Identify the product formed ( A and E )

Step 1:

(∵${\mathrm{NO}}_2$​  is m-directing group while ${\mathrm{CH}}_3$ group is​  o− and p− directing group. Thus, $\mathrm{Br}$ group attach at o− position with respect to ${\mathrm{CH}}_3$ and m-position with respect to ${\mathrm{NO}}_2$ group)

Step 2:

On reduction with $\mathrm{Sn}/\mathrm{HCl}, {\mathrm{NO}}_2^{-}$  group changes to  $-{\mathrm{NH}}_2$​ group.

Step 3:

 On treatment with ${\mathrm{NaNO}}_2+\mathrm{HCl}; {\mathrm{NH}}_2$ changes to diazo group.

Step 4:

On treatment with ${\mathrm{H}}_3{\mathrm{PO}}_2$​, only diazo group $\left({\mathrm{N}}_2\mathrm{Cl}\right)$ reacts and gets removed.

Step 5:

On oxidation with ${\mathrm{KMnO}}_4 /{\mathrm{OH}}^{-}, -{\mathrm{CH}}_3$​ group changes to $-\mathrm{COOH}$ group.