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Question #45100

Difficulty: 8/10 | 🟢 Free

The Question

Consider the electrophilic monobromination of $N,N,2,6$ -tetramethylaniline (Compound X) and 2,6-dimethylphenol (Compound Y) under standard conditions. Identify the major regioselective products formed from X and Y, respectively.

Options

A.
3-bromo- $N,N,2,6$ -tetramethylaniline and 4-bromo-2,6-dimethylphenol
B.
4-bromo- $N,N,2,6$ -tetramethylaniline and 4-bromo-2,6-dimethylphenol
C.
3-bromo- $N,N,2,6$ -tetramethylaniline and 3-bromo-2,6-dimethylphenol
D.
4-bromo- $N,N,2,6$ -tetramethylaniline and 3-bromo-2,6-dimethylphenol

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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

Steric Hindrance (B-Strain)|Ortho-Effect in Anilines|Electrophilic Substitution Activation/Deactivation

Why This Question Is Tricky

Common Mistakes

This question tests the concept of Steric Inhibition of Resonance (SIR) or the Ortho-Effect in electrophilic aromatic substitution.

Students often assume that any group with a lone pair (like $-NMe_2$ or $-OH$ ) will dominate the directing effects via a strong +M effect, directing the electrophile to the para position (C4).

However, they fail to account for the steric bulk of the $-NMe_2$ group compared to the -OH group.

In $N,N,2,6$ -tetramethylaniline, the ortho methyl groups force the $-NMe_2$ group out of coplanarity, breaking conjugation.

This converts a strong +M ortho/para director into a -I meta director, shifting the regioselectivity to the meta position (C3).

The -OH group in 2,6-dimethylphenol does not suffer from SIR and directs para normally.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Organic Chemistry – Some Basic Principles and Techniques - Class 11 Part 2
Concept	Steric Hindrance (B-Strain)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free