Question

Write chemical equations of two methods of preparation of monohydric alcohol and also write chemical equation of the reactions of ethyl alcohol with conc. sulphuric acid (H$_2$SO$_4$) at different temperatures.

Hint:

Remember the temperature effect: $140^\circ C \to$ ether, $170^\circ C \to$ ethene, higher $T \to$ oxidation products.

Answer 1

Part 1: Two methods of preparation of monohydric alcohol.
Method 1: By hydration of alkenes.
Ethene reacts with water in presence of conc. H$_2$SO$_4$ or phosphoric acid catalyst to give ethanol: \[ CH_2=CH_2 + H_2O \;\xrightarrow{H_2SO_4}\; CH_3CH_2OH \] Method 2: By reduction of aldehydes.
Acetaldehyde on reduction with hydrogen gives ethanol: \[ CH_3CHO + H_2 \;\xrightarrow{Ni/Pt}\; CH_3CH_2OH \] Part 2: Reactions of ethyl alcohol with conc. H$_2$SO$_4$ at different temperatures.
(i) At 443 K (140$^\circ$C): Formation of diethyl ether.
\[ C_2H_5OH + C_2H_5OH \;\xrightarrow[443K]{conc. H_2SO_4}\; C_2H_5OC_2H_5 + H_2O \] (ii) At 443–463 K: Dehydration to form ethene.
\[ C_2H_5OH \;\xrightarrow[443-463K]{conc. H_2SO_4}\; CH_2=CH_2 + H_2O \] (iii) As oxidising agent: Formation of acetaldehyde.
\[ CH_3CH_2OH \;\xrightarrow{conc. H_2SO_4}\; CH_3CHO + H_2 \] Conclusion:
Monohydric alcohols can be prepared by hydration of alkenes or reduction of aldehydes. Ethanol reacts with conc. H$_2$SO$_4$ differently depending on temperature, giving ether, ethene, or acetaldehyde.