Addition Hcn Nh3 Grignard

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

Lesson

The trap first: Students confuse reactivity order in nucleophilic addition. Ketones are NOT more reactive than aldehydes — it is the reverse. Aldehydes have less steric hindrance (one H, one R) and weaker electron donation from a single alkyl group, making their carbonyl carbon more electrophilic.

Nucleophilic addition to carbonyls is the signature reaction of aldehydes and ketones. The carbonyl C=O has a polarised carbon (δ+) attacked by nucleophiles. The general mechanism: nucleophile attacks Cδ+ → π-bond breaks → oxygen becomes negatively charged → protonation gives the addition product.

Three nucleophiles NEET targets:

  1. HCN addition — produces cyanohydrins (R-CH(OH)-CN from aldehydes, R₂C(OH)-CN from ketones). Requires a trace of base (CN⁻ is the actual nucleophile). The product gains one carbon — this is a chain-lengthening reaction.

  2. NH₃ derivatives (NH₂OH, NH₂NH₂, 2,4-DNP, semicarbazide) — carbonyl + H₂N-G → C=N-G + H₂O. These are condensation reactions forming oximes, hydrazones, 2,4-DNP derivatives, and semicarbazones respectively. The C=N linkage is called a Schiff base when G = aryl/alkyl.

  3. Grignard reagent (RMgX) — the carbanion (R⁻) attacks Cδ+. HCHO → 1° alcohol; other aldehydes → 2° alcohol; ketones → 3° alcohol. Hydrolysis with dilute acid completes the reaction.

Reactivity order: HCHO > other aldehydes > ketones (steric + electronic). This ordering is the single most tested concept in NEET nucleophilic addition questions (NCERT Class 12 Chemistry Chapter 12, page 8).

Watch-out: When a question says "addition of HCN to propanone," the product is 2-hydroxy-2-methylpropanenitrile — not propanal cyanohydrin. Match the carbonyl substrate to the correct product skeleton.

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

The addition of HCN to acetaldehyde in the presence of a base produces:

MCQ 2Easy RecallPractice

The reaction of an aldehyde with hydroxylamine (NH₂OH) produces:

MCQ 3Easy RecallPractice

The reaction of formaldehyde with methylmagnesium bromide (CH₃MgBr) followed by hydrolysis gives:

MCQ 4Direct ApplicationPractice

Arrange the following in decreasing order of reactivity towards nucleophilic addition: (I) HCHO, (II) CH₃CHO, (III) (CH₃)₂CO, (IV) (C₆H₅)₂CO

MCQ 5Direct ApplicationPractice

The product formed when acetone reacts with HCN followed by hydrolysis is:

MCQ 6Direct ApplicationPractice

Which of the following Grignard reactions produces a tertiary alcohol?

MCQ 7CalculationPractice

An unknown carbonyl compound X reacts with 2,4-DNP to give an orange precipitate, but does NOT reduce Tollens' reagent. X reacts with CH₃MgBr followed by acid hydrolysis to give 2-methyl-2-butanol. Compound X is:

MCQ 8CalculationPractice

Compound A (C₃H₆O) forms a cyanohydrin B with HCN. Hydrolysis of B gives compound C (a hydroxy acid). C has a chiral centre. Compound A is:

Worked Example

  1. 1

    Given

    - Substrate: benzaldehyde (C₆H₅CHO) — an aromatic aldehyde - Nucleophiles: (a) HCN/NaCN, (b) C₆H₅MgBr then dilute H₂SO₄

  2. 2

    Required

    - Product for each reaction - Reactivity comparison with acetophenone (C₆H₅COCH₃)

  3. 3

    Concept

    Nucleophilic addition to carbonyl: Nu⁻ attacks Cδ+ → tetrahedral alkoxide → protonation. Reactivity depends on steric accessibility and electrophilicity of carbonyl carbon.

  4. 4

    Formula/Principle

    Reactivity order: Aldehyde > Ketone (less steric hindrance at carbonyl C; fewer electron-donating groups).

  5. 5

    Substitution / Application

    (a) C₆H₅CHO + CN⁻ → C₆H₅CH(O⁻)CN → protonation → C₆H₅CH(OH)CN (mandelonitrile / benzaldehyde cyanohydrin) (b) C₆H₅CHO + C₆H₅MgBr → C₆H₅CH(O⁻MgBr)(C₆H₅) → H₃O⁺ → (C₆H₅)₂CHOH (benzhydrol, a secondary alcohol)

  6. 6

    Calculation/reasoning

    No numerical calculation needed. The logic is mechanistic: in both cases, the nucleophile (CN⁻ or C₆H₅⁻) attacks the electrophilic carbonyl carbon, breaking the C=O π-bond.

  7. 7

    Final answer

    - (a) Product: mandelonitrile, C₆H₅CH(OH)CN - (b) Product: benzhydrol, (C₆H₅)₂CHOH (a 2° alcohol) - Benzaldehyde is more reactive than acetophenone because: (i) the aldehyde has only one substituent on carbonyl C vs two in ketone → less steric crowding; (ii) the methyl group in acetophenone provides +I electron donation, reducing Cδ+ character further.

  8. 8

    Common trap

    Students may invert the reactivity, assuming that acetophenone's conjugated phenyl makes it "more activated." In fact, both compounds have a phenyl ring, but the additional methyl group in acetophenone adds both steric and electronic deactivation. The relevant mistake: believing ketones are more reactive than aldehydes toward nucleophilic addition.

  9. 9

    Similar NEET-style question

    "Arrange the following in decreasing order of reactivity toward addition of HCN: (I) HCHO, (II) C₆H₅CHO, (III) CH₃COCH₃, (IV) C₆H₅COC₆H₅." Expected answer: I > II > III > IV.

Before solving, remember these

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
Key Fact

Common nucleophilic addition reactions

+HCN → cyanohydrin. +NaHSO₃ → bisulphite addition product (test for aldehydes/methyl ketones). +NH₃ derivatives → imines, oximes, hydrazones (used to identify carbonyls). +Grignard (RMgX) → 1°/2°/3° alcohol.

-- NCERT, p. 10

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.8

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