Propene when heated with chlorine at about `500^@` C forms

Options

A. `CH_2Cl.CH=CH_2`
B. `CH_3.CHCl.CH_2Cl`
C. `CH_2Cl. CHCl. CH_2Cl`
D. All of three

Explanation

To solve the question about the reaction of propene with chlorine at about 500°C, we will follow these steps:

### Step 1: Identify the Reactants
The reactant in this reaction is propene, which has the molecular formula C3H6. Its structure can be represented as:
\[ \text{CH}_2= \text{CH} - \text{CH}_3 \]

### Step 2: Understand the Reaction Conditions
The reaction occurs at high temperature, specifically around 500°C. At this temperature, chlorine (Cl2) can undergo homolytic cleavage to form chlorine radicals (Cl•).

### Step 3: Describe the Mechanism
At high temperatures, the reaction proceeds via a radical mechanism rather than an electrophilic addition. The chlorine radicals will react with the double bond in propene.

### Step 4: Formation of the Product
When a chlorine radical reacts with propene, it can abstract a hydrogen atom, leading to the formation of a new radical. The chlorine radical will then add to the carbon atom that was involved in the double bond, resulting in the formation of a stable compound.

The structure of the product formed is:
\[ \text{CH}_2\text{Cl} - \text{CH} = \text{CH}_2 \]
This compound is known as 1-chloro-2-propene.

### Final Answer
Therefore, the product formed when propene is heated with chlorine at about 500°C is:
\[ \text{CH}_2\text{Cl} - \text{CH} = \text{CH}_2 \]

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