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Question #44034

Difficulty: 8/10 | 🟢 Free

The Question

Consider the following reaction sequences for nitrobenzene ( $\ce{C6H5NO2}$ ):

Sequence I: $\ce{C6H5NO2} \xrightarrow{\ce{LiAlH4,\ dry\ ether}} \ce{P}$

Sequence II: $\ce{C6H5NO2} \xrightarrow{\ce{Sn/HCl}} \xrightarrow{\ce{NaOH(aq)}} \ce{Q} \xrightarrow{\ce{NaNO2,\ HCl,\ 273\ K}} \ce{R} \xrightarrow[\ce{pH \approx 9-10}]{\ce{Phenol}} \ce{S}$

Identify the major products $\ce{P}$ and $\ce{S}$ .

Options

A.
$\ce{P}$ = $\ce{Ph-N=N-Ph}$ ; $\ce{S}$ = $p$ - $\ce{Ph-N=N-C6H4OH}$
B.
$\ce{P}$ = $\ce{Ph-NH2}$ ; $\ce{S}$ = $p$ - $\ce{Ph-N=N-C6H4OH}$
C.
$\ce{P}$ = $\ce{Ph-NH-NH-Ph}$ ; $\ce{S}$ = $o$ - $\ce{Ph-N=N-C6H4OH}$
D.
$\ce{P}$ = $\ce{Ph-N=N(O)-Ph}$ ; $\ce{S}$ = $p$ - $\ce{Ph-N=N-C6H4NH2}$

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

Aniline Resonance Pitfalls|Reduction Selectivity (Sn/HCl vs LiAlH4)|Inductive Effect Trends

Why This Question Is Tricky

Common Mistakes

This question tests a critical 'Chemical Exception' in reduction selectivity.

Students often falsely generalize that $\ce{LiAlH4}$ , being a powerful reducing agent, will reduce all nitro groups to primary amines (as it does for aliphatic nitro compounds).

However, they overlook that in aromatic systems, the resonance-enhanced electron density and the nucleophilic nature of hydride attack lead to bimolecular condensation, halting the reduction at azobenzene.

Option B is the classic distractor designed to catch students who apply the aliphatic rule to aromatic systems.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Amines - Class 12
Concept	Aniline Resonance Pitfalls
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free