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.