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:
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.
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.
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 8, page 234).
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.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
The addition of HCN to acetaldehyde in the presence of a base produces:
Answer: B. HCN adds across the C=O of acetaldehyde to form a cyanohydrin — 2-hydroxypropanenitrile (CH₃CH(OH)CN). This is a standard nucleophilic addition (NCERT Class 12 Chemistry Chapter 8, page 234).
Why A is wrong: Acetonitrile (CH₃CN) is a nitrile but is formed by dehydration of acetaldoxime or from substitution reactions, not by HCN addition to a carbonyl.
Why C is wrong: Acetic acid would require oxidation of acetaldehyde, not nucleophilic addition of HCN.
Why D is wrong: Ethanol would require reduction of acetaldehyde (e.g., NaBH₄ or LiAlH₄), not HCN addition.
The reaction of an aldehyde with hydroxylamine (NH₂OH) produces:
Answer: D. NH₂OH reacts with the carbonyl group to form an oxime (C=N-OH) with elimination of water. This is a condensation reaction specific to hydroxylamine (NCERT Class 12 Chemistry Chapter 8, page 236).
Why A is wrong: A hydrazone is formed with hydrazine (NH₂NH₂), not hydroxylamine.
Why B is wrong: A cyanohydrin is formed by HCN addition, not by reaction with an NH₃ derivative.
Why C is wrong: A semicarbazone is formed with semicarbazide (NH₂NHCONH₂), not hydroxylamine.
The reaction of formaldehyde with methylmagnesium bromide (CH₃MgBr) followed by hydrolysis gives:
Answer: C. HCHO + CH₃MgBr → CH₃CH₂OMgBr; hydrolysis gives ethanol (a primary alcohol). Grignard addition to formaldehyde always yields a 1° alcohol with one more carbon than the Grignard reagent (NCERT Class 12 Chemistry Chapter 8, page 234).
Why A is wrong: Methanol has only one carbon — CH₃MgBr adds a methyl group to HCHO giving a two-carbon product (ethanol), not methanol.
Why B is wrong: A secondary alcohol forms when a Grignard reagent adds to an aldehyde other than HCHO (e.g., CH₃CHO + CH₃MgBr → 2-propanol). HCHO gives a 1° alcohol.
Why D is wrong: A tertiary alcohol forms when a Grignard reagent adds to a ketone (e.g., acetone + CH₃MgBr → 2-methyl-2-propanol). HCHO cannot produce a 3° alcohol.
Arrange the following in decreasing order of reactivity towards nucleophilic addition: (I) HCHO, (II) CH₃CHO, (III) (CH₃)₂CO, (IV) (C₆H₅)₂CO
Answer: A. Reactivity decreases with increasing steric bulk and electron donation to carbonyl carbon. HCHO (no alkyl) > CH₃CHO (one methyl) > acetone (two methyls) > benzophenone (two bulky phenyl groups with +M effect). This is the standard NCERT ordering (Class 12 Chemistry Chapter 8, page 234).
Why B is wrong: This is the exact reverse order — it would apply if ketones were more reactive than aldehydes, which contradicts the steric + electronic argument. Aldehydes are more reactive than ketones (trap: mistake: carbonyl ketone more reactive).
Why C is wrong: Placing CH₃CHO above HCHO is incorrect. HCHO has zero steric hindrance and no +I alkyl group, making its carbonyl carbon the most electrophilic.
Why D is wrong: Placing acetone above CH₃CHO reverses the aldehyde-vs-ketone reactivity. One R group (aldehyde) donates less electron density and offers less steric shielding than two R groups (ketone).
The product formed when acetone reacts with HCN followed by hydrolysis is:
Answer: C. Acetone + HCN → acetone cyanohydrin ((CH₃)₂C(OH)CN). Hydrolysis of the -CN group (→ -COOH) gives 2-hydroxy-2-methylpropanoic acid. The two methyl groups from acetone are retained on the central carbon (NCERT Class 12 Chemistry Chapter 8, page 234).
Why A is wrong: Lactic acid (2-hydroxypropanoic acid) has only one methyl group on the α-carbon — it would form from acetaldehyde + HCN + hydrolysis, not acetone.
Why B is wrong: 2-Methylpropanoic acid lacks the -OH group. The cyanohydrin intermediate retains the hydroxyl from the nucleophilic addition step.
Why D is wrong: Malonic acid (propanedioic acid) has two -COOH groups and would not form from a single HCN addition to a monoketone.
Which of the following Grignard reactions produces a tertiary alcohol?
