Author Archives: Rohit

INTEXT Questions

Question 1.
Sleeping pills are recommended by doctors to the patients suffering from sleeplessness but it is not advisable to take its doses without consultation with the doctor. Why ?
Solution:
Most of the dru gs taken in doses higher than recommended may cause harmful effect and act as poison. Therefore, a doctor should always be consulted before taking medicine.

Question 2.
With reference to which classification has the statement, “ranitidine is an antacid” been given?
Solution:
This statement refers to the classification according to pharmacological effect of the drug because any drug which will be used to counteract the effect of excess acid in the stomach will be called antacid.

Question 3.
Why do we require artificial sweetening agents?
Solution:
Those people who have diabetes or who need to control intake of calories, they cannot take sugar. They need its substitute. These substitutes are called artificial sweeteners. These are excreted from the body in urine unchanged.

Question 4.
Write the chemical equation for preparing sodium soap from glyceryloleate and glyceryl palmitate. Structural formulae of these compounds are given below.

1. (C₁₅H₃₁COO)₃C₃H₅ – Glyceryl palmitate
2. (C₁₇H₃₃COO)₃C₃H₅ – Glyceryl oleate.

Solution:

Question 5.
Following type of non-ionic detergents are present in liquid detergents, emulsifying agents and wetting agents. Label the hydrophilic and hydrophobic parts in the molecule. Identify the functional group(s) present in the molecule.

Solution:

The functional groups present in the molecule are alcoholic group and ethereal linkage (—O—).

NCERT Exercises

Question 1.
Why do we need to classify drugs in different ways?
Solution:
In a general way, the drug may be defined as a substance used in the prevention, diagnosis, treatment or cure of diseases. We must classify the drug so as to understand its action on our body. Drugs are classified on the basis of (a) pharmacological effect, (b) drug action, (c) chemical structure and (d) molecular targets.

Question 2.
Explain the term, target molecules or drug targets as used in medicinal chemistry.
Solution:
Drugs usually interact with biomolecules, such as carbohydrates, lipids, proteins and nucleic acids. These are called target molecules or drug targets. These perform various functions in the body. For example, proteins which perform the role of biological catalysts in the body are called enzymes and those which are crucial to communication system in the body are called receptors. Nucleic acids have coded genetic information for the cell. Lipids and carbohydrates are structural parts of the cell membrane.

Question 3.
Name the macromolecules that are chosen as drug targets.
Solution:
Macromolecules of biological origin such as carbohydrates, lipids, proteins and nucleic acids.

Question 4.
Why should not medicines be taken without consulting doctors?
Solution:
Drugs are designed to interact with specific targets so that these have the least chance of affecting other targets. Only a doctor can diagnose a disease properly and prescribe the correct medicine in proper dose because excess of medicines may have harmful effects on our body. So we should not take medicines without consulting doctors.

Question 5.
Define the term chemotherapy.
Solution:
Chemotherapy (literally means chemical treatment) is the science in which chemicals are used for the treatement of diseases. Chemotherapy is defined as the use of chemicals (drugs) to injure or destroy infectious microorganisms without causing any injury to the host. Chemotherapy has developed into a vast subject today and efforts are being continuously made to search new drugs as to free human beings from various types of diseases.

Question 6.
Which forces are involved in holding the drugs to the active site of enzymes?
Solution:
Substrates bind to the active site of the enzyme through a variety of interactions such as ionic bonding, hydrogen bonding, van der Waals interaction or dipol-dipole interaction.

Question 7.
While antacids and antiallergic drugs interfere with the function of histamines, why do these not interfere with the function of each other?
Solution:
Histamine stimulates the secretion of pepsin and hydrochloric acid in the stomach. The drug cimetidine (antacid) was designed to prevent the interaction of histamine with the receptors present in the stomach wall. This resulted in release of lesser amount of acid. Antacid and antiallergic drugs work on different receptors.

Question 8.
Low level of noradrenaline is the cause of depression. What type of drugs are needed to cure this problem? Name two drugs.
Solution:
Noradrenaline is one of the Neurotransmitters that plays a role in mood changes. If the level of noradrenalines remains low for some reason, then the signal¬sending activity becomes low and the person suffers from depression. In such situations, antidepressant drugs are required. These drugs inhibit the enzymes which catalyse the degradation of noradrenaline. If the enzyme is inhibited, this important neurotransmitter is slowly metabolised and can activate its receptor for longer periods of time, thus counteracting the effect of depression. Iproniazid and phenylzine are two such drugs.

Question 9.
What is meant by the term ‘broad spectrum antibiotics’? Explain.
Solution:
The range of bacteria or other microorganisms that are affected by a certain antibiotic is expressed as its spectrum of action. Antibiotics which kill or inhibit a wide range of Gram-positive and Gram-negative bacteria are said to be broad spectrum antibiotics, e.g., Chloramphenicol, Tetracycline, etc.

Question 10.
How do antiseptics differ from disinfectants? Give one example of each.
Solution:
Antiseptics and disinfectants are the chemicals which either kill or prevent the growth of microorganisms. Antiseptics are applied to the living tissues such as wounds, cuts, ulcers and diseased skin surfaces. Examples are furacine, soframicine, etc. These are not ingested like antibiotics. On the other hand, disinfectants are applied to inanimate objects such as floors, drainage system, instruments, etc.

Same substances can act as an antiseptic as well as disinfectant by varying the concentration. For example, 0.2% solution of phenol is an antiseptic while its 1% solution is disinfectant.

Question 11.
Why are cimetidine and ranitidine better antacids than sodium hydrogencarbonate or magnesium or aluminium hydroxide?
Solution:
Excessive hydrogencarbonate can make the stomach alkaline and trigger the production of even more acid. Metal hydroxides are better alternatives because being insoluble, these do not increase the pH above neutrality. These treatments control only symptoms, and not the cause. Therefore, with these metal salts, the patients cannot be treated easily.

The drugs cimetidine (Tegamet) and ranitidine (Zantac) prevent the interaction of histamine with the receptors present in the stomach wall. This resulted in release of lesser amount of acid.

Question 12.
Name a substance which can be used as an antiseptic as well as disinfectant.
Solution:
0.2% solution of phenol is an antiseptic while its 1% solution is disinfectant.

Question 13.
What are the main constituents of dettol?
Solution:
Dettol is a mixture of chloroxylenol (4-chloro-3,5-dimethylphenol) and terpineol.

Question 14.
What is tincture of iodine? What is its use?
Solution:
2-3% solution of iodine in alcohol- water mixture is known as tincture of iodine. It is a powerful antiseptic.

Question 15.
What are food preservatives?
Solution:
Food preservatives prevent spoilage of food due to microbial growth or these may be defined as the substances which are capable of inhibiting the process of fermentation, acidification or any other decomposition of food. The most commonly used preservatives include table salt, sugar, vegetable oils and sodium benzoate. Salts of sorbic acid and propanoic acid are also used as preservatives.

Question 16.
Why is use of aspartame limited to cold foods and drinks?
Solution:
Use of aspartame is limited to cold foods and soft drinks because it in unstable to heat and decomposes at cooking temperature.

Question 17.
What are artificial sweetening agents? Give two examples.
Solution:
The chemical substances which give sweetening effect to food but do not add any calorie to our body are celled artificial sweetening agents, e.g., aspartame, saccharin etc.

Question 18.
Name the sweetening agent used in the preparation of sweets for a diabetic patient.
Solution:
Saccharin is of great value of diabetic persons and people who need to control intake of calories because it is excreted from the body in urine unchanged and appears to be entirely inert and harmless when taken.

Question 19.
What problem arises in using alitame as artificial sweetener?
Solution:
Alitame is high potency sweetener (2000 times as sweet as cane sugar). The control of sweetness of food is difficult while using it.

Question 20.
How are synthetic detergents better than soaps?
Solution:
Soaps when used in hard water form insoluble precipitates which separate as scum in water and are useless as cleansing agent. In fact these are hindrance to good washing, because the precipitate adheres onto the fibre of the cloth as gummy mass. On the other hand, synthetic detergents can be used both in soft and hard water as they give foam even in hard water. Some of the detergents give foam even in ice cold water.

Question 21.
Explain the following terms with suitable examples
(i) cationic detergents
(ii) anionic detergents and
(iii) non-ionic detergents.
Solution:
(i) Cationic detergents : Cationic detergents are quarternary ammonium salts of amines with acetates, chlorides or bromides as’ anions. Cationic part possess a long hydrocarbon chain and a positive charge on nitrogen atom. Hence, these are called cationicdetergents.Cetyltrimethylammonium bromide is a popular cationic detergent and is used in hair conditioners.
Cationic detergents have germicidal properties and are expensive, therefore these are of limited use.

(ii) Anionic detergents : Anionic detergents are sodium salts of sulphonated long chain alcohols or hydrocarbons. Alkyl hydrogensulphates formed by treating long chain alcohols with concentrated sulphuric acid are neutralised with alkali to form anionic detergents. Similarly alkyl benzene sulphonates are obtained by neutralising alkyl benzene sulphonic acids with alkali.


In anionic detergents, the anionic part of the molecule is involved in the cleansing action. Sodium salts of alkylbenzenesulphonates are an important class of anionic detergents.

They are mostly used for household work. Anionic detergents are also used in toothpastes.

(iii) Non-ionic detergents : Non-ionic detergents do not contain any ion in their constitution. One such detergent is formed when stearic acid reacts with polyethyleneglycol.

Liquid dishwashing detergents are non-ionic type.

Question 22.
What are biodegradable and non-biodegradable detergents? Give one example of each.
Solution:
Biodegradable detergents : Detergents having straight hydrocarbon chains are easily degraded by microorganisms and hence are called biodegradable detergents, e.g.,

Non-biodegradable detergents : Detergents containing branched hydrocarbon chains are not easily degraded by the microorganisms and hence are called non-biodegradable detergents.

Question 23.
Why do soaps not work in hard water?
Solution:
Hard water contains calcium and magnesium ions. These ions form insoluble calcium and magnesium soaps respectively when sodium or potassium soaps are dissolved in hard water.

These insoluble soaps separate as scum in water and are useless as cleansing agent. In fact these are hindrance to good washing, because the precipitate adheres onto the fibre of the cloth as gummy mass.

Question 24.
Can you use soaps and synthetic detergents to check the hardness of water?
Solution:
Soaps and detergents can be used to check the hardness of water. Hard water forms curdy white precipitate with Ca²⁺ and Mg²⁺ ions present in hard water whereas no such precipitates are formed by detergents in hard water.

Question 25.
Explain the cleansing action of soaps.
Solution:
Soap is sodium or potassium salt of a higher fatty acid and may be represented as RCOO^– Na (e.g., sodium stearate CH₃(CH₂)₁₆ COO^– Na⁺ When dissolved in water, it dissociates into RCOO^– and Na⁺ ions. The RCOO^– ions, however, consist of two parts – a long hydrocarbon chain R (also called non-polar ‘tail’) which is hydrophobic (water repelling), and a polar group COO- (also called polar-ionic ‘head’), which is hydrophilic (water loving).

