In the reaction
`Me_(3)N overset(BrCN)toA underset((ii)Delta)overset((i)H_(2)O)toB`
B is

Options

A. `MeNH_(2)`
B. `Me_(3)NO`
C. `Me_(2)NH`
D. `Me_(2)NCOOH`

Answer

To solve the problem, we need to analyze the given reaction step by step.

### Step 1: Identify the starting compound
The starting compound is trimethylamine, represented as \( \text{Me}_3\text{N} \). It has three methyl groups attached to a nitrogen atom.

### Step 2: Understand the reaction with BrCN
In the first part of the reaction, \( \text{Me}_3\text{N} \) reacts with \( \text{BrCN} \). The bromine (Br) atom is a good leaving group, and the cyanide group (CN) can act as a nucleophile. The nitrogen atom in trimethylamine has a lone pair of electrons and can attack the carbon of the cyanide group.

### Step 3: Formation of compound A
When \( \text{Me}_3\text{N} \) reacts with \( \text{BrCN} \), the bromine atom leaves, and the nitrogen forms a bond with the carbon of the cyanide group. This results in the formation of a new compound, which can be represented as \( \text{Me}_3\text{N}^+\text{C} \equiv \text{N} \) (a quaternary ammonium salt with a cyanide group). This compound is denoted as \( A \).

### Step 4: Heating the compound A
The next step involves heating compound \( A \) under conditions denoted by \( (ii) \Delta \). Heating can lead to the hydrolysis of the cyanide group. The cyanide group can be converted into a carboxylic acid through hydrolysis.

### Step 5: Formation of compound B
When compound \( A \) is hydrolyzed, the cyanide group (\( -C \equiv N \)) will be converted into a carboxylic acid group (\( -COOH \)). Thus, the final product \( B \) will be trimethylamine with a carboxylic acid group attached, which is trimethylammonium carboxylate.

### Conclusion
The final product \( B \) is trimethylammonium carboxylate.

### Summary of the Steps:
1. Identify the starting compound as trimethylamine (\( \text{Me}_3\text{N} \)).
2. Recognize the reaction with \( \text{BrCN} \) leading to the formation of compound \( A \).
3. Understand that heating compound \( A \) leads to hydrolysis.
4. Conclude that the final product \( B \) is trimethylammonium carboxylate.