what is the major product of the following reaction ?
`H_(2)C=CH-CH_(2)-OHunderset("excess")overset("HBr")to"product"`

Options

A. `CH_(3)-overset(Br)overset(|)CH-CH_(2)-Br`
B. `H_(2)C=CH-CH_(2)-Br`
C. `CH_(3)-overset(Br)overset(|)CH-CH_(2)-OH`
D. `CH_(3)-overset(OH)overset(|)CH-CH_(2)-OH`

Explanation

To determine the major product of the reaction between 1-propen-2-ol (H₂C=CH-CH₂-OH) and excess HBr, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactant is 1-propen-2-ol (H₂C=CH-CH₂-OH) and the reagent is excess HBr.

2. **Protonation of the Alcohol**:
The hydroxyl group (-OH) in the alcohol can be protonated by HBr. This will convert the -OH group into a better leaving group (water, H₂O).

3. **Formation of Carbocation**:
The double bond (C=C) can react with HBr. The H⁺ from HBr will add to one of the carbon atoms in the double bond, leading to the formation of a carbocation. The more stable carbocation will form, which is typically the one that is more substituted.

4. **Markovnikov's Rule**:
According to Markovnikov's rule, when HX (where X is a halogen) adds to an alkene, the hydrogen atom will attach to the carbon with the greater number of hydrogen atoms already attached. In this case, the H will add to the terminal carbon (C1), and the positive charge will be on the more substituted carbon (C2).

5. **Nucleophilic Attack by Bromide Ion**:
The bromide ion (Br⁻) will then attack the carbocation formed in the previous step. This will lead to the substitution of the bromine atom at the position of the carbocation.

6. **Final Product**:
The final product of this reaction will be 1-bromo-2-propanol (CH₃-CHBr-CH₃).

### Summary of the Major Product:
The major product of the reaction between 1-propen-2-ol and excess HBr is 1-bromo-2-propanol.