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Question #44618

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The Question

Consider the reaction of (2S)-2-bromobutane with $1.0\text{ M } \ce{NaOEt}$ at $330\text{ K}$ . When performed in ethanol ( $\ce{EtOH}$ ), the activation energy for the $\text{S}_\text{N}2$ pathway is $E_{a,1}$ and for the $\text{E}2$ pathway is $E_{a,2}$ . When the solvent is switched to dimethyl sulfoxide (DMSO) under identical conditions, the activation energy for the $\text{S}_\text{N}2$ pathway is $E_{a,3}$ . Identify the major product formed in $\ce{EtOH}$ and the correct thermodynamic relationship between the activation energies.

Options

A.
trans-2-butene, $E_{a,2} < E_{a,1}$ and $E_{a,1} > E_{a,3}$
B.
(2R)-2-ethoxybutane, $E_{a,1} < E_{a,2}$ and $E_{a,1} < E_{a,3}$
C.
cis-2-butene, $E_{a,2} < E_{a,1}$ and $E_{a,1} < E_{a,3}$
D.
(2S)-2-ethoxybutane, $E_{a,1} > E_{a,2}$ and $E_{a,1} > E_{a,3}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Numerical_Precision

Improve calculation stamina and avoid rounding errors in high-stakes competitive examinations.

Concepts Behind This Question

SN2 Stereochemical Inversion|Reaction Progress Coordinate|E1/E2 Solvent and Base Effects

Why This Question Is Tricky

Common Mistakes

The core trap tests the student's ability to distinguish between the kinetic solvent effect on nucleophilicity versus basicity.

Students often memorize that 'strong nucleophiles favor SN2,' overlooking that in polar protic solvents, hydrogen bonding selectively suppresses nucleophilicity (raising SN2 activation energy) much more than basicity.

The hidden variable is the ground-state stabilization of the nucleophile, which alters the reaction coordinate, making the E2 pathway kinetically favored ( $E_{a,2} < E_{a,1}$ ) despite the presence of a good nucleophile.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Haloalkanes and Haloarenes - Class 12
Concept	SN2 Stereochemical Inversion
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free