Aldehydes, Ketones and Carboxylic Acids NEET PYQReal NEET & AIPMT 2005–2021 questions, year-tagged, with solutions
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Aldehydes, Ketones and Carboxylic Acids previous year questions
41 questions. Tap an option to check your answer.
Q1NEET 2021
What is the IUPAC name of the organic compound formed in the following chemical reaction ? Acetone —[(i) C₂H₅MgBr, dry Ether; (ii) H₂O, H⁺]→ Product
Not quite — the answer is D.
C₂H₅MgBr adds to the carbonyl carbon of acetone, and hydrolysis gives the tertiary alcohol (CH₃)₂C(OH)CH₂CH₃. Its longest chain has four carbons with OH and CH₃ on C-2: 2-methylbutan-2-ol. 2-methylpropan-2-ol would come from CH₃MgBr.
Q2NEET 2021
Match List - I with List - II. List - I | List - II (a) C₆H₆ —(CO, HCl; Anhyd. AlCl₃/CuCl)→ | (i) Hell-Volhard-Zelinsky reaction (b) R–CO–CH₃ + NaOX → | (ii) Gattermann-Koch reaction (c) R–CH₂–OH + R'COOH —(Conc. H₂SO₄)→ | (iii) Haloform reaction (d) R–CH₂COOH —((i) X₂/Red P; (ii) H₂O)→ | (iv) Esterification Choose the correct answer from the options given below.
Not quite — the answer is D.
Benzene + CO/HCl with AlCl₃/CuCl gives benzaldehyde (Gattermann-Koch). A methyl ketone with NaOX gives haloform (haloform reaction). Alcohol + carboxylic acid with conc. H₂SO₄ gives an ester (esterification). An acid with X₂/red P gives an α-halo acid (Hell-Volhard-Zelinsky).
Q3NEET 2021
The intermediate compound 'X' in the following chemical reaction is : C₆H₅–CH₃ + CrO₂Cl₂ —CS₂→ X —H₃O⁺→ C₆H₅–CHO
Not quite — the answer is A.
In the Etard reaction, chromyl chloride in CS₂ converts the methyl group of toluene into a brown chromium complex, C₆H₅CH(OCrOHCl₂)₂, which is hydrolysed to benzaldehyde. The diacetate C₆H₅CH(OCOCH₃)₂ is the intermediate of the CrO₃/acetic anhydride route, not chromyl chloride.
Q4NEET 2020 · Q107
Reaction between benzaldehyde and acetophenone in presence of dilute NaOH is known as :
Not quite — the answer is D.
Acetophenone has α-hydrogens and forms an enolate that attacks benzaldehyde, which has none; condensation between two different carbonyl compounds is a cross aldol (Claisen–Schmidt) condensation, giving benzalacetophenone. Cannizzaro reaction needs concentrated alkali and an aldehyde without α-hydrogen reacting with itself or another such aldehyde.
Q5NEET 2019 · Q78
The major product of the following reaction is: phthalic acid (benzene-1,2-dicarboxylic acid) + NH₃ —strong heating→
Not quite — the answer is B.
Phthalic acid first forms the ammonium salt, then the amide, and on strong heating the amide nitrogen attacks the second carboxyl group with loss of water to give the cyclic phthalimide. The mono-amide is only an intermediate at moderate temperature.
Q6NEET 2018 · Q67
Carboxylic acids have higher boiling points than aldehydes, ketones and even alcohols of comparable molecular mass. It is due to their
Not quite — the answer is D.
Carboxylic acids form strong intermolecular hydrogen bonds and exist as dimers even in the vapour phase, raising the boiling point. Intramolecular H-bonding would lower, not raise, the boiling point.
Q7NEET 2017 · Q44
Consider the reactions: X (C₂H₆O) —(Cu/573 K)→ A; A —([Ag(NH₃)₂]⁺, ⁻OH, Δ)→ Silver mirror observed; A —(⁻OH, Δ)→ Y; A —(NH₂–NH–CO–NH₂)→ Z. Identify A, X, Y and Z
Not quite — the answer is C.
C₂H₆O that is dehydrogenated by Cu at 573 K must be ethanol, giving ethanal as A, which gives the silver mirror with Tollens' reagent. With dilute alkali and heat ethanal undergoes aldol condensation to but-2-enal as Y, and with semicarbazide it forms the semicarbazone Z. Methoxymethane is also C₂H₆O but cannot be dehydrogenated to an aldehyde.
