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Problems · Example 9.14

Q.How will you convert ethanoic acid into benzene?

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Degrade the acid to methane (sodalime decarboxylation), couple up to a C2 unit (Wurtz), strip it down to ethyne (two dehydrohalogenations), then let three ethyne molecules cyclise to benzene in a red-hot iron tube at 873 K: CHX3COOH→CHX3COONa→CHX4→CHX3Cl→CX2HX6→CX2HX5Cl→CHX2=CHX2→CHX2BrCHX2Br→CHX2=CHBr→HC≡CH→CX6HX6\ce{CH3COOH -> CH3COONa -> CH4 -> CH3Cl -> C2H6 -> C2H5Cl -> CH2=CH2 -> CH2BrCH2Br -> CH2=CHBr -> HC#CH -> C6H6}.

The strategy

There is no one-step path from a two-carbon acid to a six-carbon aromatic ring. But this unit gives us one reaction that builds benzene directly: the cyclic polymerisation of ethyne (§9.5.4, method (i)) — three HC≡CH\ce{HC#CH} molecules passed through a red-hot iron tube at 873 K join into one benzene ring. So the whole conversion becomes: turn ethanoic acid into ethyne, then cyclise.

Step-by-step route

1. Acid → salt. Neutralise ethanoic acid: CHX3COOH+NaOH→CHX3COONa+HX2O\ce{CH3COOH + NaOH -> CH3COONa + H2O}.

2. Salt → methane. Sodalime decarboxylation removes the carboxyl carbon: CHX3COONa+NaOH→CaO,ΔCHX4+NaX2COX3\ce{CH3COONa + NaOH ->[CaO, \Delta] CH4 + Na2CO3}.

3. Methane → chloromethane. Photochemical chlorination (§9.2.3): CHX4+ClX2→hνCHX3Cl+HCl\ce{CH4 + Cl2 ->[h\nu] CH3Cl + HCl}.

4. Chloromethane → ethane. The Wurtz reaction couples two methyl groups: 2 CHX3Cl+2 Na→dry etherCHX3−CHX3+2 NaCl\ce{2CH3Cl + 2Na ->[\text{dry ether}] CH3-CH3 + 2NaCl}. This is the step that grows the carbon count from 1 to 2.

5. Ethane → chloroethane. CX2HX6+ClX2→hνCX2HX5Cl+HCl\ce{C2H6 + Cl2 ->[h\nu] C2H5Cl + HCl}.

6. Chloroethane → ethene. Dehydrohalogenation with alcoholic KOH: CX2HX5Cl→alc. KOHCHX2=CHX2+HCl\ce{C2H5Cl ->[\text{alc. KOH}] CH2=CH2 + HCl}.

7. Ethene → 1,2-dibromoethane. Addition of bromine: CHX2=CHX2+BrX2→CHX2Br−CHX2Br\ce{CH2=CH2 + Br2 -> CH2Br-CH2Br}. …

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