Consider the following reactions :
`(I) Me - underset ((A)) -= - Me overset (H_2 +) underset (Pd + BaSO_4) rarr underset ((B)) ? overset (D_2//Pd//C)rarr (C)`
`(II) Me - underset ((A))-= - Me overset (Na + Liq. NH_3) underset (+EtOH)rarr underset ((D)) ? overset (D_2//Pt//C)rarr (E)`
Which of the following statements are correct ?

Options

A. (B) is cis-but-2-ene and (D) is trans-but-2-ene.
B. (B) is trans-but-2-ene and (D) is cis-but-2-ene.
C. (c) is meso form and (E) is racemic form.
D. (C) is racemic form and (E) is meso form.

Explanation

To solve the given reactions and identify the products, let's break down the steps systematically.

### Step 1: Identify the Structure of Compound A
The compound A is represented as `Me - underset ((A)) -= - Me`. This indicates that A is but-2-yne, which can be written as:
\[ \text{CH}_3\text{C} \equiv \text{CCH}_3 \]

### Step 2: Reaction of A with H2 and Pd/BaSO4
The first reaction involves the reduction of but-2-yne (A) using hydrogen gas (H2) in the presence of palladium (Pd) and barium sulfate (BaSO4). This setup is known as Lindlar's catalyst, which selectively hydrogenates alkynes to cis-alkenes.

- **Product B**: The product formed from this reaction is cis-but-2-ene:
\[ \text{B} = \text{cis-CH}_3\text{C}=\text{CCH}_3 \]

### Step 3: Reaction of B with D2 and Pd/C
Next, we consider the reaction of product B with deuterium gas (D2) in the presence of palladium on carbon (Pd/C). This reaction leads to the formation of a meso compound because the addition of D2 occurs in a symmetrical manner.

- **Product C**: The product formed is a meso compound:
\[ \text{C} = \text{meso-2,3-dideuterobutane} \]

### Step 4: Identify the Structure of Compound D
Now, we analyze the second reaction where compound A (but-2-yne) is treated with sodium (Na) in liquid ammonia (NH3) and ethanol (EtOH). This is known as Birch reduction, which converts alkynes to trans-alkenes.

- **Product D**: The product formed from this reaction is trans-but-2-ene:
\[ \text{D} = \text{trans-CH}_3\text{C}=\text{CCH}_3 \]

### Step 5: Reaction of D with D2 and Pt/C
Finally, we look at the reaction of product D with D2 in the presence of platinum on carbon (Pt/C). This reaction will lead to the formation of a racemic mixture because the addition of D2 occurs in a non-symmetrical manner.

- **Product E**: The product formed is a racemic mixture of 2,3-dideuterobutane:
\[ \text{E} = \text{racemic-2,3-dideuterobutane} \]

### Summary of Products
- **B**: cis-but-2-ene
- **C**: meso-2,3-dideuterobutane
- **D**: trans-but-2-ene
- **E**: racemic-2,3-dideuterobutane

### Conclusion
Based on the products identified:
- B is cis-but-2-ene (correct)
- D is trans-but-2-ene (correct)
- C is a meso compound (correct)
- E is a racemic mixture (correct)