The shape of the lesson
By the end of the lesson, learners can
- Primary, secondary, tertiary: Warm each class of alcohol with acidified dichromate in the simulator.
- The tri-iodomethane test: Add alkaline iodine solution to several alcohols in the simulator and warm.
- K2Cr2O7/H+: orange to green.
- Distil for aldehyde; reflux for acid.
- Tertiary: no reaction.
- CHI3 (yellow) for CH3CH(OH)– and CH3CO–.
One picture every idea returns to
Acidified potassium dichromate(VI) is a mild oxidising agent; when it oxidises an alcohol, orange Cr2O72− is reduced to green Cr3+. A primary alcohol is oxidised to an aldehyde, which must be distilled off as it forms; with excess oxidant heated under reflux it goes on to a carboxylic acid. A secondary alcohol gives a ketone. A tertiary alcohol is not oxidised, because the C–OH carbon has no hydrogen. The tri-iodomethane test (alkaline iodine, warm) gives a pale yellow precipitate of CHI3 with alcohols containing CH3CH(OH)–.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Orange to green tells the class.. Warm an alcohol with acidified dichromate and watch the colour. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 17 min | Teach and check: Primary, secondary, tertiary | Warm each class of alcohol with acidified dichromate in the simulator. Project tab 3 of the lab (Hydroxy compounds lab, oxidation) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: The tri-iodomethane test | Add alkaline iodine solution to several alcohols in the simulator and warm. Project tab 5 of the lab (Hydroxy compounds lab, tri-iodomethane test) and let a student drive it while the class predicts. | Mix-Pair-Share |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 15 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in The tri-iodomethane test, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Mix-Pair-Share
Students mix around the room until you call pair; they pair with the nearest classmate, discuss the prompt, then mix again. Three prompts of rising difficulty, three different partners.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Distillation gives the aldehyde; reflux gives the carboxylic acid. | The colour change is orange to green. Tertiary alcohols are not oxidised because there is no hydrogen on the carbon bearing the OH. The tri-iodomethane test needs a CH 3 group next to the CH(OH), so propan-1-ol is negative. |
Oxidation of alcohols: worksheet
6 questions, 15 marks. Written in the style of the papers; no past paper question is reproduced.
Give the organic product, if any, when each is heated with acidified K2Cr2O7: (a) butan-2-ol, (b) 2-methylpropan-2-ol, (c) propan-1-ol under reflux.
Explain how the apparatus is chosen to make ethanal rather than ethanoic acid from ethanol, and why it works.
Write equations using [O] for the oxidation of ethanol to (a) ethanal and (b) ethanoic acid.
Describe tests using acidified dichromate(VI) and alkaline iodine that would distinguish butan-1-ol, butan-2-ol and 2-methylpropan-2-ol.
Explain why tertiary alcohols are not oxidised by acidified dichromate(VI).
Which of these give a positive tri-iodomethane test: ethanol, methanol, pentan-3-ol, pentan-2-ol? Explain.
Answer key and marking notes
- (a) butanone.
- (b) no reaction (tertiary).
- (c) propanoic acid.
- Use distillation apparatus and add the oxidant slowly (ethanol in excess).
- Ethanal (b.p. 21 °C) boils off as soon as it forms.
- It is removed from the oxidising agent before it can be oxidised to ethanoic acid.
- (a) CH3CH2OH + [O] → CH3CHO + H2O.
- (b) CH3CH2OH + 2[O] → CH3COOH + H2O.
- Acidified dichromate: butan-1-ol and butan-2-ol turn it green; 2-methylpropan-2-ol stays orange.
- Alkaline iodine: only butan-2-ol gives a pale yellow precipitate (CHI3).
- So butan-1-ol: green, no precipitate; butan-2-ol: green, precipitate; tertiary: no change in either.
- Oxidation removes the H from the O–H and an H from the carbon bearing the OH.
- In a tertiary alcohol that carbon has no hydrogen atom, so it cannot be oxidised without breaking C–C bonds.
