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MHT-CET Chemistry · Aldehydes, Ketones and Carboxylic Acids

Nucleophilic Addition and Condensation Reactions

The carbonyl carbon is electrophilic, so nucleophiles add to it — HCN gives a cyanohydrin, an alcohol a hemiacetal, ammonia derivatives the oximes, hydrazones and semicarbazones — and two carbonyls with α-hydrogens combine in the aldol reaction, while those without α-hydrogen disproportionate in the Cannizzaro reaction.

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Why this matters

15 PYQs, none HARD. Three ask the reagent for a semicarbazone (semicarbazide, three times), one for a cyanohydrin (HCN), one which drawing is a hemiacetal; nine are aldol and Cannizzaro — propanone to mesityl oxide, ethanal to but-2-enal, methanal plus benzaldehyde giving methanoic acid and benzyl alcohol, and what type of reaction each is. Two cards.

Concept 1 of 2

Cyanohydrins, Hemiacetals and the Ammonia Derivatives

Intuition

The nucleophile's atom ends up on the carbonyl carbon. HCN puts CN there next to a new OH (cyanohydrin). One alcohol molecule gives a carbon carrying both OH and OR (hemiacetal); a second alcohol makes the acetal. Ammonia derivatives NH₂–Z add then lose water to give C=N–Z: Z = OH → oxime, Z = NH₂ → hydrazone, Z = NHC₆H₅ → phenylhydrazone, Z = NHCONH₂ → semicarbazone.

Definition

  • CH3CHO+HCN→CH3CH(OH)CN\text{CH}_3\text{CHO} + \text{HCN} \to \text{CH}_3\text{CH(OH)CN}, acetaldehyde cyanohydrin. Reagent: HCN (not NaHSO₃, which gives the bisulphite adduct).
  • Hemiacetal: OH and OR on the SAME carbon, RCH(OH)OR’\text{RCH(OH)OR'}; acetal: two OR groups.
  • NH2OH\text{NH}_2\text{OH} → oxime; NH2NH2\text{NH}_2\text{NH}_2 → hydrazone; NH2NHC6H5\text{NH}_2\text{NHC}_6\text{H}_5 → phenylhydrazone; NH2NHCONH2\text{NH}_2\text{NHCONH}_2 (semicarbazide) → semicarbazone; 2,4-DNP → 2,4-dinitrophenylhydrazone (orange precipitate).
  • A hydroxy-ketone such as CH3CH(OH)COCH2CH3\text{CH}_3\text{CH(OH)COCH}_2\text{CH}_3 is named as the ketone: 2-hydroxypentan-3-one.

Addition then loss of water

>C=O+H2N-Z→>C(OH)-NH-Z→−H2O>C=N-Z\text{>C=O} + \text{H}_2\text{N-Z} \to \text{>C(OH)-NH-Z} \xrightarrow{-\text{H}_2\text{O}} \text{>C=N-Z}

Worked example

Name the product of propanone with (i) hydroxylamine, (ii) phenylhydrazine, (iii) HCN.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Aldehydes, Ketones and Carboxylic AcidsEASY
Identify the reagent 'R' used in the following reaction. Ketone →R\xrightarrow{R} Semicarbazone

[Q72 · 15th May Shift 1 · 2023]

Confusing hydrazine with semicarbazide

NH₂NH₂ gives a hydrazone; the semicarbazone needs the urea-like NH₂NHCONH₂. All four ammonia derivatives are offered together; match the Z group to the product name.
Drill 5 more on cyanohydrins, hemiacetals and the ammonia derivatives

Concept 2 of 2

Aldol Reaction and the Cannizzaro Reaction

Intuition

With an α-hydrogen, dilute base makes an enolate that ADDS to a second carbonyl — the aldol, a β-hydroxy carbonyl; heating then eliminates water to the α,β-unsaturated compound (aldol condensation = addition + elimination). Without an α-hydrogen (methanal, benzaldehyde), concentrated base makes two molecules DISPROPORTIONATE: one is oxidised to the acid, the other reduced to the alcohol (Cannizzaro).

Definition

  • Ethanal →dil. NaOH\xrightarrow{\text{dil. NaOH}} 3-hydroxybutanal (aldol, A) →Δ, −H2O\xrightarrow{\Delta,\ -\text{H}_2\text{O}} but-2-enal (crotonaldehyde, B).
  • Propanone →Ba(OH)2\xrightarrow{\text{Ba(OH)}_2} 4-hydroxy-4-methylpentan-2-one (A) →Δ\xrightarrow{\Delta} 4-methylpent-3-en-2-one (mesityl oxide, B).
  • Aldol formation is a nucleophilic ADDITION; the condensation is addition–elimination (nucleophilic addition-elimination).
  • Cannizzaro (no α-H, conc. NaOH): 2 HCHO → HCOOH + CH₃OH; 2 C₆H₅CHO → C₆H₅COOH + C₆H₅CH₂OH. Crossed: HCHO + C₆H₅CHO → methanoic acid + phenylmethanol (methanal is oxidised, benzaldehyde reduced). Type: disproportionation.

Aldol and Cannizzaro

2 RCH2CHO→OH−RCH2CH(OH)CHR-CHO→ΔRCH2CH=CR-CHO;2 ArCHO→conc. OH−ArCOO−+ArCH2OH2\,\text{RCH}_2\text{CHO} \xrightarrow{\text{OH}^-} \text{RCH}_2\text{CH(OH)CHR-CHO} \xrightarrow{\Delta} \text{RCH}_2\text{CH=CR-CHO};\qquad 2\,\text{ArCHO} \xrightarrow{\text{conc. OH}^-} \text{ArCOO}^- + \text{ArCH}_2\text{OH}

Worked example

Give the aldol and the condensation product of propanal, and the products of the Cannizzaro reaction of 2,2-dimethylpropanal.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Aldehydes, Ketones and Carboxylic AcidsMODERATE
Identify the product 'B' in the following sequence of reactions. Propanone →Ba(OH)2A→Δ,−H2OB\xrightarrow{\text{Ba(OH)}_2} A \xrightarrow{\Delta,-\text{H}_2\text{O}} B

[Q65 · 3rd May 2nd Shift · 2023]

Reducing the wrong aldehyde in the crossed Cannizzaro

Methanal is the stronger hydride donor, so IT is oxidised (to methanoic acid) and benzaldehyde is reduced (to benzyl alcohol). 'Methanol and benzoic acid' is the reversed, planted option.
Drill 8 more on aldol reaction and the cannizzaro reaction

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

  • Cyanohydrins, Hemiacetals and the Ammonia Derivatives

    Addition then loss of water

    >C=O+H2N-Z→>C(OH)-NH-Z→−H2O>C=N-Z\text{>C=O} + \text{H}_2\text{N-Z} \to \text{>C(OH)-NH-Z} \xrightarrow{-\text{H}_2\text{O}} \text{>C=N-Z}
  • Aldol Reaction and the Cannizzaro Reaction

    Aldol and Cannizzaro

    2 RCH2CHO→OH−RCH2CH(OH)CHR-CHO→ΔRCH2CH=CR-CHO;2 ArCHO→conc. OH−ArCOO−+ArCH2OH2\,\text{RCH}_2\text{CHO} \xrightarrow{\text{OH}^-} \text{RCH}_2\text{CH(OH)CHR-CHO} \xrightarrow{\Delta} \text{RCH}_2\text{CH=CR-CHO};\qquad 2\,\text{ArCHO} \xrightarrow{\text{conc. OH}^-} \text{ArCOO}^- + \text{ArCH}_2\text{OH}

Watch out for (2)

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