`CH_(3)CH_(2)CH_(2)CONH_(2)overset(Br_(2)//KOH)to`Product
The organic product of above reaction is.

Options

A. `CH_(3)CH_(2)CH_(2)NH_(2)`
B. `CH_(3)underset(CH_(2))underset(|)(CH)CH_(3)`
C. `CH_(3)CH_(2)CH_(2)CH_(2)NH_(2)`
D. `CH_(3)CH_(2)underset(NH_(2))underset(|)(C)HCH_(3)`

Explanation

To solve the problem, we need to analyze the reaction of butanamide (CH₃CH₂CH₂CONH₂) with bromine (Br₂) in the presence of potassium hydroxide (KOH). This reaction is known as the Hofmann bromamide degradation reaction.

### Step-by-Step Solution:

1. **Identify the Reactant**:
The reactant is butanamide, which has the structure CH₃CH₂CH₂CONH₂.

2. **Understand the Reaction**:
The Hofmann bromamide degradation involves the treatment of an amide with bromine and a strong base (KOH). The general reaction leads to the formation of a primary amine and the loss of one carbon atom from the carbon chain.

3. **Reaction Mechanism**:
- The amide group (C=O and NH₂) reacts with Br₂ and KOH.
- The nitrogen atom in the amide group will rearrange, and one carbon atom will be eliminated as carbon dioxide (CO₂).
- The remaining structure after the elimination will be a primary amine.

4. **Determine the Product**:
- In this case, the butanamide (4 carbon atoms) will lose one carbon atom during the reaction.
- The remaining structure will be propanamine (CH₃CH₂CH₂NH₂), which has 3 carbon atoms.

5. **Final Product**:
The final product of the reaction is propanamine (CH₃CH₂CH₂NH₂).

### Conclusion:
The organic product of the reaction between butanamide and bromine in the presence of KOH is propanamine.