ChemistryCBSE Class 12 · Aldehydes, Ketones and Carboxylic Acids

Nucleophilic Addition Reactions of Aldehydes and Ketones

Updated 15 October 2026 · Fact-checked

In nucleophilic addition, a nucleophile attacks the electron-poor carbon of the C=O group, forming a tetrahedral alkoxide intermediate that then gains a proton. Aldehydes react faster than ketones for steric and electronic reasons. Key reagents are HCN (cyanohydrins), $NaHSO_3$ (bisulphite adducts), Grignard reagents (alcohols), alcohols (acetals) and ammonia derivatives (oximes, hydrazones, semicarbazones).

Understand Nucleophilic Addition Reactions of Aldehydes and Ketones

Why the carbonyl group adds nucleophiles. In C=O, oxygen is more electronegative than carbon, so the carbon carries a partial positive charge and the oxygen a partial negative charge. The carbon is $s p^{2}$ and planar, open to attack from above or below the plane.

Mechanism. (i) A nucleophile approaches the carbonyl carbon roughly perpendicular to the plane of the $s p^{2}$ orbitals. As the new bond forms, the π electrons move onto oxygen, and the carbon becomes $s p^{3}$ : a tetrahedral alkoxide intermediate. (ii) This intermediate captures a proton from the medium to give the neutral addition product. Because the product carbon is tetrahedral, crowding around it matters.

Reactivity: aldehydes > ketones. (1) *Steric:* a ketone carries two alkyl groups that hinder the nucleophile's approach and crowd the tetrahedral intermediate; an aldehyde has one group and a small H. (2) *Electronic:* alkyl groups release electrons and reduce the positive charge on the carbonyl carbon; a ketone has two. So methanal > ethanal > propanone. Aromatic aldehydes such as benzaldehyde are less reactive than aliphatic aldehydes like propanal, because resonance with the ring reduces the carbon's positive charge. Electron-withdrawing groups on the ring ( $- N O_{2}$ ) increase reactivity; electron-releasing groups ( $- O C H_{3}$ , $- C H_{3}$ ) decrease it.

Important additions.

*Hydrogen cyanide.* Aldehydes and ketones add HCN to give cyanohydrins. The reaction is slow with pure HCN and is catalysed by a base, which generates the stronger nucleophile $C N^{-}$ : $C H_{3} C H O + H C N \to C H_{3} C H (O H) C N$ . Cyanohydrins are useful synthetic intermediates.

*Sodium hydrogensulphite.* $N a H S O_{3}$ adds to give a crystalline bisulphite addition compound, water soluble. The equilibrium lies well to the right for most aldehydes and less so for ketones (steric). Because dilute mineral acid or alkali regenerates the carbonyl compound, the reaction is used to separate and purify aldehydes.

*Grignard reagents.* $R M g X$ adds to the carbonyl group; hydrolysis of the adduct gives alcohols: methanal → primary alcohol, other aldehydes → secondary, ketones → tertiary.

*Alcohols.* With one equivalent of a monohydric alcohol and dry HCl, an aldehyde gives a hemiacetal, and with a second equivalent an acetal (a gem-dialkoxy compound). Ketones react with ethane-1,2-diol under similar conditions to give cyclic ethylene glycol ketals. Dry HCl protonates the carbonyl oxygen, making the carbon more electrophilic. Acetals and ketals are hydrolysed back by aqueous mineral acid, so they are used to protect a carbonyl group during other reactions.

*Ammonia and its derivatives ( $H_{2} N - Z$ ).* These add to the carbonyl group and then lose water (addition–elimination), forming $C = N - Z$ . The reaction is reversible and catalysed by acid; it works best in a weakly acidic medium. Products: hydroxylamine → oxime; hydrazine → hydrazone; phenylhydrazine → phenylhydrazone; 2,4-dinitrophenylhydrazine (2,4-DNP) → 2,4-dinitrophenylhydrazone (yellow, orange or red precipitate, used to identify aldehydes and ketones); semicarbazide → semicarbazone. In semicarbazide, $H_{2} N C O N H N H_{2}$ , only the terminal $- N H_{2}$ attached to NH reacts; the other two N atoms have lone pairs delocalised into the C=O of the amide part, so they are not nucleophilic.

