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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