`1-"Methylethylene oxide" underset(HBr)overset("excess")to X,` Product 'X' will be-

Options

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B.
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D.

Answer

To solve the problem of determining the product 'X' formed when methyl ethylene oxide is treated with excess HBr, we can follow these steps:

### Step 1: Identify the Structure of Methyl Ethylene Oxide
Methyl ethylene oxide is a cyclic ether with a three-membered ring structure. It can be represented as follows:

```
O
/ \
CH3 CH2
```

### Step 2: Understand the Reaction with HBr
When methyl ethylene oxide reacts with excess HBr, the first step involves the protonation of the ether oxygen atom. HBr dissociates into H⁺ and Br⁻ ions. The lone pair of electrons on the oxygen atom attacks the H⁺ ion, leading to the formation of a positively charged intermediate.

### Step 3: Formation of the Bromonium Ion
After the protonation, the structure becomes:

```
OH2⁺
/ \
CH3 CH2
```

This is a bromonium ion intermediate, where the oxygen is now positively charged.

### Step 4: Nucleophilic Attack by Bromide Ion
In the presence of excess HBr, the Br⁻ ion will attack the more substituted carbon atom (the one attached to the methyl group) of the bromonium ion. This results in the opening of the ring and the formation of a bromoalcohol.

### Step 5: Final Product Structure
The final product 'X' can be represented as follows:

```
Br
/
CH3 - CH - OH
```

This structure indicates that we have a bromo group (Br) and a hydroxyl group (OH) attached to the same carbon atom, which is a result of the nucleophilic attack by Br⁻.

### Conclusion
The product 'X' formed from the reaction of methyl ethylene oxide with excess HBr is 1-bromo-2-methanol.

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