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Question #44366

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The Question

An enantiomerically pure sample of $(2S)$ -2-bromobutane is treated with a solution of potassium \textit{tert}-butoxide in \textit{tert}-butanol at $350\text{ K}$ . Which of the following best describes the major product formed and its optical activity?

Options

A.
$(2R)$ -2-\textit{tert}-butoxybutane, Optically active
B.
But-1-ene, Optically inactive
C.
$(E)$ -But-2-ene, Optically inactive
D.
Racemic 2-\textit{tert}-butoxybutane, Optically inactive

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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

Optical Activity and Enantiomers|Williamson Ether Synthesis Constraints|SN2 Stereochemical Inversion

Why This Question Is Tricky

Common Mistakes

This problem tests the kinetic competition between $S_N2$ substitution and $E2$ elimination, specifically targeting the 'Williamson Ether Synthesis' trap.

Students often rote-memorize that an alkoxide plus an alkyl halide yields an ether with $S_N2$ stereochemical inversion (predicting the $(2R)$ -ether).

The 'Chemical Exception' is the use of a sterically hindered base ( $\ce{^-O^tBu}$ ) with a secondary halide, which completely shuts down $S_N2$ in favor of $E2$ .

Furthermore, the steric bulk alters the standard Zaitsev regioselectivity, leading to the achiral Hofmann product (but-1-ene) as the major product.

The transition from a chiral substrate to an achiral alkene tests deep mechanistic understanding over simplified flowcharts.

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