Reactants and Conditions:
- HBr (Hydrogen bromide)
- CH2 = CH – OCH3 (Methyl vinyl ether)
- Anhydrous conditions (without water)
- Room temperature

Reaction:
The reaction between HBr and CH2 = CH – OCH3 under anhydrous conditions at room temperature leads to the formation of BrCH2– CH2– OCH3.

Explanation:
1. Electrophilic Addition Reaction:
The reaction between HBr and CH2 = CH – OCH3 involves an electrophilic addition reaction mechanism. HBr acts as an electrophile due to the presence of a polar covalent bond between hydrogen and bromine. The double bond in CH2 = CH – OCH3 acts as a nucleophile and attacks the electrophilic hydrogen in HBr.

2. Formation of Carbocation:
The attack of the double bond on HBr leads to the formation of a carbocation intermediate. The oxygen atom in CH2 = CH – OCH3 donates electron density towards the carbon-carbon double bond, making it more susceptible to attack by the electrophilic hydrogen in HBr. As a result, the hydrogen atom in HBr gets attached to one of the carbon atoms, while the bromine atom gains a positive charge.

3. Nucleophilic Attack:
The bromide ion (Br-) acts as a nucleophile and attacks the carbocation intermediate. The bromide ion, being negatively charged, is attracted towards the positively charged carbon atom. The bromide ion donates its lone pair of electrons to form a bond with the positively charged carbon atom.

4. Formation of Product:
The final product is BrCH2– CH2– OCH3, where the bromine atom is attached to one of the carbon atoms in the double bond, and the oxygen atom remains attached to the other carbon atom. The reaction proceeds via an electrophilic addition mechanism, where the double bond in CH2 = CH – OCH3 is broken, and the bromine atom is added to the carbon atoms.

In conclusion, the correct answer is option 'C' - BrCH2– CH2– OCH3. This is the product formed when HBr reacts with CH2 = CH – OCH3 under anhydrous conditions at room temperature.