Definition
Carbonyl compounds
Contain C=O group. Aldehyde: R-CHO (1° carbonyl). Ketone: R-CO-R' (2° carbonyl). Carboxylic acid: R-COOH. Different reactivity due to neighbouring groups.
-- NCERT Class 12 Chemistry, Ch. 8, p. 228Carbonyl Nucleophilic Addition
8 MCQs9-step worked example
Source: NCERT Alcohols, Phenols, Ethers and Carbonyl CompoundsPYQ coverage: NEET 2022, 2026Official key: NTA-verifiedLast updated: 27 Sep 2026
Carbonyl Nucleophilic Addition, explained for NEET
The carbonyl group (C=O) in aldehydes and ketones is the site of nucleophilic addition — and the single most tested reaction mechanism in this chapter. The trap that costs marks: believing ketones are more reactive than aldehydes. The opposite is true.
Why aldehydes win. The carbon of C=O carries a partial positive charge (δ+) because oxygen is more electronegative. A nucleophile attacks this electrophilic carbon. Two factors favour aldehydes over ketones:
- Steric effect. Aldehydes have one R group and one H attached to the carbonyl carbon. Ketones have two R groups — more crowding blocks the incoming nucleophile.
- Electronic effect (+I). Each alkyl group donates electron density toward the carbonyl carbon via the inductive effect, reducing its δ+ character. One R in an aldehyde means less donation; two R groups in a ketone means more donation and a less electrophilic carbon.
Reactivity order: HCHO > RCHO > R₂CO (NCERT Class 12 Chemistry Chapter 8, page 236).
General mechanism (simplified):
- Nucleophile (Nu⁻) attacks the carbonyl carbon.
- The π electrons of C=O shift to oxygen, forming a tetrahedral alkoxide intermediate.
- The alkoxide is protonated to give the addition product.
Common nucleophiles tested in NEET: HCN (→ cyanohydrin), NH₃ derivatives (→ Schiff base, oxime, hydrazone), and Grignard reagents (→ alcohol after hydrolysis).
Physical properties: the boiling-point ladder. The same C=O dipole decides where aldehydes and ketones sit among compounds of similar molecular mass. NCERT Class 12 Chemistry, Chapter 8, page 235 states that their boiling points are higher than those of hydrocarbons and ethers of comparable molecular masses, because of weak molecular association arising out of dipole-dipole interactions, and lower than those of alcohols of similar molecular masses, because they have no intermolecular hydrogen bonding. Its table for molecular masses 58 and 60 runs n-butane (273 K) < methoxyethane (281 K) < propanal (322 K) < acetone (329 K) < propan-1-ol (370 K), and Example 8.2 on the same page adds that butanal is more polar than ethoxyethane. Carboxylic acids sit above all of them: NCERT Class 12 Chemistry, Chapter 8, page 249 says they are higher boiling than aldehydes, ketones and even alcohols of comparable molecular masses, because their molecules associate more extensively through intermolecular hydrogen bonding; most exist as dimers even in the vapour phase. So, at comparable molecular mass: hydrocarbon < ether < aldehyde or ketone < alcohol < carboxylic acid. When NEET asks for the increasing "polarity" of an ether, a ketone, an alcohol and an acid, the official key follows this same ladder of intermolecular attraction.
Watch-out for NEET: When a question asks you to compare the rate of nucleophilic addition between an aldehyde and a ketone with the same nucleophile, the aldehyde reacts faster. Formaldehyde (HCHO), with no alkyl groups at all, is the most reactive carbonyl compound toward nucleophilic addition (NCERT Class 12 Chemistry Chapter 8, page 236).
Can you answer these Carbonyl Nucleophilic Addition MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
In the nucleophilic addition reaction of carbonyl compounds, the nucleophile attacks:
Show answer and why every option is right or wrong
Answer: A. The carbonyl carbon bears a partial positive charge (δ+) due to the higher electronegativity of oxygen, making it the electrophilic centre that the nucleophile attacks (NCERT Class 12 Chemistry Chapter 8, page 236).
Why B is wrong: B is wrong because oxygen bears the partial negative charge (δ−) and is not the site of nucleophilic attack — nucleophiles seek electron-deficient centres, not electron-rich ones.
Why C is wrong: C is wrong because α-carbon chemistry (enolisation, aldol) is a separate reaction pathway. In simple nucleophilic addition, the nucleophile attacks the carbonyl carbon directly.
