Secretary for its work. Rajiv has experience as a personal secretary so he cannot be appointed as a secretary to Pharma India Pvt. Ltd.

2. Harshada has cleared her CS examination and intends to make her career as a Secretary at the Government department.

Question (a).
Can she apply for the post of Secretary at Government Department?
Answer:
No, a secretary of the Government department must have a clear Civil Services Examination and must be in Indian Administrative Services.

Question (b).
Can she be appointed as the Secretary of ABC Ltd.?
Answer:
Yes, she can be appointed as the secretary of ABC Ltd which is a company, and shehasclearedherCS examination.

Question (c).
As a Company Secretary will Harshada be considered as an employee?
Answer:
Yes, a company secretary is a full-time employee and gets a salary as determined by the Board of Directors.

4. Distinguish between the following.

Question 1.
Personal Secretary and Company Secretary
Answer:

Question 2.
Secretary of Co-operative Society and Secretary of Government Department
Answer:

Question 3.
Company Secretary and Secretary of a Government Department
Answer:

Question 4.
Secretary of Non-profit Association and Secretary of Co-operative Society.
Answer:

5. Answer in brief.

Question 1.
Describe any four qualities of a Secretary.
Answer:
Qualities of a Secretary are as follows:

1. Accuracy: A secretary must be accurate in his work like accuracy in office work, record keeping, drafting letters, reports, etc.
2. Loyalty: A secretary should be loyal. He is a custodian of secret information about his business. He should not disclose the secret of the business to outsiders.
3. Adaptability: There are many changes that take place within the business and outside the business. A secretary must adapt himself to every situation and should be ready for every challenge.
4. Leadership: As a leader, a secretary should be able to guide, advise and direct the subordinates. He should create team spirit among his subordinates.

Question 2.
State any four functions of a Secretary.
Answer:
The functions of a Secretary are as follows:

1. Correspondence: It is a basic function of the secretary. He drafts letters, handles inward and outward mails, types letters, files important correspondence. He must handle correspondence well so as to maintain the goodwill of the business.
2. Reception functions: A secretary courteously attends to the visitors, telephone calls, arrange meetings, fix appointments, etc.
3. Financial functions: Secretary handles the financial transactions and maintains proper books of accounts. He handles banking transactions like depositing money, withdrawing money from the bank, issuing cheques and demand drafts, etc.
4. Administrative functions: Secretary performs administrative functions like supervising and controlling the activities of the business, effective coordination, training, orientation, promotion of employees.

Question 3.
Mention any four features of a Secretary.
Answer:
Features of a Secretary are as follows:

1. Individual: Only an individual can be appointed as a secretary. A partnership firm, corporate body, or institution cannot be appointed as a secretary.
2. Employee: Secretary is a paid employee of the business. He is given a salary or honorarium for the work or task done by him.
3. Confidential officer: A secretary is the custodian of the secret or confidential information of his employer and business.
4. Appointment: A secretary can be appointed by a busy person, Co-operative society, Non-profit association, Government, Joint Stock company, etc.

6. Justify the following statements.

Question 1.
A secretary should be courteous and have a pleasing personality.
Answer:

1. A secretary should be courteous as he deals with people within and outside the organization.
2. As he gets the plans executed in the business he must be kind and polite with fellow workers.
3. As a secretary represents the business to outsiders and as a leader within the business, he must have a pleasing personality with confidence and the ability to tackle challenging situations.
4. Thus, a secretary should be courteous and have a pleasing personality.

Question 2.
A secretary is considered a paid employee.
Answer:

1. A secretary is a paid employee of the business.
2. He works as a secretary and receives a salary or honorarium in return.
3. He is not the owner but works for them.
4. Thus, a secretary is considered a paid employee.

Question 3.
Only individuals can be secretaries.
Answer:

1. A secretary is always an individual.
2. Only a person can perform the work of a secretary.
3. A corporate body, institution, or partnership firm cannot be appointed as a secretary.
4. Thus, only individuals can be secretaries.

