In the reaction, A and B are respectively:

Question

Accepted Answer

In the given reactions, we are dealing with the hydrogenation of alkynes to alkenes using different reagents, which lead to different stereochemical outcomes.

1. **Reaction with $H_2, Pd/BaSO_4$ (Lindlar's Catalyst):**
   - This reaction involves the partial hydrogenation of an alkyne to an alkene using Lindlar's catalyst, which is a poisoned palladium catalyst. The key characteristic of this catalyst is that it selectively hydrogenates alkynes to alkenes without further reducing the alkene to an alkane.
   - Importantly, Lindlar's catalyst facilitates the formation of a **cis-alkene**. This is because the hydrogenation occurs on the surface of the catalyst, and both hydrogen atoms add to the same side of the alkyne, resulting in a syn addition. Therefore, the product A is a cis-alkene.

2. **Reaction with $Na/liq~NH_3$:**
   - This reaction is known as the Birch reduction, which involves the reduction of an alkyne to an alkene using sodium in liquid ammonia. This method is distinct from the Lindlar's catalyst reaction in that it produces a **trans-alkene**.
   - The mechanism involves the formation of a radical anion intermediate, which leads to the anti addition of hydrogen atoms across the triple bond. As a result, the product B is a trans-alkene.

Given these explanations, Option 3 is correct because:
- Product A, formed using Lindlar's catalyst, is a cis-alkene.
- Product B, formed using sodium in liquid ammonia, is a trans-alkene.

**Why the other options are incorrect:**

1. **Option 1 (cis-alkene and cis-alkene):**
   - This option is incorrect because the reaction with $Na/liq~NH_3$ does not produce a cis-alkene; it produces a trans-alkene.

2. **Option 2 (trans-alkene and trans-alkene):**
   - This option is incorrect because the reaction with $H_2, Pd/BaSO_4$ (Lindlar's catalyst) does not produce a trans-alkene; it produces a cis-alkene.

3. **Option 4 (trans-alkene and cis-alkene):**
   - This option is incorrect because the stereochemistry of the products is reversed. The reaction with Lindlar's catalyst produces a cis-alkene, and the reaction with $Na/liq~NH_3$ produces a trans-alkene, not the other way around.

Thus, the correct answer is Option 3: cis-alkene and trans-alkene.