The RCOO^– ions are, therefore, present on the surface with their COO^– groups in water and the hydrocarbon chains R staying away from it and remain at the surface. But at critical micelle concentration, the anions are pulled into the bulk of the solution and aggregate to form a spherical shape with their hydrocarbon chains pointing towards the centre of the sphere with COO^– part remaining outward on the surface of the sphere. An aggregate thus formed is known as ‘ionic micelle’. These micelles may contain as many as 100 such ions.

The cleansing action of soap is due to the fact that soap molecules form micelle around the oil droplet in such a way that hydrophobic part of the stearate ions is in the oil droplet and hydrophilic part projects out of the grease droplet like the bristles. Since the polar groups can interact with water, the oil droplet surrounded by stearate ions is now pulled in water and removed from the dirty surface. Thus soap helps in emulsification and washing away of oils and fats. The negatively charged sheath around the globules prevents them from coming together and forming aggregates.

Question 26.
If water contains dissolved calcium hydrogen carbonate, out of soaps and synthetic detergents which one will you use for cleaning clothes?
Solution:
Synthetic detergents.

Question 27.
Label the hydrophilic and hydrophobic parts in the following compounds.

1. CH₃(CH₂)₁₀CH₂OSO^–₃Na⁺
2. CH₃(CH₂)₁₅ N⁺ (CH₃)₃ Br
3. CH₃(CH₂)₁₆COO(CH₂CH₂O)nCH₂CH₂OH

Solution:

INTEXT Questions

Question 1.
What are polymers ?
Solution:
Polymers are high molecular mass substances consisting of large number of repeating structural units. They are also called as macromolecules. Some examples of polymers are polythene, bakelite, rubber, nylon 6, 6 etc.

Question 2.
How are polymers classified on the basis of structure?
Solution:
On the basis of structure, the polymers are classified as below :

1. Linear polymers such as polythene, polyvinyl chloride, etc.
2. Branched chain polymers such as low density polythene.
3. Cross linked polymers such as bakelite, melamine, etc.

Question 3.
Write the names of monomers of the following polymers :

Solution:

1. Hexamethylene diamine and adipic acid
2. Caprolactam
3. Tetrafluoroethene

Question 4.
Classify the following as addition and condensation polymers : Terylene, Bakelite, Polyvinyl chloride, Polythene.
Solution:
Addition polymers : Polyvinyl chloride, Polythene.
Condensation polymer : Terylene, Bakelite.

Question 5.
Explain the difference between Buna-N and Buna-S.
Solution:
Buna-N is a copolymer of 1, 3-butadiene and acrylonitrile and Buna-S is a copolymer of 1, 3-butadiene and styrene.

Question 6.
Arrange the following polymers in increasing order of their intermolecular forces.

1. Nylon 6,6, Buna-S, Polythene
2. Nylon 6, Neoprene, Polyvinyl chloride.

Solution:
The increasing order of intermolecular forces is :

1. Buna-S, Polythene, Nylon 6, 6
2. Neoprene, Polyvinyl chloride, Nylon 6.

NCERT Exercises

Question 1.
Explain the terms polymer and monomer.
Solution:
Polymers are very large molecules having high molecular mass, which are formed by joining of repeating structural units on a large scale derived from monomers, e.g., Polythene, PVC, Nylon-6, 6 etc.

Monomer is a simple molecule capable of undergoing polymerisation and leading to the formation of the corresponding polymer e.g., Ethene, Vinyl chloride, etc.

Question 2.
What are natural and synthetic polymers? Give two examples of each type.
Solution:
Natural Polymers – These are substances of natural origin, i.e., these are found in nature mainly in plants and animals. The well known natural polymers are proteins (polymers of amino acids), polysaccharides (polymers of monosaccharides), etc.

Synthetic Polymers – These are man made polymers i.e., polymers synthesised in laboratory. These include synthetic plastics, fibres and synthetic rubber. Specific examples are polythene and dacron.

Question 3.
Distinguish between the terms homopolymer and copolymer and give an example of each.
Solution:
A polymer which is obtained from only one type of monomer molecules is known as homopolymer e.g., Polythene, Polyvinyl chloride, etc.

Question 4.
How do you explain the functionality of a monomer?
Solution:
Functionality of a monomer is defined as the number of bonding sites in the monomers.

Question 5.
Define the term polymerisation.
Solution:
The process of joining together of a large number of simple small molecules (monomers) to make very large molecules (polymer) is termed polymerisation.

Question 6.

Solution:

Question 7.
In which classes, the polymers are classified on the basis of molecular forces?
Solution:
On the basis of molecular forces present between the chains of various polymers, the polymers are classified into the following four groups

1. Elastomers
2. Fibres
3. Thermoplastic polymers and
4. Thermosetting polymers.

Question 8.
How can you differentiate between addition and condensation polymerisation?
Solution:
In addition polymerisation, the molecules of the same or different monomers add together to form a large polymer molecule without the elimination of simple molecules like H₂O, HCl etc. Condensation polymerisation is a process in which two or more bifunctional molecules undergo a series of condensation reactions with the elimination of some simple molecules like H₂O, HCl, alcohol leading to the formation of polymers.

Question 9.
Explain the term copolymerisation and give two examples.
Solution:
Copolymerisation is a process in which a mixture of more than one monomeric species is allowed to polymerise. The copolymer contains multiple units of each monomer in the chain. The examples are copolymer of 1, 3-butadiene and styrene. Another example is the copolymer of 1, 3-butadiene and acrylonitrile.

Question 10.
Write the free radical mechanism for the polymerisation of ethene.
Solution:
The polymerisation of ethene to polythene consists of heating or exposing to light a mixture of ethene with a small amount of benzoyl peroxide initiator. The process starts with the addition of phenyl free radical formed by the peroxide to the ethene double bond thus regenerating a new and larger free radical. This step is called chain initiating step. As this radical reacts with another molecule of ethene, another bigger sized radical is formed. The repetition of this sequence with new and bigger radicals carries the reaction forward and the step is termed as chain propagating step. Ultimately, at some stage the product radical thus formed reacts with another radical to form the polymerised product. This step is called the chain terminating step. The sequence of steps may be depicted as follows :
Chain initiating steps :

Chain terminating step :
For termination of the long chain, these radicals can combine in different ways to form polythene. One mode of termination of chain is shown as under :

Question 11.
Define thermoplastics and thermosetting polymers with two examples of each.
Solution:
Thermoplastic polymers : These are the linear or slightly branched long chain molecules capable of repeatedly softening on heating and hardening on cooling. These polymers possess intermolecular forces of attraction intermediate between elastomers and fibres. Some common thermoplastics are polythene, polystyrene, polyvinyls, etc.

Thermosetting polymers : These polymers are cross linked or heavily branched molecules, which on heating undergo extensive cross linking in moulds and become infusible. These cannot be reused. Some common examples are bakelite, urea-formaldelyde resins, etc.

Question 12.
Write the monomers used for getting the following polymers.

1. Polyvinyl chloride
2. Teflon
3. Bakelite

Solution:

Question 13.
Write the name and structure of one of the common initiators used in free radical addition polymerisation.
Solution:
Benzoyl peroxide

Question 14.
How does the presence of double bonds in rubber molecules influence their structure and reactivity?
Solution:
Natural rubber is a linear polymer of isoprene i.e., 2-methy l-1, 3-butadiene.

In the polymer double bonds are located between C₂ and C₃ isoprene units. The cis- polyisoprene molecule consists of various chains held together by weak van der Waals interactions and has coiled structure. This cis-configuration about double bonds does not allow the chains to come closer for effective attraction due to weak van der Waals interactions. Thus, it can be stretched like a spring and exhibits elastic properties.

Question 15.
Discuss the main purpose of vulcanisation of rubber.
Solution:
Natural rubber becomes soft at high temperature (>335 K) and brittle at low temperature (<283 K) and shows high water absorption capacity. It is soluble in non¬polar solvents and is non-resistant to attack by oxidising agents. To improve upon these physical properties, a process of vulcanisation is carried out. This process consists of heating a mixture of raw rubber with sulphur and an appropriate additive at a temperature range between 373 K to 415 K. On vulcanisation, sulphur forms cross links at the reactive sites of double bonds and thus the rubber gets stiffened.

Question 16.
What are the monomeric repeating units of Nylon-6 and Nylon-6,6?
Solution:

Question 17.
Write the names and structures of the monomers of the following polymers :

1. Buna-S
2. Buna-N
3. Dacron
4. Neoprene.

Solution:

Question 18.
Identify the monomers in the following polymeric structures.

Solution:

Question 19.
How is dacron obtained from ethylene glycol and terephthalic acid?
Solution:
Dacron is obtained from ethylene glycol and terephthalic acid by condensation polymerisation reaction.

Question 20.
What is a biodegradable polymer? Give an example of a biodegradable aliphatic polyester.
Solution:
A large number of polymers are quite resistant to the environmental degradation processes and are thus responsible for the accumulation of polymeric solid waste materials. These solid wastes cause acute environmental problems and remain undegraded for quite a long time.

Aliphatic polyesters are one of the important classes of biodegradable polymers. One such example is of PHBV.

INTEXT Questions

Question 1.
Glucose or sucrose are soluble in water, but cyclohexane or benzene (simple six membered ring compounds) are insoluble in water. Explain.
Solution:
Glucose or sucrose contain several hydroxyl groups in their molecules which form hydrogen bonding with water molecules due to which they dissolve in water. On the other hand compounds like benzene or cyclohexane cannot form hydrogen bonds with water molecules, so they are insoluble in water.

Question 2.
What products are expected when lactose is hydrolysed ?
Solution:
Lactose (C₁₂H₂₂O₁₁) on hydrolysis with dilute acid yields an equimolar mixture of D-glucose and D-galactose.

Question 3.
How do you explain the absence of aldehyde group in pentaacetate of glucose ?
Solution:
The cyclic hemiacetal form of glucose contains an OH group at C-1 which gets hydrolysed in the aqueous solution to produce the open chain aldehydic form which then reacts with NH₂OH to form the corresponding oxime. Therefore, glucose contains an aldehydic group. On the other hand, when glucose is reacted with acetic anhydride, the OH group at C-1, along with the four other OH groups at C-2, C-3, C-4 and C-6 form a pentaacetate. As the pentaacetate of glucose does not contain a free OH group at C-1, it cannot get hydrolysed in aqueous solution to produce the open chain aldehydic form and thus glucose pentaacetate does not react with NH₂OH to form glucose oxime. Hence, glucose pentaacetate does not contain the aldehdye group.