Question 4.
A secretary is a link between the staff and members.
Answer:

1. With the help of effective communication, the secretary strikes a balance between the staff and members.
2. The plans, policies, and decisions made by the top-level are communicated to the staff, which helps inefficient working of the business.
3. Thus, a secretary is a link between the staff and members.

Question 5.
A personal secretary is appointed by busy individuals.
Answer:

1. A personal secretary is appointed by busy persons to handle their day-to-day work.
2. Professionals and busy persons appoint a secretary so that routine work is handled by the secretary and they can focus on other important work.
3. Thus, a personal secretary is appointed by busy individuals.

7. Answer the following questions.

Question 1.
Define a Secretary and explain the importance of a secretary.
Answer:
According to Oxford Dictionary “A person whose work is to write for others, especially one who is employed to conduct correspondence, keep records and to transact various other business for another person or for a society or corporation or public body.”

The importance of a secretary are:
(i) Provide guidance and advice:
The secretary provides guidance to the management for efficient working and achieving goals. He also gives advice to the superior on important matters and helps them to take the correct decision.

(ii) Custodian of secret information:
The secretary is closely connected with top management and has access to all the secret and confidential information about the business. He safeguards confidential documents and decisions of the business.

(iii) Correspondent:
Secretary handles all the inward and outward mails of the business. He conducts correspondence with members, directors, registrar, bank, government authorities, etc.

(iv) Conducting meetings:
Secretary carries out all formalities before, during, and after the meeting. He prepares notice, agenda, minutes, etc., and makes all the arrangements for the meetings.

(v) Administrator:
Secretary handles office administration and routine management work efficiently. On behalf of the employer, he looks after the working of day to day work of the business.

(vi) Legal officer:
Secretary performs various statutory duties. He prepares books, reports, registers, documents as prescribed by the Companies Act, 2013.

(vii) Link between management and staff:
The secretary is a link between management and staff. He maintains a balance between management and staff through effective communication and proper coordination.

Question 2.
Explain the features of a secretary.
Answer:
The features of a secretary are as follows:

1. Individual: Only an individual can be appointed as a secretary. An organization, firm, society, or corporate body cannot be appointed as a secretary.
2. Employee: Secretary is not the owner of the business. He is a paid employee. Secretary works in the business and expects salary or honorarium for the work done by him.
3. Custodian of secret information: The secretary has custody of all the secret information of the business as he is closely connected with the top-level management.
4. Appointment: Secretary can be appointed by busy persons, professionals, non-profit associations, co-operative society, or Joint Stock companies.
5. Qualities: In order to perform his duties efficiently a secretary must possess qualities like intelligence, accuracy, loyalty, Courtesy, etc.

Question 3.
Describe the qualities of a secretary.
Answer:
The qualities of a secretary are as follows:

1. Accuracy: It means correctness. While corresponding, preparing reports and records, books of accounts, and register the secretary must be accurate in his work.
2. Adaptability: Secretary must have the ability to adjust with the change within and outside the business like change in management rules and policies, change in government policies, etc.
3. Loyalty: Secretary has access to all the secret information of the business. He must be faithful and not disclose the information to anyone.
4. Cooperative: Secretary works with the people and through the people. He should instill the feeling of team spirit and oneness.
5. Punctuality: Secretary must have a sense of time. He must complete the work on time. If the work is delayed it would affect the growth of the business.
6. Piteasing personality: Secretary must have an impressive and confident personality. With decent behaviour, a good temper, and a positive approach, he can gain respect from others.
7. Tactful: A secretary handles different people and situations. He should be able to handle situations wisely and deal with people in a sensible way.

Question 4.
Explain briefly the various types of Institutional secretaries.
Answer:
Types of Institutional secretaries:
(i) Secretary of a Non-profit Association:
The non-profit association is formed to provide service and not to earn profits. The managing committee looks, after the working of Non-profit association. The Secretary of a Non-profit association is appointed by the managing committee on a salary or honorarium basis. There is no prescribed qualifications of the secretary of the Non-profit association however, knowledge of correspondence, banking transactions, knowledge of computers are expected in these associations.

