`underset((A))(EtCOCl) overset((i)H_2 N.OH) underset ((ii) overset(Ө)O H) rarr (B) overset(EtOH) rarr (C)`
Which of the following statements are correct about the given reaction ?

Options

A. The compouds `(B)` and `( C)`, respectively, are `Et-C -= N` and `EtNH_2`.
B. The compounds `(B)` and `( C)`, respectively, are `EtN=C=O` and `EtNHCOOEt`.
C. The reaction is known as Lossen rearrangement reaction.
D. The reaction proceeds via the formation of acyl carbene `(EtCO ddot CH)` as the intermediate species.

Explanation

To analyze the given reaction step by step, we will break down the transformations occurring from compound A to compound C.

### Step 1: Reaction of Compound A with Hydroxylamine
- **Reaction**: The acid chloride (EtCOCl) reacts with hydroxylamine (H2N.OH).
- **Mechanism**:
- The nitrogen of hydroxylamine attacks the carbonyl carbon of the acid chloride.
- This results in the formation of an intermediate where the chlorine (Cl) leaves as HCl.
- The product of this step is an oxime (EtC=NOH).

**Hint**: Identify the nucleophilic attack of hydroxylamine on the carbonyl carbon of the acid chloride.

### Step 2: Reaction in Basic Medium
- **Reaction**: The oxime (EtC=NOH) is treated with a base (OH-).
- **Mechanism**:
- The hydroxide ion (OH-) abstracts a proton (H+) from the nitrogen of the oxime.
- This results in the formation of a negatively charged intermediate (EtC=N-O^-).
- A water molecule is eliminated in this process.

**Hint**: Consider the role of the base in deprotonating the nitrogen atom.

### Step 3: Rearrangement (Lawson's Rearrangement)
- **Reaction**: The negatively charged nitrogen attacks the carbonyl carbon.
- **Mechanism**:
- This step is known as Lawson's rearrangement.
- The nitrogen donates its electrons to form a double bond with the carbon, resulting in the formation of compound B (EtC=N-O^-).

**Hint**: Recognize the significance of the nitrogen's electron donation in the rearrangement process.

### Step 4: Reaction of Compound B with Ethanol
- **Reaction**: Compound B (EtC=N-O^-) reacts with ethanol (EtOH).
- **Mechanism**:
- The nitrogen in compound B can be protonated by ethanol, or the ethyl group from ethanol can attack the carbon, leading to the formation of compound C.
- The final product will have a carbonyl group (C=O) and an ethoxy group (OEt) attached to it, along with the nitrogen still bonded to one ethyl group and one hydrogen.

**Hint**: Focus on how the ethyl group from ethanol can participate in the reaction to form the final product.

### Summary of Products
- **Compound A**: Ethyl acyl chloride (EtCOCl)
- **Compound B**: Ethyl oxime (EtC=N-OH)
- **Compound C**: Ethyl carbamate or ethyl ester (C=O with OEt and NH with one ethyl)

### Conclusion
After analyzing the steps, we can conclude which statements about the reaction are correct based on the transformations and mechanisms involved. The correct statements are:
- Option B is correct (the reaction involves Lawson's rearrangement).
- Option C is correct (the reaction does not proceed via carbene).

### Final Answer
The correct statements about the given reaction are Option B and Option C.

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