H$_2$SO$_4$ C$_2$H$_5$OH      →         C$_2$H$_4$                     170$^0$C The reaction above illustrates

Analysis of the Question

The reaction provided is: C$_2$H$_5$OH $\xrightarrow{H_2SO_4, 170^\circ C}$ C$_2$H$_4$ + H$_2$O

You are asked to identify the type of chemical reaction occurring when ethanol (C$_2$H$_5$OH) is heated with concentrated tetraoxosulphate(VI) acid (H$_2$SO$_4$) at 170°C to produce ethene (C$_2$H$_4$).

Explanation of the Concept

This reaction is a classic example of an elimination reaction. Specifically, it is the removal of a water molecule from an alcohol.

• Dehydration: In organic chemistry, dehydration refers to the removal of a water molecule (H$_2$O) from a compound. When concentrated H$_2$SO$_4$ acts on ethanol at a high temperature (170°C), it acts as a dehydrating agent. It pulls the hydroxyl group (-OH) and a hydrogen atom (-H) from adjacent carbon atoms in the ethanol molecule, forcing the formation of a double bond between the carbons to produce an alkene (ethene).
• Why other options are incorrect: • Oxidation: This would involve adding oxygen or removing hydrogen to form an aldehyde or carboxylic acid (e.g., using acidified KMnO$_4$). • Esterification: This involves the reaction between an alcohol and a carboxylic acid to form an ester and water. • Polymerization: This involves joining many small monomer units (like ethene) to form a long-chain polymer (like polyethene).

Step-by-Step Solution

1. Identify the reactants: Ethanol (C$_2$H$_5$OH).
2. Identify the conditions: Concentrated H$_2$SO$_4$ at 170°C.
3. Analyze the transformation: The ethanol loses one H atom and one OH group. • C$_2$H$_5$OH $\rightarrow$ C$_2$H$_4$ + H$_2$O
4. Conclude the process: Since water is removed from the molecule to form a new product, the process is defined as dehydration.

Correct Answer

The correct option is C.

Exam Tip

Remember the "Temperature Rule" for ethanol and H$_2$SO$_4$:

• At 170°C, you get ethene (dehydration/elimination).
• At 140°C, you get ethoxyethane (diethyl ether) through a condensation reaction. Always check the temperature in the question—it is the key to distinguishing between these two common JAMB chemistry reactions!