CH2=CH2→Br2/CCl4A→alc.KOH/ΔB→excessNaNH2/ΔC→CH3−Br(2eq)DThe compound D is

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${CH}_{2} = {CH}_{2} \overset{{Br}_{2} / {CCl}_{4}}{\to} A \overset{alc. KOH / Δ}{\to}$ $B \to_{e x c e s s}^{{NaNH}_{2} / Δ} C \overset{{CH}_{3} - Br (2 e q)}{\to} D$

The compound D is

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3 - hexyne

2 - pentyne

2 - butyne

1 - pentyne

answer is A.

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Detailed Solution

${Na}^{+} : C^{Θ} \equiv C^{Θ} : {Na}^{+} \to_{2 e q}^{{CH}_{3} Br}$

${CH}_{3} - C \equiv C - {CH}_{3}$

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