- Ethanol and pentan-2-ol: both contain CH3CH(OH)–.
- Methanol and pentan-3-ol lack a CH3 group attached to the CH(OH) carbon, so are negative.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
Propan-1-ol is warmed with acidified potassium dichromate(VI). What colour change is seen?
- Aorange to green, as Cr2O72− is reduced to Cr3+
- Borange to purple, as the chromium is oxidised
- Cgreen to orange, as Cr3+ is oxidised to Cr2O72−
- Dpurple to colourless, as manganate(VII) is reduced
Use the flow chart. 2-methylpropan-2-ol, (CH3)3COH, is heated with acidified potassium dichromate(VI). What is observed and why?
- Athe solution stays orange; the alcohol is tertiary and has no hydrogen on the carbon bonded to OH
- Bthe solution turns green; the tertiary alcohol is oxidised to a ketone
- Cthe solution turns green; the alcohol is oxidised to a carboxylic acid under reflux
- Dthe solution stays orange because tertiary alcohols are too volatile to react
Ethanol is heated under reflux with an excess of acidified potassium dichromate(VI). What is the main organic product?
- Aethanoic acid, CH3COOH
- Bethanal, CH3CHO, because reflux stops further oxidation
- Cethene, C2H4, by dehydration
- Dno product, because ethanol is a tertiary alcohol
Which equation, using [O] for the oxidising agent, shows the oxidation of ethanol to ethanal?
- ACH3CH2OH + [O] → CH3CHO + H2O
- BCH3CH2OH + [O] → CH3CHO + H2
- CCH3CH2OH + 2[O] → CH3CHO + H2O
- DCH3CH2OH + [O] → CH3COOH + H2
Alkaline aqueous iodine (iodine and sodium hydroxide) is warmed separately with four alcohols. Which gives a pale yellow precipitate of tri-iodomethane, CHI3?
- Apropan-2-ol, CH3CH(OH)CH3
- Bpropan-1-ol, CH3CH2CH2OH
- Cmethanol, CH3OH
- D2-methylpropan-2-ol, (CH3)3COH
The diagram shows two ways of heating a primary alcohol with acidified dichromate(VI). Why does the left-hand method give an aldehyde while the right-hand method gives a carboxylic acid?
- Athe aldehyde has a lower boiling point than the alcohol and acid (no hydrogen bonding between its molecules), so in distillation it leaves the flask before it can be oxidised further; under reflux it is returned to the oxidant
- Bdistillation removes the dichromate from the flask, so no further oxidation is possible
- Cunder reflux the temperature is lower, which favours the carboxylic acid
- Dthe aldehyde has a higher boiling point than the alcohol, so it stays in the flask during distillation
An alcohol X, C4H10O, turns acidified dichromate(VI) from orange to green and the organic product does not react further. X also gives a pale yellow precipitate with alkaline aqueous iodine. What is X?
- Abutan-2-ol, CH3CH(OH)CH2CH3
- Bbutan-1-ol, CH3CH2CH2CH2OH
- C2-methylpropan-2-ol, (CH3)3COH
- D2-methylpropan-1-ol, (CH3)2CHCH2OH
Which pair of compounds BOTH give a positive tri-iodomethane (iodoform) test with alkaline aqueous iodine?
- Aethanol, CH3CH2OH, and ethanal, CH3CHO
- Bmethanol, CH3OH, and methanal, HCHO
- Cpropan-1-ol, CH3CH2CH2OH, and propanal, CH3CH2CHO
- Dpropan-2-ol, CH3CH(OH)CH3, and propanal, CH3CH2CHO
Quiz key
Acidified K2Cr2O7 contains orange Cr2O72− ions (chromium +6). As it oxidises the alcohol it is reduced to Cr3+ (+3), which is green. Orange to purple is a classic error, mixing up the colours of chromium and manganese chemistry; chromium is reduced here, not oxidised. Green to orange is the reverse change, which would need the alcohol to be an oxidising agent. Purple to colourless describes acidified manganate(VII), a different oxidising agent.
