Question

Which of the following cannot be prepared by the reduction of either a ketone or an aldehyde with NaBH\(_4\) in methanol?

• A. 2-Butanol
• B. 2-Methyl 2-propanol
• C. 2-Methyl 1-propanol
• D. 1-Butanol
• E. 2-Phenylethanol

Hint:

Sodium borohydride (NaBH\(_4\)) selectively reduces aldehydes and ketones to alcohols. However, the formation of a compound like 2-methyl-2-propanol requires a different starting material, as it cannot be directly synthesized from a simple aldehyde or ketone.

Answer 1

The Correct Option is B

Step 1: Sodium borohydride (NaBH\(_4\)) is a selective reducing agent that is commonly used to reduce aldehydes and ketones to their corresponding alcohols. In methanol, NaBH\(_4\) will reduce the carbonyl group (\( C=O \)) of an aldehyde or ketone to a hydroxyl group (\( -OH \)).

Step 2: Let's examine each option:

- Option (A) 2-Butanol: This can be prepared by reducing 2-butanone (a ketone) with NaBH\(_4\) in methanol, resulting in 2-butanol.

- Option (C) 2-Methyl-1-propanol: This can be prepared by reducing 2-methylpropanal (an aldehyde) with NaBH\(_4\) in methanol, resulting in 2-methyl-1-propanol.

- Option (D) 1-Butanol: This can be prepared by reducing butanal (an aldehyde) with NaBH\(_4\) in methanol, resulting in 1-butanol.

- Option (E) 2-Phenylethanol: This can be prepared by reducing acetophenone (a ketone) with NaBH\(_4\) in methanol, resulting in 2-phenylethanol.

Step 3: Option (B) 2-Methyl-2-propanol cannot be prepared by the reduction of either a ketone or an aldehyde because there is no simple aldehyde or ketone that can be reduced to 2-methyl-2-propanol. The structure of 2-methyl-2-propanol requires a tert-butyl group, which is not achievable through the reduction of simple aldehydes or ketones with NaBH\(_4\).

Thus, the correct answer is option (B).