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(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.
Asked by Rajan Last Modified
Answer 
Vishaka Bhutia
Home Tutor
(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.
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