Oxidation of Alcohols and the Tri-iodomethane Test

AS · 12 min

Oxidation is the reaction that tells primary, secondary and tertiary alcohols apart. With acidified potassium dichromate(VI), a primary alcohol gives an aldehyde and then a carboxylic acid, a secondary alcohol gives a ketone, and a tertiary alcohol does not react at all. This note explains why, shows how the choice between distillation and heating under reflux controls which product you get, builds the full ionic equations, and covers the tri-iodomethane (iodoform) test for the $\ce{CH3CH(OH)-}$ group. These reactions appear in Paper 2 synthesis and deduction questions and in Paper 3 as practical tests.

The oxidising agent

The usual reagent is potassium dichromate(VI), $\ce{K2Cr2O7}$ , acidified with dilute sulfuric acid. The dichromate(VI) ion is reduced to the chromium(III) ion:

\ce{Cr2O7^2- + 14H+ + 6e- -> 2Cr^3+ + 7H2O}

Key result

Colour change when an alcohol is oxidised: orange ( $\ce{Cr2O7^2-}$ ) to green ( $\ce{Cr^3+}$ ). Chromium is reduced from $+6$ to $+3$ .

Acidified potassium manganate(VII) also oxidises primary and secondary alcohols (purple to colourless), and is sometimes used to make carboxylic acids, but dichromate is the standard reagent because it is easier to control.

In simplified equations, the oxygen supplied by the oxidising agent is written $\ce{[O]}$ .

What gets oxidised, and why tertiary alcohols do not react

When an alcohol is oxidised to a carbonyl compound, two hydrogen atoms are removed: the H of the O–H group and an H on the carbon bearing the OH. These two H atoms combine with $\ce{[O]}$ to form water, and a C=O double bond forms.

\ce{R2CH-OH + [O] -> R2C=O + H2O}

A primary alcohol, $\ce{RCH2OH}$ , has two H atoms on that carbon. It is oxidised to an aldehyde, RCHO, which still has an H on the carbonyl carbon and can be oxidised further to a carboxylic acid, RCOOH.
A secondary alcohol, $\ce{R2CHOH}$ , has one H on that carbon. It is oxidised to a ketone, $\ce{R2CO}$ , which has no H on the carbonyl carbon and is not oxidised further.
A tertiary alcohol, $\ce{R3COH}$ , has no H on the carbon bearing the OH. It cannot form a C=O without breaking a C–C bond, so it is not oxidised: the dichromate stays orange.

Key result

alcohol	product with acidified $\ce{K2Cr2O7}$	colour of dichromate
primary	aldehyde (distil), then carboxylic acid (reflux)	orange → green
secondary	ketone	orange → green
tertiary	no reaction	stays orange

Primary alcohols: aldehyde or carboxylic acid?

The first product, the aldehyde, is easily oxidised further:

\ce{CH3CH2OH + [O] -> CH3CHO + H2O}

\ce{CH3CHO + [O] -> CH3COOH}

Overall: $\ce{CH3CH2OH + 2[O] -> CH3COOH + H2O}$ .

Which product you collect depends on the apparatus. The key fact is that the aldehyde has a much lower boiling point than the alcohol or the acid, because aldehyde molecules cannot hydrogen bond to each other (they have no O–H group).

compound	boiling point / $^\circ\text{C}$	hydrogen bonding between molecules?
ethanal, $\ce{CH3CHO}$	21	no
ethanol, $\ce{CH3CH2OH}$	78	yes
ethanoic acid, $\ce{CH3COOH}$	118	yes

Method

To make the aldehyde: distil it off as it forms

Heat the acidified dichromate in a flask fitted for distillation (a still head, thermometer and sloping condenser).
Add the primary alcohol slowly, so that the alcohol is in excess and the oxidising agent is limited.
The aldehyde, with the lowest boiling point, distils out as soon as it forms, before it can be oxidised further. Collect it in a receiver cooled in ice.

Method

To make the carboxylic acid: heat under reflux

Use an excess of acidified dichromate.
Heat the mixture in a flask with a vertical (reflux) condenser. Vapours of alcohol and aldehyde condense and drip back into the flask, so they stay in contact with the oxidising agent until fully oxidised to the acid.
After refluxing, rearrange the apparatus for distillation and distil off the carboxylic acid.

Secondary alcohols give a ketone with either apparatus; the ketone is not oxidised further.

