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Question #44925

Difficulty: 8/10 | 🟢 Free

The Question

Compound $\textbf{X}$ is prepared by treating $p$ -nitroaniline with $NaNO_2$ and aqueous $HCl$ at $273\text{ K}$ . The resulting cold solution of $\textbf{X}$ is immediately added to a reaction vessel containing an equimolar mixture of $CuCl$ and $N,N$ -dimethylaniline. The reaction medium is carefully buffered at $\text{pH} = 4.5$ and maintained at $273\text{ K}$ . Which of the following is the major product $\textbf{Y}$ formed?

Options

A.
1-chloro-4-nitrobenzene
B.
$N,N$ -dimethyl-4-[(4-nitrophenyl)diazenyl]aniline
C.
$N,N$ -dimethyl-2-[(4-nitrophenyl)diazenyl]aniline
D.
$N,N$ -dimethyl-4-nitroaniline

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

Sandmeyer vs Gattermann|Azo Coupling Regioselectivity|Electrophilic Substitution Control

Why This Question Is Tricky

Common Mistakes

The problem tests the kinetic competition between nucleophilic radical substitution (Sandmeyer) and electrophilic aromatic substitution (Azo coupling).

Students typically default to the Sandmeyer product when $CuCl$ is present.

However, the 'Hidden Variable' is the pH 4.5 buffer and the low temperature ( $273\text{ K}$ ).

At this pH, $N,N$ -dimethylaniline remains sufficiently unprotonated to act as a highly activated nucleophile.

Azo coupling is kinetically much faster (lower activation energy) at $0^\circ\text{C}$ than the $Cu(I)$ -catalyzed single-electron transfer required for the Sandmeyer reaction, subverting the expected haloarene product.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Amines - Class 12
Concept	Sandmeyer vs Gattermann
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free