Answer: D. Ketone + RMgX → tertiary alkoxide → hydrolysis gives 3° alcohol. CH₃COCH₃ + C₂H₅MgBr → 2-methyl-2-butanol (tertiary). Only ketones produce tertiary alcohols with Grignard reagents (NCERT Class 12 Chemistry Chapter 8, page 234).
Why A is wrong: HCHO + RMgX always gives a primary alcohol (one carbon added to formaldehyde's single carbon).
Why B is wrong: Aldehydes other than HCHO + RMgX give secondary alcohols. CH₃CHO + C₂H₅MgBr → 2-butanol (2°).
Why C is wrong: CO₂ + RMgX followed by hydrolysis gives a carboxylic acid (RCO₂H), not an alcohol.
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:
Answer: A. Positive 2,4-DNP + negative Tollens' → X is a ketone. Product is 2-methyl-2-butanol: (CH₃)(C₂H₅)C(OH)(CH₃). The Grignard adds CH₃ to the carbonyl carbon. Working backward: remove the added CH₃ from the tertiary carbon → the ketone was CH₃COC₂H₅ (butanone). This requires two reasoning steps: (1) identify ketone vs aldehyde, (2) reverse-engineer the Grignard product.
Why B is wrong: Acetone + CH₃MgBr → 2-methyl-2-propanol (tert-butanol), which has 4 carbons. The product stated is 2-methyl-2-butanol (5 carbons), so the ketone must have 4 carbons (butanone), not 3 (acetone).
Why C is wrong: Acetaldehyde is an aldehyde — it would give a positive Tollens' test (silver mirror), contradicting the observations.
Why D is wrong: Propanal is an aldehyde — it would reduce Tollens' reagent. Additionally, aldehyde + CH₃MgBr gives a secondary alcohol, not a tertiary alcohol.
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:
Answer: B. Molecular formula C₃H₆O with one degree of unsaturation → aldehyde or ketone. If A = propanal (CH₃CH₂CHO): cyanohydrin = CH₃CH₂CH(OH)CN; hydrolysis → CH₃CH₂CH(OH)COOH (2-hydroxybutanoic acid), which has a chiral centre at C-2 (four different groups: H, OH, COOH, C₂H₅). If A = acetone: cyanohydrin hydrolysis → 2-hydroxy-2-methylpropanoic acid — the central carbon bears two identical methyl groups → no chirality. Two steps: (1) identify carbonyl from formula, (2) test chirality of hydrolysis product.
Why A is wrong: Acetone cyanohydrin on hydrolysis gives 2-hydroxy-2-methylpropanoic acid: the α-carbon has two identical CH₃ groups → no chiral centre. The question requires chirality, ruling out acetone.
Why C is wrong: Propan-1-ol (C₃H₈O) does not match the molecular formula C₃H₆O (which has one degree of unsaturation). Alcohols do not react with HCN to form cyanohydrins.
Why D is wrong: Methoxyethane (C₃H₈O) has the wrong molecular formula and is an ether — ethers do not undergo nucleophilic addition with HCN.
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Given
• Substrate: benzaldehyde (C₆H₅CHO) — an aromatic aldehyde• Nucleophiles: (a) HCN/NaCN, (b) C₆H₅MgBr then dilute H₂SO₄
Required
• Product for each reaction• Reactivity comparison with acetophenone (C₆H₅COCH₃)
Concept
Nucleophilic addition to carbonyl: Nu⁻ attacks Cδ+ → tetrahedral alkoxide → protonation. Reactivity depends on steric accessibility and electrophilicity of carbonyl carbon.
Formula/Principle
Reactivity order: Aldehyde > Ketone (less steric hindrance at carbonyl C; fewer electron-donating groups).
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)
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.
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.
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.
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.
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. 236+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 Class 12 Chemistry, Ch. 8, p. 237More in Alcohols, Phenols, Ethers and Carbonyl Compounds: 3 exam traps and mistakes · 2 formulas · 2 question patterns from its other lessons.
5 questions from NEET 2020, 2021, 2022, 2026. Answers verified against NTA official keys.
One of the products formed in the reaction shown in the figure is
+ dry HO3 RMgX+CO Y RCOOH 2 ⎯⎯⎯→ ⎯⎯⎯→ ether What is Y in the above reaction?
Reaction between acetone and methylmagnesium chloride followed by hydrolysis will give :
All 23 past-paper questions from Alcohols, Phenols, Ethers and Carbonyl Compounds →
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
inverts aldehyde ketone reactivity
Believes ketones more reactive
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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