Ethanol reacts with concentrated tetraoxosulphate (VI) acid at a temperature above 170$^0$C to form

Analysis

The question asks for the product formed when ethanol ($C_2H_5OH$) undergoes a reaction with concentrated tetraoxosulphate(VI) acid ($H_2SO_4$) at a specific temperature (above 170°C). This is a classic organic chemistry reaction involving the dehydration of an alcohol.

Concept: Dehydration of Alcohols

Concentrated tetraoxosulphate(VI) acid acts as a dehydrating agent in this reaction. The outcome of the reaction between ethanol and concentrated $H_2SO_4$ depends heavily on the temperature:

1. At 140°C: The reaction produces an ether (ethoxyethane).
2. At 170°C and above: The reaction undergoes intramolecular dehydration, where a molecule of water is removed from a single molecule of ethanol, resulting in the formation of an alkene (ethene).

Step-by-Step Solution

1. Identify the reactants: Ethanol ($CH_3CH_2OH$) and concentrated $H_2SO_4$.
2. Identify the process: Since the temperature is above 170°C, the concentrated $H_2SO_4$ removes a water molecule ($H_2O$) from the ethanol molecule.
3. Write the chemical equation: $$CH_3CH_2OH \xrightarrow{conc. H_2SO_4, >170^\circ C} CH_2=CH_2 + H_2O$$
4. Analyze the product: The product $CH_2=CH_2$ is ethene, which is the simplest member of the alkene homologous series.

Conclusion

Comparing this to your options:

• A. Ethanone (a ketone, not formed here)
• B. Ethene (the correct alkene product)
• C. Ethyne (an alkyne, not formed here)
• D. Ethanal (an aldehyde, formed by oxidation, not dehydration)

Exam Tip

Remember the "Temperature Rule" for ethanol and $H_2SO_4$:

• 140°C = Ether (Think: "Two alcohols join to make an ether").
• 170°C = Ethene (Think: "High heat breaks it down to an alkene").