Tests for Carbonyl Compounds

AS · 11 min

A colourless liquid in a bottle could be an aldehyde, a ketone, an alcohol or something else entirely. Four simple test-tube reactions settle the question: 2,4-dinitrophenylhydrazine shows that a carbonyl group is present, Tollens' reagent and Fehling's solution show whether it is an aldehyde or a ketone, and alkaline aqueous iodine shows whether it contains a $\ce{CH3CO-}$ group. This note explains each test, why it works, the observations examiners expect and how to combine results to identify an unknown. These tests appear as observation questions in Paper 2 and as practical tests in Paper 3.

Why aldehydes and ketones behave differently

An aldehyde has a hydrogen atom on the carbonyl carbon. That C–H bond can be oxidised to C–OH, turning the aldehyde into a carboxylic acid:

\ce{RCHO + [O] -> RCOOH}

A ketone has two carbon groups on the carbonyl carbon and no hydrogen. It cannot be oxidised without breaking a C–C bond, which mild oxidising agents cannot do.

So aldehydes are reducing agents and ketones are not. Tollens' reagent, Fehling's solution and acidified dichromate are all mild oxidising agents: aldehydes react with them, ketones do not.

Test 1: 2,4-DNPH for any carbonyl group

Key result

2,4-dinitrophenylhydrazine (2,4-DNPH, Brady's reagent)

Add a few drops of the compound to 2,4-DNPH solution.

Aldehydes and ketones: an orange (yellow to orange-red) precipitate forms.

No precipitate: no aldehyde or ketone. (Carboxylic acids and esters also contain C=O but do not give a precipitate.)

The precipitate is a 2,4-dinitrophenylhydrazone, formed by a condensation reaction between the carbonyl compound and 2,4-DNPH. You do not need its structure at AS.

The precipitate has a further use. Each aldehyde or ketone gives a hydrazone with its own sharp melting point. To identify the exact compound:

Method

Identifying a carbonyl compound from its 2,4-DNPH derivative

Filter off the orange precipitate.
Recrystallise it: dissolve it in the minimum volume of hot solvent, cool, filter off the pure crystals and dry them.
Measure the melting point.
Compare it with a data table of melting points of 2,4-dinitrophenylhydrazones. A match identifies the original carbonyl compound.

This is how chemists distinguished isomers such as butanal and 2-methylpropanal before spectroscopy, since both give the same results in every other test.

Test 2: Tollens' reagent (silver mirror)

Tollens' reagent is an alkaline solution of the diamminesilver(I) ion, $\ce{[Ag(NH3)2]+}$ . It must be made fresh:

Method

Preparing and using Tollens' reagent

To aqueous silver nitrate, add a few drops of aqueous sodium hydroxide: a brown precipitate of silver(I) oxide forms.
Add dilute aqueous ammonia, drop by drop, until the precipitate just dissolves. This is Tollens' reagent.
Add a few drops of the compound and warm in a water bath (not a flame: organic compounds are flammable).
Aldehyde: a silver mirror forms on the inside of the tube (or a grey-black precipitate of silver). Ketone: no change (solution stays colourless).

The aldehyde reduces $\ce{Ag+}$ to silver metal and is itself oxidised. Because the solution is alkaline, it ends up as the carboxylate ion:

\ce{Ag+ + e- -> Ag}

\ce{RCHO + 2Ag+ + H2O -> RCOOH + 2Ag + 2H+}

In the alkaline reagent the full equation is:

\ce{RCHO + 2[Ag(NH3)2]+ + 3OH- -> RCOO- + 2Ag + 4NH3 + 2H2O}

Silver goes from $+1$ to 0 (reduced); the aldehyde carbon is oxidised.

Test 3: Fehling's solution

Fehling's solution contains copper(II) ions, kept in solution in alkali by complexing with tartrate ions. It is deep blue.

Key result

Warm the compound with Fehling's solution in a water bath.

Aldehyde: the blue solution gives a red-brown (brick-red) precipitate of copper(I) oxide, $\ce{Cu2O}$ .

Ketone: no change; the solution stays blue.

The aldehyde reduces copper(II) to copper(I):

\ce{RCHO + 2Cu^2+ + 5OH- -> RCOO- + Cu2O + 3H2O}

Copper goes from $+2$ to $+1$ .

Tip

Benedict's solution, used in biology for reducing sugars, works the same way: glucose has an aldehyde group. At AS chemistry, Fehling's solution and Tollens' reagent are the named reagents.

Test 4: acidified potassium dichromate(VI)

Warm the compound with acidified $\ce{K2Cr2O7}$ . Aldehyde: orange to green (aldehyde oxidised to carboxylic acid). Ketone: stays orange. This test also responds to primary and secondary alcohols, so it is only conclusive once you know you have a carbonyl compound (from 2,4-DNPH).

