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 Question:
We are given a multi-step synthesis starting from an unknown compound 'A' reacting with a strong base (lithium amide) to form ethynyl lithium.
We must determine the identity of the original starting material 'A'.

Step 2: Detailed Explanation:
Lithium amide ($LiNH_2$) is a very strong base. It is capable of abstracting acidic protons.
Among the given options, only terminal alkynes possess an acidic hydrogen atom.
Compound 'A' reacts with $LiNH_2$ to form ethynyl lithium ($HC \equiv C^- Li^+$).
The only hydrocarbon that can yield the ethynyl anion upon deprotonation is ethyne ($HC \equiv CH$).
The complete reaction sequence is:
1. Ethyne ($HC \equiv CH$) + $LiNH_2 \rightarrow$ Ethynyl lithium ($HC \equiv C^- Li^+$) + $NH_3$
2. Ethynyl lithium ($HC \equiv C^- Li^+$) + Bromoethane ($CH_3CH_2Br$) $\rightarrow$ But-1-yne ($HC \equiv C-CH_2-CH_3$) + $LiBr$ (via an $S_N2$ reaction).

Step 3: Final Answer:
The starting material 'A' is Ethyne, matching option (b).