Describe the formation of ethanoic acid by the oxidation of ethanol: (a) with acidified aqueous potassium manganate(VII) (b) by bacterial oxidation during vinegar production

Organic Chemistry – Carboxylic Acids

Objective

Describe how ethanoic acid (acetic acid) is formed from ethanol by (a) acidified aqueous potassium manganate(VII) and (b) bacterial oxidation during vinegar production. Use clear steps, analogies, and exam‑friendly tips. 🍋

(a) Chemical Oxidation with Acidified Potassium Manganate(VII)

When ethanol (\$\ce{CH3CH2OH}\$) is treated with acidified \$\ce{KMnO4}\$, the reaction proceeds in two stages, just like a two‑step recipe: first a sweet cake (acetaldehyde) is made, then it is baked into a sour cake (acetic acid). The manganese(VII) ion is reduced to manganese(II) while the alcohol is oxidised.

  1. Acidic conditions protonate the alcohol, making it a better leaving group.

  2. Potassium manganate(VII) (\$\ce{MnO4^-}\$) accepts electrons, being reduced to \$\ce{Mn^{2+}}\$.

  3. The alcohol is first oxidised to acetaldehyde (\$\ce{CH3CHO}\$).

  4. Further oxidation of the aldehyde gives ethanoic acid (\$\ce{CH3COOH}\$).

StepReaction
1

\$\ce{CH3CH2OH + H+ -> CH3CH2OH2+}\$ (protonated ethanol)

2

\$\ce{CH3CH2OH2+ + MnO4^- -> CH3CHO + Mn^{2+} + H2O}\$ (first oxidation)

3

\$\ce{CH3CHO + MnO4^- -> CH3COOH + Mn^{2+} + H2O}\$ (second oxidation)

Exam Tip: Remember that \$\ce{KMnO4}\$ is a strong oxidiser; the reaction is two‑step. Write the overall equation: \$\ce{CH3CH2OH + 2 KMnO4 + 2 H2SO4 -> CH3COOH + 2 MnSO4 + 2 KHSO4 + 3 H2O}\$. Show the intermediate \$\ce{CH3CHO}\$ if asked. Use the mnemonic “Ethanol → Acetaldehyde → Acetic Acid”.

(b) Bacterial Oxidation – Vinegar Production

In nature, certain bacteria (e.g., Acetobacter) act like tiny chefs that turn alcohol into vinegar. The process is a biological oxidation that also goes through an aldehyde intermediate, but the bacteria provide the enzyme alcohol dehydrogenase and acetaldehyde dehydrogenase.

  1. Alcohol dehydrogenase converts ethanol to acetaldehyde.

  2. Acetaldehyde dehydrogenase then oxidises acetaldehyde to ethanoic acid.

  3. Oxygen from the air is the ultimate oxidiser.

Analogy: Think of the bacteria as a factory line: the first worker (alcohol dehydrogenase) makes a raw material (acetaldehyde), and the second worker (acetaldehyde dehydrogenase) refines it into the final product (acetic acid). The factory runs on oxygen, just like a car needs fuel to run.

EnzymeReaction
Alcohol dehydrogenase

\$\ce{CH3CH2OH + NAD+ -> CH3CHO + NADH + H+}\$

Acetaldehyde dehydrogenase

\$\ce{CH3CHO + NAD+ + H2O -> CH3COOH + NADH + H+}\$

Exam Tip: When asked about vinegar production, mention the two enzymes and that oxygen is the final oxidiser. Write the overall reaction: \$\ce{CH3CH2OH + 1/2 O2 -> CH3COOH + H2O}\$. Highlight that the process is biological and occurs at room temperature, unlike the harsh chemical oxidation.

Key Take‑aways for the Exam

  • Both methods give the same final product: ethanoic acid.

  • In chemical oxidation, \$\ce{KMnO4}\$ is the oxidiser; in biological oxidation, bacteria and oxygen are the oxidisers.

  • Remember the intermediate \$\ce{CH3CHO}\$ in both pathways.

  • Use clear, step‑by‑step diagrams or tables to show the sequence.

  • Practice writing balanced equations for both processes.