Kerala KEAM 5 June 2024 Solved Paper

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).Quick Tip: 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.