\ce{CH3CH(OH)CH3 + [O] -> CH3COCH3 + H2O}

Practical skills

Reflux is used whenever a reaction mixture must be heated for a long time without losing volatile substances. Add anti-bumping granules to the flask for smooth boiling. Water enters the condenser at the bottom and leaves at the top, so the condenser is completely filled with cold water. Never heat a sealed system: the top of a reflux condenser is open. Organic liquids are flammable, so heat with an electric heating mantle or water bath, not a Bunsen flame. In a distillation, the thermometer bulb sits level with the side arm, to measure the temperature of the vapour that is distilling.

Full ionic equations

To write the full equation, combine the half-equation for dichromate with a half-equation for the organic compound. Organic half-equations are balanced with $\ce{H2O}$ , $\ce{H+}$ and electrons.

Ethanol to ethanal: $\ce{CH3CH2OH -> CH3CHO + 2H+ + 2e-}$

Multiply by 3 to transfer 6 electrons and add to the dichromate half-equation:

\ce{3CH3CH2OH + Cr2O7^2- + 14H+ -> 3CH3CHO + 6H+ + 2Cr^3+ + 7H2O}

Cancel $\ce{6H+}$ :

\ce{3CH3CH2OH + Cr2O7^2- + 8H+ -> 3CH3CHO + 2Cr^3+ + 7H2O}

Ethanol to ethanoic acid: $\ce{CH3CH2OH + H2O -> CH3COOH + 4H+ + 4e-}$ . The lowest common multiple of 4 and 6 is 12:

\ce{3CH3CH2OH + 2Cr2O7^2- + 16H+ -> 3CH3COOH + 4Cr^3+ + 11H2O}

Check: H $18 + 16 = 34$ left, $12 + 22 = 34$ right; O $3 + 14 = 17$ left, $6 + 11 = 17$ right; charge $-4 + 16 = +12$ left, $+12$ right.

Distinguishing primary, secondary and tertiary alcohols

Method

Classifying an unknown alcohol

Warm a few drops of the alcohol with acidified potassium dichromate(VI).
If the solution stays orange, the alcohol is tertiary.
If it turns green, the alcohol is primary or secondary. Repeat the oxidation with distillation and collect the product.
Test the distillate with Tollens' reagent (silver mirror) or Fehling's solution (red-brown precipitate). A positive result shows an aldehyde, so the alcohol was primary. A negative result shows a ketone, so the alcohol was secondary.

The tests for aldehydes and ketones are explained in Tests for carbonyl compounds.

The tri-iodomethane (iodoform) test

Key result

Test for the $\ce{CH3CH(OH)-}$ group (a methyl group on the carbon bearing the OH):

Warm the alcohol with alkaline aqueous iodine (iodine solution with aqueous sodium hydroxide added until the brown colour just fades).

Positive result: a pale yellow precipitate of tri-iodomethane, $\ce{CHI3}$ , which has an antiseptic smell.

The test works in two stages. First, alkaline iodine oxidises the $\ce{CH3CH(OH)-}$ group to a $\ce{CH3CO-}$ group (a methyl ketone, or ethanal from ethanol). Second, the $\ce{CH3CO-}$ group reacts with more alkaline iodine to give $\ce{CHI3}$ and a carboxylate ion with one carbon fewer.

So the test is positive for any compound with $\ce{CH3CH(OH)-}$ or $\ce{CH3CO-}$ . For alcohols:

alcohol	contains $\ce{CH3CH(OH)-}$ ?	tri-iodomethane test
methanol, $\ce{CH3OH}$	no	negative
ethanol, $\ce{CH3CH2OH}$	yes (R = H)	positive (the only primary alcohol that is)
propan-1-ol, $\ce{CH3CH2CH2OH}$	no	negative
propan-2-ol, $\ce{CH3CH(OH)CH3}$	yes	positive
butan-2-ol, $\ce{CH3CH(OH)CH2CH3}$	yes	positive
pentan-3-ol, $\ce{CH3CH2CH(OH)CH2CH3}$	no	negative
2-methylpropan-2-ol, $\ce{(CH3)3COH}$	no (no H on the OH carbon)	negative

The overall equations, for interest:

\ce{CH3CH2OH + 4I2 + 6OH- -> CHI3 + HCOO- + 5I- + 5H2O}

\ce{CH3CH(OH)CH3 + 4I2 + 6OH- -> CHI3 + CH3COO- + 5I- + 5H2O}

Tip

Tertiary alcohols such as 2-methylbutan-2-ol, $\ce{CH3CH2C(OH)(CH3)2}$ , have a methyl group on the OH carbon but no hydrogen on it, so they cannot be oxidised to a $\ce{CH3CO-}$ compound and give a negative test. The group needed is specifically $\ce{CH3CH(OH)-}$ , with the H.

