Year 12

AQA A level Chemistry Carbonyls Notes

Carbonyls

— Chemistry · AS Level (7405) · Crash Course —
★ Exam Tip: Carbonyls = aldehydes + ketones. Examiners love test-and-observation tables (Tollens’, Fehling’s, 2,4-DNPH, iodoform, dichromate), the nucleophilic addition mechanism with HCN (racemic mixture!) and “identify the compound” deductions. Four-carbon compounds are examiners’ favourites.

1. What are Carbonyls?

Definition: organic compounds containing a C=O bond.

Type Structure Board examples
Aldehyde R–CHO : C=O carbon is bonded to one H (always H) and R; R can be anything (alkyl, aryl, H). The CHO is always at the end of the chain (C1) — never in the middle. methanal HCHO · ethanal CH₃CHO · propanal CH₃CH₂CHO · butanal · 2-methylpropanal CH₃CH(CH₃)CHO · 2-methylbutanal · 3-methylbutanal
Ketone R–CO–R′ : C=O carbon bonded to two carbon groups (R, R′ cannot be H). Always in the middle of the chain. propanone CH₃COCH₃ · butanone CH₃COCH₂CH₃ · pentan-3-one · pentan-2-one · 3-methylpentan-2-one
OHRaldehyde: C=O at the end
OR′Rketone: C=O in the middle
Naming rules (sir): Naming: aldehyde = -al (C1, so no locant: butanal not butan-1-al). Ketone = -one with position number from the end giving the lowest number (butanone needs no number — it can only be butan-2-one). A group with higher precedence is the suffix; an –OH with a higher-precedence group (e.g. COOH, carbonyl) is written as the prefix hydroxy-. Primary alcohol group = CH₂OH.

2. Physical Properties

  • Low solubility in water (esp. longer chains): they have only temporary and permanent dipole forces — no H on O to donate H-bonds (small ones, e.g. methanal, ethanal, propanone, do dissolve a little by H-bonding to water).
  • Lower boiling points than alcohols of similar Mr: no hydrogen bonding between their own molecules (only permanent + temporary dipoles).

3. Reaction ① — Addition of HCN (Nucleophilic Addition)

Reagent: HCN with NaCN or KCN (CN⁻ catalyses it). Product: a hydroxynitrile (cyanohydrin). Name: the C bearing OH and CN: nitrile gets C1.

Carbonyl Equation Product
ethanal CH₃CHO + HCN → CH₃CH(OH)CN 2-hydroxypropanenitrile
propanone CH₃COCH₃ + HCN → (CH₃)₂C(OH)CN 2-hydroxy-2-methylpropanenitrile
butanone CH₃COC₂H₅ + HCN → CH₃C(OH)(CN)CH₂CH₃ 2-hydroxy-2-methylbutanenitrile

Mechanism: nucleophilic addition

Oδ–Cδ+H₃CH⁻:C≡N:CN⁻ attacks δ+ CO:⁻CH₃CHCNH–C≡Nδ+δ–OHC*H₃CHCN+ :C≡N⁻step 1: CN⁻ adds to C; the C=O π electrons move onto O (O⁻)step 2: O⁻ takes H⁺ from HCN, regenerating CN⁻ (the catalyst)
nucleophilic addition: C=O is polar (Cδ+, Oδ–); CN⁻ attacks Cδ+, the π bond breaks onto O, then O⁻ picks up H⁺
  • Key points to show: δ+ on C, δ– on O; arrow from the lone pair on C of CN⁻ to Cδ⁺; arrow from the C=O π bond to O; O⁻ with lone pair; arrow from O⁻ to Hδ⁺ of HCN; arrow showing H–CN breaking.
  • The C=O carbon is planar ⇒ CN⁻ can attack from above or below with equal probability.
  • Ethanal (and other aldehydes, unsymmetrical ketones) gives a product with a chiral centre* ⇒ a racemic mixture (50 : 50 of 2 optical isomers) is made, which has no optical activity overall. Propanone gives no chiral C.
Why it matters: Importance of this reaction: ① carbon chain lengthens by 1 (CN can be hydrolysed to COOH — dil. H₂SO₄/HCl reflux, as in the halogenoalkane chapter); ② the OH group gives all the reactions of alcohols.
carbonylhydroxynitrilehydroxy-acid

