Question

Mohan heated ethanol with a compound 'X' in the presence of a few drops of conc. \( H_2SO_4 \) and observed a sweet smelling compound 'Y' is formed. When 'Y' is treated with sodium hydroxide it gives back ethanol and a compound 'Z'.
(i) Identify 'X', 'Y' and 'Z'.
(ii) Write the role of conc. \( H_2SO_4 \) in the reaction.
(iii) Write the chemical equations involved and name the reactions.

Hint:

Esterification: Alcohol + Carboxylic acid \(\xrightarrow{\text{conc. H}_2\text{SO}_4}\) Ester + Water (sweet smell) Saponification: Ester + NaOH \(\Rightarrow\) Alcohol + Sodium salt of carboxylic acid Conc. H₂SO₄ acts as catalyst and dehydrating agent.

Answer 1

Part (i): Identification of X, Y and Z
Step 1: Analyze the first reaction.
Ethanol is heated with compound X in presence of conc. H₂SO₄ to form a sweet-smelling compound Y. Sweet-smelling organic compounds are typically esters. This indicates that Y is an ester.
Step 2: Formation of ester.
Esters are formed by the reaction of a carboxylic acid with an alcohol in presence of conc. H₂SO₄ (esterification reaction). Therefore, X must be a carboxylic acid. Since ethanol is the alcohol, the ester Y will be ethyl something. The most common carboxylic acid used is ethanoic acid (acetic acid, CH₃COOH). Then Y would be ethyl ethanoate (CH₃COOC₂H₅). \[ \text{X = Ethanoic acid (CH₃COOH)} \] \[ \text{Y = Ethyl ethanoate (CH₃COOC₂H₅)} \]
Step 3: Analyze the second reaction.
When Y (ester) is treated with sodium hydroxide (NaOH), it gives back ethanol and compound Z. This is saponification (alkaline hydrolysis of ester). Esters hydrolyze to give back alcohol and salt of carboxylic acid. \[ \text{Ester (Y) + NaOH \(\Rightarrow\) Alcohol (ethanol) + Sodium salt of carboxylic acid (Z)} \] Therefore, Z is the sodium salt of the carboxylic acid X. Since X is ethanoic acid, Z is sodium ethanoate (sodium acetate, CH₃COONa). \[ \text{Z = Sodium ethanoate (CH₃COONa)} \]
Step 4: Final identification.
\[ \boxed{\text{X = Ethanoic acid (CH₃COOH)}
\boxed{\text{Y = Ethyl ethanoate (CH₃COOC₂H₅)}
\boxed{\text{Z = Sodium ethanoate (CH₃COONa)}} \] Part (ii): Role of conc. H₂SO₄ in the reaction
Step 1: Understand the reaction.
The reaction between ethanol and ethanoic acid to form ester is reversible and slow.
Step 2: Functions of conc. H₂SO₄.
Concentrated sulphuric acid plays multiple roles:

• Dehydrating agent: It removes water formed during esterification, shifting the equilibrium towards product formation (Le Chatelier's principle).
• Catalyst: It speeds up the reaction by protonating the carboxylic acid, making it more susceptible to nucleophilic attack by alcohol.
• Absorbs water: It absorbs the water produced, preventing the reverse reaction (hydrolysis of ester).

Step 3: Final answer.
Conc. H₂SO₄ acts as a dehydrating agent and catalyst. It removes water and speeds up the esterification reaction.}} Part (iii): Chemical equations and naming of reactions
Step 1: First reaction - Esterification.
Ethanol reacts with ethanoic acid in presence of conc. H₂SO₄ to form ethyl ethanoate (ester) and water. \[ \boxed{\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{conc. H}_2\text{SO}_4} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}} \] Reaction name: \textit{Esterification} or \textit{Fischer esterification}
Step 2: Second reaction - Saponification (Alkaline hydrolysis).
Ethyl ethanoate reacts with sodium hydroxide to give back ethanol and sodium ethanoate. \[ \boxed{\text{CH}_3\text{COOC}_2\text{H}_5 + \text{NaOH} \xrightarrow{\Delta} \text{CH}_3\text{COONa} + \text{C}_2\text{H}_5\text{OH}} \] Reaction name: \textit{Saponification} or \textit{Alkaline hydrolysis of ester}