Question

Identify the product 'B' in following reaction.
$\text{Ethyl phenyl ketone} \xrightarrow{H_2N-NH_2} A \xrightarrow{KOH, HO-(CH_2)_2-OH} B$

• A. Phenylhydrazone
• B. Ethyl benzene
• C. n-Propyl benzene
• D. Isopropyl benzene

Hint:

Logic Tip: The Wolff-Kishner and Clemmensen reductions are the two most famous ways to strip a carbonyl oxygen and replace it with two hydrogens. The carbon skeleton remains completely unchanged. Ethyl phenyl ketone ($Ph-CO-Et$) becomes $Ph-CH_2-Et$, which is simply a straight 3-carbon chain attached to benzene (n-propyl benzene).

Answer 1

The Correct Option is C

Concept:
The given two-step sequence represents the Wolff-Kishner reduction. This reaction completely reduces the carbonyl group ($-C=O$) of an aldehyde or ketone into a methylene group ($-CH_2-$) to form an alkane.
Step 1: Identify the starting material and the first step.
The starting material is Ethyl phenyl ketone, which has the structure $C_6H_5 - C(=O) - CH_2CH_3$. In the first step, the ketone reacts with hydrazine ($H_2N-NH_2$). Water is eliminated, and a carbon-nitrogen double bond is formed. $$C_6H_5-C(=O)-CH_2CH_3 + H_2N-NH_2 \rightarrow C_6H_5-C(=N-NH_2)-CH_2CH_3 + H_2O$$ The product 'A' is a hydrazone (specifically, ethyl phenyl ketone hydrazone).
Step 2: Identify the second step and the final product.
In the second step, the hydrazone 'A' is heated with a strong base (like $KOH$) in a high-boiling solvent like ethylene glycol ($HO-CH_2CH_2-OH$). Under these conditions, nitrogen gas ($N_2$) is expelled, and the $=N-NH_2$ group is replaced by two hydrogen atoms. $$C_6H_5-C(=N-NH_2)-CH_2CH_3 \xrightarrow{KOH, \Delta} C_6H_5-CH_2-CH_2CH_3 + N_2$$
Step 3: Name the final product 'B'.
The resulting alkane chain attached to the benzene ring has 3 carbon atoms arranged in a straight line: $-CH_2-CH_2-CH_3$. This straight 3-carbon chain is a normal propyl (n-propyl) group. Therefore, the IUPAC name of product 'B' is n-propyl benzene.