Why D is wrong: D is wrong because while the π electrons are displaced during the reaction, the nucleophile targets the electrophilic carbon atom, not the electron cloud itself. The π electrons shift to oxygen as the tetrahedral intermediate forms.
MCQ 2Easy RecallPractice
Which intermediate is formed during nucleophilic addition to a carbonyl group before protonation?
Show answer and why every option is right or wrong
Answer: C. When the nucleophile attacks the carbonyl carbon, the C=O π bond breaks and the electron pair shifts to oxygen, forming a tetrahedral alkoxide intermediate (NCERT Class 12 Chemistry Chapter 8, page 236).
Why A is wrong: A is wrong because no carbon loses an electron to become positively charged during nucleophilic addition — the mechanism is heterolytic addition to the C=O, not ionisation.
Why B is wrong: B is wrong because nucleophilic addition to C=O is an ionic (polar) mechanism, not a radical one. No homolytic bond cleavage occurs.
Why D is wrong: D is wrong because the carbon gains a bond to the nucleophile and does not acquire a negative charge. The negative charge resides on oxygen in the alkoxide intermediate.
MCQ 3Easy RecallPractice
The correct order of reactivity toward nucleophilic addition is:
Show answer and why every option is right or wrong
Answer: B. Formaldehyde has no alkyl groups (least steric hindrance, least +I effect), followed by aldehydes (one R), then ketones (two R groups). Both steric and electronic factors favour this order (NCERT Class 12 Chemistry Chapter 8, page 236).
Why A is wrong: A is wrong because it inverts the correct order — ketones are the least reactive due to two bulky, electron-donating R groups. This is the classic trap of assuming more substitution means more reactivity (mistake: mistake: carbonyl ketone more reactive).
Why C is wrong: C is wrong because it places ketones above aldehydes. Two alkyl groups on a ketone provide greater steric hindrance and stronger +I donation than the single alkyl group on an aldehyde, making ketones less reactive.
Why D is wrong: D is wrong because it places a monosubstituted aldehyde above formaldehyde. HCHO has zero alkyl substituents, so it has the least steric crowding and the most electrophilic carbonyl carbon.
MCQ 4Direct ApplicationPractice
Among the following, which compound undergoes nucleophilic addition most readily?
Show answer and why every option is right or wrong
Answer: A. A is correct. Formaldehyde has the smallest groups (two H atoms) on the carbonyl carbon — minimal steric hindrance and no +I-donating alkyl groups — making it the most electrophilic and therefore the most reactive toward nucleophilic addition (NCERT Class 12 Chemistry Chapter 8, page 236).
Why B is wrong: B is wrong because although acetaldehyde is more reactive than acetone, it still has one methyl group donating electron density and adding steric bulk, making it less reactive than HCHO.
Why C is wrong: C is wrong because acetone has two methyl groups providing both steric hindrance and +I effect, reducing the electrophilicity of the carbonyl carbon compared to aldehydes and formaldehyde.
Why D is wrong: D is wrong because benzophenone has two bulky phenyl rings creating significant steric hindrance, and the phenyl groups also stabilise the carbonyl through resonance, further reducing electrophilicity. It is the least reactive among the four.
MCQ 5Direct ApplicationPractice
When HCN adds to acetaldehyde (CH₃CHO), the product formed is:
Show answer and why every option is right or wrong
Answer: D. HCN adds across the C=O bond: CN⁻ attacks the carbonyl carbon, and H⁺ protonates the alkoxide oxygen, forming the cyanohydrin CH₃CH(OH)CN. This is a textbook nucleophilic addition (NCERT Class 12 Chemistry Chapter 8, page 236).
Why A is wrong: A is wrong because ethanol would require reduction of the carbonyl group (e.g., by NaBH₄ or LiAlH₄), not nucleophilic addition of HCN. HCN introduces a –CN group, not –H.
Why B is wrong: B is wrong because propionitrile would require replacement of the oxygen entirely and reduction — this is not what nucleophilic addition achieves. The –OH group is retained in the cyanohydrin product.
Why C is wrong: C is wrong because acetic acid is an oxidation product of acetaldehyde (e.g., with KMnO₄), not an addition product. HCN adds to C=O; it does not oxidise the aldehyde.
MCQ 6Direct ApplicationPractice
Acetaldehyde reacts with hydroxylamine (NH₂OH) to form:
Show answer and why every option is right or wrong
Answer: D. Hydroxylamine is an ammonia derivative that undergoes nucleophilic addition-elimination with aldehydes. The –NH₂ of NH₂OH adds to the carbonyl carbon; after loss of water, the product is an oxime, CH₃CH=NOH (NCERT Class 12 Chemistry Chapter 8, page 236).