Nucleophilic Addition Reactions of Aldehydes and Ketones: formulas and key facts

General mechanism: $> C = O + N u^{-} \to> C (O^{-}) N u H^{+} > C (O H) N u$; Planar $s p^{2}$ carbon becomes tetrahedral $s p^{3}$ .
Reactivity order: $H C H O > C H_{3} C H O > C H_{3} C O C H_{3}; aliphatic aldehyde > C_{6} H_{5} C H O$; Steric hindrance and +I effect of alkyl groups; resonance in aromatic aldehydes.
Cyanohydrin formation: $C H_{3} C H O + H C N O H^{-} C H_{3} C H (O H) C N$; Base generates $C N^{-}$ .
Acetals and ketals: $R C H O R^{'} O H, dry HCl R C H (O H) O R^{'} R^{'} O H R C H (O R^{'})_{2}; R_{2} C O + H O C H_{2} C H_{2} O H \to cyclic ketal$; Hydrolysed back by aqueous acid.
Addition–elimination with H₂N–Z: $> C = O + H_{2} N - Z ⇌> C = N - Z + H_{2} O$; Z = OH (oxime), $N H_{2}$ (hydrazone), $N H C_{6} H_{5}$ (phenylhydrazone), $N H C O N H_{2}$ (semicarbazone), 2,4-DNP.
Grignard addition: $> C = O + R M g X \to> C (R) O M g X H_{2} O > C (R) O H$; HCHO → 1°, RCHO → 2°, $R_{2} C O$ → 3° alcohol.

How to solve Nucleophilic Addition Reactions of Aldehydes and Ketones questions

Typical questions: write a product, show the mechanism, arrange compounds by reactivity, explain a reagent's behaviour, or identify an unknown from derivatives. Work from the carbonyl carbon.

1Identify the carbonyl compound (aldehyde or ketone; aliphatic or aromatic) and the reagent's nucleophilic atom.
2For simple additions (HCN, $N a H S O_{3}$ , $R M g X$ ), attach the nucleophile to the carbonyl carbon and H (or MgX, then H) to oxygen.
3For $H_{2} N - Z$ reagents, replace the C=O oxygen by =N–Z and remove one $H_{2} O$ .
4For reactivity order, count alkyl/aryl groups on the carbonyl carbon and look at ring substituents: more or bulkier electron-releasing groups → less reactive; $- N O_{2}$ → more reactive.
5For mechanisms, show (i) nucleophile attack giving the tetrahedral alkoxide, (ii) protonation; mention acid or base catalysis where used.
6For identification, a 2,4-DNP precipitate confirms a carbonyl group; other tests (Tollens', iodoform) then separate aldehydes from ketones.
7Check whether the product has a new chiral centre (for example the cyanohydrin of an unsymmetrical carbonyl compound).

Nucleophilic Addition Reactions of Aldehydes and Ketones in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Bisulphite addition for purification

Suppose a perfume lab has benzaldehyde, which smells of almonds, contaminated with toluene left over from its synthesis. A chemist shakes the mixture with saturated aqueous sodium hydrogensulphite and crystals separate.

What are the crystals, why does toluene not form them, and how is pure benzaldehyde recovered?

Try it first, then show the solutionSolution

Benzaldehyde adds $N a H S O_{3}$ across its C=O group, forming a crystalline bisulphite addition compound: $C_{6} H_{5} C H (O H) S O_{3}^{-} N a^{+}$ .
Toluene has no carbonyl group, so no nucleophilic addition occurs; it stays in the liquid.
The crystals are filtered and washed.
Warming them with dilute mineral acid or alkali reverses the addition and regenerates pure benzaldehyde.

Answer: The crystals are the bisulphite adduct of benzaldehyde; toluene has no C=O; dilute acid or alkali regenerates the aldehyde.

Memory hook: Bisulphite grabs carbonyls into crystals and lets them go again with acid or alkali.

Tricks to remember Nucleophilic Addition Reactions of Aldehydes and Ketones

Short memory aids. Say each one aloud once, then use it on a question the same day.

O-H-P-S-D for ammonia derivatives
Oxime (hydroxylamine), Hydrazone (hydrazine), Phenylhydrazone, Semicarbazone, DNP derivative: '–ne' products all have C=N.
Use it when: Naming products of $H_{2} N - Z$ reactions.
Fewer neighbours, faster attack
HCHO has no alkyl neighbours, aldehydes one, ketones two: more neighbours, more crowding, less δ+.
Use it when: Reactivity orders.
Acetal = aldehyde's armour
Turn C=O into an acetal or ketal to protect it; take the armour off with aqueous acid.
Use it when: Questions on protecting groups and acetal formation.

