The racemic `-2,3-`dibromopentane on reaction with
`KI` in acetone give `:`

Options

A. `E-`Pent `-2-`ene
B. `Z-`Pent `-2-2`ene
C. `Both `(a)` and `(b)`
D. None

Explanation

To solve the problem of what racemic 2,3-dibromopentane gives when reacted with KI in acetone, we can follow these steps:

### Step 1: Understand the Structure of Racemic 2,3-Dibromopentane
Racemic 2,3-dibromopentane has two stereoisomers due to the presence of two chiral centers. The structure can be represented as follows:

- CH3-CHBr-CHBr-CH2-CH3

This molecule exists as a racemic mixture, meaning it contains equal amounts of both enantiomers.

### Step 2: Reaction with KI in Acetone
When 2,3-dibromopentane reacts with potassium iodide (KI) in acetone, a nucleophilic substitution reaction occurs. The bromine atoms are replaced by iodine atoms.

### Step 3: Mechanism of Reaction
1. **Nucleophilic Attack**: The iodide ion (I-) from KI acts as a nucleophile and attacks the carbon atom bonded to bromine, leading to the displacement of bromine.
2. **Stereochemistry**: Since the reaction involves a chiral center, the stereochemistry of the product will depend on which enantiomer of 2,3-dibromopentane is reacting.

### Step 4: Formation of Alkenes
The reaction leads to the formation of alkenes through elimination. The two possible products from the elimination of HBr are:
- **Cis isomer**: When the groups on the double bond are on the same side (cis).
- **Trans isomer**: When the groups on the double bond are on opposite sides (trans).

### Step 5: Identify the Products
1. **Cis-2-pentene (Z)**: This is formed when the two substituents (ethyl and methyl groups) are on the same side of the double bond.
2. **Trans-2-pentene (E)**: This is formed when the two substituents are on opposite sides of the double bond.

### Conclusion
The reaction of racemic 2,3-dibromopentane with KI in acetone produces both cis-2-pentene (Z) and trans-2-pentene (E).

### Final Answer
The products are **cis-2-pentene (Z)** and **trans-2-pentene (E)**.

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