Question

\(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C} \mathrm{MgBr}\) can be prepared by the reaction of
(A) \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{Br}\) with \(\mathrm{MgBr}_{2}\)
(B) \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CH}\) with \(\mathrm{MgBr}_{2}\)
(C) \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CH}\) with \(\mathrm{KBr}\) and \(\mathrm{Mg}\) metal
(D) \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CH}\) with \(\mathrm{CH}_{3} \mathrm{MgBr}\)

Answer 1

Ans: (D)
 \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{H}\) is a terminal alkyne, which has acidic hydrogen.
\(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{H}+\mathrm{CH}_{3} \mathrm{MgBr} \rightarrow \mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CMg} \mathrm{Br}+\mathrm{CH}_{4}\)

Answer 2

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Answer 3

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