Nucleophilic Addition Reactions of Aldehydes and Ketones: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Most reactive carbonyl · 1 mark · MCQBoard level

Which compound is the most reactive towards nucleophilic addition?
(a) Propanone (b) Benzaldehyde (c) Ethanal (d) Methanal

Try it first, then show the solutionSolution

Methanal has two H atoms on the carbonyl carbon: least hindrance and no electron-releasing alkyl groups.

Answer: (d) Methanal.

Example 2: Two simple products · 2 marks · Very short answerBoard level

Write the structures of the products when propanone reacts with (a) HCN (b) hydroxylamine in weakly acidic medium.

Try it first, then show the solutionSolution

(a) Cyanohydrin: $(C H_{3})_{2} C (O H) C N$ , 2-hydroxy-2-methylpropanenitrile (1 mark).
(b) Oxime: $(C H_{3})_{2} C = N O H$ , propanone oxime (1 mark).

Answer: (a) $(C H_{3})_{2} C (O H) C N$ . (b) $(C H_{3})_{2} C = N - O H$ .

Example 3: Mechanism of cyanohydrin formation · 3 marks · Short answerHarder board level

Write the mechanism of the base-catalysed addition of HCN to ethanal. Why is a base added?

Try it first, then show the solutionSolution

Base generates the nucleophile: $H C N + O H^{-} ⇌ C N^{-} + H_{2} O$ (½ mark).
$C N^{-}$ attacks the carbonyl carbon; the π electrons move to oxygen, giving the tetrahedral intermediate $C H_{3} C H (O^{-}) C N$ (1 mark).
The alkoxide takes a proton from HCN or water: $C H_{3} C H (O H) C N$ (ethanal cyanohydrin), regenerating $C N^{-}$ or $O H^{-}$ (1 mark).
HCN is a weak acid and a poor nucleophile; base converts it into the strong nucleophile $C N^{-}$ , speeding up the reaction (½ mark).

Answer: $C N^{-}$ attack → alkoxide → protonation to $C H_{3} C H (O H) C N$ ; base produces $C N^{-}$ .

Example 4: Ordering reactivity · 3 marks · Short answerHarder board level

Arrange in increasing order of reactivity towards nucleophilic addition and justify: (a) methanal, propanone, ethanal, pentan-3-one (b) benzaldehyde, 4-methoxybenzaldehyde, 4-nitrobenzaldehyde.

Try it first, then show the solutionSolution

(a) Pentan-3-one < propanone < ethanal < methanal (1 mark).
Reason: ketones have two alkyl groups (more steric hindrance and +I); pentan-3-one has two bulkier ethyl groups; aldehydes have fewer, methanal none (½ mark).
(b) 4-Methoxybenzaldehyde < benzaldehyde < 4-nitrobenzaldehyde (1 mark).
Reason: $- O C H_{3}$ donates electrons (+R) and reduces the carbonyl carbon's δ+; $- N O_{2}$ withdraws electrons and increases it (½ mark).

Answer: (a) Pentan-3-one < propanone < ethanal < methanal. (b) 4-Methoxy < unsubstituted < 4-nitro benzaldehyde.

Example 5: Semicarbazones, pH and protection · 5 marks · Long answerHOTS / competency

(a) Write the equation for the reaction of propanone with semicarbazide. (b) Semicarbazide has two $- N H_{2}$ groups, but only one reacts. Explain. (c) Why is the reaction carried out in a weakly acidic medium and not in strongly acidic solution? (d) How can the C=O group of a ketone be protected using ethane-1,2-diol, and how is it recovered?

