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

31 PYQs, one HARD. Of the fifteen set from 2021 to 2025, 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

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

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

Addition then loss of water

\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.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 15th May Shift 1 · Q74Easy

Example 1 · Aldehydes, Ketones and Carboxylic Acids · Nucleophilic Addition and Condensation Reactions

Identify the reagent 'R' used in the following reaction. Ketone

\xrightarrow{R}

Semicarbazone

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

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 $\xrightarrow{\text{dil. NaOH}}$ 3-hydroxybutanal (aldol, A) $\xrightarrow{\Delta,\ -\text{H}_2\text{O}}$ but-2-enal (crotonaldehyde, B).
Propanone $\xrightarrow{\text{Ba(OH)}_2}$ 4-hydroxy-4-methylpentan-2-one (A) $\xrightarrow{\Delta}$ 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\,\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.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 3rd May Shift 1 · Q65Moderate

Example 2 · Aldehydes, Ketones and Carboxylic Acids · Nucleophilic Addition and Condensation Reactions

Identify the product 'B' in the following sequence of reactions. Propanone

\xrightarrow{\text{Ba(OH)}_2} A \xrightarrow{\Delta,-\text{H}_2\text{O}} B

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
$\text{>C=O} + \text{H}_2\text{N-Z} \to \text{>C(OH)-NH-Z} \xrightarrow{-\text{H}_2\text{O}} \text{>C=N-Z}$
Aldol Reaction and the Cannizzaro Reaction
Aldol and Cannizzaro
$2\,\text{RCH}_2\text{CHO} \xrightarrow{\text{OH}^-} \text{RCH}_2\text{CH(OH)CHR-CHO} \xrightarrow{\Delta} \text{RCH}_2\text{CH=CR-CHO};\qquad 2\,\text{ArCHO} \xrightarrow{\text{conc. OH}^-} \text{ArCOO}^- + \text{ArCH}_2\text{OH}$

Watch out for (2)

Confusing hydrazine with semicarbazide
→ Cyanohydrins, Hemiacetals and the Ammonia Derivatives
Reducing the wrong aldehyde in the crossed Cannizzaro
→ Aldol Reaction and the Cannizzaro Reaction

Test yourself on Aldehydes, Ketones and Carboxylic Acids

20 past MHT-CET questions from this chapter, timed at 18 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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