In the Hofmann-Bromamide rearrangement intermediate
compounds are :

Options

A. `R - CONHBr`
B. `[ R - overset(overset(O)(||))C - bar(N) - Br ] Na^(+)`
C. `R - N = C = O`
D. All of these

Answer

To solve the question regarding the intermediate compounds in the Hofmann-Bromamide rearrangement, we will outline the process step by step.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The Hofmann-Bromamide rearrangement involves the conversion of a primary amide into a primary amine with the loss of one carbon atom. The general structure of the starting material is R-C(=O)-NH2.

2. **Initial Reaction with Bromine**:
The primary amide (R-C(=O)-NH2) reacts with bromine (Br2) in the presence of a base such as sodium hydroxide (NaOH). This leads to the formation of an N-bromoamide intermediate:
\[ R-C(=O)-NH2 + Br2 + NaOH \rightarrow R-C(=O)-NBr + HBr + H2O \]
Here, the first intermediate is R-C(=O)-NBr.

3. **Formation of Isocyanate**:
The N-bromoamide can further react with hydroxide ions (OH-) to form an isocyanate intermediate:
\[ R-C(=O)-NBr + OH- \rightarrow R-N=C=O + Br- + H2O \]
This is the second intermediate, which is the alkyl isocyanate (R-N=C=O).

4. **Hydrolysis of Isocyanate**:
The isocyanate (R-N=C=O) can undergo hydrolysis in the presence of water to yield the primary amine:
\[ R-N=C=O + H2O \rightarrow R-NH2 + CO2 \]
This results in the formation of the final product, which is the primary amine (R-NH2).

5. **Summary of Intermediates**:
The intermediates formed during the Hofmann-Bromamide rearrangement are:
- N-bromoamide (R-C(=O)-NBr)
- Alkyl isocyanate (R-N=C=O)

### Final Answer:
The intermediate compounds in the Hofmann-Bromamide rearrangement are:
1. N-bromoamide (R-C(=O)-NBr)
2. Alkyl isocyanate (R-N=C=O)