Page No 379: - Chapter 12 Aldehydes Ketones & Carboxylic Acids Exercise Solutions class 12 ncert solutions Chemistry - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question 12.17:

Complete each synthesis by giving missing starting material, reagent or products

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(xi)

Answer:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(xi)

Question 12.18:

Give plausible explanation for each of the following:

(i) Cyclohexanone forms cyanohydrin in good yield but 2, 2, 6 trimethylcyclohexanone does not.

(ii) There are two −NH₂ groups in semicarbazide. However, only one is involved in the formation of semicarbazones.

(iii) 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.

Answer:

(i) Cyclohexanones form cyanohydrins according to the following equation.

In this case, the nucleophile CN⁻ can easily attack without any steric hindrance. However, in the case of 2, 2, 6 trimethylcydohexanone, methyl groups at α-positions offer steric hindrances and as a result, CN⁻ cannot attack effectively.

For this reason, it does not form a cyanohydrin.

(ii) Semicarbazide undergoes resonance involving only one of the two −NH₂ groups, which is attached directly to the carbonyl-carbon atom.

Therefore, the electron density on −NH₂ group involved in the resonance also decreases. As a result, it cannot act as a nucleophile. Since the other −NH₂ group is not involved in resonance; it can act as a nucleophile and can attack carbonyl-carbon atoms of aldehydes and ketones to produce semicarbazones.

(iii) Ester along with water is formed reversibly from a carboxylic acid and an alcohol in presence of an acid.

If either water or ester is not removed as soon as it is formed, then it reacts to give back the reactants as the reaction is reversible. Therefore, to shift the equilibrium in the forward direction i.e., to produce more ester, either of the two should be removed.

Question 12.19:

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.

Answer:

% of carbon = 69.77 %

% of hydrogen = 11.63 %

% of oxygen = {100 − (69.77 + 11.63)}%

= 18.6 %

Thus, the ratio of the number of carbon, hydrogen, and oxygen atoms in the organic compound can be given as:

Therefore, the empirical formula of the compound is C₅H₁₀O. Now, the empirical formula mass of the compound can be given as:

5 × 12 + 10 ×1 + 1 × 16

= 86

Molecular mass of the compound = 86

Therefore, the molecular formula of the compound is given by C₅H₁₀O.

Since the given compound does not reduce Tollen’s reagent, it is not an aldehyde. Again, the compound forms sodium hydrogen sulphate addition products and gives a positive iodoform test. Since the compound is not an aldehyde, it must be a methyl ketone.

The given compound also gives a mixture of ethanoic acid and propanoic acid.

Hence, the given compound is  Pentan-2-one.

The given reactions can be explained by the following equations:

Question 12.20:

Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why?

Answer:

Resonance structures of phenoxide ion are:

It can be observed from the resonance structures of phenoxide ion that in II, III and IV, less electronegative carbon atoms carry a negative charge. Therefore, these three structures contribute negligibly towards the resonance stability of the phenoxide ion. Hence, these structures can be eliminated. Only structures I and V carry a negative charge on the more electronegative oxygen atom.

Resonance structures of carboxylate ion are:

In the case of carboxylate ion, resonating structures I′ and II′ contain a charge carried by a more electronegative oxygen atom.

Further, in resonating structures I′ and II′, the negative charge is delocalized over two oxygen atoms. But in resonating structures I and V of the phexoxide ion, the negative charge is localized on the same oxygen atom. Therefore, the resonating structures of carboxylate ion contribute more towards its stability than those of phenoxide ion. As a result, carboxylate ion is more resonance-stabilized than phenoxide ion. Hence, carboxylic acid is a stronger acid than phenol.