Test 5: alkaline aqueous iodine (tri-iodomethane test)

Key result

Test for the $\ce{CH3CO-}$ group (a methyl ketone, or ethanal), and also for $\ce{CH3CH(OH)-}$ in alcohols.

Warm the compound with alkaline aqueous iodine (iodine solution with aqueous sodium hydroxide).

Positive: a pale yellow precipitate of tri-iodomethane, $\ce{CHI3}$ (antiseptic smell).

The $\ce{CH3}$ of the $\ce{CH3CO-}$ group becomes $\ce{CHI3}$ , and the rest of the molecule becomes a carboxylate ion with one carbon fewer:

\ce{CH3COCH3 + 3I2 + 4OH- -> CHI3 + CH3COO- + 3I- + 3H2O}

For a general methyl ketone: $\ce{CH3COR + 3I2 + 4OH- -> CHI3 + RCOO- + 3I- + 3H2O}$ .

compound	$\ce{CH3CO-}$ ?	result
ethanal, $\ce{CH3CHO}$	yes (the only aldehyde)	positive
propanal, $\ce{CH3CH2CHO}$	no	negative
propanone, $\ce{CH3COCH3}$	yes	positive
butanone, $\ce{CH3COCH2CH3}$	yes	positive
pentan-3-one, $\ce{CH3CH2COCH2CH3}$	no	negative

Putting the tests together

Key result

test	aldehyde	ketone
2,4-DNPH	orange precipitate	orange precipitate
Tollens' reagent, warm	silver mirror	no change
Fehling's solution, warm	blue → red-brown precipitate	stays blue
acidified $\ce{K2Cr2O7}$ , warm	orange → green	stays orange
alkaline $\ce{I2}$ , warm	pale yellow precipitate only for ethanal	pale yellow precipitate if $\ce{CH3CO-}$
$\ce{NaBH4}$	reduced to primary alcohol	reduced to secondary alcohol

Method

Deducing the nature of an unknown carbonyl compound

2,4-DNPH: orange precipitate confirms an aldehyde or ketone.
Tollens' (or Fehling's): positive means aldehyde; negative means ketone.
Alkaline iodine: positive means a $\ce{CH3CO-}$ group (a methyl ketone, or ethanal).
Use the molecular formula to list the possible structures that fit all the results.
If isomers remain, use the melting point of the 2,4-DNPH derivative (or a spectrum) to decide.

Worked examples

Routine: propanal and propanone

Give the observations when propanal and propanone are separately tested with (a) 2,4-DNPH, (b) Tollens' reagent, (c) Fehling's solution, (d) alkaline aqueous iodine.

Solution

test	propanal	propanone
(a) 2,4-DNPH	orange precipitate	orange precipitate
(b) Tollens'	silver mirror	no change
(c) Fehling's	blue to red-brown precipitate	stays blue
(d) alkaline iodine	no precipitate	pale yellow precipitate

Routine: what happens in the silver mirror test

Ethanal is warmed with Tollens' reagent. (a) State what is seen. (b) Write a half-equation for the change in silver and an equation using $\ce{[O]}$ for the organic change. (c) Name the type of reaction the ethanal undergoes.

Solution

(a) A silver mirror forms on the inside of the test tube.

(b) $\ce{Ag+ + e- -> Ag}$ ; $\ce{CH3CHO + [O] -> CH3COOH}$ (in the alkaline reagent it is present as ethanoate ions).

Standard: the isomers of C4H8O

There are three aldehydes and ketones with molecular formula $\ce{C4H8O}$ . Name them and predict the result of each with Tollens' reagent and with alkaline aqueous iodine. Explain how the two that give the same results could be told apart.

Solution

Butanal, $\ce{CH3CH2CH2CHO}$ : Tollens' positive (silver mirror); iodine negative.

2-methylpropanal, $\ce{(CH3)2CHCHO}$ : Tollens' positive; iodine negative.

Butanone, $\ce{CH3COCH2CH3}$ : Tollens' negative; iodine positive (pale yellow precipitate, with propanoate ions).

Butanal and 2-methylpropanal give the same results. React each with 2,4-DNPH, recrystallise the orange precipitate, measure its melting point and compare with data-book values: the two derivatives have different melting points.

Standard: deducing a structure

Compound E, $\ce{C5H10O}$ , has a straight carbon chain. It gives an orange precipitate with 2,4-DNPH, no change with Fehling's solution, and a pale yellow precipitate with alkaline aqueous iodine. Identify E and give the organic products of the iodine reaction.

