Reaction I `Ph - overset(overset(O)(||))C - NH_2 overset( overset(oplus)(O) D, Br_2 ) (to) A`
Reaction II `Ph -overset(overset(O)(||)) C - ND_2 overset(overset(oplus)(O) D, Br_2) (to) B`
Products A and B are :

Options

A. `Ph - NH_2 and Ph - ND_2`
B. `Ph - ND_2 and Ph - NH_2`
C. Both `Ph - NH_2`
D. Both `Ph - ND_2`

Explanation

To solve the problem, we need to analyze the two reactions step by step and determine the products formed in each case.

### Step 1: Identify the Reactants
- **Reaction I**: The reactant is `Ph - C(=O) - NH2` (where Ph represents a phenyl group).
- **Reaction II**: The reactant is `Ph - C(=O) - ND2`.

### Step 2: Understand the Reaction Conditions
Both reactions are treated with `O D-` (deuterated hydroxide) and `Br2`. The presence of `O D-` suggests that deuterium will be involved in the reactions, affecting the products.

### Step 3: Mechanism of Reaction I
1. **Deprotonation**: The `O D-` abstracts a hydrogen atom from the amine group (`NH2`), forming `Ph - C(=O) - NH-`.
2. **Bromination**: The bromine (`Br2`) reacts with the nitrogen, resulting in `Ph - C(=O) - NBr`.
3. **Further Deprotonation**: Another `O D-` abstracts the remaining hydrogen, leading to a negatively charged nitrogen (`Ph - C(=O) - N-`).
4. **Elimination**: The nitrogen can lose a bromine atom, forming a nitrene intermediate.
5. **Migration**: The phenyl group migrates to the nitrogen, forming an isocyanate intermediate.
6. **Final Product Formation**: The reaction with `O D-` leads to the formation of `Ph - NH - C(=O) - O D` and eventually to `Ph - NH2` after tautomerization and decarboxylation.

### Step 4: Mechanism of Reaction II
1. **Deprotonation**: The `O D-` abstracts a deuterium from the amine group (`ND2`), forming `Ph - C(=O) - N D-`.
2. **Bromination**: The bromine (`Br2`) reacts with the nitrogen, resulting in `Ph - C(=O) - N DBr`.
3. **Further Deprotonation**: Another `O D-` abstracts the remaining deuterium, leading to a negatively charged nitrogen (`Ph - C(=O) - N-`).
4. **Elimination**: Similar to Reaction I, the nitrogen can lose a bromine atom, forming a nitrene intermediate.
5. **Migration**: The phenyl group migrates to the nitrogen, forming an isocyanate intermediate.
6. **Final Product Formation**: The reaction with `D2O` leads to the formation of `Ph - N D - C(=O) - O D` and eventually to `Ph - N D2` after tautomerization and decarboxylation.

### Step 5: Identify Products A and B
- **Product A** from Reaction I: `Ph - NH2`
- **Product B** from Reaction II: `Ph - N D2`

### Conclusion
The products of the reactions are:
- A = `Ph - NH2`
- B = `Ph - N D2`

Thus, the correct answer is option A: `Ph - NH2` and `Ph - N D2`.

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