Question 4.
The melting points and solubility in water of amino acids are higher than those of the corresponding haloacids. Explain.
Solution:
The amino acids exist as zwitter ions, H₃N⁺ OHR – COO^–. Because of this dipolar salt like character they have strong dipole- dipole attractions. So, their melting points are higher than halo acids which do not have sail like character. Moreover, due to this salt like character, they interact strongly with H₂O. Thus, solubility of amino acids in water is higher than that of the corresponding halo acids which do not have salt like character.

Question 5.
Where does the water present in the egg go after boiling the egg.
Solution:
The boiling of an egg is a common example of denaturation of proteins present in the white portion of an egg.

The albumin present in the white of an egg gets coagulated when the egg is boiled hard. The soluble globular protein present in it is denatured resulting in the formation of insoluble fibrous protein.

Question 6.
Why vitamin C cannot be stored in our body?
Solution:
Vitamin C is a water-soluble vitamin. Water-soluble vitamins when supplied regularly in the diet cannot be stored in out body because they are readily excreted in urine.

Question 7.
What products would be formed when a nucleotide from DNA containing thymine is hydrolysed ?
Solution:
When a nucleotide from DNA containing thymine is completely hydrolysed, the products obtained are :

1. 2-deoxy-D(-)ribose.
2. two pyrimidine i.e., guanine (G) and adenine (A).
3. two purines, i.e., thymine (T) and cytosine (C) and
4. phosphoric acid.

Question 8.
When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained ? What does this fact suggest about the structure of RNA ?
Solution:
A DNA molecule has two strands in which the four complementary bases pair each other, viz. cytosine (C) always pairs with guanine (G) while thymine (T) always pairs with adenine (A). Therefore, when a DNA molecule is hydrolysed the molar amount of cytosine is always equal to that of guanine and that of adenine is always equal to that of thymine RNA also contains four bases, the first three are same as in DNA but the fourth one is uracil (U).

As in RNA there is no relationship between the quantities of four bases (C, G, A and U) obtained, therefore, the base-pairing principle, viz., (A) pairs with (U) and (C) pairs with (G) is not followed. So, unlike DNA, RNA has a single strand.

NCERT Exercises

Question 1.
What are monosaccharides?
Solution:
A carbohydrate that cannot be hydrolysed further to give simpler unit of polyhydroxy aldehyde or ketone is called a monosaccharide. With a few exceptions they have general formula, C_,,H₂„O„. About 20 monosaccharides are known to occur in nature. Some common examples are glucose, fructose, ribose, etc.

Question 2.
What are reducing sugars?
Solution:
All those carbohydrates which reduce Fehling’s solution and Tollen’s reagent are referred to as reducing sugars. All monosaccharides whether aldose or ketose are reducing sugars.

Question 3.
Write two main functions of carbohydrates in plants.
Solution:
Two main functions of carbohydrates are

1. Cell wall of bacteria and plants is made up of a polysaccharide, cellulose.
2. Starch is the major food reserve material in plants.

Question 4.
Classify the following into monosaccharides and disaccharides. Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose.
Solution:
Monosaccharides : Ribose, 2-deoxyribose, galactose and fructose
Disaccharides : Maltose and Lactose

Question 5.
What do you understand by the term glycosidic linkage?
Solution:
Disaccharides on hydrolysis with dilute acids or enzymes yield two molecules of either the same or different monosaccharides. The two monosaccharides are joined together by an oxide linkage formed by the loss of a water molecule. Such a linkage between two monosaccharides units through oxygen atom is called glycosidic linkage.

Question 6.
What is glycogen? How is it different from starch ?
Solution:
The carbohydrates are stored in animal body as glycogen. It is also known as animal starch because its structure is similar to amylopectin. It is present in liver, muscles and brain. When the body needs glucose, enzymes break the glycogen down to glucose. Glycogen is also found in yeast and fungi.

Starch is the main storage polysaccharide of plants. It is the most important dietary source for human being. High content of starch is found in cereals, roots, tubers and some vegetables. It is a polymer of two components-amylose (15-20%) which is water soluble and amylopectin(80-85%) which is water insoluble.

Question 7.
What are the hydrolysis products of (i) sucrose and (ii) lactose ?
Solution:

1. Sucrose on hydrolysis gives one unit of glucose and one unit of fructose.
   
2. Lactose on hydrolysis with dilute acids yields an equimolar mixture of D-glucose and D-galactose.

Question 8.
What is the basic structural difference between starch and cellulose ?
Solution:
The basic structural difference between starch and cellulose is of linkage between the glucose units. In starch, there is a-D-glycosidic linkage. Both the components of starch-amylose and amylopectine are polymer of α-D-glucose. On the other hand, cellulose is a linear polymer of β-D-glucose in which C₁ of one glucose unit is connected to C₄ of the other through β-D-glycosidic linkage.

Question 9.
What happens when D-glucose is treated with the following reagents?

1. Hl
2. Bromine water
3. HNO₃

Solution:

Question 10.
Enumerate the reactions of D-glucose which cannot be explained by its open chain structure.
Solution:
The open chain structure of D-glucose OHC – (CHOH)₄ – CH₂OH fails to explain the following reactions :

(i) Though it contains the aldehyde (-CHO) group, glucose does not give

2,4-DNP test, Schiff’s test and it does not form the hydrogen sulphite addition product with NaHSO₃.

(ii) The pentaacetate of glucose does not T react with hydroxylamine (NH₂OH) to form ’ the oxime indicating the absence of free -CHO group.

(iii) The formation of two anomeric methyl glycosides by glucose on reaction with CH₃OH and dry HCl can be explained in terms of the cyclic structure. The equilibrium mixture of a-and (3-glucose react separately with methanol in the presence of dry HCl gas to form the corresponding methyl D-glucosides.

These pentaacetates donot have a free -OH group at C₁ and hence are not hydrolysed in aqueous solution to produce the open chain aldehyde form and hence do not react with NH₂OH to form glucose oxime.

(v) The existence of glucose in two crystalline forms termed as a and β-D-glucose can again be explained on the basis of cyclic structure of glucose and not by its open chain structure. It was proposed that one of the -OH groups may add to – CHO group and form a cyclic hemiacetal structure. It was found that glucose forms a 6-membered ring in which -OH at C – 5 is involved in ring formation. This explains the absence of -CHO group and also existence of glucose in two forms as shown below. These two forms exist in equilibrium with open chain structure.

Question 11.
What are essential and non-essential amino acids? Give two examples of each type.
Solution:
There are about 20 amino acids which make up the bio-proteins. Out of these 10 amino acids (non-essential) are synthesised by our bodies and rest are essential in the diet (essential amino acids) and supplied to our bodies by food which we take because they cannot be synthesised in our body.
e.g. Essential amino acid – Valine and Leucine
Non-essential amino acid – Glycine and Alanine

Question 12.
Define the following as related to proteins

1. Peptide linkage
2. Primary structure
3. Denaturation.

Solution:
(i) Peptide Linkage : Proteins are the polymers of a-amino acids which are connected to each other by peptide bond or peptide linkage. Chemically, peptide linkage is an amide formed between -COOH group and -NH2 group. The reaction between two molecules of similar or different amino acids, proceeds through the combination of the amino group of one molecule with the carboxyl group of the other. This results in the elimination of a water molecule and formation of a peptide bond -CO-NH-. The product of the reaction is called a dipeptide because it is made up of two amino acids. For example, when carboxyl group of glycine combines with the amino group of alanine we get a dipeptide, glycylalanine.

(ii) Primary Structure : Proteins may have one or more polypeptide chains. Each polypeptide in a protein has amino acids linked with each other in a specific sequence and it is this sequence of amino acids that is said to be the primary structure of that protein. Any change in this primary structure i.e., the sequence of amine acids creates a different protein.

(iii) Denaturation : Protein found in a biological system with a unique three-dimensional structure and biological activity is called a native protein. When a protein in its native form, is subjected to physical change like change in temperature or chemical change in pH, the hydrogen bonds are disturbed. Due to this, globules unfold and helix get uncoiled and protein loses its biological activity. This is called denaturation of protein. During denaturation 2° and 3° structures are destroyed but 1° structure remains intact. The coagulation of egg white on boiling is a common example of denaturation. Another example is curdling of milk which is caused due to the formation of lactic acid by the bacteria present in milk.

Question 13.
What are the common types of secondary structure of proteins?
Solution:
The secondary structure of protein refers to the shape in which a long polypeptide chain can exist. They are found to exist in two different types of structures viz, a-helix and P-pleated sheet structure. These structures arise due to the regular folding of the backbone of the polypeptide chain due to hydrogen bonding between

Question 14.
What type of bonding helps in stabilising the α-helix structure of proteins?
Solution:
α-Helix is one of the most common ways in which a polypeptide chain forms all possible hydrogen bonds by twisting into a right handed screw (helix) with the -NH group of each amino acid residue hydrogen bonded to the >C = O of an adjacent turn of the helix.

Question 15.
Differentiate between globular and fibrous proteins.
Solution:
Characteristic differences between globular and fibrous proteins can be given as :

Globular Proteins

1. These are cross linked proteins and are condensation product of acidic and basic amino acids.
2. These are soluble in water, mineral acids and bases.
3. These proteins have three dimensional folded structure. These are stabilised by internal hydrogen bonding, e.g., egg albumin enzymes.

Fibrous Proteins

1. These are linear condensation polymer
2. These are insoluble in water but soluble in strong acids and bases.
3. These are linear polymers held together by intermolecular hydrogen bonds. e.g., hair, silk.

Question 16.
How do you explain the amphoteric behaviour of amino acids?
Solution:
Due to dipolar or Zwitter ionic structure, amino acids are amphoteric in nature. The acidic character of the amino acids is due to the N⁺H₃ group while the basic character is due to the COO^– group.

Question 17.
What are enzymes?
Solution:
Life is possible due to the coordination of various chemical processes in living organisms. An example is the digestion of food, absorption of appropriate molecules and ultimately production of energy. This process involves a sequence of reactions and all these reactions occur in the body under very mild conditions. This occurs with the help of certain biocatalysts called enzymes. Almost all the enzymes are globular proteins. Enzymes are specific for a particular reaction and for a particular substrate. They are generally named after the compound or class of compounds upon which they work. For example, the enzyme that catalyses hydrolysis of maltose into glucose is named as maltose.

Question 18.
What is the effect of denaturation on the structure of proteins?
Solution:
Proteins are very sensitive to the action of heat, mineral acids, alkalies etc. On heating or on treatment with mineral acids, soluble forms of proteins such as globular proteins often undergo coagulation or precipitation to give fibrous proteins which are insoluble in water. This coagulation also results in the loss of the biological activity of the protein. That is why the coagulated proteins so formed are called denatured proteins.. Chemically, denaturation does not change the primary structure but brings about changes in the secondary and tertiary structure of proteins.