Why A is wrong: A is wrong because alcohol formation requires a reducing agent (NaBH₄, LiAlH₄) or catalytic hydrogenation. Hydroxylamine is a nitrogen nucleophile, not a hydride source.
Why B is wrong: B is wrong because hydrazones form when hydrazine (NH₂NH₂) or substituted hydrazines react with carbonyl compounds. Hydroxylamine has –OH on nitrogen, not –NH₂, leading to the oxime product.
Why C is wrong: C is wrong because Schiff bases form when primary amines (RNH₂) — not hydroxylamine — react with aldehydes or ketones. The key distinction is the substituent on nitrogen: –OH in hydroxylamine gives an oxime, not a Schiff base.
MCQ 7Concept TrapPractice
Ketones are less reactive than aldehydes toward nucleophilic addition primarily because:
Show answer and why every option is right or wrong
Answer: B. The two alkyl groups in ketones create greater steric crowding around the carbonyl carbon AND donate electron density through the +I effect, reducing the partial positive charge on carbon. Both factors lower the reactivity toward nucleophilic attack compared to aldehydes with only one alkyl group (NCERT Class 12 Chemistry Chapter 8, page 236).
Why A is wrong: A is wrong because the C=O bond strength is not the primary factor governing reactivity differences between aldehydes and ketones. The difference arises from steric and electronic effects of the substituents, not from intrinsic bond strength.
Why C is wrong: C is wrong because the electronegativity of oxygen does not change between aldehydes and ketones — it is the same oxygen atom. The difference lies in the substituents attached to the carbonyl carbon.
Why D is wrong: D is wrong because ketones do form tetrahedral intermediates during nucleophilic addition — they simply do so more slowly than aldehydes due to steric and electronic factors. The mechanism itself is the same.
MCQ 8CalculationPractice
An unknown carbonyl compound X reacts faster with HCN than acetone but slower than formaldehyde. When treated with a Grignard reagent (CH₃MgBr) followed by hydrolysis, X gives a secondary alcohol. Compound X is most likely:
Show answer and why every option is right or wrong
Answer: C. Step 1: Reactivity between formaldehyde and acetone means X is an aldehyde (RCHO). Step 2: Grignard addition to an aldehyde (RCHO + R'MgBr → RCH(OH)R' after hydrolysis) gives a secondary alcohol. Acetaldehyde fits both criteria: it is more reactive than acetone (one R group vs two) and less reactive than HCHO (one R group vs none), and CH₃CHO + CH₃MgBr → CH₃CH(OH)CH₃ (isopropanol, a secondary alcohol).
Why A is wrong: A is wrong because formaldehyde is more reactive than X (given in the stem). Additionally, HCHO + CH₃MgBr → CH₃CH₂OH (a primary alcohol), not a secondary alcohol.
Why B is wrong: B is wrong because acetone is less reactive than X (given in the stem). Also, while acetone + CH₃MgBr does give a tertiary alcohol (not secondary), the reactivity condition alone eliminates it.
Why D is wrong: D is wrong because although benzaldehyde is an aldehyde and would give a secondary alcohol with CH₃MgBr, the bulky phenyl group and its resonance stabilisation of the carbonyl make benzaldehyde less reactive than a simple aliphatic aldehyde like acetaldehyde, likely placing it below or near acetone in reactivity — inconsistent with being faster than acetone.
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How do you solve a Carbonyl Nucleophilic Addition question? A worked example
- 1
Given
Four carbonyl compounds: formaldehyde (HCHO), acetaldehyde (CH₃CHO), acetone (CH₃COCH₃), and acetophenone (C₆H₅COCH₃). Nucleophile: HCN.
- 2
Required
Decreasing order of reactivity toward nucleophilic addition.
- 3
Concept
Reactivity of carbonyl compounds toward nucleophilic addition depends on the electrophilicity of the carbonyl carbon and the steric environment around it. Factors: (a) number and size of substituents (steric hindrance), (b) electron-donating (+I) or electron-withdrawing (−I, −M) nature of substituents.
- 4
Formula / Principle
No single formula applies. The governing principle is:• Fewer and smaller substituents → more reactive (less steric hindrance, less +I donation).• Alkyl groups donate via +I; phenyl groups donate via resonance (−M from carbonyl is offset by +M of ring into carbonyl, but net effect is that phenyl stabilises the C=O, reducing electrophilicity more than a simple alkyl).
