`CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-underset(OH)underset(|)(CH)-CH_(3)overset(HBr)rarrA` (predominant) A is

Options

A. `(CH_(3))_(3)C-overset(Br)overset(|)(CH)-CH_(3)`
B. `(CH_(3))_(2)overset(Br)overset(|)(C)-CH(CH_(3))_(2)`
C. Both
D. none of the above

Answer

To solve the problem, we need to determine the product (A) formed when the given compound reacts with HBr. The compound can be represented as follows:

**Structure of the compound:**
```
CH3
|
CH3 - C - OH
|
CH3
```

### Step-by-Step Solution:

**Step 1: Protonation of the Alcohol**
- The first step in the reaction with HBr involves the protonation of the hydroxyl group (-OH) in the alcohol. The lone pair of electrons on the oxygen atom will attack a proton (H+) from HBr, forming a better leaving group (water).

**Reaction:**
```
CH3 - C(OH) - CH3 + HBr → CH3 - C(OH2+) - CH3 + Br-
```

**Step 2: Formation of Carbocation**
- After protonation, water (H2O) will leave, leading to the formation of a carbocation. The carbocation formed here is a secondary carbocation.

**Intermediate:**
```
CH3 - C+ - CH3
```

**Step 3: Carbocation Rearrangement**
- Secondary carbocations can rearrange to form more stable tertiary carbocations. In this case, a methyl group (CH3) can shift to the carbocation center, resulting in a more stable tertiary carbocation.

**Rearrangement:**
```
CH3 - C+ - CH3 → CH3 - C(CH3) - CH3
```

**Step 4: Nucleophilic Attack by Bromide Ion**
- The bromide ion (Br-) will now attack the more stable tertiary carbocation to form the final product.

**Final Product:**
```
CH3 - C(Br) - CH3
```

### Final Structure of Product A:
The final product A can be represented as:
```
CH3
|
CH3 - C - Br
|
CH3
```

### Conclusion:
The predominant product A formed from the reaction of the given compound with HBr is 2-bromo-2-methylpropane.

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