Q.Identify the compound Y in the following reaction.

The key idea is the Sandmeyer reaction: the diazonium group ($-N_2^+$) is replaced by a chlorine atom using a cuprous chloride catalyst.

Reasoning:

1. Aniline reacts with $\mathrm{NaNO_2 + HCl}$ at low temperature ($273\text{-}278\,\mathrm{K}$) to form benzenediazonium chloride, $\mathrm{C_6H_5N_2^+Cl^-}$.
2. This diazonium salt is then treated with $\mathrm{Cu_2Cl_2}$ (cuprous chloride in HCl). The Sandmeyer reaction substitutes the diazonium group with a chlorine atom, releasing $\mathrm{N_2}$ gas.
3. The product is chlorobenzene ($\mathrm{C_6H_5Cl}$). No further substitution occurs under these conditions.

The key to this question is recognising the Sandmeyer reaction — a classic method for replacing the diazonium group ($-\mathrm{N_2^+}$) with a halogen using a copper(I) halide. Let’s walk through the chemistry step by step.

1. First step: Diazotisation Aniline ($\mathrm{C_6H_5NH_2}$) reacts with $\mathrm{NaNO_2}$ and $\mathrm{HCl}$ at low temperature (273–278 K). This converts the amino group into a diazonium group:

The product is benzenediazonium chloride, a key intermediate in aromatic substitution. The low temperature is critical — diazonium salts decompose above about 5°C.

2. Second step: The Sandmeyer reaction The diazonium salt is then treated with $\mathrm{Cu_2Cl_2}$ (copper(I) chloride). This is the classic Sandmeyer reaction, where the diazonium group is replaced by a chlorine atom. The mechanism involves a single-electron transfer from Cu(I) to the diazonium ion, generating an aryl radical, which then abstracts chlorine from Cu(II) to form the aryl chloride.

The nitrogen gas ($\mathrm{N_2}$) bubbles off, driving the reaction forward.

3. What about the options?
   • (i) Chlorobenzene — This is the direct product of the Sandmeyer reaction with $\mathrm{Cu_2Cl_2}$.
   • (ii) Benzene — This would require reduction of the diazonium group (e.g., with $\mathrm{H_3PO_2}$), not with $\mathrm{Cu_2Cl_2}$.
   • (iii) 1,3-Dichlorobenzene and (iv) 1,4-Dichlorobenzene — These would require two chlorine substitutions, but the reaction conditions only introduce one chlorine. No further chlorination occurs here.

4. Confirming the product The reaction is clean: one diazonium group, one chlorine atom replaces it, and nitrogen is lost. The product is chlorobenzene, $\mathrm{C_6H_5Cl}$.

Concept: Sandmeyer Reaction

The Sandmeyer reaction is a method to replace the diazonium group ($-N_2^+$) with a halogen (Cl, Br, I) or a cyano group ($-CN$) using a copper(I) halide or copper(I) cyanide as a catalyst.

Method: Sandmeyer Reaction for Chlorination

Step 1: Identify the starting material and the reagent.

• Aniline ($\mathrm{C_6H_5NH_2}$) is first converted to benzenediazonium chloride ($\mathrm{C_6H_5N_2^+Cl^-}$) at low temperature (273–278 K) using $\mathrm{NaNO_2 + HCl}$.

Step 2: Apply the Sandmeyer reaction condition.

• The benzenediazonium chloride is treated with $\mathrm{Cu_2Cl_2}$ (copper(I) chloride).

Step 3: Write the reaction.

• The diazonium group ($-N_2^+$) is replaced by a chlorine atom ($-Cl$), and nitrogen gas ($N_2$) is released.

Step 4: Identify the product Y.

• The product is chlorobenzene ($\mathrm{C_6H_5Cl}$).

Final Answer

Y = Chlorobenzene ($\mathrm{C_6H_5Cl}$) → Option (i)

Common Mistakes in This Diazonium Reaction Problem

This question tests your understanding of the Sandmeyer reaction — specifically the replacement of the diazonium group ($-\mathrm{N}_2^+$) with chlorine using $\mathrm{Cu_2Cl_2}$.

✗ Mistake 1: Thinking $\mathrm{Cu_2Cl_2}$ gives substitution on the ring

Why students make it:

They see $\mathrm{Cu_2Cl_2}$ and assume it chlorinates the benzene ring directly (like electrophilic substitution), producing dichlorobenzenes.

How to avoid:

Remember: $\mathrm{Cu_2Cl_2}$ in the Sandmeyer reaction replaces the diazonium group ($-\mathrm{N}_2^+$) with a chlorine atom at the same position. It does not add extra chlorines to the ring.

Correct result: Only one chlorine replaces the $-\mathrm{N}_2^+$ group → chlorobenzene ($\mathrm{C_6H_5Cl}$).

✗ Mistake 2: Choosing benzene ($\mathrm{C_6H_6}$)

Why students make it:

They recall that diazonium salts can be reduced to benzene using $\mathrm{H_3PO_2}$ (hypophosphorous acid) or ethanol, and confuse the reagent.

How to avoid:

Memorise the reagent–product mapping:

Here, $\mathrm{Cu_2Cl_2}$ cannot give benzene — it gives chlorobenzene.

✗ Mistake 3: Forgetting that $\mathrm{N_2}$ gas is released

Why students make it:

They focus only on the product structure and ignore the stoichiometric clue.

How to avoid:

The equation shows $\mathrm{N_2}$ is evolved. This means the diazonium group ($-\mathrm{N}_2^+$) leaves completely. The only thing that can replace it is a single atom or group from the reagent — here, $\mathrm{Cl}$ from $\mathrm{Cu_2Cl_2}$.

✓ Quick Summary Table

Final correct answer: $\boxed{\text{(i) Chlorobenzene}}$