1. The reaction sequence shown is not visible in the text, but from the answer choices we can infer:

   • Compound $A$ is either $n$-pentane or $2$-pentyne.
   • Compound $B$ is either cis-2-butene or trans-2-butene.

2. A very standard organic conversion is:

   • Dehydrohalogenation / alkylation steps giving an alkyne such as $2$-pentyne.
   • Reduction of an alkyne to an alkene.

3. To decide $B$:

   • Partial reduction of an alkyne with dissolving metal conditions such as $\mathrm{Na/NH_3(l)}$ gives the trans-alkene.
   • Partial reduction with Lindlar catalyst gives the cis-alkene.

4. Since the choices pair an alkyne precursor with a butene product, the chemically consistent intermediate is $A = 2$-pentyne, not $n$-pentane.

5. Further, the standard selective reduction of an alkyne under dissolving metal conditions gives: $\text{2-butyne} \xrightarrow[\text{liq. } NH_3]{Na} \text{trans-2-butene}$ Hence $B$ should be trans-2-butene.

6. Therefore the matching option is: $A = 2\text{-pentyne}, \quad B = \text{trans-2-butene}$

So, the correct option is D.