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Question: Answered & Verified by Expert
$\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C} \mathrm{MgBr}$ can be prepared by the
reaction of
ChemistryHydrocarbonsWBJEEWBJEE 2018
Options:
  • A $\mathrm{CH}_{3}-\mathrm{C}=\mathrm{C}-\mathrm{Br}$ 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 $\mathrm{CH}_{3} \mathrm{MgBr}$
Solution:
2877 Upvotes Verified Answer
The correct answer is: $\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CH}$ with $\mathrm{CH}_{3} \mathrm{MgBr}$
$\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CMg}$ Br can be prepared by the reaction of $\mathrm{CH}_{3}-\mathrm{C}=\mathrm{CH}$ with $\mathrm{CH}_{3} \mathrm{MgBr}$. This
method is used in preparation of higher alkynes from lower alkynes. The chemical equation for the formation of $\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CMgBr}$ is given below
$$
\begin{array}{r}
\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CH}+\mathrm{CH}_{3} \mathrm{MgBr} \longrightarrow \\
\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CMg} X
\end{array}
$$

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