Which of the following compounds will undergo racemisation when solution of alcoholic KOH ?

Options

A.
B. `H_(3)C-overset(CH_(3))overset("| ")"CH"-CH_(2)Cl`
C.
D.

Answer

To determine which of the given compounds will undergo racemization when treated with an alcoholic KOH solution, we need to analyze the structure of each compound and the mechanism of the reaction. Here’s a step-by-step solution:

### Step 1: Understand Racemization
Racemization is the process by which an optically active compound is converted into a racemic mixture, which contains equal amounts of two enantiomers (mirror-image isomers). For racemization to occur, a chiral center must be formed or altered during the reaction.

**Hint:** Look for compounds that can form a chiral center during the reaction.

### Step 2: Identify the Role of Alcoholic KOH
Alcoholic KOH provides hydroxide ions (OH-) that can act as nucleophiles. In a nucleophilic substitution reaction, the nucleophile can attack a carbon atom that is bonded to a leaving group (like Cl), leading to the formation of a new bond.

**Hint:** Identify the leaving group in each compound and the carbon atom it is attached to.

### Step 3: Analyze Each Compound
1. **Compound 1**: If the compound has a carbon with a leaving group and does not lead to a chiral center upon substitution, racemization will not occur.

2. **Compound 2**: Similar analysis as Compound 1. If the substitution does not create a chiral center, racemization will not occur.

3. **Compound 3**: If the compound has a carbon with a leaving group that, upon substitution, leads to the formation of a carbocation where the nucleophile (OH-) can attack from either side, this can create a chiral center and thus lead to racemization.

4. **Compound 4**: If the leaving group is part of a stable aromatic system (like a phenyl group), the formation of a carbocation may not occur easily, and thus racemization is unlikely.

**Hint:** Focus on the third compound where the nucleophile can attack from two different sides after the leaving group departs.

### Step 4: Conclusion
After analyzing the compounds, we find that **Compound 3** is the one that will undergo racemization. This is because it can form a chiral center when the nucleophile attacks the carbocation formed after the leaving group departs.

**Final Answer:** The compound that will undergo racemization when treated with alcoholic KOH is **Compound 3**.