Forming C=O needs a hydrogen on the C-OH carbon. In (CH3)3COH that carbon is bonded to three CH3 groups and OH, so there is no hydrogen to remove and no reaction occurs; the dichromate stays orange. Saying a tertiary alcohol is oxidised to a ketone is the classic error: a ketone forms from a secondary alcohol, whose C-OH carbon carries one H. A carboxylic acid forms only from a primary alcohol, which has two H atoms on the C-OH carbon. Volatility has nothing to do with it; 2-methylpropan-2-ol (b.p. 82 °C) can be heated with the reagent but still does not react.
Ethanol is a primary alcohol. It is first oxidised to ethanal, but under reflux the volatile ethanal condenses and returns to the flask, where excess dichromate oxidises it further to ethanoic acid: CH3CH2OH + 2[O] → CH3COOH + H2O. Thinking reflux gives the aldehyde is the classic error; to stop at ethanal you must distil it off as soon as it forms, with the alcohol in excess. Dehydration to ethene needs hot concentrated acid or Al2O3, not an oxidising agent. Ethanol is primary (its C-OH carbon has two H atoms), not tertiary.
Oxidation to the aldehyde removes the H from O-H and one H from the C-OH carbon; these combine with one [O] to form water, so the equation is balanced with a single [O]. Hydrogen gas is never a product of dichromate oxidation; the H atoms end up in water. Using 2[O] leaves one O atom unbalanced; 2[O] is used for the full oxidation to ethanoic acid (CH3CH2OH + 2[O] → CH3COOH + H2O). The last equation is unbalanced and names the acid, not ethanal.
A positive tri-iodomethane test needs the CH3CH(OH)- group (or CH3CO-). Propan-2-ol has it: the alkaline iodine first oxidises it to propanone, CH3COCH3, then the CH3 is tri-iodinated and split off as CHI3. In propan-1-ol the OH carbon is bonded to H and C2H5, not CH3, so the test is negative; choosing it is the classic error. Methanol has no C-C bond, so there is no CH3 next to the C-OH carbon. 2-methylpropan-2-ol has CH3 groups on its C-OH carbon but no H there, so it cannot be oxidised to a methyl ketone.
Ethanal molecules cannot hydrogen bond to each other (no O-H group), so ethanal boils at 21 °C, far below ethanol (78 °C) and ethanoic acid (118 °C). With the alcohol in excess and the apparatus set for distillation, the aldehyde vaporises and is condensed into the receiver as soon as it forms, escaping further oxidation. In reflux the vertical condenser returns every vapour to the flask, so the aldehyde meets more oxidant and becomes the acid. Dichromate is an involatile ionic solid and stays in the flask in both methods. Reflux does not lower the temperature; it lets the mixture boil for a long time without loss. The aldehyde is the most volatile substance present, not the least.
The colour change shows X is oxidised, and a product that cannot be oxidised further is a ketone, so X is secondary. Butan-2-ol is the only secondary C4H10O alcohol, and it contains CH3CH(OH)-, so it also gives CHI3; it is oxidised to butanone, CH3COCH2CH3. Butan-1-ol and 2-methylpropan-1-ol are primary: their aldehydes can be oxidised further to acids, and neither has a CH3 on the C-OH carbon, so both fail the iodine test. 2-methylpropan-2-ol is tertiary, so dichromate stays orange, and it gives no CHI3.
Ethanol has CH3CH(OH)- (the OH carbon carries a CH3 and an H); the iodine oxidises it to ethanal, CH3CHO, which has CH3CO-, so both give CHI3. Ethanol is the only primary alcohol and ethanal the only aldehyde to give the test. Methanol and methanal contain only one carbon, so there is no CH3 joined to the functional carbon. Propan-1-ol and propanal have an ethyl group, not a methyl group, attached to that carbon, so both are negative. Propan-2-ol is positive, but propanal is not, so that pair fails.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.