`Ph-CH=CH_2overset(IC l)to P`, Identify major product 'P' is :

Options

A. `Ph-undersetunderset(I)(|)CH-CH_2-Cl`
B. `Ph-undersetunderset(Cl)(|)CH-CH_2-I`
C. `Ph-undersetunderset(Cl)(|)CH-CH_2-Cl`
D. `Ph-undersetunderset(I)(|)CH-CH_2-I`

Answer

To solve the problem, we need to identify the major product 'P' formed when phenylpropene (Ph-CH=CH2) reacts with iodine monochloride (ICL). Let's break down the reaction step by step.

### Step 1: Identify the Reactants
The reactant is phenylpropene, which can be represented as Ph-CH=CH2. The other reactant is iodine monochloride (ICL).

### Step 2: Mechanism of Reaction
1. **Electrophilic Attack**: The double bond in phenylpropene acts as a nucleophile and attacks the iodine in ICL. This results in the formation of a carbocation intermediate. The iodine atom becomes negatively charged (I-) and the chlorine atom leaves as a chloride ion (Cl-).

**Intermediate**: Ph-CH^+ -CH2-I

2. **Formation of a Three-Membered Ring**: The iodine ion (I-) has lone pairs and can stabilize the carbocation by forming a three-membered ring with the carbocation. This results in a cyclic intermediate where iodine is bonded to the carbocation carbon.

**Intermediate**: A three-membered ring with Iodine.

### Step 3: Nucleophilic Attack
1. **Nucleophilic Attack by Cl-**: The chloride ion (Cl-) can attack one of the carbons in the three-membered ring. Since the carbocation is more stable at the carbon adjacent to the phenyl group (due to resonance), the Cl- will preferentially attack this carbon.

2. **Formation of Products**:
- **Major Product**: When Cl- attacks the carbon adjacent to the phenyl group, we get the major product:
- Ph-CH(Cl)-CH2-I
- **Minor Product**: If Cl- were to attack the other carbon (the one bonded to iodine), we would get a minor product:
- Ph-CH-I-CH2Cl

### Step 4: Identify the Major Product
From the above analysis, the major product 'P' formed from the reaction of phenylpropene with ICL is:
- **Major Product**: Ph-CH(Cl)-CH2-I

### Conclusion
The major product 'P' is Ph-CH(Cl)-CH2-I.

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