Solution

Orange precipitate: carbonyl compound. Fehling's negative: ketone. Iodine positive: $\ce{CH3CO-}$ group, so a methyl ketone. Straight chain, five carbons: $\ce{CH3COCH2CH2CH3}$ , pentan-2-one. (The other methyl ketone, 3-methylbutan-2-one, is branched; pentan-3-one has no $\ce{CH3CO-}$ .)

Products: tri-iodomethane, $\ce{CHI3}$ , and butanoate ions, $\ce{CH3CH2CH2COO-}$ :

\ce{CH3COCH2CH2CH3 + 3I2 + 4OH- -> CHI3 + CH3CH2CH2COO- + 3I- + 3H2O}

Exam-hard: identifying four colourless liquids

Four unlabelled bottles contain propan-1-ol, propan-2-ol, propanal and propanone. Devise a sequence of tests, using as few reagents as possible, to identify each. Give the observation in each case.

Solution

Test each with 2,4-DNPH: propanal and propanone give orange precipitates; the two alcohols do not.

Test the two carbonyl compounds with Tollens' reagent, warm: propanal gives a silver mirror; propanone does not.

Test the two alcohols with alkaline aqueous iodine, warm: propan-2-ol (contains $\ce{CH3CH(OH)-}$ ) gives a pale yellow precipitate; propan-1-ol does not.

Three reagents identify all four. (Alkaline iodine alone would pick out both propanone and propan-2-ol; that is why the 2,4-DNPH step comes first.)

Exam-hard: a quantitative silver mirror

$0.440\ \text{g}$ of an aldehyde J is warmed with excess Tollens' reagent. The silver formed is collected, washed and dried: its mass is $2.16\ \text{g}$ . J gives a pale yellow precipitate with alkaline aqueous iodine. Identify J. ( $A_r$ : Ag 107.9)

Solution

$n(\ce{Ag}) = 2.16 / 107.9 = 0.0200\ \text{mol}$ .

Each aldehyde molecule reduces two $\ce{Ag+}$ ions ( $\ce{RCHO + 2Ag+ + H2O -> RCOOH + 2Ag + 2H+}$ ), so $n(\textbf{J}) = 0.0100\ \text{mol}$ .

$M_r(\textbf{J}) = 0.440 / 0.0100 = 44.0$ . Aldehydes $\ce{C_{n}H_{2n}O}$ : $14n + 16 = 44$ , $n = 2$ : $\ce{C2H4O}$ , ethanal, $\ce{CH3CHO}$ .

Confirmation: ethanal is the only aldehyde with a $\ce{CH3CO-}$ group, so it is the only aldehyde that gives a positive tri-iodomethane test.

Watch out

2,4-DNPH colour. The precipitate is orange (accept yellow/orange or orange-red). The reagent tests for aldehydes and ketones only; carboxylic acids and esters do not react.
Fehling's result. "Red-brown precipitate" or "brick-red precipitate", not "turns red" or "goes orange". The starting colour is blue.
Silver mirror with ketones. Ketones give no change with Tollens' and Fehling's. Only aldehydes reduce them.
Heating with a flame. Warm Tollens' and Fehling's tests in a hot water bath.
Iodoform test as an aldehyde test. It tests for $\ce{CH3CO-}$ . Most aldehydes are negative; ethanal is the exception.
The leftover product of the iodoform test. The carboxylate ion has one carbon fewer than the original compound.

Exam tip

Observation questions need the reagent, the conditions (warm) and the observation for both a positive and a negative result where asked: "Tollens': aldehyde gives silver mirror; ketone, no change".
Questions often give a molecular formula and test results and ask for a structure. Write down each deduction separately; marks are often awarded for the reasoning ("negative with Fehling's, so a ketone").
When asked for "a test to distinguish X from Y", choose a test where one gives a positive result and the other does not, and state both observations.
In Paper 3 you may be asked to prepare Tollens' reagent from silver nitrate, sodium hydroxide and ammonia. Add ammonia until the precipitate just dissolves.

Summary

2,4-DNPH: orange precipitate with aldehydes and ketones; melting point of the recrystallised derivative identifies the compound.
Aldehydes have H on the carbonyl carbon, so they are oxidised (to carboxylic acids); ketones are not.
Tollens' ( $\ce{[Ag(NH3)2]+}$ , warm): aldehyde gives silver mirror ( $\ce{Ag+}$ reduced to Ag); ketone no change.
Fehling's (blue $\ce{Cu^2+}$ , warm): aldehyde gives red-brown $\ce{Cu2O}$ precipitate; ketone stays blue.
Acidified dichromate: aldehyde orange → green; ketone stays orange.
Alkaline aqueous iodine: pale yellow $\ce{CHI3}$ with $\ce{CH3CO-}$ compounds (methyl ketones and ethanal) and $\ce{CH3CH(OH)-}$ alcohols.