Q8NEET 2017 (Gujarati paper) · Q9
If the rate of the reaction: R–CO–Z + Nu⁻ → R–CO–Nu + Z⁻ is fastest, then Z is:
Not quite — the answer is A.
Nucleophilic acyl substitution is fastest when Z⁻ is the best leaving group, i.e. the weakest base. Cl⁻ (conjugate base of the strong acid HCl) leaves most easily, so acid chlorides are most reactive; acetate is next, while ethoxide and especially NH₂⁻ are poor leaving groups.
Q9NEET 2017 (Gujarati paper) · Q22
Consider the following sequence of reactions: Benzene —(CO, HCl, anhydrous AlCl₃/CuCl)→ A —('B', OH⁻, 293 K)→ C₆H₅–CH=CH–CO–C₆H₅. The substance 'B' is:
Not quite — the answer is B.
The Gattermann–Koch reaction converts benzene into benzaldehyde, the intermediate A. A base-catalysed cross aldol (Claisen–Schmidt) condensation of benzaldehyde with acetophenone, C₆H₅COCH₃, then gives benzalacetophenone, C₆H₅CH=CHCOC₆H₅. Acetone would give C₆H₅CH=CHCOCH₃, not the diphenyl product.
Q10NEET 2017 (Gujarati paper) · Q33
Among the following acids, the strongest acid is:
Not quite — the answer is C.
Electron-withdrawing groups on the α-carbon stabilise the carboxylate ion. Three fluorine atoms, the most electronegative, withdraw electrons most strongly, making trifluoroacetic acid the strongest (pKa about 0.2), ahead of trichloroacetic acid and the single-substituent acids.
Q11NEET 2016 Phase 2 · Q87
The correct structure of the product A formed in the reaction cyclohex-2-en-1-one —(H₂ (gas, 1 atmosphere), Pd/carbon, ethanol)→ A is
Not quite — the answer is B.
H₂ with Pd/C at 1 atm hydrogenates the C=C double bond of the α,β-unsaturated ketone but does not reduce the C=O group under these mild conditions, so cyclohexanone is formed. Reducing the carbonyl to an alcohol while keeping the C=C would need a hydride reagent such as NaBH₄ with CeCl₃.
Q12NEET 2016 Phase 1 · Q54
The correct statement regarding a carbonyl compound with a hydrogen atom on its alpha-carbon, is:
Not quite — the answer is D.
An α-hydrogen can shift to the carbonyl oxygen, giving an enol that is in rapid equilibrium with the keto form; this proton-shift isomerism is keto-enol tautomerism. Carbonylation means introducing a CO group, which is a different reaction altogether.
Q13NEET 2016 Phase 1 · Q83
The product formed by the reaction of an aldehyde with a primary amine is:
Not quite — the answer is A.
A primary amine adds to the carbonyl group and then loses water to form an imine, R–CH=N–R', known as a Schiff base. The reaction is a condensation, not an oxidation, so no acid or ketone is formed.
Q14AIPMT 2015
An organic compound 'X' having molecular formula C₅H₁₀O yields phenyl hydrazone and gives negative response to the Iodoform test and Tollens' test. It produces n-pentane on reduction. 'X' could be:
Not quite — the answer is D.
Forming a phenylhydrazone shows a C=O group; a negative Tollens' test rules out pentanal; a negative iodoform test rules out the CH₃CO– group of 2-pentanone. Reduction to n-pentane needs an unbranched chain, so X is 3-pentanone (CH₃CH₂COCH₂CH₃).
Q15AIPMT 2014
Which one is most reactive towards Nucleophilic addition reaction?
Not quite — the answer is C.
Aldehydes are more reactive than ketones, and an electron-withdrawing –NO₂ at the para position makes the carbonyl carbon more electrophilic. Electron-donating –CH₃ lowers reactivity, so p-nitrobenzaldehyde is most reactive.
Q16NEET 2013
Reaction by which benzaldehyde cannot be prepared
Not quite — the answer is D.
Etard reaction (toluene + CrO₂Cl₂), Rosenmund reduction (benzoyl chloride + H₂/Pd–BaSO₄) and Gattermann–Koch reaction (benzene + CO + HCl/AlCl₃) all give benzaldehyde. Zn/Hg with conc. HCl (Clemmensen conditions) does not convert benzoic acid into benzaldehyde.
