Carbonyl Nucleophilic Addition

8 MCQs9-step worked example
Source: NCERT BiomoleculesPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

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:

  1. 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.
  2. 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 12, page 2).

General mechanism (simplified):

  1. Nucleophile (Nu⁻) attacks the carbonyl carbon.
  2. The π electrons of C=O shift to oxygen, forming a tetrahedral alkoxide intermediate.
  3. 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).

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 12, page 8).

Practice 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:

MCQ 2Easy RecallPractice

Which intermediate is formed during nucleophilic addition to a carbonyl group before protonation?

MCQ 3Easy RecallPractice

The correct order of reactivity toward nucleophilic addition is:

MCQ 4Direct ApplicationPractice

Among the following, which compound undergoes nucleophilic addition most readily?

MCQ 5Direct ApplicationPractice

When HCN adds to acetaldehyde (CH₃CHO), the product formed is:

MCQ 6Direct ApplicationPractice

Acetaldehyde reacts with hydroxylamine (NH₂OH) to form:

MCQ 7Concept TrapPractice

Ketones are less reactive than aldehydes toward nucleophilic addition primarily because:

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:

Worked Example

  1. 1

    Given

    Four carbonyl compounds: formaldehyde (HCHO), acetaldehyde (CH₃CHO), acetone (CH₃COCH₃), and acetophenone (C₆H₅COCH₃). Nucleophile: HCN.

  2. 2

    Required

    Decreasing order of reactivity toward nucleophilic addition.

  3. 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. 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. 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. 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. 7

    Final answer

    **Decreasing order of reactivity:** HCHO > CH₃CHO > CH₃COCH₃ > C₆H₅COCH₃

  8. 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. 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.) ---

Before solving, remember these

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, p. 2
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, p. 8

Formulas

pKa of carboxylic acid

Stronger acid than phenol due to more effective resonance over two equivalent oxygens. EWG substituents (Cl, NO2) increase acidity.

SymbolQuantitySI Unit
pKa-log Ka-

Valid when

  • Aqueous solution
  • α-substituent effects strongest

pKa of phenol vs aliphatic alcohol

Phenols ~10⁶× more acidic than aliphatic alcohols due to resonance stabilisation of phenoxide ion. Electron-withdrawing substituents lower pKa further.

SymbolQuantitySI Unit
pKa-log Ka-

Valid when

  • Aqueous solution
  • Substituent effects shift values

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

Trap

Oxidation product depends on reagent strength

Category: Organic Reaction Conditions

1° alcohol: PCC/PDC → aldehyde (stops). KMnO4/K2Cr2O7 in acidic → carboxylic acid (continues). 2° alcohol: any oxidiser → ketone. 3° alcohol: not oxidised by ordinary reagents.

When it triggers

Question gives 1° alcohol oxidation with specified reagent.

How to avoid

PCC, PDC, Swern, DMP: mild → stop at aldehyde. KMnO4, K2Cr2O7, CrO3, jones: strong → carboxylic acid. Reagent choice matches desired product.

Past Year Questions

20 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.

NEET 2024Revised key

Given below are two statements: Statement I : Aniline does not undergo Friedel-Crafts alkylation reaction. Statement II : Aniline cannot be prepared through Gabriel synthesis . In the light of the above statements, choose the correct answer from the options given below:

1Both statement I and Statement II are true
2Both Statement I and Statement II are false
3Statement I is correct but Statement II is false
4Statement I is incorrect but Statement II is true
NTA Answer: Option 1(revised_final)
NEET 2024Revised key

Fehling’s solution ‘A’ is

1aqueous copper sulphate
2alkaline copper sulphate
3alkaline solution of sodium potassium tartrate (Rochelle’s salt)
4aqueous sodium citrate
NTA Answer: Option 1(revised_final)
NEET 2022

Given below are two statements: Statement I: In Lucas test, primary, secondary and tertiary alcohols are distinguished on the basis of their reactivity with conc. HCl + ZnCl , known as Lucas Reagent. 2 Statement II: Primary alcohols are most reactive and immediately produce turbidity at room temperature on reaction with Lucas Reagent. In the light of the above statements, choose the most appropriate answer from the options given below:

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 4(final)
NEET 2021

Match List-I with List-II List-I List-II CO, HCI (a) (i) Hell-Volhard-Zelinsky Anhyd. AlCl/ Cu3Cl reaction O R — C — CH + (b) (ii) Gattermann-Koch 3 NaOX reaction R — CH — OH (c) (iii) Haloform reaction 2 +R′ COOH Conc. HSO 2 4 (d) R — CH COOH (iv) Esterification 2 ( II) X /Red P 2 (ii) HO 2 Choose the correct answer from the options given below.

1(a) - (ii), (b) - (iii), (c) - (iv), (d) - (i)
2(a) - (iv), (b) - (i), (c) - (ii), (d) - (iii)
3(a) - (iii), (b) - (ii), (c) - (i), (d) - (iv)
4(a) - (i), (b) - (iv), (c) - (iii), (d) - (ii)
NTA Answer: Option 1(final)

How NEET usually asks this

Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.

Sources

NCERT refs: Class 12 Chemistry Chapter 12, p.2 | Class 12 Chemistry Chapter 12, p.8

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