Question

A can be

• (a)
• (b)
• (c)
• (d)

Hint:

\(\mathrm{LiAlH_4}\) attacks less substituted carbon of unsymmetrical epoxides.

Answer 1

The Correct Option is A

The reaction in question involves the reduction of an epoxide using Lithium Aluminium Hydride (LiAlH₄) followed by hydrolysis. LiAlH₄ is a strong reducing agent and can open the epoxide ring to form an alcohol.

The given epoxide is:

The structure of the epoxide can be simplified as CH₃-CH(CH₂)-O. When treated with LiAlH₄, the following reaction takes place:

1. The LiAlH₄ attacks the less hindered carbon in the epoxide, leading to the opening of the ring.
2. The resulting alkoxide ion is then protonated during the acidic workup (H⁺), forming an alcohol.

As a result, the product A can be described as:

\(CH_3CH_2CH(OH)CH_3\)

This corresponds to option (a): CH₃CH(OH)CH₂CH₃, which is actually a mistake in labeling. Option (a) should be CH₃CH(CH₃)OH, which corresponds effectively to 2-butanol.

Therefore, the correct answer is option (a) which represents 2-butanol, formed from the reduction and ring opening of the given epoxide.