JEE Chemistry 2024 Question with Solution

Given: Benzene reacts first with $\text{HNO}_3$ and $\text{H}_2\text{SO}_4$, and then the product reacts with $\text{Sn/HCl}$.

Find: The identities of products A and B.

The reaction of benzene with concentrated nitric acid and sulfuric acid is nitration of benzene. Therefore, product A is nitrobenzene:

$$\text{C}_6\text{H}_6 \xrightarrow[\text{}]{\text{HNO}_3/\text{H}_2\text{SO}_4} \text{C}_6\text{H}_5\text{NO}_2$$

So,

$$A = \text{C}_6\text{H}_5\text{NO}_2$$

Nitrobenzene on reduction with $\text{Sn/HCl}$ gives aniline, because the nitro group is reduced to an amino group:

$$\text{C}_6\text{H}_5\text{NO}_2 \xrightarrow[]{\text{Sn/HCl}} \text{C}_6\text{H}_5\text{NH}_2$$

Hence,

$$B = \text{C}_6\text{H}_5\text{NH}_2$$

Therefore, the correct option is A. The products are $\text{C}_6\text{H}_5\text{NO}_2$ and $\text{C}_6\text{H}_5\text{NH}_2$.

Given:

• Starting compound is benzene.
• Reagents in the first step are $\text{HNO}_3$ and $\text{H}_2\text{SO}_4$.
• Reagent in the second step is $\text{Sn/HCl}$.

Find: Which compounds correspond to A and B.

Step 1: Benzene undergoes electrophilic substitution with the nitrating mixture $\text{HNO}_3/\text{H}_2\text{SO}_4$. This introduces a nitro group on the benzene ring.

$$\text{Benzene} \rightarrow \text{Nitrobenzene}$$

Thus, product A is nitrobenzene:

$$A = \text{C}_6\text{H}_5\text{NO}_2$$

Step 2: Nitrobenzene is reduced by $\text{Sn/HCl}$. In this reduction, the $-\text{NO}_2$ group is converted into $-\text{NH}_2$.

$$\text{Nitrobenzene} \rightarrow \text{Aniline}$$

Hence, product B is aniline:

$$B = \text{C}_6\text{H}_5\text{NH}_2$$

the solution marks the correct option as B, but the extracted working clearly gives A = $\text{C}_6\text{H}_5\text{NO}_2$ and B = $\text{C}_6\text{H}_5\text{NH}_2$, which matches option A in the given option order. Therefore, the defensible correct option is A.