A keto ester $\left(A\right)$ with molecular formula ${\mathrm{C}}_6{\mathrm{H}}_{10}{\mathrm{O}}_3$ on treatment with $\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}+{\mathrm{I}}_2$ does not give iodoform but on boiling with dilute KOH gives a compound $\left(B\right)$ with molecular formula ${\mathrm{C}}_4{\mathrm{H}}_5{\mathrm{O}}_3\mathrm{K}$ which upon acidification followed by heating undergoes decarboxylation to give acetone. The keto ester $\left(A\right)$ is

The decreasing order of ease of alkaline hydrolysis for the following esters is

In the following reactions, products $\mathrm{A}$ and $\mathrm{B}$ are:

The increasing order of the reactivity of the following with $\text{L}\text{i}\text{A}\text{l}{\text{H}}_4$ is:

$\left(A\right)$

$\left(B\right)$

$\left(C\right)$

$\left(D\right)$

The major product of the following reaction is:

The major product obtained in the following reaction is:

How will you bring about the following conversion in not more than two steps? 

Benzoic acid to m-Nitrobenzyl alcohol

Mention the reagents used for the following conversions:

(m) ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CH}}_3\stackrel{\mathrm{A}}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{CHO}$

(n) ${\mathrm{CH}}_3\mathrm{CHO}\stackrel{\mathrm{B}}{\to }{\mathrm{CH}}_3{\mathrm{COOC}}_2{\mathrm{H}}_5$

(o) ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{CHO}\stackrel{\mathrm{C}}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COCH}\left(\mathrm{OH}\right){\mathrm{C}}_6{\mathrm{H}}_5$

(p) ${\mathrm{CH}}_3{\mathrm{COCH}}_3\stackrel{\mathrm{D}}{\to }{\left({\mathrm{CH}}_3\right)}_2\mathrm{C}\left(\mathrm{OH}\right){\mathrm{CH}}_2{\mathrm{COCH}}_3$

(q) ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{COOH}\stackrel{\mathrm{E}}{\to }{\mathrm{CH}}_3\mathrm{CH}\left(\mathrm{Br}\right)\mathrm{COOH}$

(r) ${\mathrm{CH}}_3{\mathrm{COCH}}_3\stackrel{\mathrm{F}}{\to }{\mathrm{CH}}_3{\mathrm{COCH}}_2\mathrm{Br}$