Q17AIPMT 2012 Mains
Consider the following reaction: C₆H₅COCl —(H₂, Pd–BaSO₄)→ 'A' The product 'A' is
Not quite — the answer is A.
This is the Rosenmund reduction: hydrogen over poisoned Pd–BaSO₄ reduces an acid chloride only as far as the aldehyde, so benzoyl chloride gives benzaldehyde, C₆H₅CHO. The poisoned catalyst prevents further reduction to the alcohol.
Q18AIPMT 2012 Mains
Consider the reaction: RCHO + NH₂NH₂ → RCH=N–NH₂. What sort of reaction is it?
Not quite — the answer is D.
The nucleophilic nitrogen of hydrazine adds to the electrophilic carbonyl carbon, and the intermediate then loses water to form the hydrazone C=N bond. This is nucleophilic addition followed by elimination, not an electrophilic or free-radical process.
Q19AIPMT 2012 Prelims
CH₃CHO and C₆H₅CH₂CHO can be distinguished chemically by
Not quite — the answer is C.
CH₃CHO contains a CH₃CO– group and gives a yellow iodoform precipitate with I₂/NaOH, while C₆H₅CH₂CHO has no methyl ketone/aldehyde group and does not. Both are aldehydes with α-hydrogen, so both reduce Tollens' reagent and (being aliphatic-type aldehydes) Fehling's and Benedict's solutions, which therefore cannot distinguish them.
Q20AIPMT 2012 Prelims
Acetone is treated with excess of ethanol in the presence of hydrochloric acid. The product obtained is:
Not quite — the answer is A.
In dry HCl, acetone adds one ethanol to give a hemiketal, which with excess ethanol loses water and adds a second ethanol to give the ketal 2,2-diethoxypropane, (CH₃)₂C(OC₂H₅)₂. The hemiketal (CH₃)₂C(OH)(OC₂H₅) is only the intermediate, and no C–C bond forming product like a higher ketone is formed.
Q21AIPMT 2012 Prelims
Which of the following acids does not exhibit optical isomerism?
Not quite — the answer is B.
Maleic acid (cis-HOOC–CH=CH–COOH) is planar with no chiral carbon, so it shows geometrical but not optical isomerism. Lactic acid, tartaric acid and α-amino acids (except glycine) have chiral carbons.
Q22AIPMT 2012 Prelims
The correct order of decreasing acid strength of trichloroacetic acid (A), trifluoroacetic acid (B), acetic acid (C) and formic acid (D) is:
Not quite — the answer is B.
Electron-withdrawing (−I) groups stabilise the carboxylate ion. F is more electronegative than Cl, so CF₃COOH > CCl₃COOH; HCOOH is stronger than CH₃COOH because CH₃ is electron-releasing. Order: trifluoroacetic > trichloroacetic > formic > acetic.
Q23AIPMT 2012 Prelims
Predict the products in the given reaction: m-Chlorobenzaldehyde (m-Cl–C₆H₄–CHO) —(50% KOH)→ ?
Not quite — the answer is D.
m-Chlorobenzaldehyde has no α-hydrogen, so with concentrated (50%) KOH it undergoes the Cannizzaro reaction: one molecule is reduced to m-chlorobenzyl alcohol and another oxidised to m-chlorobenzoate. The aryl C–Cl bond is not hydrolysed under these conditions, so products with –OH replacing Cl are wrong.
Q24AIPMT 2011 Mains
Match the compounds given in List-I with List-II and select the suitable option using the code given below. List-I: a. Benzaldehyde, b. Phthalic anhydride, c. Phenyl benzoate, d. Methyl salicylate List-II: (i) Phenolphthalein, (ii) Benzoin condensation, (iii) Oil of wintergreen, (iv) Fries rearrangement
Not quite — the answer is B.
Benzaldehyde undergoes benzoin condensation; phthalic anhydride condenses with phenol to give phenolphthalein; phenyl benzoate undergoes Fries rearrangement to hydroxybenzophenones; methyl salicylate is oil of wintergreen. Only one code gives all four pairs correctly.
Q25AIPMT 2011 Mains
The order of reactivity of phenyl magnesium bromide (PhMgBr) with the following compounds: (I) CH₃CHO, (II) (CH₃)₂C=O and (III) Ph₂C=O is
Not quite — the answer is B.