Practice

Question

Name the reagent used to show that a compound contains a carbonyl group, and give the positive observation.
Give two reagents that distinguish an aldehyde from a ketone, with the observation for each in both cases.
Explain why aldehydes are oxidised by Fehling's solution but ketones are not.
Which of these give a pale yellow precipitate with alkaline aqueous iodine: ethanal, propanal, propanone, butanone, pentan-3-one, ethanol?
Write a half-equation for the reduction of silver ions in Tollens' reagent, and state the change in oxidation number of copper in the Fehling's test.
Explain why Tollens' and Fehling's tests are carried out in a water bath rather than over a Bunsen flame.
Describe how the 2,4-DNPH derivative is used to identify a particular ketone.
Compound F, $\ce{C4H8O}$ , gives an orange precipitate with 2,4-DNPH but does not react with Tollens' reagent. Identify F and predict its result with alkaline aqueous iodine.
An aldehyde M, $\ce{C5H10O}$ , is reduced by $\ce{NaBH4}$ to an alcohol that contains a chiral centre. Identify M, showing why the other aldehydes of formula $\ce{C5H10O}$ do not fit.
$1.29\ \text{g}$ of a straight-chain aldehyde is warmed with excess Tollens' reagent, giving $3.24\ \text{g}$ of silver. (a) Calculate the $M_r$ of the aldehyde and identify it. (b) Predict its result with alkaline aqueous iodine and with 2,4-DNPH. ( $A_r$ : H 1.0, C 12.0, O 16.0, Ag 107.9)

Answers

2,4-dinitrophenylhydrazine (2,4-DNPH, Brady's reagent): an orange precipitate.
Tollens' reagent, warm: aldehyde gives a silver mirror; ketone, no change. Fehling's solution, warm: aldehyde gives a red-brown precipitate (blue solution changes); ketone, stays blue. (Acidified dichromate is also acceptable: aldehyde orange to green; ketone stays orange.)
An aldehyde has a hydrogen atom on the carbonyl carbon, which can be oxidised (the aldehyde becomes a carboxylic acid/carboxylate); it acts as a reducing agent and reduces $\ce{Cu^2+}$ to $\ce{Cu+}$ ( $\ce{Cu2O}$ ). A ketone has no H on the carbonyl carbon and cannot be oxidised without breaking a C–C bond, so it does not reduce Fehling's.
Ethanal, propanone, butanone and ethanol.
$\ce{Ag+ + e- -> Ag}$ . Copper: $+2$ to $+1$ .
Organic compounds are flammable, so naked flames must be avoided; a water bath also gives gentle, even heating.
React the ketone with 2,4-DNPH, filter off the orange precipitate, recrystallise it (dissolve in the minimum of hot solvent, cool, filter, dry), measure its melting point, and compare with a table of melting points of 2,4-DNPH derivatives of known ketones.
Carbonyl compound but not an aldehyde, so a ketone with four carbons: butanone, $\ce{CH3COCH2CH3}$ . It contains $\ce{CH3CO-}$ , so it gives a pale yellow precipitate (of $\ce{CHI3}$ ) with alkaline aqueous iodine.
Aldehydes $\ce{C5H10O}$ : pentanal (reduces to pentan-1-ol, not chiral); 3-methylbutanal (reduces to 3-methylbutan-1-ol; C3 carries two $\ce{CH3}$ , not chiral); 2,2-dimethylpropanal (reduces to 2,2-dimethylpropan-1-ol, not chiral); 2-methylbutanal, $\ce{CH3CH2CH(CH3)CHO}$ , reduces to 2-methylbutan-1-ol, $\ce{CH3CH2CH(CH3)CH2OH}$ , in which C2 carries H, $\ce{CH3}$ , $\ce{C2H5}$ and $\ce{CH2OH}$ : chiral. M is 2-methylbutanal.
(a) $n(\ce{Ag}) = 3.24 / 107.9 = 0.0300\ \text{mol}$ ; $n(\text{aldehyde}) = 0.0150\ \text{mol}$ ; $M_r = 1.29 / 0.0150 = 86.0$ . $14n + 16 = 86$ , so $n = 5$ : $\ce{C5H10O}$ ; straight chain: pentanal, $\ce{CH3CH2CH2CH2CHO}$ . (b) Alkaline iodine: no precipitate (no $\ce{CH3CO-}$ group). 2,4-DNPH: orange precipitate.

Why aldehydes and ketones behave differently

Test 1: 2,4-DNPH for any carbonyl group

Test 2: Tollens' reagent (silver mirror)

Test 3: Fehling's solution

Test 4: acidified potassium dichromate(VI)

Test 5: alkaline aqueous iodine (tri-iodomethane test)

Putting the tests together

Worked examples

Practice

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