Question

(D) $\underset{\left(ii\right)C_2{H}_{5}ONa}{\overset{\left(i\right)N{H}_{2}N{H}_2}{\to }}$ CH${}_3$CH${}_2$CH${}_3$

$D$ in the given reaction is:

• A. Iso-propyl alcohol
• B. Butanaldehyde
• C. Acetone
• D. None of these

Answer 1

Answer: (C)

Acetone

This is Wolff-Kishner reduction process, where reduction of an aldehyde or ketone carbonyl group $(C=O)$ into a methylene group $\left(C{H}_2\right)$ takes place.

The aldehyde or ketone is condensed with hydrazine to form a hydrazone. This hydrazone is then reduced by reaction with a strong base like sodium ethoxide to form a carbanion intermediate, which is subsequently protonated.

Among all options, acetone has three carbon (as in product) and having $-C=O$ group, hence undergoes given reaction and gives propane.