Nucleophilic addition of PhMgBr gets slower as electron-donating and bulky groups around the carbonyl carbon increase. Acetaldehyde (one H, one CH₃) is most reactive, acetone (two CH₃) is next, and benzophenone (two bulky phenyls with resonance) is least reactive: I > II > III.
Q26AIPMT 2011 Prelims
Clemmensen reduction of a ketone is carried out in the presence of which of the following?
Not quite — the answer is C.
Clemmensen reduction converts C=O to CH₂ using zinc amalgam and concentrated HCl. Hydrazine with KOH in ethylene glycol is the Wolff–Kishner reduction, and LiAlH₄ reduces ketones only to alcohols.
Q27AIPMT 2010 Prelims · Q89
In a set of reactions, ethyl benzene yielded a product D: C₆H₅CH₂CH₃ —(KMnO₄, KOH)→ B —(Br₂, FeCl₃)→ C —(C₂H₅OH, H⁺)→ D 'D' would be
Not quite — the answer is A.
Alkaline KMnO₄ oxidises the whole ethyl side chain to –COOH, so B is benzoic acid. –COOH is meta-directing, so Br₂/FeCl₃ gives m-bromobenzoic acid as C. Esterification with ethanol/H⁺ then gives D, ethyl m-bromobenzoate. Products that keep a CH₂ side chain ignore the side-chain oxidation.
Q28AIPMT 2010 Prelims · Q96
Which of the following reactions will not result in the formation of carbon-carbon bonds?
Not quite — the answer is C.
The Cannizzaro reaction is a disproportionation of an aldehyde into an alcohol and a carboxylate by hydride transfer; no new C–C bond is formed. Friedel–Crafts acylation, Reimer–Tiemann (ring –CHO) and Wurtz coupling all make new C–C bonds.
Q29AIPMT 2010 Prelims · Q99
Among the given compounds, the most susceptible to nucleophilic attack at the carbonyl group is
Not quite — the answer is A.
Acyl chlorides are the most reactive acid derivatives: Cl is strongly electron-withdrawing and gives almost no resonance donation to C=O, and Cl⁻ is an excellent leaving group. Anhydrides, esters and amides (least) are progressively less reactive.
Q30AIPMT 2009 Prelims · Q175
Propionic acid with Br₂/P yields a dibromo product. Its structure would be:
Not quite — the answer is C.
In the Hell–Volhard–Zelinsky reaction, Br₂/P replaces the α-hydrogens (on the carbon next to –COOH). Propionic acid has two α-H, so the dibromo product is CH₃–CBr₂–COOH. Substitution at the β-carbon or on both carbons does not occur.
Q31AIPMT 2008 Prelims · Q87
A strong base can abstract an α-hydrogen from
Not quite — the answer is D.
The α-hydrogen of a ketone is acidic (pKa about 19–20), because the carbanion formed is stabilised as a resonance enolate by the C=O group. Alkanes, alkenes and amines have no such electron-withdrawing group to stabilise the carbanion.
Q32AIPMT 2008 Prelims · Q90
The relative reactivities of acyl compounds towards nucleophilic substitution are in the order of
Not quite — the answer is B.
Reactivity in nucleophilic acyl substitution follows how easily the leaving group departs: Cl⁻ > RCOO⁻ > RO⁻ > NH₂⁻, the weaker the base the better. So acyl chloride > acid anhydride > ester > amide. Placing ester above anhydride ignores that carboxylate is a better leaving group than alkoxide.
Q33AIPMT 2008 Prelims · Q97
Acetophenone when reacted with a base, C₂H₅ONa, yields a stable compound which has the structure?
Not quite — the answer is B.
Ethoxide removes an α-hydrogen from the CH₃ group, and the enolate attacks a second acetophenone (aldol addition). The β-hydroxy ketone then dehydrates to the conjugated α,β-unsaturated ketone C₆H₅C(CH₃)=CH–CO–C₆H₅, which is stabilised by conjugation. The diols are pinacol-type reduction products, not products of base-catalysed condensation.
Q34AIPMT 2007 Prelims · Q63
Consider the following compounds. (i) C₆H₅COCl (ii) O₂N–C₆H₄–COCl (para) (iii) H₃C–C₆H₄–COCl (para) (iv) OHC–C₆H₄–COCl (para) The correct decreasing order of their reactivity towards hydrolysis is
Not quite — the answer is C.
