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Question #44414
The Question
Compound \textbf{P} is methyl 4,5-epoxy-5-methylhexanoate. \textbf{P} undergoes the following reaction sequence:
- \textbf{P} is treated with excess \ce{LiAlH4} in dry THF followed by acidic workup to yield compound \textbf{Q}.
- \textbf{Q} is reacted with 1 equivalent of acetyl chloride (\ce{CH3COCl}) in the presence of pyridine at to yield the major product \textbf{R}.
- \textbf{R} is treated with Lucas reagent (\ce{ZnCl2} / conc. \ce{HCl}) at to yield compound \textbf{S}.
Identify the IUPAC name of compound \textbf{S}.
Options
- A.
5-chloro-5-methylhexyl acetate
- B.
4-chloro-5-methylhexyl acetate
- C.
6-chloro-2-methylhexan-2-yl acetate
- D.
Methyl 5-chloro-5-methylhexanoate
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Correct Answer
Step-by-Step Solution
Explanation
How the Examiner Thinks
Exception_Exposure
Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.
Concepts Behind This Question
Lucas Test Logic|Esterification Steric Hindrance|Reduction Selectivity (Hydrides)
Why This Question Is Tricky
Common Mistakes
This question tests the opposing regioselectivities of two common reactions.
In esterification, primary alcohols react much faster than tertiary alcohols due to steric hindrance.
Conversely, in the Lucas test (SN1), tertiary alcohols react much faster due to carbocation stability.
The problem forces the student to correctly assign these competing reactivities in a single molecule.
It also traps students who misapply the regioselectivity of epoxide opening (acidic vs basic conditions) or fail to recognize that \ce{LiAlH4} reduces esters, unlike \ce{NaBH4}.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Alcohols, Phenols, and Ethers - Class 12 |
| Concept | Lucas Test Logic |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |