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Question #44043

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The Question

Consider the optically active alcohol $\textbf{P}$ , which is $(2R, 3S)$ -3-phenylbutan-2-ol. $\textbf{P}$ undergoes two distinct reaction sequences to form primary amines $\textbf{R}$ and $\textbf{T}$ :

$\textbf{Sequence 1:}$

$\textbf{P} \xrightarrow{\text{Phthalimide, } PPh_3\text{, DEAD}} \textbf{Q}$
$\textbf{Q} \xrightarrow{NH_2NH_2\text{, EtOH, }\Delta} \textbf{R}$

$\textbf{Sequence 2:}$

$\textbf{P} \xrightarrow{SOCl_2\text{, Pyridine}} \textbf{S}$
$\textbf{S} \xrightarrow{\text{Excess } NH_3\text{ (aq), sealed tube, } 373\text{ K}} \textbf{T}$ (Major primary amine)

Which of the following statements correctly describes the stereochemical relationship between $\textbf{R}$ and $\textbf{T}$ ?

Options

A.
$\textbf{R}$ is a single enantiomer with $(2S, 3S)$ configuration, while $\textbf{T}$ is a racemic mixture of $(2S, 3S)$ and $(2R, 3R)$ isomers.
B.
$\textbf{R}$ is a single enantiomer with $(2S, 3S)$ configuration, while $\textbf{T}$ is a single enantiomer with $(2R, 3S)$ configuration.
C.
Both $\textbf{R}$ and $\textbf{T}$ are the same single enantiomer with $(2S, 3S)$ configuration.
D.
$\textbf{R}$ is a racemic mixture of $(2S, 3S)$ and $(2R, 3R)$ isomers, while $\textbf{T}$ is a single enantiomer with $(2R, 3S)$ configuration.

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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

Gabriel Phthalimide Synthesis|Ammonolysis Complexity

Why This Question Is Tricky

Common Mistakes

This problem tests the critical distinction between the stereospecific Gabriel-Mitsunobu synthesis and the complex realities of direct ammonolysis.

The 'Hidden Variable' is Neighboring Group Participation (NGP) by the phenyl ring.

Students often assume ammonolysis will proceed via a simple $S_N2$ inversion (which would incorrectly yield the $(2R, 3S)$ product).

However, the intermediate $(2S, 3S)$ -2-chloro-3-phenylbutane (\textit{threo}) forms a $C_2$ -symmetric phenonium ion, leading to complete racemization.

This perfectly illustrates why Gabriel synthesis is synthetically superior for producing pure primary amines, prioritizing stereochemical selectivity over atom economy.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Amines - Class 12
Concept	Gabriel Phthalimide Synthesis
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free