- 5
Substitution / Analysis
| Compound | Substituents on C=O | Steric bulk | +I / Resonance effect |
|---|---|---|---|
| HCHO | H, H | Minimal | None |
| CH₃CHO | CH₃, H | Low | One +I (CH₃) |
| CH₃COCH₃ | CH₃, CH₃ | Moderate | Two +I (CH₃ × 2) |
| C₆H₅COCH₃ | C₆H₅, CH₃ | High (phenyl is bulky) | +I (CH₃) + resonance stabilisation (C₆H₅) | - 6
Reasoning
HCHO: no alkyl groups, no steric hindrance — most electrophilic carbon. Most reactive.
CH₃CHO: one methyl — slight +I, slight steric effect. Second most reactive.
CH₃COCH₃: two methyls — greater +I and steric effect than acetaldehyde. Third.
C₆H₅COCH₃: phenyl is bulkier than methyl AND provides resonance stabilisation of the carbonyl (delocalisation of the lone pair on oxygen into the ring stabilises the C=O, making the carbon less δ+). Least reactive. - 7
Final answer
Decreasing order of reactivity:
HCHO > CH₃CHO > CH₃COCH₃ > C₆H₅COCH₃ - 8
Common trap
The common distractor inverts aldehyde-ketone reactivity (mistake: students believe ketones are more reactive because they are "more substituted" — confusing substitution with reactivity in nucleophilic addition, where substitution decreases reactivity). Another trap: treating phenyl as equivalent to methyl, missing the additional resonance stabilisation that makes acetophenone less reactive than acetone.
- 9
Similar NEET-style question
"Which of the following undergoes nucleophilic addition with NaHSO₃ most readily: (a) benzaldehyde, (b) acetaldehyde, (c) acetone, (d) benzophenone?" (Answer: acetaldehyde — aliphatic aldehyde with minimal steric and electronic deactivation. Note: benzaldehyde, though an aldehyde, has the phenyl group reducing reactivity relative to acetaldehyde.)
---
What to remember before solving Carbonyl Nucleophilic Addition questions
Key Fact
Nucleophilic addition to C=O
Mechanism: (1) nucleophile attacks electrophilic C; (2) tetrahedral intermediate; (3) protonation of O. Aldehydes more reactive than ketones (less steric hindrance, less +I from one R).
-- NCERT Class 12 Chemistry, Ch. 8, p. 236The boiling points of aldehydes and ketones are higher than hydrocarbons and ethers of comparable molecular masses, due to weak molecular association arising out of dipole-dipole interactions. They are lower than those of alcohols of similar molecular masses due to absence of intermolecular hydrogen bonding.
-- NCERT Class 12 Chemistry, Ch. 8, p. 235Carboxylic acids are higher boiling liquids than aldehydes, ketones and even alcohols of comparable molecular masses, due to more extensive association of carboxylic acid molecules through intermolecular hydrogen bonding.
-- NCERT Class 12 Chemistry, Ch. 8, p. 249Where do students lose marks on Carbonyl Nucleophilic Addition?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Root cause: concept gap
Correction
Aldehydes more reactive: less steric hindrance, less +I from one R group. Order: HCHO > RCHO > R₂CO.
More in Alcohols, Phenols, Ethers and Carbonyl Compounds: 2 exam traps and mistakes · 2 formulas · 2 question patterns from its other lessons.
Carbonyl Nucleophilic Addition questions from past NEET papers
2 questions from NEET 2022, 2026. Answers verified against NTA official keys.
NEET 2026
1A < B < C < D
2C < A < D < B
3C < A < B < D
4A < C < B < D
NTA Answer: Option 4(final)
NEET 2022
1Statement I is incorrect but Statement II is correct
2Both Statement I and Statement II are correct
3Both Statement I and Statement II are incorrect
4Statement I is correct but Statement II is incorrect
NTA Answer: Option 2(final)
All 23 past-paper questions from Alcohols, Phenols, Ethers and Carbonyl Compounds →
How does NEET ask about Carbonyl Nucleophilic Addition?
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
Predict product of nucleophilic addition to aldehyde/ketone. Aldehydes more reactive than ketones.
InterpretationMedium
Common distractors
inverts aldehyde ketone reactivity
Believes ketones more reactive
Sources
NCERT refs: Class 12 Chemistry Chapter 8, p.236 | Class 12 Chemistry Chapter 8, p.235 | Class 12 Chemistry Chapter 8, p.249
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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