(ii) Secretary of a Co-operative Society:
It is a voluntary organization formed for the welfare of its members. Managing committee handles or manage the working of the society. One of the members of the managing committee is appointed as a secretary. Secretary of Co-operative society can be appointed as a full-time employee. Secretary must have full knowledge of all the provisions of the Co-operative Societies Act.

(iii) Secretary of a Government Department:
The Secretary of a Government department is appointed by the Central or State Government. Secretary handles the work of various government departments on behalf of the ministers and act as an administrative head of the department. The secretary of the government department should be a graduate and be in Indian Administrative Services [IAS].

(iv) Secretary of a Joint Stock Company:
The Secretary of a Joint Stock company holds a key managerial position and handles the business on behalf of the members. Secretary of a Joint Stock Company is appointed by the Board of Directors. According to the Companies Act, 2013 the Companies Secretary must be a member of the Institute of Company Secretaries of India (ICSI). Joint Stock Company works on a large scale, hence, full-time secretary handles day-to-day work on behalf of the Board of directors.

Activity 1. (Textbook Page No. 9)

Find out the following:

1. Company Secretary of Mahindra and Mahindra Ltd.
2. Principal Secretary of Prime Minister of India.
3. Chief Secretary, Maharashtra State.
4. Secretary, School Education Department, Government of Maharashtra.

Answer:

1. Narayan Shankar
2. Pramod Kumar Mishra
3. Sitaram Kunte
4. Vishal Solanki (IAS)

Activity 2. (Textbook Page No. 9)

Atmaram Bendre is a Secretary of Gokulghar Housing Society, Enlist his duties.
Answer:
Duties of a Secretary performed by Atmaram Bendre residing at Gokulghar Housing Society, Malad (W), Mumbai are as follows:

1. To hold a monthly meeting of Managing Committee.
2. To prepare notice along with agenda and issue to members of the managing committee.
3. To record and maintain minutes book of the managing committee.
4. To assist the chairman before, during, and after the meeting.
5. To issue a Share certificate to members of the Co-operative Housing Society.
6. To transfer Share certificate within a given period of time.
7. To maintain books relating to the resignation and transfer of members of society.
8. To examine the documents related to transferring flats.
9. To take appropriate actions against default members by giving legal notice, if any.
10. To maintain accounting records required for auditing.
11. To finalize the audit report.
12. To utilize the sinking fund of the society with prior approval of members of society for repair, maintenance internal leakage problem, etc.
13. To give No Objection Certificate for flat agreement of sale or any kind of loans on flat.
14. To abide bye-laws as guided by Co-operative Housing Society Act, 1960.
15. To maintain all correspondence records, files, documents, etc.
16. To issue notice to the member who breaches the bye-laws of the society.
17. To maintain records for the societies such as account records, statutory records, and membership records.
18. To produce records as and when required by the government authorities with the consent of members.
19. To issue allotment or cancellation letter as required by the members of the society.
20. To take contingency meetings as and when required in the event of an emergency.

Balbharti Yuvakbharati English 11th Digest Chapter 4.3 Extracts of Drama – (A) A Midsummer – Night’s Dream Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 11 English Yuvakbharati Solutions Chapter 4.3 Extracts of Drama – A Midsummer – Night’s Dream

11th English Digest Chapter 4.3 Extracts of Drama – A Midsummer – Night’s Dream Textbook Questions and Answers

Characters:

1. Choose the odd one out.

Question (i)
Bottom, Moth, Mustard seed, Cobweb
Answer:
Bottom

Question (ii)
Flute, Snug, Quince, Cobweb
Answer:
Cobweb

Question 2.
Match the columns.

Answer:

Question 3.
Draw a character sketch of Oberon as an enemy of his wife but a friend of lovers.
Answer:
Oberon, the king of fairies, is shown having two faces. On the one hand, he is shown hatching a plan with the help of Puck by using ‘Love in Idleness’ ensuring that the proper lovers end up loving each other. He is also shown sympathising Helena on seeing Demetrius’ cold behaviour towards her. He brings blessings, good health and peace to all the newly married couples towards the end.

