Question 12 of 32intermediate🔧 ApplyShort Answer2 marks

How will you convert ethanoic acid into benzene?

Correct Answer

Ethanoic acid is first converted to methane, which is then converted to ethyne, and finally polymerised to benzene.

First, ethanoic acid (CH3COOH\text{CH}_3\text{COOH}) is treated with NaOH(aq)\text{NaOH(aq)} to form sodium ethanoate (CH3COONa\text{CH}_3\text{COONa}), which on heating with soda lime yields methane (CH4\text{CH}_4).

Next, methane is converted to CH3Cl\text{CH}_3\text{Cl} (using Cl2\text{Cl}_2), which undergoes Wurtz reaction with Na/dry ether to form ethane (C2H6\text{C}_2\text{H}_6). Ethane is then converted to C2H5Cl\text{C}_2\text{H}_5\text{Cl} (using Cl2\text{Cl}_2) and then to ethene (CH2=CH2\text{CH}_2=\text{CH}_2). Ethene reacts with Br2\text{Br}_2 to form CH2Br−CH2Br\text{CH}_2\text{Br}-\text{CH}_2\text{Br}, which on treatment with alc. KOH gives CH2=CHBr\text{CH}_2=\text{CHBr}. This is further treated with NaNH2\text{NaNH}_2 to form ethyne (CH≡CH\text{CH}\equiv\text{CH}).

Finally, ethyne undergoes cyclic polymerization by passing through a red hot iron tube at 873 K to form benzene.

Exercise: Problem 9.14 | Q: 9.14 | (Page Page 24)
For More Understanding

Explanation

The conversion requires moving from a 2-carbon carboxylic acid to a 6-carbon aromatic ring. The textbook context outlines the step-by-step pathway: decarboxylation to methane, chain elongation via Wurtz reaction to ethane, halogenation and dehydrohalogenation to ethyne, and finally cyclic polymerization of ethyne to benzene.

Solution Steps

  1. Step 1: CH3COOH→NaOH(aq)CH3COONa→Soda limeCH4\text{CH}_3\text{COOH} \xrightarrow{\text{NaOH(aq)}} \text{CH}_3\text{COONa} \xrightarrow{\text{Soda lime}} \text{CH}_4

  2. Step 2: CH4→Cl2CH3Cl→Na/dry etherC2H6→Cl2C2H5Cl→alc. KOHCH2=CH2\text{CH}_4 \xrightarrow{\text{Cl}_2} \text{CH}_3\text{Cl} \xrightarrow{\text{Na/dry ether}} \text{C}_2\text{H}_6 \xrightarrow{\text{Cl}_2} \text{C}_2\text{H}_5\text{Cl} \xrightarrow{\text{alc. KOH}} \text{CH}_2=\text{CH}_2

  3. Step 3: CH2=CH2→Br2CH2Br−CH2Br→alc. KOHCH2=CHBr→NaNH2CH≡CH\text{CH}_2=\text{CH}_2 \xrightarrow{\text{Br}_2} \text{CH}_2\text{Br}-\text{CH}_2\text{Br} \xrightarrow{\text{alc. KOH}} \text{CH}_2=\text{CHBr} \xrightarrow{\text{NaNH}_2} \text{CH}\equiv\text{CH}

  4. Step 4: CH≡CH→Red hot iron tube, 873 KBenzene\text{CH}\equiv\text{CH} \xrightarrow{\text{Red hot iron tube, 873 K}} \text{Benzene}