Which alkene will give optically inactive product with `Br_2//"CCl"_4` ?

Options

A. 1- butene
B. Propene
C. cis - 2 - butene
D. trans - 2 - butene

Answer

To determine which alkene will give an optically inactive product when reacted with \( \text{Br}_2 \) in the presence of \( \text{CCl}_4 \), we need to analyze the structures of the given alkenes and the products formed during the reaction.

### Step-by-Step Solution:

1. **Identify the Alkenes**:
We need to look at the four alkene options provided. Let's denote them as:
- Alkene 1: 1-butene
- Alkene 2: Propene
- Alkene 3: Cis-butene
- Alkene 4: Trans-butene

2. **Reaction with \( \text{Br}_2 \) in \( \text{CCl}_4 \)**:
The reaction of alkenes with bromine in the presence of carbon tetrachloride (\( \text{CCl}_4 \)) typically results in a vicinal dibromide. The addition of bromine occurs across the double bond.

3. **Analyze Products for Chirality**:
- **Alkene 1 (1-butene)**:
- Product: \( \text{CH}_3-\text{CH}(\text{Br})-\text{CH}_2-\text{Br} \)
- This product has a chiral center, making it optically active.

- **Alkene 2 (Propene)**:
- Product: \( \text{CH}_3-\text{C}(\text{Br})-\text{CH}_2-\text{Br} \)
- This product also has a chiral center, making it optically active.

- **Alkene 3 (Cis-butene)**:
- Product: \( \text{CH}_3-\text{C}(\text{Br})-\text{CH}_3 \) with bromines on adjacent carbons.
- This can form enantiomers, thus it is optically active.

- **Alkene 4 (Trans-butene)**:
- Product: \( \text{CH}_3-\text{C}(\text{Br})-\text{CH}_3 \) with bromines on adjacent carbons.
- This product has a plane of symmetry, resulting in a meso compound, which is optically inactive.

4. **Conclusion**:
The only alkene that will give an optically inactive product when reacted with \( \text{Br}_2 \) in \( \text{CCl}_4 \) is **Trans-butene**.

### Final Answer:
**Trans-butene** will give an optically inactive product with \( \text{Br}_2 \) in \( \text{CCl}_4 \).

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