The final product $[D]$ of the reaction is
$CHCl_3$ $\xrightarrow[hv]{O_2} [A]$ $\xrightarrow{EtOH} [B]$ $\xrightarrow{2 \ mole \ MeMgBr} [C]$ $\xrightarrow{CaOCl_2} [D]$ $\xrightarrow{\Delta} CH_3-C(=O)-CH_3$

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.

The increasing order of the acidity of the following carboxylic acids is:

The general formula for monocarboxylic acids is

Consider the following sequence of reactions to give the major product $(X)$: $(i)$ $CH_3Cl$ / anhydrous $AlCl_3$, (ii) $Cl_2$ / $FeCl_3$, (iii) $K_2Cr_2O_7$ / $H_2SO_4$. $P$ g of the major product $(X)$ formed is reacted with $NaHCO_3$ solution to liberate a gas which occupied $11.2 \ dm^3$ at $STP$. $P = \text{ . . . . . . }$ g.

The ester among the following is