Hydrolysis is a nucleophilic attack at the carbonyl carbon, faster when electron-withdrawing groups make it more positive. –NO₂ withdraws more strongly than –CHO, H is neutral and –CH₃ donates, so (ii) > (iv) > (i) > (iii).
Q35AIPMT 2007 Prelims · Q66
Reduction of aldehydes and ketones into hydrocarbons using zinc amalgam and conc. HCl is called
Not quite — the answer is D.
Clemmensen reduction uses Zn–Hg and concentrated HCl to convert C=O into CH₂. Wolff–Kishner reduction does the same with hydrazine and KOH in ethylene glycol.
Q36AIPMT 2007 Prelims · Q68
Which of the following represents the correct order of the acidity in the given compounds?
Not quite — the answer is C.
Electron-withdrawing halogens stabilise the carboxylate ion, more so with higher electronegativity: F > Cl > Br, and unsubstituted acetic acid is weakest. So FCH₂COOH > ClCH₂COOH > BrCH₂COOH > CH₃COOH.
Q37AIPMT 2007 Prelims · Q87
Which one of the following on treatment with 50% aqueous sodium hydroxide yields the corresponding alcohol and acid?
Not quite — the answer is A.
The Cannizzaro reaction (disproportionation into alcohol and acid salt with concentrated alkali) needs an aldehyde with no α-hydrogen, such as benzaldehyde. Butanal and phenylacetaldehyde have α-H and undergo aldol condensation instead.
Q38AIPMT 2007 Prelims · Q98
The product formed in Aldol condensation is
Not quite — the answer is A.
The aldol addition of an enolisable aldehyde or ketone gives a β-hydroxy aldehyde or β-hydroxy ketone (which may then dehydrate). The OH ends up on the β-carbon, not the α-carbon, and no ester or acid is formed.
Q39AIPMT 2006 Prelims · Q71
A carbonyl compound reacts with hydrogen cyanide to form cyanohydrin which on hydrolysis forms a racemic mixture of α-hydroxy acid. The carbonyl compound is :
Not quite — the answer is A.
Acetaldehyde adds HCN to give CH₃CH(OH)CN, whose central carbon has four different groups; attack from either face is equally likely, so a racemic cyanohydrin and, on hydrolysis, racemic lactic acid result. Formaldehyde, acetone and diethyl ketone give cyanohydrins with two identical groups, which are achiral.
Q40AIPMT 2006 Prelims · Q74
Nucleophilic addition reaction will be most favoured in :
Not quite — the answer is D.
Aldehydes are more reactive than ketones towards nucleophiles because they have less steric hindrance and only one electron-releasing alkyl group. Among the aldehydes, CH₃CHO has a smaller alkyl group than CH₃CH₂CHO, so it reacts most readily. Both ketones are less reactive.
Q41AIPMT 2005 Prelims · Q58
In a set of reactions, acetic acid yielded a product D. CH₃COOH —(SOCl₂)→ A —(Benzene, Anhyd. AlCl₃)→ B —(HCN)→ C —(HOH)→ D The structure of D would be :
Not quite — the answer is A.
CH₃COOH → (SOCl₂) CH₃COCl → (C₆H₆, AlCl₃) acetophenone C₆H₅COCH₃ → (HCN) cyanohydrin C₆H₅C(OH)(CH₃)CN → (hydrolysis) C₆H₅C(OH)(CH₃)COOH. The cyanohydrin is only C, and no CH₂ is inserted between ring and carbon.
How often Aldehydes, Ketones and Carboxylic Acids came up
Questions from this chapter in each paper on this site.
- NEET 2021: 3
- NEET 2020: 1
- NEET 2019: 1
- NEET 2018: 1
- NEET 2017: 1
- NEET 2017 (Gujarati paper): 3
- NEET 2016 Phase 2: 1
- NEET 2016 Phase 1: 2
- AIPMT 2015: 1
- AIPMT 2014: 1
- NEET 2013: 1
- AIPMT 2012 Mains: 2
- AIPMT 2012 Prelims: 5
- AIPMT 2011 Mains: 2
- AIPMT 2011 Prelims: 1
- AIPMT 2010 Prelims: 3
- AIPMT 2009 Prelims: 1
- AIPMT 2008 Prelims: 3
- AIPMT 2007 Prelims: 5
- AIPMT 2006 Prelims: 2
- AIPMT 2005 Prelims: 1
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