Give plausible explanation for each of the following:
Cyclohexanone forms cyanohydrin in good yield but 2,2,6-trimethyl cyclohexanone does not.
These are two −NH2 groups in semicarbazide. However, only one is involved in the formation of semicarbazones.
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.
Give plausible explanation for each of the following:
Cyclohexanone forms cyanohydrin in good yield but 2,2,6-trimethyl cyclohexanone does not.
These are two −NH2 groups in semicarbazide. However, only one is involved in the formation of semicarbazones.
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.
In 2,2,6 trimethyl cyclohexanone, three methyl groups are presents at α− position with respect to the ketonic (>C=O) group. Therefore, these groups cause steric hindrance during the nucleophilic attack of CN−ion so cyanohydrin is not formed. Due to the absence of methyl groups in cyclohexanone, there is no steric hindrance and cyanohydrin is formed.
Resonance in semicarbazide
Semicarbazide has two amino (NH2) groups, out of which one is involved in resonance as shown above. Electron-density on this (NH2) decreases and it does not act as a nucleophile.
But the other (NH2) group (attached o NH) has a lone pair of electrons which are not involved in resonance. So, this pair is available for the nucleophilic attack on the carbonyl group (>C=O) of aldehydes and/or ketones.
Esterification is a reversible reaction.
RCOOH+R′OHH2SO4⇌RCOOR′Ester+H2O
When the sufficient amount of products is formed, the rate of forwarding reaction decreases and the reverse reaction begins. To avoid this condition i.e., in order to shift the equilibrium in the forward direction, the concentration of products (ester and/or water) should be drecreased. (Le-Chatelier's principle). So, water should be removed from time to time.
In 2,2,6 trimethyl cyclohexanone, three methyl groups are presents at α− position with respect to the ketonic (>C=O) group. Therefore, these groups cause steric hindrance during the nucleophilic attack of CN−ion so cyanohydrin is not formed. Due to the absence of methyl groups in cyclohexanone, there is no steric hindrance and cyanohydrin is formed.
Resonance in semicarbazide
Semicarbazide has two amino (NH2) groups, out of which one is involved in resonance as shown above. Electron-density on this (NH2) decreases and it does not act as a nucleophile.
But the other (NH2) group (attached o NH) has a lone pair of electrons which are not involved in resonance. So, this pair is available for the nucleophilic attack on the carbonyl group (>C=O) of aldehydes and/or ketones.
Esterification is a reversible reaction.
RCOOH+R′OHH2SO4⇌RCOOR′Ester+H2O
When the sufficient amount of products is formed, the rate of forwarding reaction decreases and the reverse reaction begins. To avoid this condition i.e., in order to shift the equilibrium in the forward direction, the concentration of products (ester and/or water) should be drecreased. (Le-Chatelier's principle). So, water should be removed from time to time.
