JEE Chemistry 2025 Question with Solution

Given: The reaction sequence involves treatment of [A] with $\text{NaNO}_2$ and $\text{HCl}$ at $273\text{–}278 \, \text{K}$, followed by reaction with $\text{KI}$, and then with $2\text{Na}$ in dry ether.

Find: Identify [A], [B], and [C].

The first step is diazotization. An aromatic primary amine forms a diazonium salt under these conditions, so [A] must be aniline, $\text{C}_6\text{H}_5\text{NH}_2$.

The diazonium salt then reacts with $\text{KI}$. This replaces the diazonium group by iodine, so [B] is iodobenzene, $\text{C}_6\text{H}_5\text{I}$.

In the final step, aryl halide with $2\text{Na}$ in dry ether undergoes coupling to form biphenyl. Therefore [C] is biphenyl, $\text{C}_6\text{H}_5\text{-C}_6\text{H}_5$.

Therefore, the correct option is C.

Given: Recognize the reactions involved: diazotization, Sandmeyer reaction, and coupling reaction.

Find: The identities of [A], [B], and [C].

Step 1: The first reaction involves the treatment of [A] with $\text{NaNO}_2$ and $\text{HCl}$ at $273\text{–}278 \, \text{K}$.

This is a typical diazotization reaction where an aromatic amine (aniline) is converted into a diazonium salt. Thus, [A] is aniline, $\text{C}_6\text{H}_5\text{NH}_2$.

Step 2: The diazonium salt formed reacts with $\text{KI}$.

This is a Sandmeyer reaction where the diazonium group is replaced by iodide to form an iodobenzene compound. Hence, [B] is iodobenzene, $\text{C}_6\text{H}_5\text{I}$.

Step 3: The compound undergoes a reaction with $2\text{Na}$ in dry ether.

This reaction is a typical Wurtz reaction where iodobenzene reacts with sodium to form biphenyl. Therefore, [C] is biphenyl, $\text{C}_6\text{H}_5\text{-C}_6\text{H}_5$.

Thus, the correct answer is: Aniline ([A]) $\rightarrow$ Iodobenzene ([B]) $\rightarrow$ Biphenyl ([C]). Hence, the correct option is C.