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Question #44932

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The Question

Consider the substrate \textbf{X}, ethyl 4-(bromomethyl)benzoate ( $p\text{-BrCH}_2\text{-C}_6\text{H}_4\text{-COOCH}_2\text{CH}_3$ ). Three different synthetic pathways are employed to convert the bromomethyl group into a primary amine:

\textbf{Pathway 1:} \textbf{X} is reacted with potassium phthalimide in DMF, followed by refluxing with excess hydrazine ( $\text{NH}_2\text{NH}_2$ ) in ethanol to yield major product \textbf{P}.

\textbf{Pathway 2:} \textbf{X} is reacted with $\text{NaN}_3$ in DMF, followed by catalytic hydrogenation ( $\text{H}_2$ , $\text{Pd/C}$ ) at 298 K to yield major product \textbf{Q}.

\textbf{Pathway 3:} \textbf{X} is reacted with $\text{AgNO}_2$ to isolate the major aliphatic nitro compound, which is subsequently treated with excess $\text{LiAlH}_4$ in THF followed by aqueous workup to yield major product \textbf{R}.

Identify the major products \textbf{P}, \textbf{Q}, and \textbf{R}.

Options

A.
\textbf{P}: $p\text{-H}_2\text{N-CH}_2\text{-C}_6\text{H}_4\text{-CONHNH}_2$ ; \textbf{Q}: $p\text{-H}_2\text{N-CH}_2\text{-C}_6\text{H}_4\text{-COOCH}_2\text{CH}_3$ ; \textbf{R}: $p\text{-H}_2\text{N-CH}_2\text{-C}_6\text{H}_4\text{-CH}_2\text{OH}$
B.
\textbf{P}: $p\text{-H}_2\text{N-CH}_2\text{-C}_6\text{H}_4\text{-COOCH}_2\text{CH}_3$ ; \textbf{Q}: $p\text{-H}_2\text{N-CH}_2\text{-C}_6\text{H}_4\text{-COOCH}_2\text{CH}_3$ ; \textbf{R}: $p\text{-H}_2\text{N-CH}_2\text{-C}_6\text{H}_4\text{-COOCH}_2\text{CH}_3$
C.
\textbf{P}: $p\text{-H}_2\text{N-CH}_2\text{-C}_6\text{H}_4\text{-COOCH}_2\text{CH}_3$ ; \textbf{Q}: $p\text{-H}_2\text{N-CH}_2\text{-C}_6\text{H}_4\text{-CONHNH}_2$ ; \textbf{R}: $p\text{-HONH-CH}_2\text{-C}_6\text{H}_4\text{-CH}_2\text{OH}$
D.
\textbf{P}: $p\text{-H}_2\text{N-CH}_2\text{-C}_6\text{H}_4\text{-COOH}$ ; \textbf{Q}: $p\text{-H}_2\text{N-CH}_2\text{-C}_6\text{H}_4\text{-COOCH}_2\text{CH}_3$ ; \textbf{R}: $p\text{-H}_2\text{N-CH}_2\text{-C}_6\text{H}_4\text{-COOCH}_2\text{CH}_3$

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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

Reduction Selectivity (Sn/HCl vs LiAlH4)|Gabriel Phthalimide Synthesis|Ammonolysis Complexity

Why This Question Is Tricky

Common Mistakes

The core trap lies in 'Functional Group Compatibility' and 'Chemoselectivity'.

Students often memorize reagent-to-product transformations (e.g., Gabriel Phthalimide = primary amine, LiAlH4 = reduction to amine) without considering collateral damage to other sensitive functional groups.

The 'Hidden Variable' here is the ester group.

In Pathway 1, the hydrazine used for phthalimide cleavage is highly nucleophilic and destroys the ester, forming a hydrazide.

In Pathway 3, LiAlH4 acts as a 'chemical sledgehammer', indiscriminately reducing both the nitro group and the ester.

Only the azide pathway (Pathway 2) offers the chemoselectivity required to preserve the ester.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Amines - Class 12
Concept	Reduction Selectivity (Sn/HCl vs LiAlH4)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free