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Question #43979

Difficulty: 8/10 | 🟢 Free

The Question

Compound $P$ is 1,1,1-trichloro-2-(4-chlorophenyl)-3-nitropropane. \begin{itemize} \item \textbf{Reaction 1:} $P$ is treated with $Sn/HCl$ followed by basification to yield major product $X$ . \item \textbf{Reaction 2:} $P$ is treated with excess $LiAlH_4$ in dry THF followed by aqueous workup to yield major product $Y$ . \end{itemize} Identify $X$ and $Y$ .

Options

A.
$X$ = 3,3,3-trichloro-2-(4-chlorophenyl)propan-1-amine ; $Y$ = 2-(4-chlorophenyl)propan-1-amine
B.
$X$ = 3,3,3-trichloro-2-(4-chlorophenyl)propan-1-amine ; $Y$ = 2-phenylpropan-1-amine
C.
$X$ = 2-(4-chlorophenyl)propan-1-amine ; $Y$ = 3,3,3-trichloro-2-(4-chlorophenyl)propan-1-amine
D.
$X$ = 3,3,3-trichloro-2-phenylpropan-1-amine ; $Y$ = 2-(4-chlorophenyl)propan-1-amine

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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

DDT and Freon Chemistry|Reduction Selectivity (Sn/HCl vs LiAlH4)|Carboxylic Acid Strength

Why This Question Is Tricky

Common Mistakes

This question tests the chemoselectivity of reducing agents on polyhalogenated nitro compounds.

The trap is the assumption that metallic reducers ( $Sn/HCl$ ) and hydride donors ( $LiAlH_4$ ) have identical reactivity profiles towards halogens.

Students often forget that $LiAlH_4$ readily reduces aliphatic $C-Cl$ bonds (like in DDT analogs) to alkanes via hydrogenolysis, but cannot reduce aryl $C-Cl$ bonds.

Conversely, $Sn/HCl$ selectively reduces the nitro group without affecting any halogens.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Haloalkanes and Haloarenes - Class 12
Concept	DDT and Freon Chemistry
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free