\text{CH}\_3\text{-CH}2\text{-C} \equiv \text{CH} \rightleft... | Chemistry - Alkynes (Reactions and Isomerization) | SolveeIt | Solveeit

Today, August 19

Question

$\text{CH}_3\text{-CH}_2\text{-C} \equiv \text{CH} \rightleftharpoons^{\text{A}}_{\text{B}} \text{CH}_3\text{C} \equiv \text{C-CH}_3$

Visual representation for Chemistry

A

alcoholic KOH and NaNH2

B

NaNH2 and alcoholic KOH

C

NaNH2 and Lindlar

D

Lindlar and NaNH2

Answer

NaNH2 and alcoholic KOH

Explanation

Solution

The terminal alkyne (1‐butyne) is first deprotonated by a strong base. NaNH₂ (a very strong base) effectively removes the acidic terminal hydrogen, forming an acetylide ion that rearranges to form the internal alkyne (2‐butyne). In the reverse reaction, alcoholic KOH (a milder base) is used, thereby making the process reversible under controlled conditions.