Question

4 -nitrotoluene is treated with bromine to get compound ' $P$ ' which is reduced with Sn and HCl to get compound ' $Q$ '''Q ' is diazotised and the product is treated with phosphinic acid to get compound ' $R$ '' $R$ ' is oxidised with alkaline solutionKMnO ${}_{4_1}$ to get the final product. Identify the finăl product.

• A. $2$-Bromo-$4$-hydroxybenzoic acid
• B. Benzoic acid
• C. $4$-Bromobenzoic acid
• D. $3$-Bromobenzoic acid
• E. $2$-Bromobenzoic acid

Answer 1

Answer: (E)

Given,

Step 1.

($\because N{O}_2.$ is $m$ -directing group while $C{H}_3$ group iso $-$ 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.