On the other hand, in his dealings with his wife Titania, Oberon is potrayed as a wicked man. At the beginning of the play, Oberon is shown fighting with Titania over the custody of an Indian boy. Here, he is however, trying to prove his authority as a male and win the boy. He also tricks his wife by casting a spell upon her that leaves her fall in love with Bottom. He then releases the spell off her when he gets what he has wanted – The boy.

Question 4.
Comment on the loving pair of Lysander and Helena from the point of view of developing their character sketch.
Answer:
Hermia and Lysander loved each other and they eloped in the forest to get married. Helena loved Demetrius but it was the effect of the love potion which was applied to Lysander’s eyes when he slept. Therefore, as he woke up he saw Helena and he started developing feelings for her. But Helena loved Demetrius. On her part Helena was very right. At the end both found their right companions Helena – Demetrius and Lysander – Hermia.

Setting:

1. Correct the given sentences with justification.

Question (i)
The play is restricted to only a part of the woods.
Answer:
The play is not restricted to only a part of the woods because the references, Quince’s cottage, in the another part of the wood etc., are there in the play.

Question (ii)
Since there is a reference to the Indian boy, there are some scenes from India too.
Answer:
The reference to the Indian boy is found in Act II scene I as Puck says “A lovely boy, stolen from an Indian king” and when Titania said that his mother was a votaress of my order and in the spiced Indian air. The reference of Titania’s Indian friend is also found.

Question 2.
The characters are a part of the stage setting. How does this reflect when the characters of the play range from the Duke and the Indian boy to the fairies?
Answer:
Characters are a part of stage setting. This reflects in ‘A Midsummer Night’s Dream’ from the Duke in the palace as he instructs Philostrate to arrange for the celebration with great revelry. However, in the another part of the wood Oberon and Titania struggle for the custody of the Indian boy. Oberon uses love potion on the eyelids of Titania when she sleeps. As the setting of the stage changes the characters change accordingly. Because of the presence of the fairies and an Indian boy, the characters in the play range from real to imaginary.

Question 3.
What changes in the stage setting would you suggest.
Answer:
The characters are the part of stage setting and the play starts with Theseus, Duke of Athens as he wins the queen of Amazons, Hippolyta. But later, in their marriage celebration there are the king and the queen of fairies, Oberon and Titania as well as other fairies and elves as imaginary characters. Characters are arranged from real to imaginary. I suggest that all characters should be real humans.

Question 4.
Comment on the versality and the aptness of the stage setting, as per the requirement of the play ‘A Midsummer Night’s Dream’.
Answer:
As per the requirement of the play ‘A Midsummer Night’s Dream’ the versality and the aptness of the stage setting go hand in hand. The play starts in the palace and then the stage of the play develops in two different parts of the wood. The stage-setting has been arranged according to the need of the characters and the plot. The characters are set accordingly onto the stage.

Plot:

Question 1.
State whether the following statements are True/False.

1. Lysander and Demetrius fall in love with Helena as a result of the love potion.
2. Oberon transforms Bottom’s head into that of an ass.
3. Titania falls in love with an ass.
4. Both Demetrius and Lysander fight for Helena.

Answer:

1. True
2. False
3. False
4. True

2. Give reasons:

Question (i)
Oberon and Titania fight for the custody of the Indian boy because –
(a) Oberon wants the custody of the Indian boy so that Titania would give him a lot time along with the boy.
(b) Titania wants the Indian boy because of her love and duty towards the small boy.

Question 3.
The consequences of Oberon ‘s jealousy for Titania are comic rather than tragic, comment.
Answer:
Oberon was jealous of his queen Titania because she refused to give the boy to him . She fell in love with Bottom. Oberon decided to seek revenge on his queen. It was Oberon himself who made Titania to fall in love with Bottom who wa§ having the head of a donkey. Comic incident was created by that scene in the drama.

