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Question #45093

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The Question

Consider the reaction of 4-methylpent-3-en-2-one (Mesityl oxide) under two different sets of conditions:

Reaction I: 1. LDA (1.05 eq), THF, $-78 ^\circ\text{C}$ ; 2. TMSCl Reaction II: 1. KH (1.05 eq), THF, $25 ^\circ\text{C}$ ; 2. TMSCl

Identify the major products P and Q formed in Reaction I and Reaction II, respectively.

Options

A.
P = 4-methyl-2-(trimethylsiloxy)penta-1,3-diene; Q = 2-methyl-4-(trimethylsiloxy)penta-1,3-diene
B.
P = 2-methyl-4-(trimethylsiloxy)penta-1,3-diene; Q = 4-methyl-2-(trimethylsiloxy)penta-1,3-diene
C.
P = 4-methyl-1-(trimethylsilyl)pent-3-en-2-one; Q = 4-methyl-3-(trimethylsilyl)pent-4-en-2-one
D.
P = 4-methyl-2-(trimethylsiloxy)penta-1,3-diene; Q = 4-methyl-5-(trimethylsilyl)pent-3-en-2-one

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

Bond Cleavage Energetics|Enolate Regioselectivity|Resonance Energy and Stability

Why This Question Is Tricky

Common Mistakes

The question tests the differentiation between kinetic and thermodynamic enolates in extended conjugated systems.

The primary trap is failing to recognize that thermodynamic control favors the linearly conjugated $\gamma$ -dienolate over the cross-conjugated $\alpha$ -enolate due to superior resonance stabilization.

A secondary trap exploits the HSAB principle: students often incorrectly predict C-alkylation (typical for alkyl halides) instead of O-silylation (required for hard silyl chlorides), which is tested via distractors C and D.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Organic Chemistry – Some Basic Principles and Techniques - Class 11 Part 2
Concept	Bond Cleavage Energetics
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free