Organic Reactions

The given compound is CH₃–CH=CH–CH₂–CHO, an α,β-unsaturated aldehyde (4-pentenal).

Step 1: Reaction with LiAlH₄
Lithium aluminium hydride (LiAlH₄) is a strong reducing agent. It reduces aldehydes (–CHO) to primary alcohols (–CH₂OH) but does not reduce C=C double bonds.

The reduction of CH₃–CH=CH–CH₂–CHO gives CH₃–CH=CH–CH₂–CH₂–OH.

Step 2: Acidic workup with H₃O⁺
This step neutralizes the reaction mixture and protonates the alkoxide intermediate to yield the final alcohol.

Therefore, the overall reaction is:

CH₃–CH=CH–CH₂–CHO ⟶ (i) LiAlH₄ / (ii) H₃O⁺ ⟶ CH₃–CH=CH–CH₂–CH₂–OH

Hence, the product is:
CH₃–CH=CH–CH₂–CH₂–OH

Conversion of Ethanol to Ethyl Bromide (A): Ethanol reacts with bromine in the presence of red phosphorus to form ethyl bromide. So, Ethyl Bromide .
Conversion of Ethyl Bromide to Propanenitrile (B): Ethyl bromide reacts with alcoholic potassium cyanide to form propanenitrile. So, Propanenitrile .
Conversion of Propanenitrile to Propylamine (C): Propanenitrile undergoes reduction with lithium aluminum hydride in ether to form propylamine. So, Propylamine .

The Cannizzaro reaction is a classic redox process where non-enolizable aldehydes, which lack an -hydrogen, are treated with a strong base ( ) to undergo disproportionation. This reaction results in the simultaneous formation of an alcohol and a carboxylate salt.
Consider the reaction with methanal (formaldehyde):

The given compound is butanone . When treated with hydrogen in the presence of a nickel catalyst and heat , the reaction involves the reduction of the carbonyl group into a hydroxyl group, resulting in the formation of a secondary alcohol. The reaction proceeds as follows: Thus, the correct product is a secondary alcohol, which is .
The correct answer is (B).