4. Reaction ② — Reduction

  • Aldehydes are reduced to primary alcohols; ketones to secondary alcohols.
  • Reagents: LiAlH₄ in dry ether · NaBH₄ (aq) · H₂ + Ni catalyst + heat (≈200 °C). Show hydride as [H] in equations.
CH₃CHO + 2[H] → CH₃CH₂OHCH₃COCH₃ + 2[H] → CH₃CH(OH)CH₃ (board writes “+ [H]” — balance needs 2[H])
Careful: Both LiAlH₄ and NaBH₄ reduce C=O only — they leave C=C alone (MCQ 26: hept-4-enal + NaBH₄ keeps C=C, gives the alkenol). H₂/Ni reduces both C=C and C=O (CH₃(CH₂)₅CH₂OH ⇒ A). Therefore if you see “either H₂/Ni or NaBH₄” the answer depends on which — read carefully.

5. Reaction ③ — Oxidation (and the Tests)

  • Aldehydes are oxidised to carboxylic acids.
  • Ketones NEVER oxidise (no H on the C=O carbon).
Reagent Aldehyde Ketone Note
Acidified K₂Cr₂O₇ orange → green no change (stays orange) also 1° and 2° alcohols
Tollens’ reagent silver mirror / silver–grey precipitate (Ag⁺ → Ag) no reaction warm; “[Ag(NH₃)₂]⁺ in NaOH”
Fehling’s reagent brick-red precipitate (Cu²⁺ → Cu₂O) no reaction warm; Cu²⁺ blue → red ppt
Sir: You don’t need the chemistry of Tollens’/Fehling’s — just the reagent → observation. Tollens’ distinguishes an aldehyde from a ketone (MCQ 8: CH₃CHO vs CH₃COCH₃ ⇒ A). “Does not give a silver mirror” ⇒ not an aldehyde. Mark schemes accept “silver precipitate”.
1° alcoholaldehydecarboxylic acid[O][O]2° alcohol → ketone and stopsketones: no further oxidation
the oxidation ladder: 1° alcohol → aldehyde → acid; 2° alcohol → ketone and stops

6. Reaction ④ — 2,4-DNPH (Brady’s test)

Test for aldehydes AND ketones (i.e. for the C=O group): both give a yellow/orange precipitate with 2,4-dinitrophenylhydrazine. (Mark schemes accept “orange” or “yellow”.)

ArNH–NH₂+ O=C(R)(R′)ArNH–N=C(R)(R′)+ H₂OAr = 2,4-dinitrophenyl ring (NO₂ on 2 and 4)O of C=O + 2 H from NH₂ leave as H₂O
addition–elimination = a CONDENSATION reaction
Definition: Condensation reaction = two large molecules join, releasing a small molecule such as H₂O. 2,4-DNPH does not distinguish aldehyde from ketone — use Tollens’/Fehling’s for that. Both a C=O in a ketone/aldehyde give a positive; carboxylic acids/esters do not.

7. Reaction ⑤ — Triiodomethane (Iodoform) Test

Reagent: I₂(aq) + NaOH(aq) (alkaline aqueous iodine), warm. A yellow precipitate of CHI₃ (triiodomethane) with:

  • methyl ketones: CH₃–CO–R;
  • secondary methyl alcohols: CH₃–CH(OH)–R (oxidised to the methyl ketone first);
  • also ethanal (CH₃CHO) and ethanol (CH₃CH₂OH) — the only aldehyde and 1° alcohol that do.
CH₃CH₂COCH₃ + 3I₂ + 4NaOH → CHI₃ + CH₃CH₂COONa + 3NaI + 3H₂OCH₃CH(OH)CH₂CH₂CH₃ + 4I₂ + 6NaOH → CHI₃ + CH₃CH₂CH₂COONa + 5NaI + 5H₂Oboard writes the skeleton: ketone + I₂ + NaOH → CHI₃ + RCO₂Na + NaI + H₂O (balance as above)
Look for the “methyl-carbonyl” unit: The CH₃CO– carbon is turned into CHI₃; the rest becomes a carboxylate with one fewer C than the carbonyl. Check the structure has CH₃–C(=O)– or CH₃–CH(OH)–. Butanone gives a positive; pentan-3-one does not (no CH₃CO).

