Question:

A can be

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\(\mathrm{LiAlH_4}\) attacks less substituted carbon of unsymmetrical epoxides.
Updated On: Apr 23, 2026
  • (a)
  • (b)
  • (c)
  • (d)
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The Correct Option is A

Solution and Explanation

Step 1: Formula / Definition}
\[ \mathrm{LiAlH_4} \rightarrow \mathrm{H^-} \text{ nucleophile} \]
Step 2: Calculation / Simplification}
Unsymmetrical epoxide opening follows \(S_N2\) at less hindered carbon.
\(\mathrm{H^-}\) attacks terminal \(\mathrm{CH_2}\) carbon.
Ring opens forming alkoxide on secondary carbon.
Protonation yields \(\mathrm{CH_3CH(OH)CH_3}\) (Propan-2-ol).
Step 3: Final Answer
\[ \mathrm{CH_3CH(OH)CH_3} \]
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