Question 4.
There were some reasons why Theseus was initially against but later gave consent for the marrige of Hermia with Lysander. Explain.
Answer:
The reasons why Theseus was initially against but later gave consent for the marrige of Hermia with Lysander because Hermia’s father’s wish was that she should marry with Demetrius and she disobeyed her father. The Duke wants that Hermia should follow the rules and decision of her father. But she eloped with Lysander. It was Hermia’s true love for Lysander that made Theseus to change his decision and agreed that Hermia should marry with Lysander.

Form:

1. Select the correct options.

Question (i)
A Midsummers Night’s Dream is a _______
(a) poetic drama
(b) comedy of errors
(c) comedy based on fantasy
(d) a character play
(e) a revenge tragedy
(f) belongs to realm of dreams.
Answer:
A Midsummer Night’s Dream is a comedy of errors.

Question 2.
Find 2/4 expressions of humour from the extract.
Answer:
Expressions of humour from the extract are:
1. Helena: You do advance your cunning more and more –
Lysander: I had no judgement when to her I swore.
2. Hermia: What’s this to my Lysander? Where is he? Ah, good Demetrius, wilt thou give him me ?
Demetrius: I had rather give his carcass to my bounds.
Hermia: Out dog / out cur/ thou drivest me past the bounds.

Question 3.
‘A Midsummer Night’s Dream’ is one of the best examples of Shakespeare’s comedy of errors. Comment.
Answer:
A Midsummer Night’s Dream is one of the best examples of Shakespear’s comedy of errors because it has all the features of comedy of errors like conflict between characters, resolution over the problem, cleared confusions between the love triangle, reunion and marriage. It all seem very funny and interesting.

Theme:

Question 1.
Shakespeare is acknowledged as the greatest writer because he understood human nature better than anyone else. Explain the statement in context of the of play.
Answer:
Many authors have tried to potray love as human nature, however, none could display it better than Shakespeare does in his plays. In this play – ‘A Midsummer Night’s Dream’ he explains human nature of ‘control’ through love. He has given a clear picturisation of how love controls a person in the character of Theseus as he forgives Hermia and allows her to marry her love Lysander. It reveals that true love prevails and finds solace.

Shakespeare also explains human nature of conflicts between family and friends. He gives a clear picturisation of conflict when the magic potion ‘Love in Idleness’ was wrongly used on Lysander and he falls in love with Helena because of which Hermia and Helena start a quarrel. Another conflict potrayed, is between the family, when Hermia is given three choices by her father – marry Demetrius, become a nun or die to which she rebels against her father and decides to elope to the woods to marry her love Lysander. Hence, Shakespeare is acknowledged as the greatest writer of human nature.

Question 2.
Prove with the theme of the play/ extract that the deeper human emotion which profoundly interested Shakespeare, was jealousy.
Answer:
The theme of the play ‘A Midsummer Night’s Dream’ is love. Shakespeare portrays how people fall in love with those who appear beautiful to them. However, if one rejects other’s true love, then jealousy starts in the minds of the lovers. The same situation arises if one does not spend quality time with the other. In this play the example of Oberon and Titania is apt to describe human emotion.

Language:

Question 1.
Interpret the following lines in Simple English.
Puck: I’ll follow you.
Bottom: The finch, the sparrow.
Answer:
The song is sung by Bottom to try to keep his courage up as Puck has just turned his head into that of an ass. This transformation scared everyone around him and he is alone in the forest. But his song awakens Titania who falls in love with him immediately and the use of appropriate words and songs are there for the other parts of the wood also.

2. Comment on the literary device, used in the following lines.

Question (i)
Titania: Be kind and courteous to this gentleman
Answer:
Tautology. The words kind and courteous denote the same meaning to give dramatic effect.

Question (ii)
Titania: Come wait upon him: lead him to my bower.
Answer:
Repetition. The word him is repeated two times to give dramatic effect.
Apostrophe. Time has been addressed by Titania to other fairies.

