Question

How many molecules of methyl iodide are required to obtain tetramethyl ammonium iodide from dimethyl amine?

• A. 1
• B. 3
• C. 2
• D. 4

Hint:

Simply calculate the difference in the number of methyl groups between your reactant and your target product. Dimethyl amine has 2 methyl groups, and tetramethyl ammonium iodide has 4. To bridge this gap, you must add exactly $4 - 2 = 2$ methyl groups via methyl iodide.

Answer 1

The Correct Option is C

Step 1: Understand the question.
We start with dimethyl amine, $(\text{CH}_3)_2\text{NH}$, and want to reach tetramethyl ammonium iodide. We must count how many methyl iodide ($\text{CH}_3\text{I}$) molecules are needed.

Step 2: Count carbons on nitrogen at start and end.
Dimethyl amine has 2 methyl groups on nitrogen. The final salt $(\text{CH}_3)_4\text{N}^+\text{I}^-$ has 4 methyl groups on nitrogen. So we must add $4 - 2 = 2$ new methyl groups.

Step 3: Know how methyl iodide adds groups.
Each $\text{CH}_3\text{I}$ molecule attaches exactly one methyl group to the nitrogen. So we will need one molecule per methyl group added.

Step 4: First addition.
Dimethyl amine plus one $\text{CH}_3\text{I}$ gives trimethyl amine (now 3 methyl groups).
\[ (\text{CH}_3)_2\text{NH} + \text{CH}_3\text{I} \rightarrow (\text{CH}_3)_3\text{N} + \text{HI} \]

Step 5: Second addition.
Trimethyl amine plus a second $\text{CH}_3\text{I}$ gives the quaternary salt (now 4 methyl groups).
\[ (\text{CH}_3)_3\text{N} + \text{CH}_3\text{I} \rightarrow (\text{CH}_3)_4\text{N}^+\text{I}^- \]

Step 6: Count the molecules used.
We used 2 molecules of methyl iodide in total, which is option 3.
\[ \boxed{2} \]