Question 19.
How are vitamins classified? Name the vitamin responsible for the coagulation of blood.
Solution:
Vitamins are classified into two groups depending upon their solubility in water or fat.
(i) Fat soluble vitamins : Vitamins which are soluble in fats and oils but insoluble in water are kept in this group. These are vitamins A, D, E and K. They are stored in liver and adipose (fat storing) tissues.
(ii) Water soluble vitamins : B group vitamins and vitamin C are soluble in water so they are grouped together. Water soluble vitamins must be supplied regularly in diet
because they are readily excreted in urine and cannot be stored (except vitamin B₁₂) in our body.
Vitamin K is responsible for coagulation of blood.

Question 20.
Why are vitamin A and vitamin C essential to us? Give their important sources.
Solution:
Deficiency of vitamin A causes Xerophthalmia (hardening of cornea of the eye) and night blindness. So its use is essential to us. It is available in fish liver oil, carrots, butter and milk. It promotes growth and increases resistance to diseases. Vitamin C is very essential to us because its deficiency causes Scurvy (bleeding of gums) and pyorrhea (loosening and bleeding of teeth). Vitamin C increases resistance of the body towards diseases. Maintains healthy skin and helps cuts and abrasions to heat properly. It is soluble in water. It is present in citrus fruits, e.g.,oranges, lemons, amla, tomato. green vegetables (Cabbage) chillies, sprou pulses and germinated grains.

Question 21.
What are nucleic acids? Mention their two important functions.
Solution:
Nucleic acids : They constitute an important class of biomolecules which are found in the nuclei of all living cells in the form of nucleoproteins (i.e., proteins containing nucleic acid as the prosthetic group). Nucleic acids are the genetic materials of the cells and are responsible for transmission of hereditary effect from one generation to the other and also carry out the biosynthesis of proteins. Nucleic acids are biopolymers (i.e., polymers present in the living system). The genetic information coded in nucleic acids controls the structure of all proteins including enzymes and thus governs the entire metabolic activity in the living organism.
Two important functions of nucleic acids are :

1. Replication : The process by which a single DNA molecule produces two identical copies of itself is called replication.
2. Protein Synthesis : DNA may be regarded as the instrument manual for the synthesis of all proteins present in the cell.

Question 22.
What is the difference between a nucleoside and a nucleotide?
Solution:
Nucleoside : A nucleoside contains only two basic components of nucleic acids, i.e., a pentose sugar and a nitrogenous base. It may be represented as Sugar-base. Depending upon the type of sugar present, nucleosides are of two types :

1. Ribonucleosides and
2. Deoxyribonucleosides.

Nucleotides : A nucleotide contains all the three basic components of nucleic acids, i.e., a phosphoric acid group, a pentose sugar and a nitrogenous base. In other words, nucleotides are nucleoside monophosphates.

Depending upon the type of sugar present, nucleotides like nucleosides are of two types :

1. Ribonucleotides and
2. Deoxyribonucleotides.

Question 23.
The two strands in DNA are not identical but are complementary. Explain.
Solution:
Two nucleic acid chains are wound about each other and held together by hydrogen bonds between pairs of bases. The trands are complementary to each other because the hydrogen bonds are formed between specific pairs of bases. Adenine forms hydrogen bonds with thymine whereas cytosine forms hydrogen bonds with guanine.

Question 24.
Write the important structural and functional differences between DNA and RNA.
Solution:
Difference between DNA and RNA.

Question 25.
What are the different types of RNA found in the cell?
Solution:
RNA molecules are of three types and they perform different functions. They are named as messenger RNA (m-RNA), ribosomal RNA (r-RNA) and transfer RNA (t-RNA).

INTEXT Questions

Question 1.
Classify the following amines as primary, secondary and tertiary :

Amines class 12 NCERT Solutions:

(a) Primary
(b) Tertiary
(c) Primary
(d) Secondary

Question 2.
(a) Write structures of different isomeric amines corresponding to the molecular formula, C₄H₁₁N.
(b) Write IUPAC names of all the isomers.
(c) What type of isomerism is exhibited by different pairs of amines?
Amines class 12 NCERT Solutions:

Question 3.
How will you convert: (i) Benzene into aniline, (ii) Benzene into N,N-dimethylaniline, (iii) Cl(CH₂)₄Cl into hexan-1,6-diamine?
Solution:

Question 4.
Arrange the following in increasing order of their basic strength.

(a) C₂H₅NH₂, C₆H₅NH₂, NH₃, C₆H₅CH₂NH₂ and (C₂H₅)₂NH
(b) C₂H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N, C₆H₅NH₂
(c) CH₃NH₂< (CH₃)₂NH, (CH₃)₃N, C₆H₅NH₂> C₆H₅CH₂NH₂.

Amines class 12 NCERT Solutions:
(a) The order of basicity is C₆H₅NH₂ < NH₃ < C₆H₅CH₂NH₂ < C₂H₅NH₂ < (C₂H₅)₂NH
The basic nature of amine arises from their ability to donate the lone pair of electrons on N to electrophiles. The availability of this l.p. depends on two factors :
(i) Electron donating/withdrawing effect of the alkyl groups attached to the N atom.
(ii) Steric hindrance posed by alkyl groups around N.
The presence of alkyl groups on the N atom increases the electron density and makes the l.p. more available for electrophiles. This happens due to the +1 effect of the alkyl groups. But, if the alkyl groups are too bulky or too many in number, they tend to sterically hinder the incoming proton and the basic strength decreases. These two factors working in opposing directions, tend to balance out each other in 2° amines, making them most basic and the basic strength follows the order :

Question 5.
Complete the following acid-base reactions and name the products :

1. CH₃CH₂CH₂NH₂ + HCl →
2. (C₂H₅)₃N + HCl →

Solution:

Question 6.
Write the reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution.
Solution:

Question 7.
Write the chemical reaction of aniline with benzoyl chloride and write the name of the product obtained.
Solution:

Question 8.
Write structures of different isomers corresponding to the molecular formula, C₃H₉N. Write IUPAC name of the isomers which will liberate nitrogen gas on treatment with nitrous acid.
Solution:

Question 9.
Convert

1. 3-Methylaniline into 3-nitrotoluene
2. Aniline into 1,3,5-tribromobenzene.

Solution:

Amines class 12 NCERT Solutions – NCERT Exercises

Question 1.
Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

1. (CH₃)₂CHNH₂
2. CH₃(CH₂)₂NH₂
3. CH₃NHCH(CH₃)₂
4. (CH₃)₃CNH₂
5. C₆H₅NHCH₃
6. (CH₃CH₂)₂NCH₃
7. m-BrC₆H₄NH₂

Solution:

1. Propan-2-amine (primary),
2. Propan-l-amine (primary),
3. N-Methylpropan-2-amine (secondary),
4. 2-Methylpropan-2-amine (primary),
5. N-Methylbenzenamine or N-Methylaniline (secondary),
6. N-Ethyl-N-methylethanamine (tertiary),
7. 3-Bromobenzenamine or 3-Bromoaniline (primary).

Question 2.
Give one chemical test to distinguish between the following pairs of compounds.

1. Methylamine and dimethylamine
2. Secondary and tertiary amine
3. Ethylamine and W-methylaniline
4. Aniline and benzylamine
5. Aniline and N-methylaniline

Solution:

Question 3.
Account for the following :
(i) pK_b of aniline is more than that of methylamine.
(ii) Ethylamine is soluble in water whereas aniline is not.
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
(iv) Although amino group is o- and p-directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.
(v) Aniline does not undergo Friedel-Crafts reaction.
(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
(vii) Gabriel phthalimide synthesis is preferred .for synthesising primary amines.
Solution:
(i) If the pK_b value of any base or compound is higher than that of another, it implies that the former is a weaker base than the latter. In aniline, the N-atom is attached to the benzene ring and therefore the lone pair on N is delocalised over the entire benzene ring. As a result, it cannot accept a proton or any other electrophile.

This is why it has a lower K_b value (lower basic strength) and high corresponding pK_b value.

In methylamine, CH₃NH₂, the electron density on nitrogen is greater than that in case of aniline. This is because -CH₃ group in methylamine, by virtue of its +1 effect, increases electron density on N, which is more available for protonation.

(ii) Any compound capable of forming hydrogen bonds with water, dissolves in it. Ethylamine is able to do the same and hence its solubility.

However, in aniline, the bulky hydrocarbon part – C₆H₅ prevents the formation of effective hydrogen bonding and therefore it is not soluble.

(iii) The formation of hydrated ferric oxide may be understood by taking into consideration the basic strength of CH₃NH₂. In presence of CH₃NH₂, water hydrolyses as

(iv) During nitration, the nitration mixture used (cone. HNO₃ and cone. H₂SO₄) protonates the NH₂ group to produce anilinium ion as

But when this reaction is carried out with aniline, no electrophile generation takes place. The reason being the presence of aniline as a base.

Aniline is a Lewis base, reacts with AlCl₃ and hence deactivates it. The Lewis acid is therefore no more available for electrophile generation and hence reaction does not take place.

(vi) Diazonium salts carry a N atom with a positive charge. This positive charge is well dispersed in aromatic diazonium salts through resonance as shown below :

Such a charge delocalisation is not possible in aliphatic amines and hence they are less stable.

(vii) Gabriel phthalimide reaction gives pure primary amines without any contamination of secondary and tertiary amines. Therefore, it is preferred for synthesising primary amines.

Question 4.
Arrange the following :
(i) In decreasing order of the pK_b values: C₂H₅NH₂, C₆H₅NHCH₃, (C₂H₅)₂NH and C₆H₅NH₂
(ii) In increasing order of basic strength: C₆H₅NH₂, C₆H₅N(CH₃)₂, (C₂H₅)₂NH and CH₃NH₂
(iii) In increasing order of basic strength:
(a) Aniline, p-nitroaniline and p-toluidine
(b) C₆H₅NH₂, C₆H₅NHCH₃, C₆H₅CH₂NH₂.
(iv) In decreasing order of basic strength in gas phase : C₂H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N and NH₃
(v) In increasing order of boiling point: C₂H₅OH, (CH₃)₂NH, C₂H₅NH₂
(vi) In increasing order of solubility in water: C₆H₅NH₂, (C₂H₅)₂NH, C₂H₅NH₂.
Solution:


The availability of l.p. on N of p-nitroaniline is drastically reduced by presence of electron withdrawing -NO₂ group on it.

In contrast, presence of electron releasing -CH₃ group increases the electron density on N atom and improves basicity in p-toluidine.

(b) C₆H₅NH₂ < C₆H₅NHCH₃ < C₆H₅CH₂NH₂ Involvement of l.p. of N in resonance causes aniline to have low basicity. In II, the -Me group through its +I effect improves the electron density on N and therefore its basic strength increases. In III, the -NH₂ is farther off from benzene ring and hence l.p. is localized on it and hence the basic strength is highest.

(iv) In gas phase, basicity follows the order : (C₂H₅)₃N > (C₂H₅)₂NH > C₂H₅NH₂ > NH₃ In gas phase, the stabilization by solvation is not present and hence basic strength follows the expected order based on +I effect of alkyl groups.

