Question:

Consider the following organic compounds:
\[ C_2H_5CHO, \, CH_3COOH, \, C_2H_5OH, \, CH_3COCH_3 \]
Which one of these compounds, when heated at 443 K in excess conc. \(\text{H}_2\text{SO}_4\), forms ethene as its major product? State the role of conc. \(\text{H}_2\text{SO}_4\) in the reaction. Also, write the chemical equation for the chemical reaction involved.

Updated On: Jun 6, 2025
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Solution and Explanation

Step 1: Identify the compound that forms ethene
The compound that forms ethene as its major product when heated at 443 K in excess concentrated sulfuric acid (\( \text{H}_2\text{SO}_4 \)) is ethanol (\( C_2H_5OH \)).
This reaction is an example of a dehydration reaction, where ethanol undergoes dehydration to form ethene.

Step 2: Role of concentrated sulfuric acid
Concentrated sulfuric acid acts as both a catalyst and a dehydrating agent. It removes water from the ethanol molecule, causing it to break down into ethene and water.

Step 3: Chemical equation for the reaction
The chemical equation for the dehydration of ethanol to form ethene is as follows:
\[ C_2H_5OH \xrightarrow{\text{conc.} \, \text{H}_2\text{SO}_4, 443 \, \text{K}} C_2H_4 + \text{H}_2\text{O} \]

Step 4: Explanation of the reaction
- When ethanol is heated with concentrated sulfuric acid at 443 K, it undergoes elimination of water (\( \text{H}_2\text{O} \)) to form ethene (\( C_2H_4 \)). 
- The concentrated sulfuric acid helps in removing water and facilitates the formation of the double bond in ethene.

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