Question

The product obtained, heating ethanol with conc. ${{H}_{2}}S{{O}_{4}}$at ${{165}^{{\mathrm O}}}-{{170}^{{\mathrm O}}}$, is
A. ${{({{C}_{2}}{{H}_{5}})}_{2}}S{{O}_{4}}$
B. $C{{H}_{2}}=C{{H}_{2}}$
C. $C{{H}_{3}}COOH$
D. ${{C}_{2}}{{H}_{5}}HS{{O}_{4}}$

Answer 1

Hint: Concentrated sulphuric acid $({{H}_{2}}S{{O}_{4}})$ acts as a dehydrating agent. When alcohol reacts with concentrated sulphuric acid at high temperatures produces alkene as a major product after the removal of water molecules. Here we also have alcohol, ethanol reacts with conc. ${{H}_{2}}S{{O}_{4}}$ produces an alkene through dehydration.

Complete Step by Step Answer:
When ethyl alcohol or ethanol reacts with concentrated sulphuric acid at a temperature range ${{165}^{{\mathrm O}}}-{{170}^{{\mathrm O}}}C$ and intramolecular dehydration occurs to form ethane.

The overall mechanism of acid hydration of alcohol proceeds through three steps. In the first step oxygen of alcohol is protonated.

In the second step, a carbocation is formed by the loss of water molecules.

In the third step or final step, an alkene is formed by an elimination reaction with carbocation rearrangement. Here an alkene is formed with the migration of hydride shift from adjacent carbon.

The overall mechanism proceeds through ${{E}_{2}}$ mechanism as ${{E}_{1}}$ pathway is not favoured here. This is because when primary alcohol reacts with the acid, a primary carbocation is generated which is extremely unstable. Through ${{E}_{2}}$elimination, the transition state will be lower in energy and therefore a proton is abstracted from adjacent carbon by a weak nucleophile, $HSO_{4}^{-}$, leading to the formation of alkene.

Therefore when ethanol is heated with conc. ${{H}_{2}}S{{O}_{4}}$at ${{165}^{{\mathrm O}}}-{{170}^{{\mathrm O}}}C$, ethane is formed.
Thus, option (B) is correct.

Note: When hydrogen halide reacts with alcohol a substitution reaction occurs but when Sulphuric acid reacts with alcohol an elimination reaction occurs. Because $HSO_{4}^{-}$is a weak nucleophile and is stabilised through resonance. The negative charge on oxygens delocalized over the molecule and hence is less available for substitution reaction.