What type of reaction is involved the reaction `CH_(2)=CH-CH_(3) + HBr to CH_(3)-CH(Br)-CH_(3)`?

Options

A. Nucleophllic addition
B. Electrophilic substation
C. Electrophilic addition
D. Free radical addition

Explanation

To determine the type of reaction involved in the reaction of propene (CH₂=CH-CH₃) with hydrogen bromide (HBr) to form 2-bromopropane (CH₃-CH(Br)-CH₃), we can follow these steps:

### Step 1: Identify the reactants and products
The reactants are propene (CH₂=CH-CH₃) and hydrogen bromide (HBr). The product formed is 2-bromopropane (CH₃-CH(Br)-CH₃).

### Step 2: Analyze the reaction mechanism
In this reaction, HBr adds across the double bond of propene. The double bond is a site of high electron density, making it susceptible to electrophilic attack.

### Step 3: Apply 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 already attached, while the halogen (Br) will attach to the carbon with fewer hydrogen atoms. In this case, the H from HBr will add to the terminal carbon (CH₂), and Br will add to the middle carbon (CH).

### Step 4: Determine the type of reaction
Since HBr is adding across the double bond of propene, this reaction is classified as an electrophilic addition reaction. In this type of reaction, an electrophile (H⁺ from HBr) attacks the nucleophilic double bond, leading to the formation of a more stable product.

### Conclusion
The reaction of propene with HBr to form 2-bromopropane is an **electrophilic addition reaction**.