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 regarding the compound that reacts with hypobromite anion to produce bromoform (CHBr₃), we need to identify the correct precursor compound that can undergo a haloform reaction. Here’s a step-by-step solution:

### Step 1: Understand the Reaction
The reaction involves a hypobromite anion (OBr⁻) reacting with a compound to produce bromoform (CHBr₃). This indicates that we are dealing with a haloform reaction.

**Hint:** Recall that haloform reactions typically involve compounds that can generate a methyl ketone or a compound that can be oxidized to a methyl ketone.

### Step 2: Identify the Haloform Reaction
The haloform reaction is characterized by the reaction of a methyl ketone or certain alcohols with halogens in the presence of a base (like sodium hydroxide). The products are a haloform (like bromoform) and a carboxylate salt.

**Hint:** Methyl ketones and some alcohols (like ethanol and secondary alcohols) can undergo this reaction.

### Step 3: Analyze the Given Options
We need to evaluate the options provided to see which compound can give rise to bromoform upon reaction with hypobromite.

1. **Isopropyl Alcohol (CH₃-CHOH-CH₃)**: This is a secondary alcohol. Upon oxidation, it can form acetone (a methyl ketone), which can then react to form bromoform.

2. **Isobutyl Alcohol (CH₃-CH(CH₃)-CH₂OH)**: This is a primary alcohol and does not lead to a methyl ketone upon oxidation, thus it cannot give bromoform.

3. **Neopentyl Alcohol (C(CH₃)₄OH)**: This is also a primary alcohol and does not lead to a methyl ketone, so it cannot give bromoform.

4. **N-Butanol (CH₃-CH₂-CH₂-CH₂OH)**: This is another primary alcohol and cannot produce a methyl ketone, thus it cannot give bromoform.

**Hint:** Focus on the structure of each compound and the type of alcohol (primary, secondary) to determine if they can form a methyl ketone.

### Step 4: Conclusion
After evaluating the options, we find that only isopropyl alcohol can undergo oxidation to form acetone, which can then react with hypobromite to yield bromoform.

**Final Answer:** The compound is **Isopropyl Alcohol (CH₃-CHOH-CH₃)**.