2-methyl-2-phenyl propanamide when reacted with bromine and NaOH forms.

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Answer

To solve the problem of what product is formed when 2-methyl-2-phenylpropanamide reacts with bromine and NaOH, we can follow these steps:

### Step 1: Understand the Structure of 2-methyl-2-phenylpropanamide
2-methyl-2-phenylpropanamide has the following structure:
- It contains a propanamide backbone, which means it has three carbon atoms (C-C-C) with an amide functional group (CONH2) attached to the second carbon.
- The second carbon also has a methyl (–CH3) group and a phenyl (–C6H5) group attached.

**Hint:** Draw the structure to visualize the molecule clearly.

### Step 2: Identify the Reaction Conditions
The reaction involves bromine (Br2) and sodium hydroxide (NaOH). This combination is indicative of the Hofmann degradation reaction, which typically involves the conversion of primary amides to amines with the loss of one carbon atom.

**Hint:** Recall that Hofmann degradation leads to the removal of one carbon atom from the amide.

### Step 3: Mechanism of Hofmann Degradation
1. **Deprotonation:** The NaOH acts as a base and abstracts a proton from the amide nitrogen, forming a negatively charged intermediate.
2. **Bromination:** This negatively charged nitrogen can then react with bromine, leading to the formation of an N-bromoamide.
3. **Formation of an Isocyanate:** The N-bromoamide undergoes rearrangement to form an isocyanate intermediate.
4. **Hydrolysis:** The isocyanate is hydrolyzed to form a carbamic acid, which subsequently decomposes to release carbon dioxide (CO2) and yield the final amine product.

**Hint:** Follow the sequence of reactions carefully, noting the intermediates formed at each step.

### Step 4: Determine the Final Product
After the Hofmann degradation, the carbonyl carbon (C=O) of the amide is lost, resulting in the formation of an amine. Since the original molecule had a phenyl group and a methyl group, the final product will be:
- **Final Product:** 2-phenylpropylamine (C6H5-CH(CH3)-NH2)

**Hint:** Remember that the product will have one less carbon than the original amide.

### Conclusion
When 2-methyl-2-phenylpropanamide reacts with bromine and NaOH, it undergoes Hofmann degradation to yield 2-phenylpropylamine.

**Final Product:** 2-phenylpropylamine