2-Phenyl-2-chloropropane on treatment with alc. KOH gives mainly

Options

A. 2-Phenylpropene
B. 3-Phenylpropene
C. 1-Phenylpropan-2-ol
D. 1-Phenylpropan-3-ol

Explanation

To solve the question regarding the reaction of 2-Phenyl-2-chloropropane with alcoholic KOH, we will follow these steps:

### Step 1: Understand the Structure of 2-Phenyl-2-chloropropane
2-Phenyl-2-chloropropane has the following structure:
- It consists of a propane chain (3 carbon atoms) with a chlorine atom and a phenyl group (C6H5) attached to the second carbon.
- The structure can be represented as:
```
Cl
|
CH3- C - CH3
|
C6H5
```

### Step 2: Identify the Reactant and the Reagent
The reactant is 2-Phenyl-2-chloropropane, and the reagent is alcoholic KOH (potassium hydroxide in alcohol). The alcohol typically used in the lab is ethanol (C2H5OH).

### Step 3: Determine the Reaction Mechanism
Alcoholic KOH acts as a base. In this case, it will facilitate an elimination reaction (E2 mechanism) because there is an alpha hydrogen available for elimination.

### Step 4: Elimination Reaction
In the E2 mechanism, the base (OH-) will abstract a hydrogen atom from the carbon adjacent to the carbon bearing the chlorine atom. The chlorine atom will leave as a chloride ion (Cl-), leading to the formation of a double bond.

- The reaction can be summarized as:
```
CH3- C(Cl)- CH3 + OH- → CH3- C=CH2 + Cl- + H2O
```

### Step 5: Identify the Product
The product formed from this elimination reaction is 2-Phenylpropene. The double bond forms between the second and third carbon atoms, resulting in the following structure:
```
C6H5
|
CH3- C = CH2
```
This compound is known as 2-Phenylpropene.

### Conclusion
Thus, the main product obtained from the reaction of 2-Phenyl-2-chloropropane with alcoholic KOH is **2-Phenylpropene**.

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