In the preparation of alkyl halide from alkene and halogen which of the following reaction is involved

Options

A. Free radical substitution
B. Nuclephilic addition
C. Electrophilie substitution
D. Nucleophilic substitution

Answer

### Step-by-Step Solution:

1. **Identify the Reactants**:
- We start with an alkene, for example, propene (C3H6), which has the structure CH2=CH-CH3.
- We will react this alkene with a halogen, such as chlorine (Cl2).

2. **Reaction Setup**:
- The reaction involves treating propene with chlorine gas (Cl2) in the presence of heat. The reaction can be represented as:
\[ \text{CH}_2=\text{CH-CH}_3 + \text{Cl}_2 \xrightarrow{\text{heat}} \text{3-chloropropene} \]

3. **Mechanism of the Reaction**:
- The reaction proceeds through a free radical mechanism.
- The Cl2 molecule undergoes homolytic cleavage upon heating, generating two chlorine radicals (Cl•):
\[ \text{Cl}_2 \xrightarrow{\text{heat}} 2 \text{Cl}• \]

4. **Radical Attack**:
- One of the chlorine radicals attacks the propene. The chlorine radical will preferentially abstract a hydrogen atom from the propene, leading to the formation of a radical on the carbon:
\[ \text{CH}_2=\text{CH-CH}_3 + \text{Cl}• \rightarrow \text{CH}_2•-\text{CH-CH}_3 + \text{HCl} \]
- This results in the formation of a CH2 radical and HCl.

5. **Formation of Alkyl Halide**:
- The generated CH2 radical can now react with another chlorine molecule (Cl2). The chlorine radical will again undergo homolytic cleavage:
\[ \text{CH}_2• + \text{Cl}_2 \rightarrow \text{CH}_2\text{Cl}-\text{CH-CH}_3 \]
- This results in the formation of 3-chloropropene, which is the alkyl halide.

6. **Conclusion**:
- The overall reaction is a free radical substitution reaction, where a chlorine radical substitutes a hydrogen atom in the alkene to form an alkyl halide.

### Final Answer:
The reaction involved in the preparation of alkyl halide from alkene and halogen is **free radical substitution**.

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