Question 3.
Shakespeare’s poetry has come to be valued for its own sake on the stage. Comment with reference to the play ‘A Midsummer Night’s Dream’.
Answer:
Shakespeare’s poetry, in his ‘A Midsummer Night’s Dream’ is found outstanding and valued for it’s own sake on the stage. The expressions of love through poetry made his drama lively and interesting. Through the poetic language Shakespeare underlined the truth of human nature and through poetic expressions made the critical moments easier.

Yuvakbharati English 11th Digest Chapter 4.3 Extracts of Drama – A Midsummer – Night’s Dream Additional Important Questions and Answers

Question 1.
Then will two at once woo one;
That must needs be sport alone:
Answer:
Reference : These lines are taken from William Shakespeare’s famous comedy ‘A Midsummer Night’s Dream’. Act III scene II.
Context: The lines are said by Puck as he applied love potion on the eyelids of Demetrius with negligence. Explanation : In ‘A Midsummer Night’s Dream’, a comic character Puck was told by Oberon, the king of fairies to use the love potion on the eyelides of Demetrius so that he wakes up and sees Helena and will fall in love with her.
But Puck uses the love potion on the eyelids of Lysander, supposing him to be Demetrius. Helena sees the first person Lysander after waking up and falls in love with her instantly. A misunderstanding of Puck made a chaotic situation between the lovers.

Question 2.
Shakespeare’s poetry has come to be valued for its own sake on the stage. Comment with reference to the play ‘A midsummer Night’s Dream’
Answer:
In midsummer night’s dream, Shakespeare’s used theatrical circle simply as fantasy. In order to made it more effective he used poetic lines for enlightening the beauty of the actions in the drama. These poetic lines revealed the nature of characters and developed poetic structure of the drama.

Extracts of Drama – A Midsummer – Night’s Dream Summary in English

In the Palace:

Theseus, Duke of Athens wins Hippolyta in war and they are to be married. So, he instructs Philostrate to arrange for the celebration.

Egeus, the father of Hermia wants his daughter to be married with Demetrius. But she refuses to marry Demetrius as she is in love with Lysander. The Duke Theseus urges her to obey her father and gives her three alternatives: (1) marry Demetrius (2) become a nun or (3) suffer a death sentence. Hermia has time to decide until Theseus, The Duke of Athen’s marriage.

Hermia and Lysander decide to elope to the woods and get married but, Demetrius who used to love Helena, now rejected her love and is interested in Hermia. Hermia tries to win back the love of Demetrius for Helena. As Hermia elopes with Lysander, Demetrius follows her and Helena follows Demetrius.

At the Quinces Cottage:

The workmen from Athens wish to perform a play ‘Pyramus and Thisby’ at the Duke’s wedding. They all plan to meet in the forest for the rehearsal.

In the Woods:

A different world has been seen in the woods. The king of fairies, Oberon and Queen of fairies, Titania are having a fight over the custody of an Indian boy. To teach a lesson to her, Oberon, with the help of Puck, a mischievous spirit, plans to use the magic potion ‘Love in Idleness’, the juice if poured on the eyelids of a sleeping person, makes his/her fall madly in love with the first person he/she sees after waking up. He thus plans to madden Titania and get the custody of the Indian boy.

Oberon then sees Demetrius’ cold behaviour towards Helena and tries to help them by developing feelings in the heart of Demetrius for her. Therefore, he asks Puck to squeeze the magic juice on the eyelids of Demetrius Taking Lysander for Demetrius, Puck squeezes the magic juice on his eyelids.

Helena who enters the wood following Demetrius, proves to be the first person Lysander sees after waking up and falls in love with her instantly. Titania found too much interested in Bottom, one of the workmen of Athens. Puck bewitches him by transforming his head into that of an ass and it bewildered Titania.

In the Another Part of the Wood:

Puck wins the heart of Oberon for punishing Titania and Bottom. But when Oberon realises Puck’s mistake of using magic potion for Lysander instead of Demetrius, he himself squeezes it on the eyelids of Demetrius and orders Puck to fetch Helena as he wakes up to restore the love of Demetrius for Helena. He also corrects the relations of Lysander with Hermia as well as orders Puck to restore Bottom’s head and feels sorry for Titania.