(v) (CH₃)₂NH < C₂H₅NH₂ < C₂H₅OH

(vi) C₆H₅NH₂ < (C₂H₅)₂NH < C₂H₅NH₂

Amines can form hydrogen bonds with water and are therefore soluble in it. However, the solubility decreases if the mass of the hydrocarbon part increases.

Question 5.
How will you convert :

1. Ethanoic acid into methanamine
2. Hexanenitrile into 1-aminopentane
3. Methanol to ethanoic acid
4. Ethanamine into methanamine
5. Ethanoic acid into porpanoic acid
6. Methanamine into ethanamine
7. Nitromethane into dimethylamine
8. Propanoic acid into ethanoic acid?

Solution:

Question 6.
Describe a method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.
Solution:
1°,2° and 3° amines can be distinguished by Hinsberg’s reagent.

Question 7.
Write short notes on the following :

1. Carbylamine reaction
2. Diazotisation
3. Hofmann’s bromamide reaction
4. Coupling reaction
5. Ammonolysis
6. Gabriel phthalimide synthesis
7. Acetylation

Solution:
(i) Carbylamine reaction : Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form isocyanides or carbylamines which are foul smelling substances. Secondary and tertiary amines do not show this reaction. This reaction is known as carbylamine reaction or isocyanide test and is used as a test for primary amines.

(ii) Diazotisation : The conversion of primary aromatic amines into diazonium salts is known as diazotisation.

The conversion is brought about by reacting the amine with HNO₂ which is prepared in situ.

(iii) Hofmann’s bromamide reaction : Primary amides when heated with Br₂ and (aqueous or ethanoic solution of) NaOH lose a carbon atom and are converted to the corresponding amines. It is an example of step-down reaction.

(iv) Coupling reaction : The reaction of diazonium salts with phenols and aromatic amines to form azo compounds having an extended conjugate system with both aromatic rings joined through the — N = N — bond, is called coupling reaction. In this reaction, the nitrogen atoms of the diazo group are retained in the product. The coupling with phenols takes place in mildly alkaline medium while that with amines occurs under faintly acidic conditions. For example,


Coupling generally occurs at the p-position with respect to the hydroxyl or the amino group, if free, otherwise it takes place at the o-position.

(v) Ammonolysis : The process of cleavage of the C — X bond in alkyl halides by ammonia molecule is called ammonolysis. 1° amine thus obtained behaves as a nucleophile and further reacts with alkyl halide to form 2°, 3° and finally quaternary ammonium salt.

(vi) Gabriel phthalimide synthesis : In this reaction phthalimide is converted into its potassium salt by treating it with alcoholic potassium hydroxide. Then potassium phthalimide is heated with an alkyl halide to yield an N-alkylpthalimide which is hydrolysed to phthalic acid and primary amine by alkaline hydrolysis.


This synthesis is very useful for the preparation of pure aralkyl and aliphatic primary amines. However, aromatic primary amines cannot be prepared by this method.

Question 8.
Accomplish the following conversions :

1. Nitrobenzene to benzoic acid
2. Benzene to m-bromophenol
3. Benzoic acid to aniline
4. Aniline to 2,4,6-tribromofluorobenzene
5. Benzyl chloride to 2-phenylethanamine
6. Chlorobenzene to p-chloroaniline
7. Aniline to p-bromoaniline
8. Benzamide to toluene
9. Aniline to benzyl alcohol.

Solution:

Question 9.
Give the structures of A, B and C in the following reactions :


Solution:

Question 10.
An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’which on heating with Br₂ and KOH forms a compound ‘C’ of molecular formula C₆H₇N. Write the structures and IUPAC names of compounds A, B and C.
Solution:

Question 11.
Complete the following reactions :

Solution:

Question 12.
Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?
Solution:
(i) Gabriel phthalimide reaction involves the nucleophilic attack of the phthalimide on the alkyl halide.

(ii) Such a nucleophilic substitution reaction is not possible if the substrate is an aryl halide.
(iii) The reason for it can be explained on the basis of
(a) Partial double bond character of C — X bond in aryl halide. Consider the following structures :

From the resonance structures we see that the C — X bond has a double bond character in structures II, III and IV and this makes cleavage of C — X bond difficult.

(b) Also the steric hindrance by the bulky aryl group prevents the incoming nucleophile.

Question 13.
Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.
Solution:
Amines do not directly react with nitrous acid, rather they react with a mixture of dil. HCl and NaNO₂ and HNO₂ is produced in situ.
The reactions are :

Question 14.
Give plausible explanation for each of the following :
(i) Why are amines less acidic than alcohols of comparable molecular masses?
(ii) Why do primary amines have higher boiling point than tertiary amines?
(iii) Why are aliphatic amines stronger bases than aromatic amines?
Solution:
(i) Loss of a proton from an amine gives RNH^– ion while loss of a proton from alcohol gives RO- ion as shown below :

As O is more electronegative than N,RO^– can accommodate the negative charge more easily than the RNH^– can.

As, RO^– is more stable than RNH^– the former is formed more. As a result, amines are less acidic than alcohols.

(ii) (a) At boiling point, the molecules in a compound break free from their inter molecular forces and escape into the vapour phase.

Weaker the inter-molecular forces, lower will be the boiling point.

(b) In 1 ° amines, there is strong H-bonding that binds the amine molecules together. Whereas in 3° amine absence of H on N atom prevents hydrogen bonding completely.

This is why 1° amines have higher boiling point.

(iii) (a) The basic nature of amines is a result of the presence of l.p. of electron on the N atom. Also the electron density is increased on N due to the +I effect of alkyl group.
(b) In aryl amines the l.p. on N is involved in resonance with the benzene ring and hence less available for protonation.

(c) In aliphatic amines there is no such delocalisation and hence it is more basic.

INTEXT Questions

Question 1.
Write the structures of the following compounds.

1. α-Methoxypropionaldehyde
2. 3-Hydroxybutanal
3. 2-Hydroxycyclopentanecarbaldehyde
4. 4-Oxopentanal
5. Di-sec.butyl ketone
6. 4-Fluoroacetophenone

Solution:

Question 2.
Write the structures of the products of the following reactions :

Solution:

Question 3.
Arrange the following compounds in increasing order of their boiling points.
CH₃CHO, CH₃CH₂OH, CH₃OCH₃, CH₃CH₂CH₃
Solution:
CH₃CH₂CH₃ < CH₃OCH₃ < CH₃CHO < CH₃CH₂OH

Question 4.
Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions.

1. Ethanal, Propanal, Propanone, Butanone.
2. Benzaldehyde, p-Tolualdehyde,

p – Nitrobenzaldehyde, Acetophenone. Hint: Consider steric effect and electronic effect.
Solution:

1. Butanone < Propanone < Propanal < Ethanal
2. Acetophenone < p-Tolualdehyde, Benzaldehyde < p-Nitrobenzaldehyde

Question 5.
Predict the products of the following reactions :

Solution:

Question 6.
Give the IUPAC names of the following compounds :

Solution:

1. 3-Phenylpropanoic acid
2. 3-Methylbut-2-enoic acid
3. 2-Methylcyclopentanecarboxylic acid
4. 2,4,6-Trinitrobenzoic acid

Question 7.
Show how each of the following compounds can be converted to benzoic acid.

1. Ethylbenzene
2. Acetophenone
3. Bromobenzene
4. Phenylethene (Styrene)

Solution:

Question 8.
Which acid of each pair shown here would you expect to be stronger ?
(i) CH₃CO₂H or CH₂FCO₂H
(ii) CH₂FCO₂H or CH₂ClCO₂H
(iii) CH₂FCH₂CH₂CO₂H or CH₃CHFCH₂CO₂H

Solution:
(i) H₂CFCOOH will be stronger of the two. The presence of electronegative F atom at the α-C causes electron withdrawal from the COOH and facilitates the release of H+.
(ii) CH₂FCO₂H is a stronger acid for the same reason as stated above. F is more electronegative than Cl, so it withdraws electrons from the carboxyl group to a greater extent.
(iii) CH₃CHFCH₂COOH is stronger. Although both the givenacidshaveFatomin them, it is the proximity of F in CH₃CHFCH₂COOH to the COOH group which makes it more acidic.

NCERT Exercises

Question 1.
What is meant by the following terms? Give an example of the reaction in each case.

1. Cyanohydrin
2. Acetal
3. Semicarbazone
4. Aldol
5. Hemiacetal
6. Oxime
7. Ketal
8. Imine
9. 2,4-DNP derivative
10. Schiff’s base

Solution:

Question 2.
Name the following compounds according to IUPAC system of nomenclature :

1. CH₃CH(CH₃)CH₂CH2CHO
2. CH₃CH₂COCH(C₂H₅)CH₂CH₂Cl
3. CH₃CH=CHCHO
4. CH₃COCH₂COCH₃
5. CH₃CH(CH₃)CH₂C(CH₃)₂COCH₃
6. (CH₃)₃CCH₂COOH
7. OHCC₆H₄CHO-p

Solution:

1. 4-Methylpentanal
2. 6-Chloro-4-ethylhexan-3-one
3. But-2-en-1-al
4. Pentane-2, 4-dione
5. 3, 3, 5-Trimethylhexan-2-one
6. 3, 3-Dimethylbutanoic acid
7. Benzene-1, 4-dicarbaldehyde

Question 3.
Draw the structures of the following compounds.

1. 3-Methylbutanal
2. p-Nitropropiophenone
3. p-Methylbenzaldehyde
4. 4-Methylpent-3-en-2-one
5. 4-Chloropentan-2-one
6. 3-Bromo-4-phenylpentanoic acid
7. p, p’-Dihydroxybenzophenone
8. Hex-2-en-4-ynoic acid

Solution:

Question 4.
Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names.

Solution:

Question 5.
Draw structures of the following derivatives.

1. The 2,4-dinitrophenylhydrazone of benz- aldehyde
2. Cyclopropanone oxime
3. Acetaldehydedimethylacetal
4. The semicarbazone of cyclobutanone
5. The ethylene ketal of hexan-3-one
6. The methyl hemiacetal of formaldehyde

Solution:

Question 6.
Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents.

1. PhMgBr and then H₃O⁺
2. Tollens’reagent
3. Semicarbazide and weak acid
4. Excess ethanol and acid
5. Zinc amalgam and dilute hydrochloric acid

Solution:

Question 7.
Which of the following compounds would undergo aldol condensation, which the Cannizzaro reaction and which neither? Write the structures of the expected products of aldol condensation and Cannizzaro reaction.

1. Methanal
2. 2-Methylpentanal
3. Benzaldehyde
4. Benzophenone
5. Cyclohexanone
6. 1-Phenylpropanone
7. Phenylacetaldehyde
8. Butan-1-ol
9. 2,2-Dimethylbutanal

Solution:
Aldol condensation is shown by those aldehydes or ketones which have at least one α-H atom while Cannizzaro reaction is undergone by aldehydes that have no α-H atom.

Question 8.
How will you convert ethanal into the following compounds?

