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Question #44420

Difficulty: 8/10 | 🟢 Free

The Question

Consider the following reaction sequence starting from 2,2-dimethylhexane-1,5-diol (Compound $\textbf{A}$ ):

$\textbf{A} + \ce{CH3COOH}$ (1 equivalent) $\xrightarrow{\ce{H2SO4} \text{ (cat.)}, 298 \text{ K}}$ $\textbf{P}$ (Major Product)
$\textbf{P} \xrightarrow{\ce{CrO3}, \ce{H2SO4}, \ce{H2O}, \text{ acetone}, 0^\circ\text{C}}$ $\textbf{Q}$
$\textbf{Q} \xrightarrow{1. \ce{KOH} \text{ (aq, excess), } \Delta \quad 2. \text{ dil. } \ce{HCl}, \text{ gentle warming}}$ $\textbf{R}$ (Major Product)

Identify the correct IUPAC name for the final major product $\textbf{R}$ .

Options

A.
3,3,6-trimethyltetrahydro-2H-pyran-2-one
B.
2,5,5-trimethyltetrahydro-2H-pyran-2-ol
C.
5-hydroxy-2,2-dimethylhexanoic acid
D.
6,6-dimethyltetrahydro-2H-pyran-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

Esterification Steric Hindrance|Nucleophilic Addition Sterics|Nucleophilic Acyl Substitution

Why This Question Is Tricky

Common Mistakes

The question tests the limits of the '1° > 2° > 3°' heuristic for alcohol reactivity in Fischer esterification.

Students often fail to account for the severe steric penalty of a neopentyl group in the sp³ tetrahedral intermediate of an acyl substitution.

If they blindly follow the heuristic, they acetylate the 1° alcohol, leaving the 2° alcohol free.

Jones oxidation would then yield a ketone, which upon hydrolysis and cyclization forms a cyclic hemiketal (Option B).

Recognizing that the unhindered 2° alcohol is kinetically favored correctly routes the synthesis through a carboxylic acid intermediate, culminating in a Thorpe-Ingold accelerated lactonization to a delta-lactone.

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