8. Summary — Test Table

Reagent Observation Positive for
Acidified K₂Cr₂O₇ (warm) orange → green 1° alcohol, 2° alcohol, aldehyde
Tollens’ / Fehling’s silver mirror / brick-red ppt aldehydes only
2,4-DNPH orange/yellow ppt aldehydes and ketones (any C=O)
I₂ / NaOH yellow ppt CHI₃ CH₃CO– or CH₃CH(OH)– (methyl ketone, 2° methyl alcohol, ethanal, ethanol)
Br₂(aq) brown → colourless alkenes
Na effervescence (H₂) –OH (alcohol, acid)
PCl₅ steamy HCl fumes –OH
NaHCO₃ effervescence (CO₂) carboxylic acid

9. Past-Paper MCQs (board)

Q Question Working / answer
1 HCN product with a chiral centre CH₃CHO + HCN ⇒ CH₃CH(OH)CN: C bonded to H, OH, CH₃, CN ⇒ chiral ⇒ A ✓. B is pentan-3-one (two identical ethyl groups ⇒ product not chiral); methanal gives CH₂(OH)CN (no chiral C)
4 ethanal → 3-carbon acid, intermediate? HCN adds, then hydrolysis ⇒ intermediate CH₃CH(OH)CN ⇒ D ✓
5 burnt-sugar compound CH₃COCHO tested with 2,4-DNPH and Fehling’s 2,4-DNPH: both C=O (Q ketone + R aldehyde); Fehling’s: aldehyde only (R) ⇒ C ✓
6 propanone + HCN, then acid hydrolysis: final product (CH₃)₂C(OH)CN → (CH₃)₂C(OH)CO₂H ⇒ D ✓
8 pair distinguished by Tollens’ test one aldehyde + one non-aldehyde ⇒ CH₃CHO and CH₃COCH₃ ⇒ A ✓
9 P: cis–trans + red-brown ppt with Fehling’s needs a CHO (Fehling’s) and a C=C with different groups on each carbon ⇒ D ✓ (board circles D); B is a ketone (no Fehling’s)
10 X: orange → green with Cr₂O₇²⁻; 1 mol X reacts with 2 mol HCN 1° alcohol / 2° alcohol / aldehyde (oxidised) and 2 C=O groups to add HCN ⇒ dialdehyde ⇒ D ✓ (OHCCH₂CH₂CHO)
12 glycol HOCH₂CH₂OH — oxidation product that won’t react with Na no OH left ⇒ OHC–CHO = C₂H₂O₂ ⇒ A ✓
Q Question Working / answer
17 reaction forming ibuprofen (aryl–CH(CH₃)CO₂H) oxidise 1° alcohol with warm acidified K₂Cr₂O₇ ⇒ B ✓
19 alcohol → product that reacts with 2,4-DNPH but NOT Fehling’s ketone ⇒ 2° alcohol ⇒ butan-2-ol ⇒ B ✓
20 P (ketoester) + phenylhydrazine ⇒ Q condensation at the ketone C=O: C=N–NH–Ph ⇒ D ✓
22 malic acid in two steps from NCCH₂CH(OH)CN step 1 HCN/NaCN adds (carbonyl → hydroxynitrile), step 2 dil. H₂SO₄ hydrolyses nitriles ⇒ B ✓
26 hept-4-enal + H₂/Ni or NaBH₄ H₂/Ni reduces both C=C and C=O ⇒ A ✓ CH₃(CH₂)₅CH₂OH
30 carvone + hot conc. MnO₄⁻ ⇒ CO₂ + X (9 C); max 2,4-DNPH per X X has 3 C=O (ketones/aldehydes) ⇒ C (3) ✓
33 civetone: no reaction with? ketone + alkene: reacts with 2,4-DNPH ✓, HBr ✓, NaBH₄ ✓; not Fehling’s (not an aldehyde) ⇒ B ✓
36 tartaric acid in 2 steps from OHC–CHO HCN/NaCN then H₂SO₄(aq) ⇒ B ✓
Q Question Working / answer
42 Y + warm dichromate ⇒ product with 2,4-DNPH but no Tollens’ ketone product ⇒ 2° alcohol ⇒ butan-2-ol ⇒ B ✓
47 ethanal → HCN, NaCN → F → hot dil. H₂SO₄ → G → CH₃OH/conc. H₂SO₄ → H CH₃CH(OH)CN → CH₃CH(OH)COOH → methyl ester CH₃CH(OH)COOCH₃ ⇒ B ✓
64 pineapple ester from butanal (NaBH₄ → X; Cr₂O₇²⁻ → Y) X = butan-1-ol, Y = butanoic acid ⇒ butyl butanoate CH₃(CH₂)₂CO₂(CH₂)₃CH₃ ⇒ B ✓
80 citral + excess H₂/Ni → V; hot conc. MnO₄⁻ reflux → W V = saturated alcohol; oxidised to the carboxylic acid ⇒ B ✓ (the CH₂OH → COOH; no C=C left)
92 G: chiral, positive iodoform, no Tollens’ 3-hydroxybutan-2-one: CH₃COCH(OH)CH₃ — chiral C (H, OH, CH₃, COCH₃) and CH₃CO ⇒ D ✓

