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 $-NH_2$ 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 presence of an acid catalyst,the water or the ester should be removed as soon as it is formed.

(N/A) $(i)$ Cyclohexanone forms cyanohydrin because the nucleophile $CN^{-}$ can easily attack the carbonyl carbon without significant steric hindrance. In $2,2,6-$trimethylcyclohexanone,the methyl groups at the $\alpha-$positions create significant steric hindrance,preventing the nucleophilic attack of $CN^{-}$.
$(ii)$ Semicarbazide $(H_2N-NH-CO-NH_2)$ exhibits resonance where the lone pair of electrons on one of the $-NH_2$ groups is delocalized towards the carbonyl group. This reduces the electron density and nucleophilicity of that specific $-NH_2$ group. The other $-NH_2$ group,which is not involved in resonance,remains nucleophilic and attacks the carbonyl carbon of aldehydes and ketones to form semicarbazones.
$(iii)$ The esterification reaction between a carboxylic acid and an alcohol is a reversible process: $RCOOH + R'OH \rightleftharpoons RCOOR' + H_2O$. According to Le Chatelier's principle,removing the products (water or ester) as they are formed shifts the equilibrium in the forward direction,thereby increasing the yield of the ester.