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Question #44422

Difficulty: 8/10 | 🟢 Free

The Question

Consider the reactions of 2,3-dimethylbutan-2-ol ( $\textbf{X}$ ) under three different sets of conditions:

I. $\textbf{X} \xrightarrow[\Delta]{\ce{CH3COOH, H2SO4 (cat.)}} \textbf{P} \text{ (Major)}$

II. $\textbf{X} \xrightarrow{\ce{Na}} \xrightarrow{\ce{CH3I}} \textbf{Q} \text{ (Major)}$

III. $\textbf{X} \xrightarrow{\ce{HBr}} \xrightarrow[\Delta]{\ce{CH3ONa}} \textbf{R} \text{ (Major)}$

Identify the major organic products $\textbf{P}$ , $\textbf{Q}$ , and $\textbf{R}$ .

Options

A.
$\textbf{P}$ : 2,3-dimethylbut-2-ene; $\textbf{Q}$ : 2-methoxy-2,3-dimethylbutane; $\textbf{R}$ : 2,3-dimethylbut-2-ene
B.
$\textbf{P}$ : 2,3-dimethylbutan-2-yl acetate; $\textbf{Q}$ : 2-methoxy-2,3-dimethylbutane; $\textbf{R}$ : 2-methoxy-2,3-dimethylbutane
C.
$\textbf{P}$ : 2,3-dimethylbut-2-ene; $\textbf{Q}$ : 2,3-dimethylbut-2-ene; $\textbf{R}$ : 2-methoxy-2,3-dimethylbutane
D.
$\textbf{P}$ : 2,3-dimethylbutan-2-yl acetate; $\textbf{Q}$ : 2,3-dimethylbut-2-ene; $\textbf{R}$ : 2,3-dimethylbut-2-ene

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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

Esterification Steric Hindrance|Williamson Ether Synthesis Constraints|E2 Regioselectivity (Saytzeff vs Hoffmann)

Why This Question Is Tricky

Common Mistakes

The problem tests the student's ability to evaluate steric constraints across three different reaction mechanisms.

Students often mistakenly assume Fischer esterification works universally, overlooking that tertiary alcohols are too sterically hindered for the $B_{Ac}2$ mechanism and instead undergo E1 dehydration under acidic reflux.

Furthermore, it contrasts the Williamson ether synthesis conditions: a bulky tertiary alkoxide can successfully undergo $S_N2$ with a methyl halide because $\ce{CH3I}$ lacks $\beta$ -hydrogens, preventing E2.

Conversely, reacting a tertiary halide with a primary alkoxide completely shifts the pathway to E2 elimination due to steric hindrance at the electrophilic carbon.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Alcohols, Phenols, and Ethers - Class 12
Concept	Esterification Steric Hindrance
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free