$HC \equiv CH$ $\xrightarrow[20\% \ H_2SO_4]{1\% \ HgSO_4} A$ $\xrightarrow{CH_3MgX} B$ $\xrightarrow{[O]} ?$
What is the final product of the above reaction sequence?

An unknown compound $A$ has a molecular formula $C_4H_6$. When $A$ is treated with an 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

When $CH_3CH_2CH_2CHCl_2$ is treated with $2 \ g$ equivalent of $NaNH_2$,the product formed is:

$CH_3-C \equiv C-CH_3$ $\xrightarrow{H^{+}/KMnO_4, \Delta} A$ $\xrightarrow{(1) LiAlH_4, (2) H^{+}} B$
$B$ will be

The correct order for acid strength of compounds $CH \equiv CH$,$CH_3 - C \equiv CH$,and $CH_2 = CH_2$ is as follows:

The correct order of acidic nature of the following alkynes is:
$I$. $HC \equiv CH$
$II$. $CH_3-C \equiv C-CH_3$
$III$. $CH_3-C \equiv CH$