1. Butane-1,3-diol
2. But-2-enal
3. But-2-enoic acid

Solution:

Question 9.
Write structural formulas and names of four possible aldol condensation products from propanal and butanal. In each case, indicate which aldehyde acts as nucleophile and which as an electrophile
Solution:
The possible products of aldol condensation from propanal and butanal are

Question 10.
An organic compound with the molecular formula C₉H₁₀O forms 2,4-DNP derivative, reduces Tollens’ reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1, 2-benzenedicarboxylic acid. Identify the compound.
Solution:
(i) It is given that the compound reduces Tollens’ reagent. This proves that the compound is an aldehyde. Further, the fact that it undergoes Cannizzaro reaction shows that it lacks an α-H atom.

(ii) On oxidation it yields 1,2-benzenedicarboxylic acid. This shows that it is an o-substituted benzaldehyde. The only possible structure for the compound is :

Question 11.
An organic compound (A) (molecular formula C₈H₁₆O₂) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid produced (8). (C) on dehydration gives but-1-ene. Write equations for the reactions involved.
Solution:

Question 12.
Arrange thefollowing compounds in increasing order of their property as indicated :

1. Acetaldehyde, Acetone, Di-tert-butyl ketone, Methyl fert-butyl ketone (reactivity towards HCN)
2. CH₃CH₂CH(Br)COOH,CH₃CH(Br)CH₂COOH, (CH₃)₂CHCOOH, CH₃CH₂CH₂COOH (acid strength)
3. Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength)

Solution:

Question 13.
Give simple chemical tests to distinguish between the following pairs of compounds.

1. Propanal and Propanone
2. Acetophenone and Benzophenone
3. Phenol and Benzoic acid
4. Benzoic acid and Ethyl benzoate
5. Pentan-2-one and Pentan-3-one
6. Benzaldehyde and Acetophenone
7. Ethanal and Propanal

Solution:
The given set of compounds may be distinguish by the following reaction.

Question 14.
How will you prepare the following compounds from benzene? You may use any inorganic reagent and any organic reagent having not more than one carbon atom

1. Methyl benzoate
2. m-Nitrobenzoic acid
3. p-Nitrobenzoic acid
4. Phenylacetic acid
5. p-Nitrobenzaldehyde

Solution:

Question 15.
How will you bring about the following conversions in not more than two steps?

1. Propanone to Propene
2. Benzoic acid to Benzaldehyde
3. Ethanol to 3-Hydroxybutanal
4. Benzene to m-Nitroacetophenone
5. Benzaldehyde to Benzophenone
6. Bromobenzene to 1 -Phenylethanol
7. Benzaldehyde to 3-Phenyipropan-1 -ol
8. Benzaldehyde to a-Hydroxyphenylacetic acid
9. Benzoic acid to m-Nitrobenzyl alcohol

Solution:

Question 16.
Describe the following :

1. Acetylation
2. Cannizzaro reaction
3. Cross-aldol condensation
4. Decarboxylation

Solution:

Question 17.
Complete each synthesis by giving missing starting material, reagents or products


Solution:

Question 18.
Give plausible explanation for each of the following :

1. Cyclohexanone forms cyanohydrin in good yield but 2, 2,6 trimethylcyclohexa- none does not.
2. There are two – NH₂ groups in semicarbazide. However, only one is involved in the formation of semicarbazones.
3. During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, the water or the ester should be removed as soon as it is formed.

Solution:

Question 19.
An organic compound contains 69.77% carbon, 11.63% hydrogen and rest, oxygen. The molecular mass of the compound is 86. It does not reduce Tollens’ reagent but forms an addition compound with sodium hydrogensulphite and give positive iodoform test. On vigorous oxidation it gives ethanoic and propanoic acid. Write the possible structure of the compound.
Solution:
(a) The given compound does not reduce Tollens’ reagent which implies that it is not an aldehyde.

Question 20.
Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why?
Solution:
(i) Phenoxide ion has the following resonating structures :



(iii) The negative charge that rests on the electronegative O atom in carboxylate ion. We know that the presence of negative charge on an electronegative atom makes the ion more stable. For the same reason RCOO^– is more stable than the phenoxide ion where the oxygen has no negative charge on it. For the above two reasons carboxylate ion is more stable and has higher acidity than phenol.

INTEXT Questions

Question 1.
Classify the following as primary, secondary and tertiary alcohols :

Solution:
Primary alcohols : (i), (ii) and (iii)
Secondary alcohols : (iv) and (v)
Tertiary alcohols : (vi)

Question 2.
Identify allylic alcohols in the above examples.
Solution:
Allvlir alcohols : (ii) and (vi).

Question 3.
Name the following compounds according to IUPAC system.


Solution:

1. 3-Chloromethyl-2-isopropylpentan-l-ol
2. 2, 5-Dimethylhexane-1,3-diol
3. 3-Bromocyclohexanol
4. Hex-l-en-3-ol
5. 2-Bromo-3-methylbut-2-en-1 -ol

Question 4.
Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent on methanal.

Solution:

Question 5.
Write structures of the products of the following reactions :

Solution:

Question 6.
Give structures of the products you would expect when each of the following alcohol reacts with (a) HCl – ZnCl₂ (b) HBr and (c) SOCl₂.

1.  Butan-1-ol
2. 2-Methylbutan-2-ol

Solution:

Question 7.
Predict the major product of acid catalysed dehydration of

1. 1-methylcyclohexanol and
2. butan-1-ol

Solution:

Question 8.
Ortho and para nitrophenols are more acidic than phenol. Draw the resonance structures of the corresponding phenoxide ions.
Solution:
The attachment of the -NO₂ group to the phenol molecule at 0- and p-positions decreases the electron density on oxygen atom. This causes the oxygen atom to pull the bond pair of electrons of the O – H bond towards itself thereby facilitating the release of H as H⁺.

The resonance structures of the phenoxide ions are :

Question 9.
Write the equations involved in the following reactions :

1. Reimer-Tiemann reaction
2. Kolbe’s reaction

Solution:

Question 10.
Write the reactions of Williamson synthesis of 2-ethoxy-3-methylpentane starting from ethanol and 3-methylpentan-2-ol.
Solution:

Question 11.
Which of the following is an appropriate set of reactants for the preparation of 1 -methoxy-4- nitrobenzene and why?

Solution:

Question 12.
Predict the products of the following reactions :

Solution:

NCERT Exercises

Question 1.
Write IUPAC names of the following compounds :


Solution:

1. 2,2,4-Trimethylpentan-3-ol
2. 5-Ethylheptan-2,4-diol
3. Butane-2,3-diol
4. Propane-1,2,3-triol
5. 2-Methylphenol
6. 4-Methylphenol
7. 2,5-Dimethylphenol
8. 2,6-Dimethylphenol
9. l-Methoxy-2-methylpropane
10. Ethoxybenzene
11. 1-Phenoxyheptane
12. 2-Ethoxybutane

Question 2.
Write structures of the compounds whose IUPAC names are as follows :

1. 2-Methylbutan-2-ol
2. 1-Phenylpropan-2-ol
3. 3,5-Dimethylhexane-1,3,5-triol
4. 2,3-Diethylphenol
5. 1-Ethoxypropane
6. 2-Ethoxy-3-methylpentane
7. Cyclohexylmethanol
8. 3-Cydohexylpentan-3-ol
9. Cyclopent-3-en-1 -ol
10. 3-Chloromethylpentan-1-ol

Solution:

Question 3.

1. Draw the structures of all isomeric alcohols of molecular formula C₅H₁₂O and give their IUPAC names.
2. Classify the isomers of alcohols in question 3 (i) as primary, secondary and tertiary alcohols.

Solution:
The isomeric alcohols with molecular formula C₅H₁₂O are :

Question 4.
Explain why propanol has higher boiling point than that of the hydrocarbon, butane.
Solution:
The boiling point of any compound depends on the strength of inter-molecular forces. Stronger is the inter-molecular attraction, higher is the boiling point.

In butane, the molecules interact with each other through weak van der Waals forces. These weak forces can be easily overcome by supplying small amount of heat energy. Thus, they have low boiling point.

In propanol, the molecules are held together by strong hydrogen bonding. These attractive forces operating between molecules are more difficult to break and therefore higher amount of heat needs to be supplied, therefore, the higher boiling point.

Question 5.
Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.
Solution:
Organic compounds are soluble in water if they are able to form hydrogen bonds with it. Alcohols are able to establish this interaction by the virtue of their OH group and are therefore soluble in water. On the other hand, other hydrocarbons of comparable mass do not dissolve in water since they cannot form hydrogen bonds.

Question 6.
What is meant by hydroboration-oxidation reaction? Illustrate it with an example.
Solution:
Hydroboration-oxidation is a method of preparation of alcohols from alkenes. The main advantage of this method is the high yield of alcohol obtained. During hydroboration, diborane (BH₃)₂ is made to react with an alkene to form an addition product. This product is then treated with hydrogen peroxide in the presence of sodium hydroxide to give alcohol.
For example,

Question 7.
Give the structure and IUPAC names of monohydric phenols of molecular formula, C₇H₈O.
Solution:

Question 8.
While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason.
Solution:
During steam distillation, it is the lower boiling compound which distills out first. Between ortho- and para-nitrophenol it is the ortho-isomer which will be steam volatile since it has a lower boiling point.

The difference in boiling point between the two isomers can be understood based on the structural difference. In ortho-isomer intra¬molecular hydrogen bonding takes place while in the pttra-isomer, inter-molecular hydrogen bonding takes place.

As a result of the strong forces operating between the molecules of p-isomer, the boiling point is higher and it is not steam volatile.

Question 9.
Give the equations of reactions for the preparation of phenol from cumene.
Solution:

Question 10.
Write chemical reaction for the preparation of phenol from chlorobenzene.
Solution:

Question 11.
Write the mechanism of hydration of ethene to yield ethanol.
Solution:
The acid catalysed hydration of ethene may be represented as :

Question 12.
You are given benzene, cone. H₂SO₄ and NaOH. Write the equations for the preparation of phenol using these reagents.
Solution:
Using the given reagents, phenol may be prepared as :

Question 13.
Show how will you synthesise :

1. 1 -Phenylethanol from a suitable alkene.
2. Cyclohexylmethanol using an alkyl halide by an S_N² reaction.
3. Pentan-1-ol using a suitable alkyl halide.

Solution:

Question 14.
Give two reactions that show the acidic nature of phenol. Compare acidity of phenol with that of ethanol.
Solution:


The resonance stabilization provided by the contributing structures (I)-(V) more than compensates for the bond breakage energy of O – H bond and thus causes phenol to be acidic in nature.

No such resonance structures are possible for ethoxide ion and therefore the conversion of ethanol to ethoxide is not favoured under normal conditions. Therefore, ethanol is less acidic than phenol.

