When cis-but-2-ene is treated with `Br_(2)` in `"CCl"_(4)` medium the product formed will be

Options

A. (2R, 3S) dibromobutane
B. (2R, 3R) dibromobutane
C. (2S, 3S) dibromobutane
D. Mixture of (2R, 3R) and (2S, 3S) dibromobutane

Answer

To determine the product formed when cis-but-2-ene is treated with Br₂ in CCl₄, we will follow these steps:

### Step 1: Identify the structure of cis-but-2-ene
Cis-but-2-ene has the following structure:
```
CH3
|
CH3-C=C-CH2
|
H
```
In this structure, both methyl (CH₃) groups are on the same side of the double bond.

### Step 2: Understand the reaction with Br₂
When cis-but-2-ene is treated with bromine (Br₂), the double bond will react with bromine to form a vicinal dibromide. The reaction occurs via an electrophilic addition mechanism.

### Step 3: Formation of the bromonium ion
1. The π bond of the double bond attacks one of the bromine atoms, leading to the formation of a bromonium ion. This intermediate has a positive charge on one of the carbons.
2. The other bromine atom is released as Br⁻.

### Step 4: Nucleophilic attack by Br⁻
1. The Br⁻ ion will attack the more substituted carbon of the bromonium ion from the opposite side (backside attack), leading to anti-addition.
2. This results in the formation of two stereoisomers due to the possibility of the nucleophile attacking from either side of the bromonium ion.

### Step 5: Draw the products
The products formed will be:
1. (2R,3S)-2,3-dibromobutane
2. (2S,3R)-2,3-dibromobutane

These two products are enantiomers of each other.

### Final Answer
The product formed when cis-but-2-ene is treated with Br₂ in CCl₄ is a mixture of (2R,3S)-2,3-dibromobutane and (2S,3R)-2,3-dibromobutane.

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