IB Chemistry SLTopic 6 — Electron TransferPaper 1 & 2Organic~12 min read
Oxidation of Alcohols
Three alcohols, the same oxidising agent, three completely different outcomes. Which one you get is decided before the reaction starts, by a single feature of the carbon carrying the OH group.
📚 What you need to know
Alcohols are primary, secondary or tertiary according to how many carbons are attached to the carbon bearing the OH.
The oxidising agent is acidified potassium dichromate(VI), K2Cr2O7 with dilute H2SO4.
The colour change is orange to green, as Cr2O72– is reduced to Cr3+.
Distil to collect the aldehyde as it forms; reflux with excess oxidising agent to reach the carboxylic acid.
Write the oxidising agent as [O] in equations, and still balance them.
Three kinds of alcohol
Find the carbon holding the OH group and count how many other carbons are bonded to it. That number is the classification, and it also decides how many hydrogens are left on that carbon — which is what really matters.
Read the blue hydrogens, not the green R groups. Oxidation removes a hydrogen from that carbon, so the number of blue labels is exactly the number of times the alcohol can be oxidised.
That is the whole explanation for tertiary alcohols, and it is worth more than memorising the rule. Oxidising an alcohol means removing the hydrogen from the carbon bearing the OH, so it can join the removed oxygen as water. A tertiary alcohol has no hydrogen on that carbon — three carbons have taken all the other places — so there is nothing to remove and the mixture stays orange.
You built that half-equation two pages ago. For the dichromate to oxidise anything it must itself be reduced, and that reduction needs H+ ions — which is why the mixture is acidified, normally with dilute sulfuric acid. The orange-to-green change is your evidence that a reaction has taken place at all.
Where each alcohol ends up
Only one arrow in this diagram needs a decision from you: whether to distil or to reflux the primary alcohol. Everything else follows from the structure.
The choice of apparatus is the choice of product, and it turns on one physical fact: an aldehyde has no hydrogen bonding between its molecules, so it boils at a lower temperature than the alcohol it came from.
Same flask, same reagents, condenser turned through ninety degrees. In distillation the vapour escapes; in reflux it has nowhere to go but back down.
Reflux needs one more thing on the list of apparatus: the top of the condenser is left open. Sealing a heated flask would let pressure build up dangerously. Add anti-bumping granules too, so the mixture boils smoothly.
WORKED EXAMPLE
Classify each alcohol as primary, secondary or tertiary, and state the organic product when each is refluxed with excess acidified potassium dichromate(VI). (a) butan-1-ol (b) butan-2-ol (c) 2-methylpropan-2-ol
(a) butan-1-olThe OH is on the end carbon, which is attached to just one other carbon.primary → butanoic acid(b) butan-2-olThe OH carbon has a carbon on each side.secondary → butanone(c) 2-methylpropan-2-olThe OH carbon carries three methyl groups and no hydrogen.tertiary → no reactionIn (a) the aldehyde, butanal, forms first but is oxidised straight on because the mixture is refluxed with excess oxidising agent.
WORKED EXAMPLE
Ethanol is oxidised with acidified potassium dichromate(VI). Write equations, using [O], for the formation of (a) ethanal and (b) ethanoic acid, and state the apparatus needed for each.
(a) stopping at the aldehydeCH₃CH₂OH + [O] → CH₃CHO + H₂Odistillation apparatusEthanal boils at a lower temperature than ethanol, so it distils off as soon as it forms and cannot be oxidised again.(b) going all the wayCH₃CH₂OH + 2[O] → CH₃COOH + H₂Oreflux apparatus, excess oxidising agentTwo [O] because two oxidation steps have happened. Count the oxygens and hydrogens to check the equation balances.
WORKED EXAMPLE
Three unlabelled bottles contain propan-1-ol, propan-2-ol and 2-methylpropan-2-ol. Describe a test that would identify the tertiary alcohol, and explain the result.
The testWarm a sample of each with acidified potassium dichromate(VI).The resultstwo turn from orange to greenone stays orange — the tertiary alcoholThe explanation2-methylpropan-2-ol has no hydrogen on the carbon carrying the OH group, so it cannot be oxidised and the dichromate is not reduced.This test cannot separate the primary from the secondary alcohol — both give the same colour change.
💡 Exam tip
Classify the alcohol first. Everything else follows from that one decision.
Name the reagent in full: acidified potassium dichromate(VI), and quote orange to green.
Use [O] for the oxidising agent, and count how many you need: one for an aldehyde or ketone, two for an acid.
Balance the equation, including the water produced.
State the apparatus explicitly: distil for an aldehyde, reflux with excess for a carboxylic acid.
Justify distillation with the aldehyde’s lower boiling point, caused by the absence of hydrogen bonding.
⚠️ Common mix-up
Classifying by the number of hydrogens on the OH group. Count the carbons attached to the carbon holding it.
Oxidising a ketone further. It cannot be done with this reagent.
Forgetting the water in the equation, which then does not balance.
Writing one [O] for the carboxylic acid route instead of two.
Refluxing when you want the aldehyde. Reflux is precisely the apparatus that stops it escaping.
Saying a tertiary alcohol “reacts slowly”. It does not react at all with this reagent.
Up next: Reducing Carboxylic Acids, Aldehydes and Ketones — every arrow on that diagram can be run backwards with the right reagent, and the same structural logic decides where you end up.
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