The reaction of acetaldehyde with conc. $KMnO_4$ gives

An optically active alkyl bromide $C_4H_9Br$ reacts with ethanolic $KOH$ to form major compound $[A]$,which reacts with bromine to give compound $[B]$. Compound $[B]$ reacts with ethanolic $KOH$ and sodamide to give compound $[C]$. One molecule of water adds to compound $[C]$ on warming with mercuric sulphate and dilute sulphuric acid at $333 \text{ K}$ to form compound $[D]$. The functional group in compound $[D]$ will be confirmed by:

Match the compounds given in List-$I$ with List-$II$ and select the suitable option using the code given below.

List-$I$	List-$II$
$A$. Benzaldehyde	$i$. Phenolphthalein
$B$. Phthalic anhydride	$ii$. Benzoin condensation
$C$. Phenyl benzoate	$iii$. Oil of wintergreen
$D$. Methyl salicylate	$iv$. Fries rearrangement

Which reagent is suitable for the following transformation?

Reactant $(A)$ is:

The reaction shown above is a Baeyer-Villiger rearrangement of an asymmetric ketone with magnesium monoperoxyphthalate hexahydrate (in the drawing,$Mg^{+2}$ is omitted for clarity). Identify the major product.