MCQ
$\mathrm{CH}_3-\mathrm{C} \equiv \mathrm{C} \ \text{MgBr}$ 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 $\mathrm{MgB}_2$
- C$\mathrm{CH}_2-\mathrm{C} \equiv \mathrm{CH}$ with $\text{KBr}$ and $\text{Mg}$ metal
- ✓$\mathrm{CH}_3-\mathrm{C} \equiv \mathrm{CH}$ with $\mathrm{CH}_3 \mathrm{MgBr}$