10. Long CIE-style Questions (board)

CIE Q1 · compounds A–E test tableA CH₃CH₂CH₂OH · B CH₃CH₂CH(OH)CH₃ · C CH₃CHO · D CH₃COCH₃ · E CH₃CH=CHCH₃
Reagent / observation Compounds
acidified K₂Cr₂O₇ boiling — green A, B, C
hot acidified KMnO₄ ⇒ ethanoic acid C, E (E gives 2 CH₃COOH; C oxidised to CH₃COOH)
H₂ / Pt — hydrogen absorbed C, D, E (C=O and C=C)
Fehling’s — brown-red ppt C
2,4-DNPH — orange ppt C, D
Br₂ in inert solvent decolourised E
CIE Q3 · lactic acid from ethanal2 stages

(a)(i) CH₃CHO + HCN → CH₃CH(OH)CN (displayed). (ii) nucleophilic addition. (iii) mechanism: δ+ C, δ– O; arrow from CN⁻ lone pair to Cδ⁺; π bond → O; O⁻ attacks Hδ⁺ of HCN; CN⁻ regenerated. (b)(i) CH₃CH(OH)CN + ½H₂SO₄ + 2H₂O → CH₃CH(OH)CO₂H + ½(NH₄)₂SO₄ · (ii) hydrolysis.

CIE Q4 · compound H (Mr 72)C 66.7 %, H 11.1 %, O 22.2 %; Br₂ decolourised, Na ⇒ H₂, Cr₂O₇²⁻ ⇒ green; 2,4-DNPH on the product: none
C : H : O = 66.7/12 : 11.1/1 : 22.2/16= 5.558 : 11.1 : 1.3875÷ 1.3875 ⇒ 4 : 8 : 1 ⇒ C₄H₈On = 72/72 = 1 ⇒ C₄H₈O

(b)(i) Br₂ test ⇒ alkene; (ii) Na test ⇒ alcohol (or carboxylic acid). (c)(i) a positive result in test 6 would show a carbonyl group; (ii) tests 5 + 6 together: the product of oxidation has no C=O reaction with 2,4-DNPH ⇒ it is a carboxylic acid, so H is a primary alcohol (an aldehyde or ketone product would have given a positive). (d) cis–trans: C=C with different groups on each carbon; the π bond prevents rotation. (e) H = but-2-en-1-ol CH₃CH=CHCH₂OH — draw cis and trans, displayed.

