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Question #44362

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The Question

An optically pure sample of (1S)-1-bromo-2,2-dimethyl-1-phenylpropane is heated with a concentrated solution of \ce{NaOCH3} in \ce{CH3OH}. The reaction yields 1-methoxy-2,2-dimethyl-1-phenylpropane. Which of the following correctly describes the stereochemical outcome of this product and the underlying kinetic rationale?

Options

A.
Complete inversion of configuration yielding the (1R)-isomer; the absence of $\beta$ -hydrogens completely suppresses E2 elimination, allowing $S_N2$ to proceed with 100% stereospecificity.
B.
Racemization yielding a nearly 1:1 mixture of (1R) and (1S) isomers; extreme steric hindrance from the \textit{tert}-butyl group blocks the $S_N2$ trajectory, forcing a unimolecular $S_N1$ mechanism.
C.
Complete retention of configuration yielding the (1S)-isomer; the \ce{CH3OH} solvent forms a tight ion-pair with the leaving group, directing front-side attack via an $S_Ni$ mechanism.
D.
Racemization yielding a nearly 1:1 mixture of (1R) and (1S) isomers; the strong base abstracts the benzylic proton to form a planar carbanion, which subsequently undergoes substitution.

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Mechanism_Trap

Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.

Concepts Behind This Question

Optical Activity and Enantiomers|SN2 Stereochemical Inversion|Steric Hindrance in SN2

Why This Question Is Tricky

Common Mistakes

The problem tests the competition between kinetic accessibility and thermodynamic feasibility in sterically hindered chiral systems.

The 'Hidden Variable' is the extreme steric hindrance of the tert-butyl group on a secondary carbon.

Students typically apply the rote rule: 'Secondary halide + Strong Nucleophile = SN2 (Inversion)'.

Furthermore, noticing the absence of beta-hydrogens reinforces their false confidence in a pure SN2 pathway.

The trap is failing to recognize that the neopentyl-like steric bulk completely blocks the SN2 trajectory, forcing the molecule to undergo an SN1 reaction (racemization) despite the presence of a strong nucleophile.

The relief of B-strain further facilitates this unexpected SN1 pathway.

Quick Facts About This Question

Question Information

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