The reaction sequence is given as:
$CH_3-CH=CH-CH_3$ $\xrightarrow{Br_2/CCl_4} \text{Intermediate}$ $\xrightarrow[(ii) NaNH_2]{(i) alc. KOH} (A)$
Product $(A)$ is:

What product is obtained by the oxymercuration of $1$-butyne using $HgSO_4 + H_2SO_4$?

The major product of the following reaction is $CH_3-C\equiv CH \xrightarrow[(ii) \ DI]{(i) \ DCl \ (1 \ equiv)}$

The product $(C)$ of the above reaction is:
Maleic anhydride $\xrightarrow{H_3O^+}$ $(A)$ $\xrightarrow{NaOH}$ $(B)$ $\xrightarrow{\text{Kolbe electrolysis}}$ $(C)$

The ozonolysis $(O_3, H_2O)$ of $(CH_3)_2CH-C \equiv C-CH_3$ gives:

In order to complete the reaction $1-Pentyne$ $\xrightarrow{a} 4-Octyne$ $\xrightarrow{b} cis-4-Octene$,$a$ and $b$ will be:
$A) NaNH_2; CH_3CH_2CH_2Br : H_2, (1 \ mole) \text{ Lindlar catalyst}$
$B) NaNH_2; CH_3CH_2CH_2Br : H_2, (2 \ moles) \text{ Pd or Ni}$
$C) NaNH_2; CH_3CH_2CH_2Br : H_2, (1 \ mole) \text{ Pd or Ni}$
$D) NaNH_2; CH_3CH_2CH_2Br : BH_3, H_2O_2, OH^-$