Try it first, then show the solutionSolution

(a) $(C H_{3})_{2} C = O + H_{2} N - N H - C O - N H_{2} \to (C H_{3})_{2} C = N - N H - C O - N H_{2} + H_{2} O$ : propanone semicarbazone (1 mark).
(b) The $- N H_{2}$ and NH next to C=O have their lone pairs delocalised onto the carbonyl oxygen by resonance (amide character), so they are poor nucleophiles. The terminal $- N H_{2}$ on the hydrazine N is not involved in resonance and attacks the carbonyl carbon (1½ marks).
(c) Mild acid protonates the carbonyl oxygen, making the carbon more electrophilic, and helps remove water. In strong acid the nucleophile's $- N H_{2}$ is itself protonated to $- N H_{3}^{+}$ and loses its lone pair, so it cannot attack (1 mark).
(d) Heat the ketone with ethane-1,2-diol and dry HCl to form a cyclic ketal, which is stable to bases, Grignard reagents and other nucleophiles. Afterwards, aqueous mineral acid hydrolyses the ketal back to the ketone (1½ marks).

Answer: (a) Propanone semicarbazone + $H_{2} O$ . (b) Two N lone pairs are delocalised into C=O. (c) Strong acid protonates the nucleophile. (d) Cyclic ketal with diol/dry HCl; remove with aqueous acid.

Where students lose marks in Nucleophilic Addition Reactions of Aldehydes and Ketones

The slips examiners see most often, and what to write instead.

Mistake: Forgetting to eliminate water with ammonia derivatives
Wrong working
$(C H_{3})_{2} C (O H) N H O H$ as the final product
Right working
$(C H_{3})_{2} C = N O H + H_{2} O$ (oxime)
Do this: $H_{2} N - Z$ reactions are addition followed by elimination of water.
Usually costs: 1 mark.
Mistake: Reversing aldehyde–ketone reactivity
Do this: Aldehydes > ketones; give both steric and electronic reasons.
Usually costs: 1 mark.
Mistake: Putting CN on oxygen in cyanohydrins
Do this: The nucleophile always bonds to the carbonyl carbon; oxygen gets H.
Usually costs: 1 mark.
Mistake: Saying strong acid speeds up oxime formation
Do this: Only weak acid helps; strong acid protonates the nucleophile and stops the reaction.
Usually costs: 1 mark in reasoning questions.

Exam tips

Show the tetrahedral alkoxide intermediate in every mechanism.
Name products systematically (cyanohydrin, oxime, hydrazone, semicarbazone, acetal, ketal).
For reactivity orders, mention both steric and electronic effects.
Use 2,4-DNP as the general test for C=O in identification questions.
In conversions, Grignard addition to carbonyl compounds is the standard way to make 1°, 2° and 3° alcohols.

