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Question #44024

Difficulty: 8/10 | 🟢 Free

The Question

Consider the organic compound \textbf{X}, which is 3-(dimethylamino)benzoic acid. \textbf{X} is subjected to two separate reaction sequences:\n\nReaction I: \textbf{X} is treated with benzenediazonium chloride ( $PhN_2^+Cl^-$ ) in a weakly acidic buffer (pH $4-5$ ) at $273\text{ K}$ to yield the major product \textbf{P}.\n\nReaction II: \textbf{X} is treated with acetyl chloride ( $CH_3COCl$ , 1 equivalent) in the presence of excess anhydrous $AlCl_3$ , followed by careful aqueous workup, to yield the major product \textbf{Q}.\n\nIdentify \textbf{P} and \textbf{Q}.

Options

A.
\textbf{P} = 5-(dimethylamino)-2-(phenyldiazenyl)benzoic acid; \textbf{Q} = 3-(dimethylamino)benzoic acid
B.
\textbf{P} = 3-(dimethylamino)-4-(phenyldiazenyl)benzoic acid; \textbf{Q} = 2-acetyl-5-(dimethylamino)benzoic acid
C.
\textbf{P} = 5-(dimethylamino)-2-(phenyldiazenyl)benzoic acid; \textbf{Q} = 2-acetyl-5-(dimethylamino)benzoic acid
D.
\textbf{P} = 3-(dimethylamino)benzoic acid; \textbf{Q} = 4-acetyl-3-(dimethylamino)benzoic acid

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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

Friedel-Crafts on Carboxylic Acids|Azo Coupling Regioselectivity|Aniline Resonance Pitfalls

Why This Question Is Tricky

Common Mistakes

This problem tests the critical distinction between electrophilic aromatic substitutions that require Lewis acid catalysis (Friedel-Crafts) and those that do not (Azo coupling).

The hidden trap lies in the Lewis acid-base interaction.

Students often mechanically apply the +M directing rules of the $-N(CH_3)_2$ group to both reactions.

However, they must recognize that under Friedel-Crafts conditions, the anhydrous $AlCl_3$ complexes with the basic nitrogen lone pair, converting the activating $-N(CH_3)_2$ group into a strongly deactivating $-N^+(CH_3)_2(AlCl_3^-)$ adduct.

This renders the ring inert.

Conversely, Azo coupling proceeds without a Lewis acid, allowing the +M effect to successfully drive the reaction.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Aldehydes, Ketones, and Carboxylic Acids - Class 12
Concept	Friedel-Crafts on Carboxylic Acids
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free