Question:medium

In the following reaction, the major product (H) is

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Using a simple frame or just bolding for the box Key Points: Grignard reagents (RMgX) add to ketones/aldehydes to form alkoxides. Alkoxides are strong nucleophiles. Intramolecular S$_N$2 reactions can occur if a nucleophile and a suitable leaving group are present in the same molecule, appropriately positioned (favored for 5/6-membered rings). The sequence is: 1) Grignard addition, 2) Intramolecular cyclization.
Updated On: Nov 28, 2025
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The Correct Option is C

Solution and Explanation

The reaction unfolds in two main stages: (A) Grignard Reaction: The nucleophilic methyl group from MeMgBr attacks the electrophilic carbonyl carbon of 5-chloro-2-pentanone. This forms a tertiary alkoxide intermediate as the C=O \(\pi\)-bond breaks.

Cl-CH2-CH2-CH2-C(=O)-Me + MeMgBrCl-CH2-CH2-CH2-C(O-MgBr+)(Me)-Me (Alkoxide Intermediate)
(B) Intramolecular SN2 Cyclization: The negatively charged oxygen (alkoxide) acts as an internal nucleophile and attacks the carbon bonded to chlorine (an electrophilic site). Chlorine acts as the leaving group, forming a stable five-membered ring (tetrahydrofuran derivative). This intramolecular reaction is faster than intermolecular reactions or protonation by trace protic sources before workup, especially for forming 5- or 6-membered rings.

Intramolecular attack leading to ring closure:
O--CH2-CH2-CH2-C(Me)(Me)-ClO-CH2-CH2-CH2-C(Me)(Me) + Cl-
(C) Workup: The acidic workup (H3O+) neutralizes remaining reagents/byproducts but does not alter the already formed cyclic ether product (H). The final major product (H) is 2,2-dimethyltetrahydrofuran, corresponding to option (C).
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