Question 15.
Explain why is ortho nitrophenol more acidic than methoxyphenol?
Solution:

Question 16.
Explain how does the -OH group attached to a carbon of benzene ring activates it towards electrophilic substitution.
Solution:
In an electrophilic substitution reaction, an electron deficient species attacks the benzene ring which is e^– rich.

When an -OH group is attached to the benzene ring, by the virtue of its electron releasing nature increases the e^– density of the ring and thus activates it, i.e., makes it a welcome site for electrophiles.

The increase in electron density can be visualised as :

From structures (I) – (V), we find that the attachment of hydroxyl group to benzene has increased the electron density (-ve charge) on the ring carbon atoms (especially C-2, C-4 and C-6). It is therefore said to have activated the ring towards electrophiles which are attracted to the increased electron density.

Question 17.
Give equations of the following reactions :

1. Oxidation of propan-1 -ol with alkaline KMnO₄ solution.
2. Bromine in CS₂ with phenol.
3. Dilute HNO₃ with phenol.
4. Treating phenol with chloroform in presence of aqueous NaOH.

Solution:

Question 18.
Explain the following with an example.
(i) Kolbe’s reaction
(ii) Reimer-Tiemann reaction
(iii) Williamson ether synthesis
(iv) Unsymmetrical ether.
Solution:
(i) Kolbe’s reaction :
The fact that phenoxide ion is even more reactive than phenol towards incoming electrophiles is made use of in this reaction. Sodium phenoxide is reacted with CO₂ followed by acid treatment to yield o-hydroxybenzoic acid as the major product.

(ii) Reimer-Tiemann reaction :
Treatment of phenol with chloroform in the presence of aqueous alkali introduces a CHO group at the ortho position. Acidification yields salicylaldehyde.

(iii) Williamson synthesis :
In this method, an alkyl halides is reacted with sodium alkoxide.
R-X + R’ – ONa → R-O-R’ + NaX
The reaction involves SN2 attack of an alkoxide ion on 1° RX.

(iv) Unsymmetrical ethers :
Unsymmetrical ethers are organic compounds where the ethereal oxygen atom is attached to two different alkyl or aryl groups, e.g.,
C₂H₅ – O – CH₃, C₆H₅O – C₂H₅, etc.

Question 19.
Write the mechanism of acid dehydration of ethanol to yield ethene.
Solution:

Question 20.
How are the following conversions carried out?

1. Propene → Propan-2-ol
2. Benzyl chloride → Benzyl alcohol
3. Ethyl magnesium chloride → Propan-1-ol
4. Methyl magnesium bromide → 2-Methylpropan-2-ol.

Solution:

Question 21.
Name the reagents used in the following reactions :

1. Oxidation of primary alcohol to carboxylic acid.
2. Oxidation of primary alcohol to aldehyde.
3. Bromination of phenol to 2,4,6-tribromophenol.
4. Benzyl alcohol to benzoic acid.
5. Dehydration of propan-2-ol to propene.
6. Butan-2-one to butan-2-ol.

Solution:

1. Alkaline KMnO₄
2. Pyridinium chlorochromate in chloromethane (CH₂Cl₂)
3. Br₂/H₂O
4. Alkaline KMnO₄
5. Cone. H₂SO₄ or H₃PO₄ at 433-443 K
6. H₂/Ni or NaBH₄ or LiAlH₄

Question 22.
Give reason for the higher boiling point of ethanol in comparison to methoxymethane.
Solution:
The higher boiling point of ethanol may be attributed to the presence of intermolecular hydrogen bonding in it.

Due to such extensive bonding, more energy needs to be supplied to ethanol to break these bonds and move it into the vapour phase.

Question 23.
Give IUPAC names of the following ethers :

Solution:

1. 1 -Ethoxy-2-methylpropane
2. 2-Chloro-l-methoxyethane
3. 4-Nitroanisole
4. 1-Methoxypropane
5. 1-Ethoxy-4,4-dimethylcyclohexane
6. Ethoxybenzene

Question 24.
Write the names of reagents and equations for the preparation of the following ethers by Williamson’s synthesis :

1. 1-Propoxypropane
2. Ethoxybenzene
3. 2-Methyl-2-methoxypropane
4. 1-Methoxyethane

Solution:

Question 25.
Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of ethers.
Solution:
The main limitation of Williamson’s ether synthesis lies in its unemployability for preparation of unsymmetrical ethers where the compound contains secondary or tertiary alkyl groups.

e.g., reaction between tert-butyl bromide and sodium methoxide yields an alkene.

This is because the competing elimination reaction predominates over S_N2 and alkene is formed.

Question 26.
How is 1-propoxypropane synthesised from propan-1 -ol? Write mechanism of this reaction.
Solution:

Question 27.
Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.
Solution:
Consider the reaction between propan- 2-ol molecules in the presence of acid.

If an ether is to be formed, another alcohol molecule must carry out a nucleophilic attack on the carbocation as

However, this does not happen because of
(a) the steric hindrance around the carbocation, and
(b) bulky size of the nucleophile which would further cause crowding.
As a result, the carbocation prefers to lose a proton and forms an alkene.

For the same reason 3° alcohols in the presence of acid do not form ethers since 3° alcohols are even more sterically hindered than 2° alcohols.

Question 28.
Write the equation of the reaction of hydrogen iodide with :

1. 1-propoxypropane
2. methoxybenzene; and
3. benzyl ethyl ether.

Solution:

Question 29.
Explain the fact that in aryl alkyl ethers (i) the alkoxy group activates the benzene ring towards electrophilic substitution and (ii) it directs the incoming substituents to ortho and para positions in benzene ring.
Solution:
Consider the following resonance structures of aryl alkyl ethers :

(i) From the above structures we find that the presence of the OR group has increased the electron density on the benzene ring and therefore the ring is said to have been activated towards incoming electrophiles.

(ii) From structures (II), (III) and (IV) we find that and the electron density has increased on C-2, C-4 and C-6, i.e., at the ortho and para positions. As a result the electrophile (E ) attaches itself to these e- rich sites and the -OR group is said to have directed the E® to ortho and para positions.

Question 30.
Write the mechanism of the reaction of Hl with methoxymethane.
Solution:
The reaction between methoxymethane and Hl is :

Question 31.
Write equations of the following reactions :

1. Friedel-Crafts reaction – alkylation of anisole.
2. Nitration of anisole.
3. Bromination of anisole in ethanoic acid medium.
4. Friedel-Craft’s acetylation of anisole.

Solution:

Question 32.
Show how would you synthesise the following alcohols from appropriate alkenes.

Solution:

Question 33.
When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place :

Give a mechanism for this reaction.
[Hint : The secondary carbocation formed in step II rearranges to a more stable tertiary carbocation by a hydride ion shift from 3rd carbon atom.]
Solution:

 INTEXT Questions

Question 1.
Write the structures of the following compounds :

(i) 2-Chloro-3-methylpentane
(ii) 1-Chloro-4-ethylcydohexane
(iii) 4-terf-Butyl-3-iodoheptane
(iv) 1, 4-Dibromobut-2-ene
(v) 1 -Bromo-4-sec-butyl-2-methylbenzene

Solution:

Question 2.
Why is sulphuric acid not used during the reaction of alcohols with Kl ?
Solution:
H₂SO₄ is a strong oxidising agent. Therefore, when it is used in presence of KI, it tends to convert KI to HI and finally oxidises it to I₂.

Question 3.
Write structures of different dihalogen derivatives of propane.
Solution:
The structures of all possible dihalogen derivatives of propane are

 Question 4.
Among the isomeric alkanes of molecular formula C₅H₁₂, identify the one that on photochemical chlorination yields
(i) A single monochloride
(ii) Three isomeric monochlorides
(iii) Four isomeric monochlorides.
Solution:
An alkane with molecular formula C₅H₁₂ can exist in the following isomeric forms :

Question 5.
Draw the structures of major monohalo products in each of the following reactions :


Solution:
The major haloderivatives formed in the given reactions are

Question 6.
Arrange each set of compounds in order of increasing boiling points.
(i) Bromomethane, bromoform, chloromethane, dibromomethane.
(ii) 1 – Chloropropane, isopropyl chloride, 1 – chlorobutane.
Solution:
The boiling points of organic compounds depend on the strength of the intermolecular forces in them. These forces are :
(a) van der Waals forces and
(b) dipole-dipole interactions These forces are dependent on the
(i) molecular mass and
(ii) surface area of the molecules

(i) As the molecular mass of the compound increases, the boiling point also increases. Therefore the correct order is
chloromethane < bromomethane < dibromomethane < bromoform

(ii) Amongst molecules with same mass, it is the size of the molecule that determines the boiling point. Branched compounds are more compact and therefore have less surface area as compared to their straight chain counterparts and therefore lower boiling point. The order of boiling point is
iso-propyl chloride < 1-chloropropane < 1-chlorobutane

Question 7.
Which alkyl halide from the following pairs would you expect to react more rapidly by an S_N² mechanism? Explain your answer.

Haloalkanes And Haloarenes Class 12 NCERT Solutions:
We know that S_N² mechanism involves a transition state wherein both, the incoming nucleophile as well as the leaving group are present around the carbon atom. There are 5 atoms simultaneously bonded to it.

Thus, for such a transition state to be possible, there should be minimum steric hindrance. Hence, 1° alkyl halides are most reactive towards S_N² followed by 2° and finally 3°.
1° RX > 2° RX > 3° RX

Question 8.
In the following pairs of halogen compounds, which compound undergoes faster S_N¹ reaction?

Haloalkanes And Haloarenes Class 12 NCERT Solutions:
S_N¹ reaction proceeds via the formation of a carbocation intermediate. This intermediate is formed by the cleavage of the C — X bond. More stable is the resultant carbocation faster is the S_N¹ reaction.
Order of stability of carbocation is
3° carbocation > 2° carbocation >1° carbocation

Question 9.
Identify A, B, C, D, E, R and R’ in the following.

Solution:

10. A hydrocarbon C₅H₁₀ does not react with chlorine in dark but gives a single monochloro compound C₅H₉Cl in bright sunlight. Identify the hydrocarbon.
Ans. The hydrocarbon with molecular formula C₅H₁₀ can either a cycloalkane or an alkene.Since the compound does not react with Cl₂ in the dark, therefore it cannot be an alkene but must be a cycloalkane. Since the cycloalkane reacts with Cl₂ in the presence of bright sunlight to give a single monochloro compound, C₅H₉Cl, therefore, all the ten hydrogen atoms of the cycloalkanes must be equivalent. Thus, the cycloalkane is cyclopentane.

