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Question #44922

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The Question

Aniline ( $\ce{C6H5NH2}$ ) is treated with $\ce{NaNO2}$ and aqueous $\ce{HCl}$ at $298\text{ K}$ . The resulting reaction mixture is immediately divided into two equal portions, I and II. Portion I is treated with $\ce{CuCl}$ and $\ce{HCl}$ . Portion II is treated with $\ce{Cu}$ powder and $\ce{HCl}$ . Let the major organic products isolated from Portion I and Portion II be $P_1$ and $P_2$ , respectively. Identify $P_1$ and $P_2$ .

Options

A.
$P_1 = \text{Chlorobenzene}, P_2 = \text{Chlorobenzene}$
B.
$P_1 = \text{Phenol}, P_2 = \text{Phenol}$
C.
$P_1 = \text{Chlorobenzene}, P_2 = \text{Phenol}$
D.
$P_1 = \text{Phenol}, P_2 = \text{Chlorobenzene}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Mechanism_Trap

Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.

Concepts Behind This Question

Sandmeyer vs Gattermann|Diazotization Temperature Sensitivity|Aniline Resonance Pitfalls

Why This Question Is Tricky

Common Mistakes

The question tests the 'Diazotization Temperature Sensitivity' trap.

Students often memorize reagent sequences (e.g., $\ce{NaNO2/HCl}$ followed by $\ce{CuCl/HCl} \rightarrow$ Sandmeyer) and focus on comparing the catalytic efficiency of $\ce{Cu(I)}$ salts versus $\ce{Cu(0)}$ powder.

However, they overlook the hidden variable: temperature.

Because the initial diazotization is performed at $298\text{ K}$ instead of the strictly required $273-278\text{ K}$ , the diazonium intermediate undergoes rapid thermal decomposition (hydrolysis) to phenol before any halogenation can occur.

The reaction conditions (temperature) hierarchically override the choice of catalyst.

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