Question

Identify 'A' in the following reaction: A + Lithium amide $\rightarrow$ Ethynyl lithium $\rightarrow$ Bromoethane $\rightarrow$ But-1-yne

• A. Ethene
• B. Ethyne
• C. But-1-ene
• D. But-2-ene

Hint:

Terminal alkynes (like ethyne) are acidic because the $sp$-hybridized carbon atom is highly electronegative. The greater s-character (50%) strongly stabilizes the negative charge formed after deprotonation.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Ethynyl lithium is a salt derived from ethyne (acetylene) by replacing an acidic hydrogen with lithium.

Step 2: Formula Application:
$A + LiNH_2 \rightarrow HC \equiv CLi + NH_3$.

Step 3: Explanation:
Lithium amide ($LiNH_2$) is a strong base that deprotonates terminal alkynes. Ethyne ($C_2H_2$) reacts with $LiNH_2$ to form Ethynyl lithium. This nucleophile then reacts with Bromoethane via $S_N2$ to form But-1-yne ($CH_3CH_2C \equiv CH$). Therefore, 'A' must be Ethyne.

Step 4: Final Answer:
The starting compound 'A' is Ethyne.