HBr reacts with `CH_2=CH-OCH_3` under anhydrous condditions at room temperature to give

Options

A. `CH_3CHO` and `CH_3Br`
B. `BrCH_2CHO` and `CH_3OH`
C. `BrCH_2-CH_2-OCH_3`
D. `H_3C-CHBr-OCH_3`

Explanation

To solve the problem of how HBr reacts with CH2=CH-OCH3 (methyl vinyl ether) under anhydrous conditions at room temperature, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactants are HBr (hydrobromic acid) and CH2=CH-OCH3 (methyl vinyl ether).

2. **Understand the Reaction Type**:
- The reaction between an alkene (in this case, methyl vinyl ether) and HBr is an electrophilic addition reaction.

3. **Mechanism of Reaction**:
- The double bond in the methyl vinyl ether acts as a nucleophile and attacks the electrophilic hydrogen (H+) from HBr. This leads to the formation of a carbocation intermediate.
- In this case, the double bond is between the two carbon atoms (C1 and C2), where C1 is attached to the OCH3 group.

4. **Formation of Carbocation**:
- The most stable carbocation will form. Since the ether (OCH3) is an electron-donating group, it stabilizes the carbocation formed at C1. Therefore, the carbocation will be formed at C2.

5. **Nucleophilic Attack by Bromide Ion**:
- The bromide ion (Br-) will then attack the carbocation formed at C2, leading to the formation of the final product.

6. **Final Product**:
- The final product of the reaction is CH3-CH(Br)-OCH3, which is 1-bromo-2-methoxyethane.

### Conclusion:
The reaction of HBr with methyl vinyl ether (CH2=CH-OCH3) under anhydrous conditions at room temperature results in the formation of 1-bromo-2-methoxyethane.