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

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Answer

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

### Step-by-Step Solution:

1. **Identify the Structure of Methyl Ethylene Oxide**:
Methyl ethylene oxide is also known as 1-methoxy-2-oxirane. It has a three-membered cyclic ether structure with a methyl group attached to one of the carbons.

2. **Reaction with HBr**:
When methyl ethylene oxide reacts with excess HBr, the oxygen atom in the epoxide ring (which is electron-rich) will abstract a proton (H+) from HBr. This results in the formation of a protonated epoxide.

3. **Formation of Carbocation**:
The protonation of the oxygen leads to a positive charge on the oxygen atom, making it a good leaving group. The bond between the oxygen and one of the carbons will break, leading to the formation of a carbocation. In this case, the bond that breaks will be the one leading to the more stable secondary carbocation.

4. **Nucleophilic Attack by Bromide Ion**:
The bromide ion (Br-) from HBr will then attack the carbocation. This will lead to the formation of a bromo alcohol.

5. **Final Product Formation**:
The final product after the nucleophilic attack will be a compound where the bromine atom is attached to the carbon that was originally part of the epoxide ring, and the hydroxyl group (OH) will be on the adjacent carbon.

Thus, the final product 'X' will be **1-bromo-2-methoxypropane**.

### Final Answer:
The product 'X' formed is **1-bromo-2-methoxypropane**.

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