Chapter 8 · Chemistry
Preparing Carboxylic Acids: Four Routes
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Four ways to make a carboxylic acid: oxidising primary alcohols and aldehydes, oxidising alkylbenzenes, hydrolysing nitriles and esters, and adding a Grignard reagent to dry ice, which adds one carbon (mechanism with curved arrows).
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Key points
- Oxidation: RCH₂OH → RCHO → RCOOH with KMnO₄ or K₂Cr₂O₇; the carbon count stays the same.
- Alkylbenzenes → benzoic acid with KMnO₄ (then H₃O⁺): the whole side chain becomes one COOH, if the carbon attached to the ring carries an H.
- Hydrolysis: R–C≡N → amide → RCOOH; esters and acid chlorides also give acids.
- RMgX + CO₂ (dry ice), then H₃O⁺ → RCOOH with one carbon more: the C–Mg bond pair attacks the carbon of CO₂.
The four routes
| Route | Starting material | Reagent | Product |
|---|---|---|---|
| Oxidation | 1° alcohol or aldehyde | KMnO₄ or K₂Cr₂O₇ (acid) | RCOOH (same C count) |
| Side-chain oxidation | alkylbenzene (toluene) | KMnO₄/OH⁻, then H₃O⁺ | benzoic acid |
| Hydrolysis | nitrile, amide, ester, acid chloride | H₂O with H⁺ or OH⁻ | RCOOH |
| Grignard | R–X → RMgX | CO₂ (dry ice), then H₃O⁺ | RCOOH (one C more) |
Grignard + CO₂, arrow by arrow
- CH₃Br + Mg in dry ether → CH₃MgBr; its C–Mg bond is polar (C δ⁻).
- The C–Mg bond pair attacks the carbon of CO₂ (new C–C bond); a C=O π pair moves onto O.
- This gives CH₃COO⁻ ⁺MgBr; H₃O⁺ gives CH₃COOH: 1 C → 2 C.
Quick quiz
5 questions. Choose an answer for each, then submit to see your score and the explanations.