The final product of the following reaction is .......
$CaO + 3C$ $\xrightarrow{\text{Heat}} A$ $\xrightarrow{H_2O} B$ $\xrightarrow{H_2SO_4, Hg^{2+}} C$

$HC \equiv CH$ $\xrightarrow[{excess}]{{CH_3MgBr}} \mathop {(A)}\limits_{\text{salt}} + \mathop {(B)}\limits_{\text{hydrocarbon}}$
$(A)$ $\xrightarrow[{excess}]{{CH_3I}} (C)$ $\xrightarrow[{(ii) H^{+} }]{{(i) NaNH_2}} (D)$

What is the end product of the following sequence of operations?
$CaC_2$ $\xrightarrow{H_2O} A$ $\xrightarrow[Hg^{2+}]{dil. H_2SO_4} B$ $\xrightarrow[H_2]{Ni} C$

An unknown compound $(A)$ has a molecular formula $C_4H_6$. When $(A)$ is treated with excess of $Br_2$, a new substance $(B)$ with formula $C_4H_6Br_4$ is formed. $(A)$ forms a white precipitate with ammoniacal silver nitrate solution. $(A)$ may be:

$CH_3-C \equiv C-CH_3$ $\xrightarrow[(1) \ O_3]{(2) \ H_2O} X$ $\xrightarrow[\Delta]{NaOH + CaO} Y$; $Y$ is

In the reaction,$CH_3-C \equiv C-CH_3 \xrightarrow{(i) X, (ii) Zn/H_2O} CH_3-C(=O)-C(=O)-CH_3$,$X$ is: