Yearly Archives: 2021

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.