Worked examples

Routine: products of oxidation

Give the organic product when each alcohol is warmed with acidified potassium dichromate(VI), and state the colour change: (a) propan-1-ol, with the product distilled off immediately; (b) propan-1-ol, heated under reflux with excess oxidising agent; (c) butan-2-ol; (d) 2-methylbutan-2-ol.

Solution

(a) Propanal, $\ce{CH3CH2CHO}$ : $\ce{CH3CH2CH2OH + [O] -> CH3CH2CHO + H2O}$ . Orange to green.

(b) Propanoic acid, $\ce{CH3CH2COOH}$ : $\ce{CH3CH2CH2OH + 2[O] -> CH3CH2COOH + H2O}$ . Orange to green.

(d) Tertiary: no reaction; the solution stays orange.

Routine: explaining the choice of apparatus

Explain why ethanal can be obtained from ethanol by distillation but not by heating under reflux.

Solution

Ethanal ( $21\ ^\circ\text{C}$ ) has a much lower boiling point than ethanol ( $78\ ^\circ\text{C}$ ) or ethanoic acid ( $118\ ^\circ\text{C}$ ), because its molecules cannot hydrogen bond to each other. In a distillation apparatus, ethanal evaporates and leaves the flask as soon as it forms, before it can be oxidised further. Under reflux, the ethanal vapour condenses and returns to the flask, where it stays in contact with the oxidising agent and is oxidised to ethanoic acid.

Standard: an ionic equation

Construct the full ionic equation for the oxidation of propan-2-ol to propanone by acidified dichromate(VI) ions.

Solution

Organic half-equation: $\ce{CH3CH(OH)CH3 -> CH3COCH3 + 2H+ + 2e-}$ . (Check: left $\ce{C3H8O}$ , right $\ce{C3H6O}$ + 2H.)

Dichromate: $\ce{Cr2O7^2- + 14H+ + 6e- -> 2Cr^3+ + 7H2O}$ .

Multiply the organic half by 3 and add, cancelling $\ce{6H+}$ :

\ce{3CH3CH(OH)CH3 + Cr2O7^2- + 8H+ -> 3CH3COCH3 + 2Cr^3+ + 7H2O}

Charge: left $-2 + 8 = +6$ ; right $+6$ . Balanced.

Standard: using the tri-iodomethane test

Three bottles contain butan-1-ol, butan-2-ol and 2-methylpropan-2-ol. Use acidified potassium dichromate(VI) and alkaline aqueous iodine to identify each.

Solution

Acidified dichromate, warm: 2-methylpropan-2-ol (tertiary) stays orange; the other two turn green.

Alkaline aqueous iodine, warm, on the two that turned green: butan-2-ol contains $\ce{CH3CH(OH)-}$ and gives a pale yellow precipitate of $\ce{CHI3}$ ; butan-1-ol gives no precipitate.

Exam-hard: identifying an alcohol from several tests

Alcohol Z, $\ce{C5H12O}$ , turns acidified dichromate(VI) green, and the organic product does not react with Tollens' reagent. Z gives a pale yellow precipitate with alkaline aqueous iodine and has a chiral centre. When Z is dehydrated, three alkenes form, two of which are geometrical isomers. Identify Z, explaining each step.

Solution

Turns dichromate green, product not an aldehyde: Z is a secondary alcohol, oxidised to a ketone.

Positive tri-iodomethane test: Z contains $\ce{CH3CH(OH)-}$ . So Z is $\ce{CH3CH(OH)R}$ with R = $\ce{C3H7}$ : either pentan-2-ol, $\ce{CH3CH(OH)CH2CH2CH3}$ , or 3-methylbutan-2-ol, $\ce{CH3CH(OH)CH(CH3)2}$ . Both have a chiral C2.

