A `overset(OBr^(-))(rarr) "CHBr"_(3)` the compound is :

Options

A. Isopropyl alcohol
B. Isobutyl alcohol
C. Neopentyl alcohol
D. n-Butanol

Answer

To solve the question, we need to identify the compound that reacts with \( \overset{OBr^{-}}{\rightarrow} \) to form \( CHBr_3 \).

### Step-by-step Solution:

1. **Understanding the Reaction**: The reaction involves a compound (let's call it A) that reacts with \( OBr^{-} \) to produce \( CHBr_3 \). This indicates that A must be a type of alcohol because \( OBr^{-} \) is a reagent that can convert alcohols to bromoalkanes.

2. **Identifying the Product**: The product \( CHBr_3 \) is known as bromoform. This suggests that the starting compound A must have a structure that allows for the formation of three bromine atoms on a carbon atom.

3. **Analyzing the Options**: The options provided are different types of alcohols. We need to determine which alcohol can lead to the formation of \( CHBr_3 \) when reacted with \( OBr^{-} \).

4. **Considering Alcohol Structures**:
- Isopropyl alcohol (2-propanol) and isobutyl alcohol (2-butanol) are branched alcohols and would not lead to the formation of \( CHBr_3 \).
- Neopentyl alcohol is also a branched alcohol and would not yield \( CHBr_3 \).
- Normal butanol (1-butanol) is a straight-chain alcohol and can react with \( OBr^{-} \) to yield \( CHBr_3 \).

5. **Conclusion**: Since normal butanol is the only straight-chain alcohol among the options, it is the compound that reacts with \( OBr^{-} \) to produce \( CHBr_3 \).

### Final Answer:
The compound is **Normal Butanol** (1-butanol).