Identify the product formed ( A and E )

Step 1:

($∵$${NO}_{2}$ is m-directing group while $${CH}_{3}$$group is o−& and& p−& directing group. Thus,& $Br$ group attach at& o−& position with respect to& ${CH}_{3}$ and& m-position with respect to& ${NO}_{2}$ group)

Step 2:

On reduction with& $Sn / HCl, {NO}_{2}^{-}$ & group changes to $$- {NH}_{2}$$ group.

Step 3:

& On treatment with& ${NaNO}_{2} + HCl; {NH}_{2}$ changes to diazo group.

Step 4:

On treatment with& $${H}_{3}{PO}_{2}$$, only diazo group $( {N}_{2}Cl )$ reacts and gets removed.

Step 5:

On oxidation with& $${KMnO}_{4} / {OH}^{−}, - {CH}_{3}$$ group changes to $- COOH$ group.