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 reactions step by step.

### Step 1: Identify the reactants
We start with trimethylamine, denoted as Me₃N, which has the structure:
\[ \text{N(CH}_3\text{)}_3 \]
This means it has three methyl groups attached to a nitrogen atom.

### Step 2: Reaction with BrCN
When trimethylamine reacts with bromine cyanide (BrCN), one of the methyl groups is replaced by a cyanide group (CN). The reaction can be represented as follows:
\[ \text{Me}_3\text{N} + \text{BrCN} \rightarrow \text{A} + \text{CH}_3\text{Br} \]
Here, product A is formed, which is dimethylsynamide (Me₂N-CN). The structure of product A can be represented as:
\[ \text{Me}_2\text{N-CN} \]

### Step 3: Hydrolysis of product A
The next step involves the hydrolysis of product A (dimethylsynamide) in the presence of water (H₂O) and heat:
\[ \text{A} + \text{H}_2\text{O} \xrightarrow{\Delta} \text{B} \]
During hydrolysis, the cyanide group (CN) is converted into a carboxylic acid group (COOH). The hydrolysis of a cyanide typically yields a carboxylic acid.

### Step 4: Determine product B
After hydrolysis, the product B will be dimethylcarbamic acid, which can be represented as:
\[ \text{B} = \text{Me}_2\text{N-COOH} \]
This means that product B is:
\[ \text{Me}_2\text{NCOOH} \]

### Conclusion
Thus, the final product B is dimethylcarbamic acid (Me₂NCOOH).

### Final Answer
B is Me₂NCOOH.

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