Question:

2-chloro-2-methylpentane on reaction with sodium methoxide in methanol yields:

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A tertiary alkyl halide with a strong base/nucleophile in a protic solvent gives both substitution (ether) and elimination products.
Updated On: Nov 11, 2026
  • All of these
  • \((a)\) and \((c)\)
  • \((c)\) only
  • \((a)\) and \((b)\)
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The Correct Option is A

Solution and Explanation

2-Chloro-2-methylpentane is a tertiary alkyl halide, \(\mathrm{CH_3-C(CH_3)Cl-CH_2CH_2CH_3}\).
In methanol with \(\mathrm{CH_3ONa}\), the carbocation/ionising pathway and the strong-base pathway both operate.
Substitution (\(\mathrm{S_N1}\)) gives the ether (a), \(\mathrm{C_2H_5CH_2C(CH_3)_2OCH_3}\).
Elimination: loss of H from a terminal \(\mathrm{CH_3}\) gives the less substituted alkene (b), \(\mathrm{C_2H_5CH_2C(CH_3)=CH_2}\); loss of H from the \(\mathrm{CH_2}\) gives the more substituted Zaitsev alkene (c), \(\mathrm{C_2H_5CH=C(CH_3)_2}\).
All three (a), (b) and (c) are formed.
Answer: (1) All of these
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