Before You Begin

Question #44627

Difficulty: 8/10 | 🟢 Free

The Question

Consider the elimination reactions of (2S,3S)-2-bromo-3-methylpentane under two different sets of conditions:

Reaction I: $t$ -BuOK in $t$ -BuOH at 323 K $\rightarrow$ Major product $P$ Reaction II: $t$ -BuOK in DMSO at 323 K $\rightarrow$ Major product $Q$

Identify the major products $P$ and $Q$ .

Options

A.
$P$ = (3S)-3-methyl-1-pentene; $Q$ = (Z)-3-methyl-2-pentene
B.
$P$ = (3S)-3-methyl-1-pentene; $Q$ = (E)-3-methyl-2-pentene
C.
$P$ = (3R)-3-methyl-1-pentene; $Q$ = (Z)-3-methyl-2-pentene
D.
$P$ = (E)-3-methyl-2-pentene; $Q$ = (3S)-3-methyl-1-pentene

Check Your Answer

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

E1/E2 Solvent and Base Effects|Basicity-Solvation Competition

Why This Question Is Tricky

Common Mistakes

Students typically rely on the heuristic that bulky bases like t-BuOK always yield the Hofmann product.

This question tests the 'Basicity-Solvation Competition' exception.

In a protic solvent (t-BuOH), hydrogen bonding heavily solvates the alkoxide, increasing its steric bulk and favoring the kinetic Hofmann product.

However, in a polar aprotic solvent (DMSO), the t-BuO- anion is unsolvated, significantly reducing its effective size and massively increasing its basicity.

This allows it to abstract the more hindered proton, shifting regioselectivity to the thermodynamic Zaitsev product.

Additionally, the problem rigorously tests stereochemical tracking during E2 anti-elimination and retention of configuration at untouched chiral centers.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Haloalkanes and Haloarenes - Class 12
Concept	E1/E2 Solvent and Base Effects
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free