Which of the statements are correct?
`R--=-R^'underset(+EtOH)overset(Na+liq. NH_3)rarr(A)overset(Br_2//C Cl_4)rarrB+C`
where (B) and (C) are:

Options

A. (a) Enantiomers if `R!=R^'`.
B. (b) Diastereomers if `R!=R^'`.
C. (c) Both are meso and hence the same compound if `R=R^'`.
D. (d) An equimolar mixture of (B) and (C) is a racemic mixture
if `R!=R^'`.

Explanation

To solve the question regarding the reactions and the nature of the products B and C, we will analyze the given reactions step by step.

### Step 1: Identify the Reactants
The reaction starts with an alkyne denoted as \( R \equiv R' \). For our analysis, we can assume:
- \( R = \text{CH}_3 \) (methyl group)
- \( R' = \text{C}_2\text{H}_5 \) (ethyl group)

### Step 2: Reaction with Sodium in Liquid Ammonia
The first step involves the reaction of the alkyne with sodium (Na) in the presence of liquid ammonia (NH₃) and ethanol (EtOH). This reaction leads to the formation of a trans-alkene due to anti-addition:
- The product formed is \( A \), which can be represented as:
\[ \text{C}_2\text{H}_5 - \text{C}(\text{H}) = \text{C}(\text{H}) - \text{CH}_3 \]
This structure has one ethyl group and one methyl group, along with a double bond.

### Step 3: Bromination of the Alkene
Next, the alkene \( A \) undergoes bromination in the presence of \( \text{Br}_2 \) and \( \text{CCl}_4 \). This reaction also results in anti-addition:
- The bromination of \( A \) produces two products \( B \) and \( C \):
- \( B \) and \( C \) will be enantiomers since they are formed from a chiral center created during the bromination.

### Step 4: Analyze the Products
Since \( R \) and \( R' \) are different (methyl and ethyl), the products \( B \) and \( C \) will be:
- Enantiomers (due to the presence of one chiral center).
- A racemic mixture (equal amounts of both enantiomers).

### Step 5: Consider the Case When \( R = R' \)
Now, if we consider the case where both \( R \) and \( R' \) are the same (e.g., both are methyl groups):
- The product \( A \) would be a trans-alkene with two identical groups.
- The bromination would yield a meso compound, which has a plane of symmetry.

### Conclusion
Based on the analysis:
- When \( R \neq R' \), \( B \) and \( C \) are enantiomers and form a racemic mixture.
- When \( R = R' \), the product is a meso compound.

### Correct Statements
1. **Statement A**: Correct (B and C are enantiomers).
2. **Statement C**: Correct (B and C form a racemic mixture).
3. **Statement D**: Correct (when R = R', the product is a meso compound).

### Final Answer
The correct statements are A, C, and D.

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