Question

The major product obtained when ethanol is heated with excess of conc. \(H_2SO_4\) at 443 K is

• A. Ethane
• B. Ethene
• C. Methane
• D. Ethyne

Answer 1

The Correct Option is B

When ethanol is heated with excess concentrated \(\text{H}_2\text{SO}_4\), it undergoes dehydration, resulting in the elimination of water molecules and the formation of an alkene. In this case, ethanol \(\text{(C}_2\text{H}_5\text{OH)}\) loses a water molecule \(\text{(H}_2\text{O)}\) to form ethene \(\text{(C}_2\text{H}_4)\)
The reaction can be represented as follows:
\(\text{C}_2\text{H}_5\text{OH} \rightarrow \text{C}_2\text{H}_4 + \text{H}_2\text{O}\)
Therefore, the major product obtained under these conditions is ethene (option B).

Answer 2

Correct answer: Ethene 

When ethanol is heated with excess of concentrated sulfuric acid (\(H_2SO_4\)) at 443 K, it undergoes an acid-catalyzed dehydration reaction to form ethene.

The reaction is: \[ \text{CH}_3\text{CH}_2\text{OH} \xrightarrow[\text{443 K}]{\text{conc. } H_2SO_4} \text{CH}_2=CH_2 + H_2O \]

Mechanism:

1. Ethanol gets protonated by \(H_2SO_4\).
2. Water is eliminated forming a carbocation.
3. A hydrogen ion is removed, forming a double bond → ethene.

This is a typical elimination (E1) reaction, and the major product is ethene.