Identify the final product `(Z)` in the following sequence of reactions. `CH_3NH_2overset(CH_3l" Excess")rarr(X) overset(AgOH)rarr(Y)overset(Delta)rarr(Z)+ CH_3OH`

Options

A. `CH_3CH_2CH_2NH_2`
B. `(CH_3)_3N`
C. `(CH_3)_2NH`
D. `(CH_3)_4overset(+)NOH^(-)`

Answer

To solve the problem step by step, we will analyze the sequence of reactions and identify the final product (Z).

### Step 1: Reaction of Methylamine with Methyl Iodide
The first reaction involves methylamine (CH₃NH₂) reacting with excess methyl iodide (CH₃I).

**Reaction:**
\[ \text{CH}_3\text{NH}_2 + \text{CH}_3\text{I} \rightarrow \text{CH}_3\text{N}^+\text{(CH}_3)_3\text{I}^- \]

In this reaction, methylamine acts as a nucleophile and attacks the electrophilic carbon in methyl iodide, resulting in the formation of a quaternary ammonium salt (X). The nitrogen atom now has four methyl groups attached and carries a positive charge.

**Hint:** Look for nucleophilic substitution reactions when amines react with alkyl halides.

### Step 2: Reaction with Silver Hydroxide
Next, the quaternary ammonium salt (X) is treated with silver hydroxide (AgOH).

**Reaction:**
\[ \text{CH}_3\text{N}^+\text{(CH}_3)_3\text{I}^- + \text{AgOH} \rightarrow \text{CH}_3\text{N}^+\text{(CH}_3)_3\text{OH}^- + \text{AgI} \]

In this step, the iodide ion (I⁻) is replaced by the hydroxide ion (OH⁻) due to the precipitation of silver iodide (AgI). The product (Y) is a quaternary ammonium hydroxide.

**Hint:** Remember that halide ions can be replaced by hydroxide ions in the presence of silver salts.

### Step 3: Heating the Quaternary Ammonium Hydroxide
The next step involves heating the quaternary ammonium hydroxide (Y).

**Reaction:**
\[ \text{CH}_3\text{N}^+\text{(CH}_3)_3\text{OH}^- \xrightarrow{\Delta} \text{CH}_3\text{N(CH}_3)_3 + \text{CH}_3\text{OH} \]

Upon heating, one of the methyl groups is eliminated along with a hydroxyl group, resulting in the formation of trimethylamine (Z) and methanol (CH₃OH).

**Hint:** Heating quaternary ammonium compounds can lead to the elimination of alkyl groups, forming tertiary amines.

### Final Product (Z)
The final product (Z) is trimethylamine, which has the structure:

\[ \text{Z} = \text{CH}_3\text{N(CH}_3)_3 \]

### Summary
The final product Z in the sequence of reactions is trimethylamine (option B).