Theseus, Hippolyta and Egeus enter and see the four Athenians and the love between them. He allows Demetrius and Helena and Hermia and Lysander to marry. At last, in the final scene, ‘Pyramus and Thisbe’ was performed for the marriage ceremony of Theseus and Hippolyta in the presence of Oberon and Titania with their fairies and elves to sing and dance and bless the newly wedded. Thus, the play ‘A Midsummer Night’s Dream’ ends on a happy note.

Glossary:

1. night-rule – dark deeds,
2. consecrated – sacred,
3. dote on – to love,
4. extremity – whole heartedly,
5. patches – clowns,
6. nuptial – marriage/ matrimonial/ moments of nuptial bliss,
7. nole – head,
8. mimic – commie actor,
9. russet-pated choughs – birds of the crow family with reddish, brown (or grey) heads,
10. sever – separate from each other,
11. stamp – a noise made by bringing one’s foot heavily on the ground but, here – thing like ‘trick’ (i.e. giving
12. Bottom an ass’s head),
13. yielders – those who yield (giving up or surrender),
14. latch’d – leached, anointed (love juice),
15. ey’d – seen,
16. antipodes – the opposite side of the earth,
17. venus – the evening star,
18. carcass – dead body,
19. mispris’d – mistaken,
20. an – and (or) even,
21. adder – circuit,
22. debt that bankrupt sleep doth sorrow owe – sleepless due to sorrow,
23. confounding – confusing and breaking,
24. Ioi tartars bow – weapon used by the Asian warriors who invaded Europe in the 13 century,
25. cupid’s archery – Roman God of love,
26. a lover’s fee – love requited,
27. fond pageant – silly behaviour,
28. tales – untrue stories,
29. sport alone – fun by itself,
30. devilish holy – a conflict between two truths,
31. nymph – nature goddess,
32. Taurus – a mountain range in Turkey,
33. superpraise – praise excessively,
34. conjure – summon/call,
35. disparage – regards being of little worth,
36. englids – brightens,
37. confederacy – plot/plan,
38. chid – rebuked/scold,
39. artificial gods – creators of works of art,
40. union in partition – two in one,
41. incorporate – united,
42. so in grace – so much in favour/in a good condition,
43. ethiope – black face, but actually a scornful reference to her dark hair,
44. canker blossom – a flower blighted by a worm lodged in the bud,
45. stealth – stealing away,
46. fond – foolish,
47. suffer her to flout – allow her to mock,
48. knot grass – a weed which creeps and makes entangling roots,
49. officious – meddlesome/self assertive enthusiastic,
50. aby – pay a heavy penalty,
51. check by jowl – closely,
52. king of shadows – fairy kings,
53. sort – occur,
54. welkin – sky,
55. Acheron – the world of dead,
56. rail thou – use violent language,
57. Aurora’s harbinger – forerunner of the dawn (the morning star),
58. wormy beds – graves,
59. Neptune – the ocean, in ancient mythology,
60. hither (archaic word) – to or towards this place,
61. constraineth (archaic word) – to impose limitations,
62. crust – (archaic words)to curse,
63. a knavish lad – a mischievous boy,
64. woe – great sorrow,
65. bedabbled – sensuous,
66. briers – wild shrubs,
67. mare – an adult female horse,
68. stalls – work-benches,
69. barren – stupid/brainless,
70. Anon – quickly,
71. the creeping fowler eye – wild geese keep a close watch on hunter who is creeping up to shoot,
72. of force she must be ey’d – inevitably (perforce) she will be seen,
73. lay breath so bitter on your bitter foe – speak so bitterly to no one but your bitter enemy,
74. may be bored – may have a hold driven right through it,
75. dead – pale and bloodless,
76. touch – feat,
77. tender – attention and care,
78. look – be sure,
79. cheer – countenance.