This conversion can be achieved by:

The treatment of $CH_3MgX$ with $CH_3-C \equiv CH$ produces

The relation between $(B)$ and $(C)$ is:
$CH_3-C \equiv CH$ $\xrightarrow{NaNH_2}$ $\xrightarrow{CH_3-I} (A)$ $\xrightarrow{Li/liq. NH_3} (B)$
$CH_3-C \equiv CH$ $\xrightarrow{NaNH_2}$ $\xrightarrow{CH_3-I} (A)$ $\xrightarrow[Pd/CaCO_3]{H_2} (C)$

In the reaction,$H_3C-C\equiv C-H$ $\xrightarrow[2. x]{1. NaNH_2, \Delta}$ $\xrightarrow{3. y} H_3C-CH=CH-CH_3$ (trans form),$x$ and $y$,respectively are

Which carbide reacts with water to produce propyne?

Which combination is best for the preparation of the compound $(A)$ shown below?