Dehydration: pentan-2-ol gives pent-1-ene and pent-2-ene, which exists as cis and trans isomers: three alkenes, two of them geometrical isomers. 3-methylbutan-2-ol gives 3-methylbut-1-ene and 2-methylbut-2-ene; neither shows cis/trans isomerism.

Z is pentan-2-ol.

Exam-hard: yield and amount of oxidant

$6.00\ \text{g}$ of propan-1-ol is heated under reflux with excess acidified potassium dichromate(VI). $5.55\ \text{g}$ of propanoic acid is obtained.

(a) Calculate the percentage yield. (b) Using the equation $\ce{3C3H7OH + 2Cr2O7^2- + 16H+ -> 3C2H5COOH + 4Cr^3+ + 11H2O}$ , calculate the minimum volume of $0.500\ \text{mol dm}^{-3}$ $\ce{K2Cr2O7}$ needed. ( $A_r$ : H 1.0, C 12.0, O 16.0)

Solution

(a) $M_r(\ce{C3H7OH}) = 60.0$ ; $n = 6.00 / 60.0 = 0.100\ \text{mol}$ . Theoretical $n(\text{acid}) = 0.100\ \text{mol}$ ; $M_r(\ce{C2H5COOH}) = 74.0$ ; theoretical mass $= 7.40\ \text{g}$ . Yield $= 5.55 / 7.40 \times 100 = 75.0\%$ .

(b) $n(\ce{Cr2O7^2-}) = 0.100 \times \dfrac{2}{3} = 0.0667\ \text{mol}$ . Volume $= 0.0667 / 0.500 = 0.133\ \text{dm}^3 = 133\ \text{cm}^3$ .

Watch out

Colour change backwards. Dichromate goes orange to green. Manganate(VII) goes purple to colourless. Writing "green to orange" or "turns orange" loses the mark.
Tertiary alcohols "oxidise to ketones". They do not react at all: there is no H on the carbon carrying the OH.
Reflux to make an aldehyde. Reflux gives the carboxylic acid. To make the aldehyde, distil it off immediately, with the alcohol in excess.
Iodoform test on any secondary alcohol. The test needs $\ce{CH3CH(OH)-}$ . Pentan-3-ol is secondary but negative. Ethanol is primary but positive.
[O] equations. Primary to acid needs $2\ce{[O]}$ and gives one $\ce{H2O}$ ; primary to aldehyde needs one $\ce{[O]}$ and gives one $\ce{H2O}$ ; aldehyde to acid needs one $\ce{[O]}$ and gives no water.

Exam tip

"State the reagents and conditions" for a primary alcohol to acid: acidified potassium dichromate(VI) (both words "acidified" and the oxidation state are often needed), heat under reflux. For the aldehyde: distil as it forms.
Observations: "orange to green". For the iodoform test: "pale yellow precipitate".
Classification questions often combine dichromate, Tollens' or Fehling's, and the tri-iodomethane test. Work through them one at a time, writing down what each result tells you.
In Paper 3, a reaction mixture is often "warmed in a hot water bath" rather than refluxed. Record colours precisely, and record "no change" when nothing happens: it is an observation too.

Summary

Acidified $\ce{K2Cr2O7}$ oxidises alcohols; orange $\ce{Cr2O7^2-}$ → green $\ce{Cr^3+}$ .
Primary → aldehyde (distil off as formed) → carboxylic acid (heat under reflux, excess oxidant).
Secondary → ketone. Tertiary → no reaction (no H on the C–OH carbon).
Aldehydes boil far lower than alcohols and acids (no hydrogen bonding), which is why distillation isolates them.
Full equations: combine $\ce{Cr2O7^2- + 14H+ + 6e- -> 2Cr^3+ + 7H2O}$ with the organic half-equation.
Tri-iodomethane test: warm with alkaline aqueous iodine; pale yellow $\ce{CHI3}$ precipitate shows $\ce{CH3CH(OH)-}$ (or $\ce{CH3CO-}$ ).

