Before You Begin

Question #44565

Difficulty: 8/10 | 🟢 Free

The Question

Consider the enolization and subsequent silylation of 2-methylcyclohexan-1-one under two distinct sets of conditions:

\textbf{Reaction I:} 1. LDA (1.05 eq), THF, $-78^\circ\text{C}$ \quad 2. $\ce{Me3SiCl}$ $\rightarrow$ Product P (Major) \textbf{Reaction II:} $\ce{Me3SiCl}$ , $\ce{Et3N}$ , DMF, $130^\circ\text{C}$ $\rightarrow$ Product Q (Major)

Let $E_{a(P)}$ and $E_{a(Q)}$ be the activation energies for the rate-determining deprotonation steps leading to the enolate precursors of P and Q, respectively. Let $G^\circ_P$ and $G^\circ_Q$ be the absolute standard free energies of these intermediate enolates.

Which of the following correctly identifies the major products P and Q, along with the thermodynamic and kinetic parameters governing their formation?

Options

A.
P = 6-methyl-1-(trimethylsiloxy)cyclohex-1-ene; Q = 2-methyl-1-(trimethylsiloxy)cyclohex-1-ene; $E_{a(P)} < E_{a(Q)}$ and $G^\circ_P > G^\circ_Q$
B.
P = 2-methyl-1-(trimethylsiloxy)cyclohex-1-ene; Q = 6-methyl-1-(trimethylsiloxy)cyclohex-1-ene; $E_{a(P)} > E_{a(Q)}$ and $G^\circ_P < G^\circ_Q$
C.
P = 6-methyl-1-(trimethylsiloxy)cyclohex-1-ene; Q = 2-methyl-1-(trimethylsiloxy)cyclohex-1-ene; $E_{a(P)} > E_{a(Q)}$ and $G^\circ_P < G^\circ_Q$
D.
P = 2-methyl-1-(trimethylsiloxy)cyclohex-1-ene; Q = 6-methyl-1-(trimethylsiloxy)cyclohex-1-ene; $E_{a(P)} < E_{a(Q)}$ and $G^\circ_P > G^\circ_Q$

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

Reaction Progress Coordinate|Enolate Regioselectivity|Alpha-Hydrogen Acidity

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to map reaction conditions (kinetic vs. thermodynamic control) directly onto a Reaction Progress Coordinate.

The primary trap lies in confusing the activation energy ( $E_a$ ) with the thermodynamic stability ( $G^\circ$ ) of the intermediates.

Students often erroneously assume that the major product always stems from the most stable intermediate, failing to account for how low-temperature, sterically hindered conditions (LDA, $-78^\circ\text{C}$ ) lock the system at the kinetic transition state, which is governed purely by the lower $E_a$ of the less hindered proton.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Chemical Kinetics - Class 12
Concept	Reaction Progress Coordinate
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free