`CH_3CONH_2` on reaction with NaOH and
`Br_2` in alcoholic medium gives :

Options

A. `CH_3 CH_2 NH_2`
B. `CH_3 CH_2 Br`
C. `CH_3 NH_2`
D. `CH_3 COONa`

Explanation

To solve the question regarding the reaction of `CH3CONH2` (acetamide) with NaOH and `Br2` in an alcoholic medium, we can break it down into several steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactants in this reaction are acetamide (`CH3CONH2`), sodium hydroxide (NaOH), and bromine (Br2) in an alcoholic medium.

2. **Understand the Reaction Conditions**:
The presence of NaOH indicates a basic medium, which is crucial for the reaction mechanism. The alcoholic medium is also important as it influences the solubility and reactivity of the reactants.

3. **Formation of the Intermediate**:
In a basic medium, NaOH will deprotonate the amide nitrogen, forming a negatively charged nitrogen species. This intermediate can then react with bromine.

4. **Bromination Step**:
The negatively charged nitrogen (from the deprotonation) attacks one of the bromine atoms in Br2, leading to the formation of a bromo-substituted amide. The product at this stage is `CH3CONBr`.

5. **Formation of Nitrene**:
The bromo-substituted amide undergoes a rearrangement where the bromine leaves, creating a nitrogen species known as nitrene. This step is crucial as it leads to the formation of an electron-deficient nitrogen.

6. **Migration of Methyl Group**:
The electron deficiency at the nitrogen allows for the migration of the methyl group (CH3) from the carbonyl carbon to the nitrogen, forming a new compound.

7. **Final Product Formation**:
The final product after the rearrangement and migration is methyl isocyanate (`CH3N=C=O`). This compound can further undergo hydrolysis in the presence of water (H2O) to yield the final product, which is methylamine (`CH3NH2`) and carbon dioxide (CO2).

### Final Answer:
The reaction of `CH3CONH2` with NaOH and `Br2` in an alcoholic medium gives methyl isocyanate, which upon hydrolysis yields methylamine.