Question

Identify the substrate 'A' in the following reaction.
\(\text{A}\rightarrow _{\text{ii)} \Delta \,}^{\text{i) Moist Ag}_2\text{O}\,}\text{CH}_3\text{CH}_2\text{N(CH}_3)_2+\text{CH}_2 = \text{CH}_2\)

• A. Diethyldimethylammonium halide
• B. Ethyltrimethylammonium halide
• C. Diethyldimethylammonium hydroxide
• D. Ethyltrimethylammonium hydroxide

Hint:

Moist Ag2O converts the quaternary ammonium halide into its hydroxide, which on heating eliminates an alkene.

Answer 1

The Correct Option is A

Step 1: Understanding the Reaction:
This is the Hofmann elimination. A quaternary ammonium halide is converted to the hydroxide using moist silver oxide, and on heating the hydroxide loses an alkene and a tertiary amine.

Step 2: Work backwards:
The products are \(\text{CH}_3\text{CH}_2\text{N(CH}_3)_2\) (ethyldimethylamine) and ethene. Together these have the atoms of a cation with the nitrogen carrying two methyl groups and two ethyl groups: \([(\text{C}_2\text{H}_5)_2\text{N(CH}_3)_2]^+\).

Step 3: Detailed Explanation:
Starting substrate A is the halide, diethyldimethylammonium halide:
\[ (\text{C}_2\text{H}_5)_2\text{N}^+(\text{CH}_3)_2\text{X}^- \xrightarrow{\text{Ag}_2\text{O}/\text{H}_2\text{O}} (\text{C}_2\text{H}_5)_2\text{N}^+(\text{CH}_3)_2\text{OH}^- \xrightarrow{\Delta} \text{CH}_3\text{CH}_2\text{N(CH}_3)_2 + \text{CH}_2\text{=CH}_2 + \text{H}_2\text{O} \]

Step 4: Why the other options are wrong.
Ethyltrimethylammonium (B, D) has only one ethyl group and could give ethene plus trimethylamine, not ethyldimethylamine. The hydroxide (C) is the intermediate formed after the treatment with \(\text{Ag}_2\text{O}\), whereas the starting substrate A has to be the halide.

Final Answer:
The substrate is diethyldimethylammonium halide, option (A). \[ \boxed{\text{Diethyldimethylammonium halide (A)}} \]