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Question #45072

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The Question

Aniline ( $\ce{C6H5NH2}$ ) is subjected to three distinct reaction pathways: (1) Treatment with $\ce{CH3Cl}$ and anhydrous $\ce{AlCl3}$ . (2) Treatment with acetic anhydride in pyridine, followed by $\ce{CH3Cl}$ /anhydrous $\ce{AlCl3}$ , and subsequent acidic hydrolysis ( $\ce{H3O+}$ , $\Delta$ ). (3) Treatment with a nitrating mixture (conc. $\ce{HNO3}$ and conc. $\ce{H2SO4}$ ) at $288\text{ K}$ . Let $\textbf{P}$ be the major species formed in Pathway 1. Let $\textbf{S}$ be the major organic product isolated from Pathway 2. Pathway 3 yields three isomeric products: $\textbf{T}$ ( $51\%$ ), $\textbf{U}$ ( $47\%$ ), and $\textbf{V}$ ( $2\%$ ). Which of the following correctly identifies $\textbf{P}$ , $\textbf{S}$ , and the mechanistic rationale for the significant yield of $\textbf{U}$ ?

Options

A.
$\textbf{P}$ : 4-Methylaniline; $\textbf{S}$ : 4-Methylaniline; Rationale for $\textbf{U}$ : Strong $-I$ effect of $-\text{NH}_2$
B.
$\textbf{P}$ : $\ce{C6H5\overset{+}{N}H2-\overline{A}lCl3}$ ; $\textbf{S}$ : 4-Methylaniline; Rationale for $\textbf{U}$ : Formation of meta-directing $-\text{NH}_3^+$
C.
$\textbf{P}$ : 3-Methylaniline; $\textbf{S}$ : 3-Methylaniline; Rationale for $\textbf{U}$ : Thermodynamic control at $288\text{ K}$
D.
$\textbf{P}$ : $N$ -Methylaniline; $\textbf{S}$ : 2-Methylaniline; Rationale for $\textbf{U}$ : Hydrogen bonding stabilization

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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

Electrophile vs Nucleophile|Electrophilic Substitution Activation/Deactivation|Aniline Resonance Pitfalls

Why This Question Is Tricky

Common Mistakes

This question tests the student's understanding of the exceptions in electrophilic aromatic substitution of aniline.

Students often blindly apply ortho/para directing rules for the -NH2 group.

The trap lies in failing to recognize that Lewis acids (like AlCl3) or Brønsted acids (like H2SO4/HNO3) react with the basic -NH2 lone pair.

This converts a strongly activating ortho/para director into a strongly deactivating meta director (-NH3+ or -NH2+-AlCl3-), completely altering the reaction's outcome or causing it to fail.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Organic Chemistry – Some Basic Principles and Techniques - Class 11 Part 2
Concept	Electrophile vs Nucleophile
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free