Giving plausible explanation for each of the following:
i. Cyclohexanone forms cyanohydrin good yield but 2,2,6-trimethylcyclohexanone does not.
ii. There are two `(-NH_(2))` 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 easter should be removed as soon as it is formed.

Explanation

(i).
Because of the presence of three methyl groups at `alpha`-positions w.r.t. the C=O groups, the nucleophilic attack by `CN^(-)` ion does not occur due to steric hindrance. Since there is no such steric hindrance in cyclohexanone, therefore, nucleophilic attach by the `CN^(-)` ion occurs readily and hence cyclohexanone cyanohydrin is obtained in good yield.
(ii).
Although semicarbazide has two `-NH_(2)` groups but one of them (ie.., which is directly attached to C=O) is involved in resonance as shown above. as a result, electron density on this `NH_(2)` group decreases and hence it does ot act as a nucleophile. in contrast, the lone pair of electrons on the other `NH_(2)` group (i.e., attached to NH) is not involved in resonance and hence is available for nucleophilic attack on the C=O group of aldehydes and ketones.
(iii) The formation of esters from a carboxylic acid and an alcohol in presence of an acid catalyst is a reversible reaction.
`underset("Carboxylic acid")(RCOOH)+underset("Alcohol")(R'OH) overset(H_(2)SO_(4))hArr underset("Ester")(RCOOR')+H_(2)O`
To shift the equilibrium in the forward direction, the water or the ester formed should be removed as fast as it is formed.