Question

4-nitrotoluene is treated with bromine to get compound '$P$' which is reduced with $Sn$ and $HCl$ to get compound '$Q$'. '$Q$' is diazotized and the product is treated with phosphinic acid to get compound '$R$'. '$R$' is oxidized with alkaline solution $KMn{O}_4$ to get the final product. Identify the final product.

• A. 2-bromo-4-hydroxybenzoic acid
• B. Benzoic acid
• C. 4-bromobenzoic acid
• D. 3-bromobenzoic acid
• E. 2-bromobenzoic acid

Answer 1

The correct option is E 2-bromobenzoic acid

Given,

Step 1.

($\because N{O}_2$ is $m$-directing group while $C{H}_3$ group is $o-$ and $p-$ directing group. Thus, $Br$ group attach at $o-$ position with respect to $C{H}_3$ and $m$-position with respect to $N{O}_2$ group)

Step 2. On reduction with $Sn/HCl,N{O}_2-$ group changes to $-N{H}_2$ group.

Step 3. On treatment with $NaN{O}_2+HCl$; $N{H}_2$ changes to diazo group (i.e. $Q$ shows diazotisation).

Step 4. On treatment with $H_{3}P{O}_2$, only diazo group $\left(N_{2}Cl\right)$ reacts and gets removed.

Step 5. On oxidation with $KMn{O}_4/O{H}^{-}-C{H}_3$ group changes to $-COOH$ group.