CIE Q5 · compound Z, C₄H₆O₂alkene + alcohol + aldehyde

(a) Br₂(aq) decolourised ⇒ alkene. (b) Na ⇒ H₂ and C₄H₅O₂Na; ethanoic acid + conc. H₂SO₄ ⇒ ester C₆H₈O₃ ⇒ alcohol. (c) 2,4-DNPH + Tollens’ silver mirror ⇒ aldehyde.

(d) Z (no cis–trans): CH₂=C(OH)–CH₂–CHO (as drawn on board). (e) Na ⇒ sodium salt; ethanoic acid ⇒ ester (O–CO–CH₃). (f) Tollens’ ⇒ CH₂=C(OH)CH₂COOH; 2,4-DNPH ⇒ the hydrazone (C=N–NH–Ar). (g) but-2-enoic acid, cis isomer: CH₃ and COOH on the same side of the C=C — draw it displayed.

CIE Q6 · paint remover (CH₂Cl₂ + CH₃OH)observations and isomers of C₂H₄O₂

(a)(i) CH₂Cl₂ + NaOH(aq) then AgNO₃: white precipitate, AgCl. (ii) CH₃OH + PCl₅: steamy fumes of HCl. (iii) CH₃OH + Na: effervescence, H₂.

(b) W: C : H : O = 40.0/12 : 6.7/1 : 53.3/16= 3.333 : 6.7 : 3.331÷ 3.331 ⇒ 1 : 2 : 1 ⇒ CH₂O

(c) further isomers of C₂H₄O₂: X = CH₃COOH, Y = HOCH=CHOH (cis/trans), Z = OHC–CH₂OH. (d) NaHCO₃: X (acid) — effervescence of CO₂; Tollens’: Z — silver mirror. (e) Y shows stereoisomerism: draw cis and trans (HOCH=CHOH).

CIE Q5 (2) · V and W (C₄H₈O)isomers with Na, Br₂, alkaline iodine, cold KMnO₄

(a) sodium ⇒ alcohol; bromine water ⇒ alkene; alkaline aqueous iodine ⇒ methyl ketone or secondary methyl alcohol.

(b) V = CH₃CH(OH)CH=CH₂ (but-3-en-2-ol: positive iodoform); W = CH₃CH=CHCH₂OH (but-2-en-1-ol: primary alcohol, no CH₃CH(OH)). (c) V shows optical isomerism (chiral C bonded to H, OH, CH₃, CH=CH₂) — draw 2 mirror images with wedge/dash. (d) X (cold dilute KMnO₄ on the C=C of both V and W) = CH₃CH(OH)CH(OH)CH₂OH, C₄H₁₀O₃ — the same neutral triol from either compound.

✓ Quick Recap — Carbonyls
  • Aldehyde R–CHO (end of chain, –al); ketone R–CO–R′ (middle, –one). Lower b.p. than alcohols (no H-bonding); little solubility.
  • HCN + NaCN/KCN: nucleophilic addition ⇒ hydroxynitrile (+1 C); CN⁻ attacks Cδ⁺ from above/below ⇒ racemic mixture (if product chiral); nitrile → COOH by acid hydrolysis.
  • Reduction: LiAlH₄/dry ether, NaBH₄, H₂/Ni 200 °C: aldehyde → 1° alcohol, ketone → 2° alcohol (NaBH₄/LiAlH₄ leave C=C).
  • Oxidation: aldehydes → acids; ketones never. K₂Cr₂O₇ orange→green; Tollens’ silver mirror; Fehling’s brick-red (aldehyde only).
  • 2,4-DNPH: orange/yellow ppt with any C=O (condensation, loses H₂O). I₂/NaOH: yellow CHI₃ with CH₃CO– / CH₃CH(OH)–.
  • Deduction questions: combine tests — Br₂ (alkene), Na (OH), Tollens’ (aldehyde), iodoform (methyl ketone), DNPH (C=O), chirality, cis–trans.