Practice

Question

State the colour change when a primary alcohol is warmed with acidified potassium dichromate(VI), and give the oxidation states of chromium before and after.
Name the organic product when each is oxidised with acidified dichromate under the conditions given: (a) butan-1-ol, distilled immediately; (b) butan-1-ol, reflux with excess oxidant; (c) pentan-3-ol; (d) 2-methylpropan-2-ol.
Explain why a tertiary alcohol is not oxidised by acidified potassium dichromate(VI).
Write equations using $\ce{[O]}$ for (a) propan-2-ol to propanone, (b) methanol to methanal, (c) ethanol to ethanoic acid.
Explain why the condenser in a reflux apparatus is vertical, and why anti-bumping granules are added.
Which of these alcohols give a positive tri-iodomethane test: methanol, ethanol, propan-1-ol, propan-2-ol, butan-2-ol, 2-methylpropan-2-ol, pentan-3-ol?
Construct the full ionic equation for the oxidation of propan-1-ol to propanal by acidified dichromate(VI) ions.
Describe how you would show that an unknown alcohol is primary rather than secondary, using acidified dichromate(VI) and one further test reagent.
Which isomer of $\ce{C4H10O}$ (an alcohol) gives a positive tri-iodomethane test? Name the organic products of the test.
Alcohol Q, $\ce{C5H12O}$ , is not oxidised by acidified dichromate(VI). (a) Identify Q. (b) Name the alkenes formed when Q is heated with concentrated phosphoric acid, and state whether either shows geometrical isomerism. (c) Explain whether Q gives a positive tri-iodomethane test.

Answers

Orange to green. Chromium changes from $+6$ (in $\ce{Cr2O7^2-}$ ) to $+3$ (in $\ce{Cr^3+}$ ).
(a) Butanal. (b) Butanoic acid. (c) Pentan-3-one. (d) No reaction (tertiary).
Oxidation to a carbonyl compound requires removal of an H atom from the carbon bonded to the OH group, along with the H of the OH. A tertiary alcohol has no H on that carbon (it is bonded to three carbons and the OH), so a C=O cannot form without breaking a C–C bond.
(a) $\ce{CH3CH(OH)CH3 + [O] -> CH3COCH3 + H2O}$ (b) $\ce{CH3OH + [O] -> HCHO + H2O}$ (c) $\ce{CH3CH2OH + 2[O] -> CH3COOH + H2O}$
The vertical condenser condenses vapours (alcohol, aldehyde, solvent) and returns them to the flask, so the mixture can be heated for a long time without losing volatile reactants or intermediates. Anti-bumping granules give small bubbles that make boiling smooth and prevent sudden violent boiling ("bumping").
Ethanol, propan-2-ol and butan-2-ol (each contains $\ce{CH3CH(OH)-}$ ).
$\ce{CH3CH2CH2OH -> CH3CH2CHO + 2H+ + 2e-}$ (×3), plus $\ce{Cr2O7^2- + 14H+ + 6e- -> 2Cr^3+ + 7H2O}$ : $\ce{3CH3CH2CH2OH + Cr2O7^2- + 8H+ -> 3CH3CH2CHO + 2Cr^3+ + 7H2O}$ .
Warm the alcohol with acidified potassium dichromate(VI), distilling the product off as it forms; the solution turns from orange to green (primary or secondary). Warm the distillate with Tollens' reagent: a silver mirror (or with Fehling's solution: a red-brown precipitate) shows an aldehyde, so the alcohol was primary. No change shows a ketone, so the alcohol was secondary.
Butan-2-ol, $\ce{CH3CH(OH)CH2CH3}$ . Products: tri-iodomethane, $\ce{CHI3}$ (pale yellow precipitate), and the propanoate ion, $\ce{CH3CH2COO-}$ (the carboxylate with one carbon fewer).
(a) Not oxidised, so tertiary. The only tertiary alcohol with formula $\ce{C5H12O}$ is 2-methylbutan-2-ol, $\ce{CH3CH2C(OH)(CH3)2}$ . (b) 2-methylbut-1-ene, $\ce{CH2=C(CH3)CH2CH3}$ , and 2-methylbut-2-ene, $\ce{(CH3)2C=CHCH3}$ . Neither shows geometrical isomerism: in each, one carbon of the C=C carries two identical groups (two H, or two $\ce{CH3}$ ). (c) Negative. Although Q has methyl groups on the carbon bearing the OH, that carbon carries no hydrogen, so Q does not contain $\ce{CH3CH(OH)-}$ and cannot be oxidised to a $\ce{CH3CO-}$ compound.

The oxidising agent

What gets oxidised, and why tertiary alcohols do not react

Primary alcohols: aldehyde or carboxylic acid?

Full ionic equations

Distinguishing primary, secondary and tertiary alcohols

The tri-iodomethane (iodoform) test

Worked examples

Practice

Console