Question

$Ph-CH_{2}-C\equiv CH\rightleftarrows_{Y}^{X}Ph-C\equiv C-CH_{3}$ .
Predict the reagents X and Y respectively.

• A. Lindlar catalyst, $NaNH_{2}$
• B. $NaNH_{2}$ and alc. $KOH$
• C. Pt catalyst, Wilkinson's catalyst
• D. Alc. $KOH$ and $NaNH_{2}$

Answer 1

Answer: (D)

Internal alkynes are more stable than terminal alkynes (because of hyperconjugation), isomerization is favoured thermodynamically. The reaction is carried out with moderately strong bases, at high temperatures (> $100^\circ C$ ), which are not able to completely deprotonate terminal alkynes.
The deciding step is the tautomerization of the acetylide anion to the propargyl anion which is stabilized by mesomerism.
On heating with alc. $KOH$ in an inert solvent, the triple bond of $1$ -alkyne is shifted towards the centre to form an isomeric $2$ -alkyne.
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The triple bond migrates from the terminal position into the $C-C$ chain.

Isomerization in the opposite direction leading to the formation of a terminal alkyne can be accomplished with strong bases, e.g. sodium amide at $150^\circ C$ , which is able to completely deprotonate terminal alkynes.



The reaction proceeds in the opposite direction because the most stable anion (acetylide) is formed under the strongly basic conditions and not the more stable hydrocarbon (internal alkyne) which is formed under less basic conditions.