UNIT 2 CHEMISTRY · PEARSON AAQ · E1.7.6
Oxidation of Primary Alcohols
FREE CHEMISTRY REVISION · TOPIC 61
Oxidation of Primary Alcohols
Pearson BTEC Level 3 AAQ Applied Science
Unit 2: Principles and Applications of Chemistry · E1.7.6
1. What you need to know
Use acidified potassium dichromate(VI), K₂Cr₂O₇, to oxidise a primary alcohol to a carboxylic acid. Predict, name and draw the acid, write a balanced equation and state the conditions. This is E1.7.6. The aldehyde stage is included only to distinguish partial oxidation from the required acid product.
2. From a primary alcohol to a carboxylic acid
A primary alcohol normally has the group R–CH₂OH: the carbon carrying OH is bonded to one other carbon. On full oxidation, this end group becomes R–C(=O)–OH, commonly written R–COOH. Methanol, CH₃OH, also oxidises to methanoic acid in the corresponding first step; the examples here use ethanol and larger primary alcohols.
The carbon chain is retained. The carbon originally carrying OH becomes the carbonyl carbon of the acid. A carboxyl group contains both a C=O double bond and a C–O–H single-bond sequence attached to the same carbon. Writing COOH is a condensed way to show that group, not an O–O bond.
RCH₂OH + 2[O] → RCOOH + H₂O
[O] represents an oxygen equivalent supplied by the oxidising agent. It is not a bottle of oxygen atoms or a claim that O₂ is the reagent. Use two [O] for full oxidation to an acid.
3. Oxidising agent and conditions
Heat the primary alcohol under reflux with excess potassium dichromate(VI) solution acidified with dilute sulfuric acid. Acidified K₂Cr₂O₇ provides the oxidising conditions; excess oxidant allows oxidation to proceed beyond the aldehyde to the carboxylic acid.
Reflux allows sustained heating while vapours condense and return to the mixture. This keeps the intermediate available for further oxidation. You need the condition and its purpose, not an apparatus procedure. Reflux alone, without suitable oxidant and sufficient oxidation, does not guarantee the acid product.
As dichromate(VI) is reduced to chromium(III), orange changes towards green. Excess orange dichromate can remain, so a mixture need not look purely green. The colour change indicates reduction of the oxidising agent; the alcohol is oxidised.
4. Partial oxidation and full oxidation
RCH₂OH + [O] → RCHO + H₂O
RCHO + [O] → RCOOH
The first stage makes an aldehyde, R–C(=O)–H. The second makes an acid. Adding these equations cancels the intermediate aldehyde and gives the overall equation with 2[O] and one H₂O.
To favour an aldehyde, use controlled partial oxidation with limited oxidant and remove the aldehyde as it forms by distillation. For the carboxylic acid required here, use excess acidified dichromate(VI) and reflux. This contrast is about conditions and products; detailed distillation or reflux techniques and reaction mechanisms are outside this lesson.
5. Worked examples: progressively larger alcohols
Worked example 1 — ethanol to ethanoic acid
CH₃CH₂OH + 2[O] → CH₃COOH + H₂O
Heat under reflux with excess acidified K₂Cr₂O₇. Ethanol has two carbons, so its acid is ethanoic acid. Atom check: left C₂H₆O₃; right C₂H₄O₂ + H₂O = C₂H₆O₃. The group CH₂OH becomes C(=O)OH.
Worked example 2 — propan-1-ol to propanoic acid
CH₃CH₂CH₂OH + 2[O] → CH₃CH₂COOH + H₂O
Retain all three carbons, including the carboxyl carbon. The acid is propanoic acid, C₃H₆O₂. The partial product would be propanal, CH₃CH₂CHO, which is not the final product under full-oxidation conditions.
Worked example 3 — butan-1-ol to butanoic acid
CH₃CH₂CH₂CH₂OH + 2[O] → CH₃CH₂CH₂COOH + H₂O
The four-carbon acid is butanoic acid. Replace only the terminal CH₂OH group by COOH. Writing CH₃CH₂COOH would lose one carbon; writing CH₃CH₂CH₂CH₂COOH would add one.
Worked example 4 — pentan-1-ol to pentanoic acid
CH₃(CH₂)₄OH + 2[O] → CH₃(CH₂)₃COOH + H₂O
Pentan-1-ol has five carbons: one CH₃ plus four CH₂ groups. Its acid has one CH₃, three CH₂ groups and one COOH carbon, still five in total. Conditions remain excess acidified dichromate(VI) and reflux.
Worked example 5 — hexan-1-ol to hexanoic acid
CH₃(CH₂)₅OH + 2[O] → CH₃(CH₂)₄COOH + H₂O
Six carbons are retained. The alcohol formula C₆H₁₄O becomes acid C₆H₁₂O₂; water accounts for the other two hydrogen atoms. Count the carbon in COOH when checking the name.
6. Work backwards and check amounts
Worked example 6 — identify the starting alcohol
Target acid: CH₃CH₂CH₂COOH, butanoic acid. Change its terminal COOH group back to CH₂OH: CH₃CH₂CH₂CH₂OH, butan-1-ol. Do not choose butan-2-ol: the OH-bearing carbon would be in a different environment and this primary-alcohol rule would not apply.
Worked example 7 — interpret the equation ratio
One mole of a primary alcohol gives at most one mole of the corresponding acid by this overall conversion. Therefore 0.0250 mol propan-1-ol has a theoretical maximum of 0.0250 mol propanoic acid and requires 0.0500 mol of [O] equivalents. This does not mean 0.0500 mol K₂Cr₂O₇: [O] equivalents and formula units of dichromate are different quantities. No dichromate mole calculation is needed here.
Common misconceptions
- Full oxidation gives an aldehyde: an aldehyde is the partial stage; the required final product is an acid.
- One [O] is sufficient for alcohol-to-acid: the overall equation needs two [O].
- COOH adds a carbon: its C is the original OH-bearing carbon.
- Acidified means add ethanoic acid: use potassium dichromate(VI) acidified with dilute sulfuric acid.
- Reflux removes the aldehyde: vapours return; removal favours stopping at the aldehyde.
- The orange-to-green change means dichromate is oxidised: it is reduced while the alcohol is oxidised.
Examination tips
- State the complete reagent: acidified potassium dichromate(VI), K₂Cr₂O₇.
- For the acid, add excess oxidant and heating under reflux.
- Show R–C(=O)–OH when asked to draw the carboxyl group explicitly.
- Keep the carbon count and change the name ending from -ol to -oic acid.
- Balance the overall equation with 2[O] and H₂O.
- Keep partial oxidation separate from the required full-oxidation answer.
Check your understanding
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