Question

A compound (X) produces an alkene when treated with H$_2$ in the presence of Na/liquid NH$_3$. Ozonolysis of this alkene gives ethanol and methanal. "X" is

• A. 1-Butyne
• B. 2-Butyne
• C. Ethyne
• D. Propyne

Hint:

In ozonolysis problems, always identify the number of carbon atoms in the products. Methanal (formaldehyde) is a strong indicator of a terminal double bond. Use this clue to reverse-engineer the parent compound.

Answer 1

The Correct Option is D

Step 1: Understand the reduction conditions.
Na in liquid NH$_3$ gives a trans-alkene from an alkyne due to partial reduction. Step 2: Analyze the ozonolysis products.
Ozonolysis giving ethanal and methanal means the double bond is between a methyl and an ethyl group: CH$_3$–CH=CH$_2$. Step 3: Work backwards.
The only alkyne that on partial reduction gives such an alkene is propyne, CH$_3$–C$\equiv$CH. Step 4: Verify by reaction:
\[ \text{Propyne} \xrightarrow[\text{liq. NH}_3]{\text{Na/H}_2} \text{CH}_3–CH=CH_2 \xrightarrow[\text{O}_3]{\text{Reductive work-up}} \text{CH}_3CHO + HCHO \] Hence, X = Propyne.