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AS 17

Carbonyl compounds

AS Level · Organic chemistry · Papers 1, 2 and 3

🎯What you need to be able to do

  • Recall how aldehydes and ketones are made by oxidising primary and secondary alcohols.
  • Describe their reduction to alcohols with NaBH4 or LiAlH4.
  • Describe the reaction with HCN (KCN catalyst, heat) to form hydroxynitriles, using ethanal and propanone, and its nucleophilic addition mechanism.
  • Use 2,4-DNPH to detect a carbonyl group.
  • Tell aldehydes from ketones with Fehling’s and Tollens’ reagents and by ease of oxidation.
  • Use the alkaline iodine test to detect a CH3CO– group.

📚The chemistry

Aldehydes and ketones both contain the carbonyl group, C=O. In an aldehyde it is at the end of the chain, with at least one H attached: RCHO. In a ketone it is inside the chain, with two alkyl groups attached: RCOR′. The carbonyl carbon is sp2 and trigonal planar, and the C=O bond is strongly polar: Cδ+=Oδ−. That δ+ carbon is the key to their chemistry — it attracts nucleophiles.

Making aldehydes and ketones

Oxidise an alcohol with acidified K2Cr2O7 or KMnO4, and distil the product off as it forms (topic 16):

\[ \mathrm{CH_3CH_2OH + [O] \rightarrow CH_3CHO + H_2O} \quad \text{primary alcohol} \rightarrow \text{aldehyde} \] \[ \mathrm{CH_3CH(OH)CH_3 + [O] \rightarrow CH_3COCH_3 + H_2O} \quad \text{secondary alcohol} \rightarrow \text{ketone} \]

Reduction

NaBH4 (sodium tetrahydridoborate, in water or ethanol) or LiAlH4 (lithium tetrahydridoaluminate, in dry ether) reduce carbonyl compounds back to alcohols. The equations use [H]:

\[ \mathrm{CH_3CHO + 2[H] \rightarrow CH_3CH_2OH} \quad \text{aldehyde} \rightarrow \text{primary alcohol} \] \[ \mathrm{CH_3COCH_3 + 2[H] \rightarrow CH_3CH(OH)CH_3} \quad \text{ketone} \rightarrow \text{secondary alcohol} \]

NaBH4 is milder and safer; LiAlH4 reacts violently with water and is used only in dry solvents, but it is strong enough to reduce carboxylic acids too.

Nucleophilic addition of HCN

Aldehydes and ketones react with hydrogen cyanide, with KCN as a catalyst and heat, to form hydroxynitriles. HCN adds across the C=O: the CN joins the carbon and the H joins the oxygen.

\[ \mathrm{CH_3CHO + HCN \rightarrow CH_3CH(OH)CN} \quad \text{2-hydroxypropanenitrile} \] \[ \mathrm{CH_3COCH_3 + HCN \rightarrow (CH_3)_2C(OH)CN} \quad \text{2-hydroxy-2-methylpropanenitrile} \]

Like the halogenoalkane reaction with KCN, this adds a carbon atom to the chain, so it is useful in synthesis. HCN is extremely toxic; KCN supplies the CN− ion that actually attacks, which is why it is the catalyst.

Mechanism (propanone)

  1. The cyanide ion is the nucleophile. Curly arrow from the lone pair on the carbon of CN− to the δ+ carbonyl carbon.
  2. At the same time, curly arrow from the C=O π bond to the oxygen: the π electrons move onto O.
  3. This gives an intermediate with a negatively charged oxygen, (CH3)2C(O−)CN.
  4. Curly arrow from a lone pair on O− to the H of an HCN molecule, and from the H–C bond of HCN to its carbon. The O takes the proton, forming the OH, and CN− is regenerated — which is why it is a catalyst.

The carbonyl carbon is planar, so CN− can attack from either side with equal probability. For an aldehyde such as ethanal, the product CH3CH(OH)CN has a chiral centre, and the two sides of attack give the two enantiomers in equal amounts.

Tests

2,4-DNPH: is there a carbonyl group?

Add 2,4-dinitrophenylhydrazine (2,4-DNPH, Brady’s reagent). Both aldehydes and ketones give an orange (yellow-orange) precipitate. Other compounds with C=O as part of another group — carboxylic acids and esters — do not. The precipitate forms by a condensation reaction; you do not need its structure.

Aldehyde or ketone?

Aldehydes are easily oxidised (to carboxylic acids), because the carbonyl carbon carries an H atom. Ketones are not oxidised by mild oxidising agents. Mild oxidising agents therefore distinguish them:

  • Tollens’ reagent (ammoniacal silver nitrate), warm: an aldehyde gives a silver mirror on the inside of the tube, as Ag+ is reduced to Ag. A ketone: no change.
  • Fehling’s solution (a deep blue copper(II) complex), warm: an aldehyde gives a brick-red precipitate of copper(I) oxide, Cu2O, as Cu2+ is reduced to Cu+. A ketone: stays blue.
  • Acidified K2Cr2O7, warm: an aldehyde turns it from orange to green; a ketone leaves it orange.

In every case the aldehyde is oxidised: RCHO + [O] → RCOOH.

The tri-iodomethane test: is there a CH3CO– group?

Warm with alkaline aqueous iodine. A compound containing CH3CO– (a methyl ketone, or ethanal) gives a pale yellow precipitate of CHI3:

\[ \mathrm{CH_3COR + 3I_2 + 4OH^{-} \rightarrow CHI_3 + RCO_2^{-} + 3I^{-} + 3H_2O} \]

Propanone and butanone give it; ethanal is the only aldehyde that does; propanal and pentan-3-one do not. The same test also detects CH3CH(OH)– in alcohols, which the iodine first oxidises to CH3CO–.