Practice questions on Nucleophilic Addition Reactions of Aldehydes and Ketones

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Aldehydes and ketones react with 2,4-dinitrophenylhydrazine to give:
(a) a colourless gas (b) a yellow, orange or red precipitate (c) a silver mirror (d) a violet colour
Hint
Test for the carbonyl group.
Answer
(b) A coloured 2,4-dinitrophenylhydrazone precipitate.
Q2 · 2 marks · Very short answerBoard level
Give two reasons why aldehydes are more reactive than ketones towards nucleophilic addition.
Hint
Steric and electronic.
Answer
Steric: ketones have two alkyl groups that hinder the nucleophile and crowd the tetrahedral intermediate. Electronic: two electron-releasing alkyl groups in ketones reduce the positive charge on the carbonyl carbon more than one group in aldehydes.
Q3 · 3 marks · Short answerHarder board level
Write the products of: (a) ethanal + $N a H S O_{3}$ (b) propanone + ethane-1,2-diol (dry HCl) (c) ethanal + excess methanol (dry HCl).
Hint
Bisulphite adduct, cyclic ketal, acetal.
Answer
(a) $C H_{3} C H (O H) S O_{3}^{-} N a^{+}$ (bisulphite addition compound). (b) 2,2-Dimethyl-1,3-dioxolane, the cyclic ethylene glycol ketal of propanone, + $H_{2} O$ . (c) $C H_{3} C H (O C H_{3})_{2}$ (1,1-dimethoxyethane, an acetal) + $H_{2} O$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Benzaldehyde is less reactive than propanal towards nucleophilic addition.
Reason (R): Resonance with the benzene ring reduces the positive charge on the carbonyl carbon of benzaldehyde.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Conjugation with the ring.
Answer
(a).
Q5 · 3 marks · Short answerHarder board level
Methylmagnesium bromide is added separately to methanal, ethanal and propanone, and each product is hydrolysed. Name the alcohols formed and classify each.
Hint
Add a methyl to the carbonyl carbon.
Answer
Methanal → ethanol (1°); ethanal → propan-2-ol (2°); propanone → 2-methylpropan-2-ol (3°).
Q6 · 2 marks · Very short answerHarder board level
Write the structures of (a) ethanal oxime (b) propanone hydrazone (c) benzaldehyde phenylhydrazone.
Hint
Replace =O by =N–Z.
Answer
(a) $C H_{3} C H = N - O H$ . (b) $(C H_{3})_{2} C = N - N H_{2}$ . (c) $C_{6} H_{5} C H = N - N H C_{6} H_{5}$ .
Q7 · 3 marks · Short answerHOTS / competency
Propanal and propanone both have the formula $C_{3} H_{6} O$ . Each is treated with HCN. Which cyanohydrin contains a chiral carbon? Explain, and predict whether the product would be optically active.
Hint
Look at the four groups on the former carbonyl carbon.
Answer
Propanal gives $C H_{3} C H_{2} C H (O H) C N$ , whose C2 carries H, OH, CN and $C_{2} H_{5}$ : chiral. Propanone gives $(C H_{3})_{2} C (O H) C N$ , with two identical $C H_{3}$ groups: achiral. The propanal cyanohydrin forms as a racemic mixture (CN⁻ attacks the planar C=O equally from both faces), so it is optically inactive.
Q8 · 4 marks · Case-basedHOTS / competency
Read and answer.
Suppose a student receives an unknown liquid X ( $C_{4} H_{8} O$ ). It gives an orange precipitate with 2,4-DNP reagent. The dried precipitate's melting point matches that reported for the 2,4-dinitrophenylhydrazone of butanone. X does not form a crystalline adduct readily with sodium hydrogensulphite.
(a) What does the 2,4-DNP test show about X? (1 mark)
(b) Write the equation for the reaction of X with 2,4-DNP, using a general formula $H_{2} N - N H - A r$ for the reagent. (1 mark)
(c) Why is a melting point of a derivative useful for identification? (1 mark)
(d) Why does X not readily form a bisulphite adduct while butanal would? (1 mark)
Hint
X is a ketone.
Answer
(a) X contains a carbonyl group (aldehyde or ketone). (b) $C H_{3} C O C H_{2} C H_{3} + H_{2} N - N H A r \to C H_{3} C (= N - N H A r) C H_{2} C H_{3} + H_{2} O$ . (c) Each carbonyl compound gives a solid derivative with a sharp, characteristic melting point, which is compared with known values. (d) X (butanone) is a ketone with two alkyl groups; steric hindrance makes the bisulphite equilibrium unfavourable, while an aldehyde like butanal adds readily.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Product of an aldehyde with HCN?Show answer
A cyanohydrin.
2.Why is $N a H S O_{3}$ used in purification?Show answer
It forms crystalline adducts that release the carbonyl compound with dilute acid or alkali.
3.Product of a ketone with ethane-1,2-diol and dry HCl?Show answer
A cyclic ethylene glycol ketal.
4.Reagent that gives an oxime?Show answer
Hydroxylamine, $N H_{2} O H$ .
5.Which is more reactive: ethanal or propanone?Show answer
Ethanal.
6.Hybridisation change at the carbonyl carbon on addition?Show answer
$s p^{2} \to s p^{3}$ .

Nucleophilic Addition Reactions of Aldehydes and Ketones: frequently asked questions

What is nucleophilic addition in Class 12 Chemistry?

A reaction in which a nucleophile adds to the electron-deficient carbon of a C=O group, forming a tetrahedral intermediate that is then protonated.

Why are aldehydes more reactive than ketones?

Ketones have two alkyl groups that crowd the carbonyl carbon and release electrons to it, making it less accessible and less positive than in aldehydes.

What is formed when aldehydes react with alcohols?

A hemiacetal with one equivalent, and an acetal with two equivalents of alcohol in the presence of dry HCl.

Why does only one NH₂ group of semicarbazide react with carbonyl compounds?

The other nitrogen atoms are attached to the C=O of the amide part and their lone pairs are delocalised by resonance, so they are not nucleophilic.

What is the 2,4-DNP test?

Aldehydes and ketones give yellow, orange or red precipitates of 2,4-dinitrophenylhydrazones with 2,4-dinitrophenylhydrazine, showing the presence of a carbonyl group.