NCERT Exercises

Question 1.
Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides :

(i) (CH₃)₂CHCH(Cl)CH₃
(ii) CH₃CH₂CH (CH₃)CH(C₂H₅)Cl
(iii) CH₃CH₂C(CH₃)₂CH₂l
(iv) (CH₃)₃CCH₂CH(Br)C₆H₅
(v) CH₃CH(CH₃)CH(Br)CH₃
(vi) CH₃C(C₂H₅)₂CH₂Br
(vii) CH₃C(CI)(C₂H₅)CH₂CH₃
(viii) CH₃CH = C(Cl)CH₂CH(CH₃)₂
(ix) CH₃CH = CHC(Br)(CH₃)₂
(x) p-ClC₆H₄CH₂CH(CH₃)₂
(xi) m-CICH₂C₆H₄CH₂C(CH₃)₃
(xii) o-BrC₆H₄CH(CH₃)CH₂CH₃

Haloalkanes And Haloarenes Class 12 NCERT Solutions:

Question 2.
Give the IUPAC names of the following compounds:
(i) CH₃CH(Cl)CH(Br) (CH₃)
(ii) CHF₂CBrClF
(iii) ClCH₂C = CCH₂Br
(iv) (CCl₃)₃CCl
(v) CH₃C(p-ClC₆H₄)₂CH(Br)CH₃
(vi) (CH₃)₃CCH = C(CI) C₆H₄ l-p
Solution:

Question 3.
Write the structures of the following organic halogen compounds.
(i) 2-Chloro-3-methylpentane
(ii) p-Bromochlorobenzene
(iii) 1 -Chloro-4-ethylcyclohexane
(iv) 2-(2-Chlorophenyl)-1 -iodooctane
(v) Perfluorobenzene
(vi) 4-tert-Butyl-3-iodoheptane
(vii) 1 -Bromo-4-sec-butyl-2-methylbenzene
(viii) 1, 4-Dibromobut-2-ene
Solution:
Structures of the given compounds are :

Question 4.
Which one of the following has the highest dipole moment?
(i) CH₂Cl₂
(ii) CHCl₃
(iii) CCl₄
Solution:

Question 5.
A hydrocarbon C₅H₁₀ does not react with chlorine in dark but gives a single monochloro compound C₅H₉Cl in bright sunlight. Identify the hydrocarbon.
Solution:
A number of structural isomers are possible for molecular formula C₅H₁₀. But, the given compound gives a single monochloro derivative when reacted with Cl₂ in sunlight suggests that, all the H-atoms in the compound are equivalent. This is possible only if the compound is a cyclic alkane.

Question 6.
Write the isomers of the compound having formula C₄H₉Br.
Solution:
The possible isomers of C₄H₉Br are

Question 7.
Write the equations for the preparation of 1 -iodobutane from
(i) 1-butanol
(ii) 1-chlorobutane
(iii) but-1-ene
Solution:

Question 8.
What are ambident nucleophiles? Explain with an example.
Solution:
Ambident nucleophiles are nucleophiles that are capable of attacking the substrate (alkyl halide) through two different atoms.

It so ‘happens due to the presence of two nucleophilic centres which arise from the contributing (resonance) structures that are possible for the ion.

e.g., In NO^–₂ ion, there is a lone pair of electrons on N and therefore makes it nucleophilic while oxygen by virtue of the negative charge acts as a nucleophile. Thus, NO^–₂ can attack via O or N atom thereby making it ambidentate.

Question 9.
Which compound in each of the following pairs will react faster in SN2 reaction with OH^– ?
(i) CH₃Br or CH₃l
(ii) (CH₃)₃CCI or CH₃Cl
Solution:
(i) Between CH₃Br and CH₃I, CH₃I will react faster via the S_N² mechanism. In S_N² mechanism, C – X bond breaks and the faster it breaks faster is the reaction.

I- is a better leaving group. Owing to its large size, the C – I bond breaks faster than the C – Br bond and reaction proceeds further at a greater rate.

(ii) The order of reactivity in an S_N² reaction depends on minimal steric hindrance around the carbon involved in the C – X bond. Lesser the steric hindrance felt by the incoming nucleophile, more reactive will be the alkyl halide towards S_N² reaction.
Based on this, CH₃Cl will react faster than (CH₃)₃CCl.

Question 10.
predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene :
(i) 1-Bromo-1-methylcyclohexane
(ii) 2-Chloro-2-methylbutane
(iii) 2, 2, 3-Trimethyl-3-bromopentane.
Solution:

Question 11.
How will you bring about the following conversions?

1. Ethanol to but-1 -yne
2. Ethane to bromoethene
3. Propeneto 1-nitropropane
4. Toluene to benzyl alcohol
5. Propene to propyne
6. Ethanol to ethyl fluoride
7. Bromomethane to propanone
8. But-1 -ene to but-2-ene
9. 1-Chlorobutane ton-octane
10. Benzene to biphenyl

Solution:

Question 12.
Explain why

1. the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride?
2. alkyl halides, though polar, are immiscible with water?
3. Grignard reagents should be prepared under anhydrous conditions?

Solution:
(i) (a) In order to understand the lower dipole moment of chlorobenzene we need to look into the contributing structures of the molecules.

(b) From the above structures we find that the C – Cl bond in chlorobenzene has a partial double bond character (structure II, III and IV). As a result, the C – Cl bond length here is shorter than the C – Cl single bond but longer than the C – Cl double bond.

(c) Also evident is the positive charge on Cl atom which reduces the partial negative (δ-) charge which it is expected to carry by the virtue of its electronegativity.

(d) Consequently, the dipole moment, which is a product of bond length and partial negative charge on Cl atom, reduces. However, in cyclohexyl chloride this does not happen. It is an alkyl halide and carbon is purely sp3 hybridised and C – Cl bond has the bond length of a single bond and 8“ appearing on Cl is also higher, thus, the greater dipole moment.

(ii) Only those compounds which can form hydrogen bonds with water are miscible with it. Alkyl halides, though polar due to the presence of electronegative halogen atom, are immiscible since they cannot form hydrogen bonds.

(iii) Grignard reagents R – Mg – X is a class of highly reactive compounds which can extract a proton even from water molecule. They thus, turn into the corresponding alkanes and render any other desired reaction ineffective.

This is why Grignard reagents are prepared in the absolute absence of water (anhydrous conditions), (e.g.,)

Question 13.
Give the uses of freon 12, DDT, carbon tetrachloride and iodoform.
Solution:
(i) Freon 12 (CCl₂F₂) is

• used in aerosol propellants
• refrigeration
• air-conditioning.

(ii) DDT (p, p’- dichlorodiphenyltrichloroethane) is

• used as an insecticide,
• mainly used against mosquitoes.

(iii) Carbontetrachloride (CCl₄) is used

• in manufacture of refrigerants and propellants for aerosol cans
• in synthesis of chlorofluorocarbons
• as degreasing agent
• as cleansing agent
• as a solvent in laboratories

(iv) Iodoform (CHI₃) is used as an antiseptic.

Question 14.
Write the structure of the major organic product in each of the following reactions :

Solution:

Question 15.
Write the mechanism of the following reaction

Solution:

Question 16.
Arrange the compounds of each set in order of reactivity towards SN2 displacement :

1. 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane
2. 1-Bromo-3-methylbutane, 2-Bromo-2- methylbutane, 3-Bromo-2-methylbutane
3. 1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methyl butane, 1-Bromo-3- methylbutane.

Solution:
S_N² reaction proceeds via the formation of transition state where the carbon atom is surrounded by 5 other atoms (groups). Thus, for such a transition state to form, the steric interactions have to be minimum. Therefore, the most favourable substrates for S_N² reactions are 1° alkyl halides followed by 2° and 3° alkyl halide. Order of reactivity towards S_N² : 1° > 2° > 3° > aryl halide.

In example (iii), although all the given alkyl halides are 1° but the steric hindrance around the carbon bearing the -Br atom decides the order of reactivity. More the number of bulky groups around this carbon lower will be its reactivity towards S_N².

Question 17.
Out of C₆ H₅ CH₂ Cl and C₆H₅ CHClC₆ H₅, which is more easily hydrolysed by aqueous KOH?
Solution:
C₆ H₅ CHClC₆ H₅ is hydrolysed faster.
(a) Hydrolysis of an alkyl halide is an example of nucleophilic substitution reaction. In case of aryl halides this follows the S_N¹ pathway i.e., via the formation of carbocation.
(b) C₆ H₅ CH₂ Cl or benzyl chloride gives

(c) Out of I & II, carbocation II is more stable. The reason is the presence of two phenyl rings attached to the carbon carrying the positive charge.
(d) As a result, the delocalisation of the +ve charge is greater and the carbocation is more stable. Due to this, (II) is formed faster and the corresponding halide is hydrolysed with greater ease as compared to benzyl chloride.

Question 18.
p-Dichlorobenzene has higher m.p. and solubility than those of o- and m-isomers. Discuss.
Solution:
The para-isomers have high melting points and solubility as compared to their ortho and meta isomers due to symmetry of para-isomers that fits into crystal lattice better than ortho and para isomers.

Question 19.
How can the following conversions be carried out?

1. Propene to propan-1 -ol
2. Ethanol to but-1 -yne
3. 1 -Bromopropane to 2-bromopropane
4. Toluene to benzyl alcohol
5. Benzene to 4-bromonitrobenzene
6. Benzyl alcohol to 2-phenylethanoic acid
7. Ethanol to propanenitrile
8. Aniline to chlorobenzene
9. 2-Chlorobutane to 3,4-dimethylhexane
10. 2-Methyl-1 -propene to 2-chloro-2- methylpropane
11. Ethyl chloride to propanoic acid
12. But-1-ene to n-butyliodide
13. 2-Chloropropaneto 1-propanol
14. Isopropyl alcohol to iodoform
15. Chlorobenzene to p-nitrophenol
16. 2-Bromopropane to 1-bromopropane
17. Chloroethane to butane
18. Benzene to diphenyl
19. ferf-Butyl bromide to isobutyl bromide
20. Aniline to phenylisocyanide

Solution:

Question 20.
The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH, alkenes are major products. Explain.
Solution:
Formation of alcohols from the reaction between alkyl chlorides and aqueous KOH is an example of simple nucleophilic substitution.

But when aqueous KOH is replaced by alcoholic KOH, alkenes are formed instead of alcohols due to elimination of HCl from an alkyl halide.

Question 21.
Primary alkyl halide C₄H₉Br (A) reacted with alcoholic KOH to give compound (B). Compound (B) is reacted with HBr to give (C) which is an isomer of (A). When (A) is reacted with sodium metal it gives compound (D). C₈H₁₈ which is different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of (A) and write the equations for all the reactions.
Solution:

Question 22.
What happens when

1. n-butyl chloride is treated with alcoholic KOH,
2. bromobenzene is treated with Mg in the presence of dry ether,
3. chlorobenzene is subjected to hydrolysis,
4. ethyl chloride is treated with aqueous KOH,
5. methyl bromide is treated with sodium in the presence of dry ether,
6. methyl chloride is treated with KCN?

Solution:


Chlorobenzene is highly unreactive towards nucleophilic substitution. However, it can be hydrolysed to phenol by heating in aqueous sodium hydroxide solution at a temperature of 623 K and 300 atm pressure. The presence of an electron withdrawing group increases the reactivity of haloarenes.