$\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C} \mathrm{MgBr}$ can be prepared by the reaction of

$\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C} \mathrm{MgBr}$ can be prepared by the reaction of
  1. $\mathrm{CH}_{3}-\mathrm{C}=\mathrm{C}-\mathrm{Br}$ with $\mathrm{MgBr}_{2}$
  2. $\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CH}$ with $\mathrm{MgBr}_{2}$
  3. $\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CH}$ with $\mathrm{KBr}$ and $\mathrm{Mg}$ metal
  4. $\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CH}$ with $\mathrm{CH}_{3} \mathrm{MgBr}$

Solution

$\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CMg}$ Br can be prepared by the reaction of $\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CH}$ 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 $\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CH}+\mathrm{CH}_{3} \mathrm{MgBr} \longrightarrow \mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CMg} X$ /

Asked in: JEE-TOPICTESTS-CHEMISTRY

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