`HBr` reacts with `CH_(2)=CH-OCH_(3)` under anhydrous conditions at room temperature to give?

Options

A. `CH_(3)CHO` and `CH_(3)OH`
B. `BrCH_(2)CHO` and `CH_(3)OH`
C. `BrCH_(2)-CH_(2)-O-CH_(2)`
D. `H_(3)C-CHBr-OCH_(3)`

Answer

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

### Step 1: Identify the Reactants
The reactants in this reaction are:
- HBr (Hydrobromic acid)
- Methyl vinyl ether (CH₂=CH-OCH₃)

### Step 2: Understand the Reaction Type
The reaction between HBr and methyl vinyl ether is an electrophilic addition reaction. In this reaction, the double bond in the vinyl ether will react with HBr.

### Step 3: Electrophilic Attack
In the presence of HBr, the double bond (C=C) in methyl vinyl ether will undergo an electrophilic attack. The H⁺ from HBr will add to one of the carbon atoms in the double bond, forming a carbocation.

### Step 4: Carbocation Formation
When H⁺ adds to the carbon atom, we can have two possible carbocations:
1. If H⁺ adds to the terminal carbon (CH₂), a primary carbocation is formed.
2. If H⁺ adds to the internal carbon (CH), a more stable secondary carbocation is formed.

In this case, the more stable secondary carbocation will be favored due to its stability.

### Step 5: Nucleophilic Attack
After the formation of the carbocation, the Br⁻ ion from HBr will act as a nucleophile and attack the positively charged carbon atom of the carbocation.

### Step 6: Product Formation
The final product of the reaction will be:
- The addition of Br to the carbon that was positively charged.
- The ether group (OCH₃) remains attached to the other carbon.

Thus, the product formed is:
**CH₃-CH(Br)-OCH₃** (2-bromo-1-methoxyethane)

### Final Answer
The reaction of HBr with methyl vinyl ether (CH₂=CH-OCH₃) under anhydrous conditions at room temperature yields **2-bromo-1-methoxyethane (CH₃-CH(Br)-OCH₃)**.

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