✏️Worked example

Compounds S and T are isomers with molecular formula C3H6O. Both give an orange precipitate with 2,4-DNPH. S gives a silver mirror with Tollens’ reagent; T gives a yellow precipitate with alkaline aqueous iodine. (a) Identify S and T. (b) Write equations for the reduction of each by NaBH4, and name the products. (c) T reacts with HCN in the presence of KCN. Name the product and describe the mechanism.

(a) Both react with 2,4-DNPH, so both are carbonyl compounds. C3H6O allows only two: propanal and propanone. S reduces Tollens’ reagent, so it is the aldehyde, S = propanal, CH3CH2CHO. T gives tri-iodomethane, so it has a CH3CO– group: T = propanone, CH3COCH3.

(b)

\[ \mathrm{CH_3CH_2CHO + 2[H] \rightarrow CH_3CH_2CH_2OH} \quad \text{propan-1-ol} \] \[ \mathrm{CH_3COCH_3 + 2[H] \rightarrow CH_3CH(OH)CH_3} \quad \text{propan-2-ol} \]

(c) The product is 2-hydroxy-2-methylpropanenitrile, (CH3)2C(OH)CN, by nucleophilic addition. A curly arrow from the lone pair on the carbon of CN− to the δ+ carbonyl carbon, and one from the C=O π bond to the oxygen, give the anion (CH3)2C(O−)CN. A lone pair on that O− then takes H+ from HCN (arrow from the lone pair to the H, and from the H–CN bond to C), forming the OH group and regenerating CN−.

Check it. Test each identification against the other results too: propanal has CH3CH2CO–, not CH3CO–, so it gives no iodoform; propanone has no H on its carbonyl carbon, so no silver mirror. The product in (c) has no chiral centre (two CH3 groups on the same carbon), whereas propanal would have given a chiral hydroxynitrile.
Starting the first curly arrow at the minus sign or at the nitrogen. In CN− the lone pair that attacks is on the carbon, because the new bond is C–C. Draw the lone pair on C and start the arrow there. And the second step needs a proton source: HCN (or H+) must appear, not just “H+ from nowhere”.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. Name the alcohol that would be oxidised to (a) butanal; (b) butanone, and state the conditions for (a).
(a) Butan-1-ol, a primary alcohol. Warm with acidified potassium dichromate(VI) and distil the butanal off as it forms, so it is not oxidised further to butanoic acid. (b) Butan-2-ol, a secondary alcohol, which gives butanone with acidified dichromate(VI) by distillation or reflux (ketones are not oxidised further).
2. Explain why the carbonyl carbon is attacked by nucleophiles, whereas the carbons of C=C in an alkene are attacked by electrophiles.
Oxygen is much more electronegative than carbon (3.5 against 2.5), so the C=O bond is polar: the π electrons are pulled towards O, leaving the carbon δ+. A δ+ carbon attracts electron-pair donors, i.e. nucleophiles. In C=C both atoms are carbon, so the bond is non-polar; the π bond is simply a region of high electron density, which attracts electron-pair acceptors, i.e. electrophiles.
3. Describe the tests, with observations, that would distinguish ethanal from propanone.
Both give an orange precipitate with 2,4-DNPH and both give tri-iodomethane with alkaline iodine (ethanal has CH3CO–), so those two tests do not distinguish them. Use a mild oxidising agent: Tollens’ reagent, warm — ethanal gives a silver mirror, propanone no change. Or Fehling’s solution, warm — ethanal gives a brick-red precipitate, propanone stays blue. Or acidified dichromate(VI) — ethanal turns it orange to green.
4. Write the equation for the reaction of ethanal with HCN, and explain why the product is formed as a mixture of two optical isomers.
CH3CHO + HCN → CH3CH(OH)CN, 2-hydroxypropanenitrile. The product carbon carries H, OH, CH3 and CN — four different groups, so it is a chiral centre. The carbonyl group of ethanal is planar, so the CN− nucleophile can attack from above or below the plane with equal probability. Attack from one side gives one enantiomer, from the other side the other, so they form in equal amounts.
5. A compound C5H10O gives an orange precipitate with 2,4-DNPH, no reaction with Fehling’s solution, and no precipitate with alkaline iodine. Identify it.
Orange precipitate: a carbonyl compound. No reaction with Fehling’s: it is a ketone, not an aldehyde. No tri-iodomethane: it has no CH3CO– group. The five-carbon ketones are pentan-2-one (CH3CO–, ruled out), 3-methylbutan-2-one (CH3CO–, ruled out) and pentan-3-one, CH3CH2COCH2CH3 — the answer.
6. Write the equation for the reaction of butanone with alkaline aqueous iodine, and name the organic ion formed.
Butanone is CH3COCH2CH3, so R = C2H5. CH3COC2H5 + 3I2 + 4OH− → CHI3 + C2H5CO2− + 3I− + 3H2O. The organic ion is propanoate, C2H5COO− — one carbon fewer than butanone, because the CH3 carbon has left as CHI3.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • Chemguide (Jim Clark) — the aldehydes and ketones section, including the nucleophilic addition of HCN drawn out
  • Royal Society of Chemistry — practical guides for the Tollens’ and Fehling’s tests and for 2,4-DNPH