`H_(3)C-CH=CH_(2)+HCl`
the intermediate of reaction is:

Options

A. `CH_(3)-CH-CH_(3)`
B. `CH_(3)-CH_(2)CH_(2)`
C. `CH_(3)-CH_(-CH_(3)`
D. `CH_(3)-CH_(2)-CH_(2)`

Explanation

To determine the intermediate of the reaction between propene (H₃C-CH=CH₂) and HCl, let's break down the steps involved in the reaction:

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactants are propene (H₃C-CH=CH₂) and hydrochloric acid (HCl).

2. **Understanding the Reaction Type**:
This reaction is an electrophilic addition reaction. In this case, HCl will add across the double bond of the alkene.

3. **Markovnikov's Rule**:
According to Markovnikov's rule, when HX (where X is a halogen) adds to an alkene, the hydrogen atom (H) will attach to the carbon with the greater number of hydrogen atoms, while the halide (Cl in this case) will attach to the carbon with fewer hydrogen atoms.

4. **Formation of the Carbocation**:
When HCl adds to propene, the hydrogen atom will add to the terminal carbon (C1), resulting in the formation of a more stable secondary carbocation at C2:
- The reaction can be represented as:
\[ H₃C-CH=CH₂ + HCl \rightarrow H₃C-CH^+-CH₃ + Cl^- \]
- Here, the carbocation (H₃C-CH^+-CH₃) is the intermediate.

5. **Nucleophilic Attack**:
The chloride ion (Cl⁻), which is a nucleophile, will then attack the carbocation. This results in the final product:
\[ H₃C-CH^+-CH₃ + Cl^- \rightarrow H₃C-CH(Cl)-CH₃ \]

6. **Final Product**:
The final product of the reaction is 2-chloropropane (H₃C-CH(Cl)-CH₃).

### Conclusion:
The intermediate formed during the reaction of propene with HCl is a secondary carbocation (H₃C-CH^+-CH₃).