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Class 12 Chemistry — questions with answers and worked explanations.
These questions are drawn from the Pariksha Sutra question bank for Aldehydes, Ketones and Carboxylic Acids, part of the Class 12 Chemistry syllabus. Each one shows the correct answer and, where a method helps, the working behind it.
Read the question, decide your answer before looking, then check the explanation — that is what turns practice into marks. If a question catches you out, the explanation is the part worth re-reading.
The longest carbon chain containing the carbonyl carbon has three carbons, so the parent hydrocarbon is propane.
The functional group is an aldehyde (–CHO), which takes the suffix “‑al” and is given the lowest possible locant, 1.
Thus the compound is named propane‑1‑al, commonly written as propanal.
Correct answer: Propanal
AlCl₃/CuCl generates the electrophile formyl cation (·CHO) from CO and HCl.
The aromatic ring undergoes electrophilic substitution with this electrophile.
Since the electrophile is a formyl group, the product is benzaldehyde.
Thus the introduced group is –CHO.
HCN adds across the carbonyl C=O of a ketone; the carbonyl carbon is electrophilic and attacks the nucleophilic cyanide ion (CN⁻).
Protonation of the resulting alkoxide gives the –OH group while the –CN remains attached to the same carbon.
The product contains both a hydroxyl and a cyano group on the former carbonyl carbon, which is the definition of a cyanohydrin.
Hence the reaction yields a cyanohydrin.
In aldol condensation a base removes an α‑hydrogen to generate an enolate ion.
The enolate then attacks the carbonyl carbon of another molecule forming the β‑hydroxy carbonyl (aldol).
If the carbonyl compound lacks an α‑hydrogen, no enolate can be formed and the reaction cannot proceed.
Hence the carbonyl must have at least one α‑hydrogen. At least one alpha-hydrogen.
Aldol condensation involves deprotonation of the α‑hydrogen of an aldehyde/ketone to form an enolate ion.
A mild, aqueous base provides the required hydroxide ions without destroying the carbonyl.
Dilute NaOH supplies OH⁻ efficiently and keeps the medium aqueous, favoring the enolate formation and subsequent C‑C coupling.
Hence the base normally used is dilute NaOH.
Acid‑sensitive groups would be destroyed by any strongly acidic conditions, so a reduction that proceeds under basic, high‑temperature conditions is required.
The Wolff‑Kishner reduction uses hydrazine and strong base (KOH) at reflux, avoiding acids altogether.
Hence it safely reduces the ketone to an alkane without harming the acid‑sensitive functionality. Wolff‑Kishner reduction.
AlCl₃ coordinates to the carbonyl oxygen of CH₃COCl, generating the acylium ion CH₃CO⁺.
The acylium ion electrophile attacks the benzene ring, forming a sigma complex that loses H⁺ to restore aromaticity.
The product retains the carbonyl group attached to the ring, giving phenyl‑acetyl (acetophenone).
Hence the reaction yields acetophenone.
Carbonyl carbon is electrophilic; its reactivity depends on how electron‑withdrawing the attached groups are.
Aldehydes have only one alkyl group, which exerts less +I effect than the two alkyl groups of a ketone, so the carbonyl is more positively polarized.
Among the given aldehydes, ethanal (CH₃CHO) has the smallest alkyl substituent, making its carbonyl carbon the most electrophilic and thus most reactive toward nucleophilic addition.
Formyl carbon in aldehydes is more electrophilic because it is attached to only one alkyl group, which has a weaker +I effect than a second alkyl group.
Hence formaldehyde (HCHO) is the most reactive toward nucleophilic addition.
In acetaldehyde (CH3CHO) the methyl group donates electron density, reducing the carbonyl’s electrophilicity relative to HCHO.
In acetone (CH3COCH3) two methyl groups donate electron density, making the carbonyl least electrophilic and thus least reactive.
Therefore the decreasing reactivity order is HCHO > CH3CHO > CH3COCH3.
In Cannizzaro reaction a strong base deprotonates the aldehyde forming a tetrahedral alkoxide intermediate. This alkoxide can transfer a hydride to another aldehyde molecule, reducing it to an alcohol. The hydride donor is therefore the tetrahedral alkoxide intermediate. The tetrahedral alkoxide intermediate.
Assign electrons to the more electronegative atom O; each C–H bond gives carbon –1, each C–C bond gives 0, and the C=O bond gives carbon +2 (since O takes both electrons).
Summing: –1 (C–H) +0 (C–C) +2 (C=O) = +1.
Thus the oxidation state of the carbonyl carbon in CH₃CHO is +1.
Electron‑donating –OCH₃ group reduces the stability of the benzoate anion by pushing electron density into the ring, decreasing acidity.
Electron‑withdrawing groups (–NO₂, –Cl) stabilize the anion by resonance/inductive effects, increasing acidity.
Hence among the para‑substituted acids, 4‑methoxybenzoic acid is the weakest acid. 4-Methoxybenzoic acid.
Identify the longest carbon chain containing the carbonyl carbon; it has three carbons → propanal.
The carbonyl carbon is at position 1, so the parent name is propanal.
A methyl group is attached to carbon 2 of this chain → 2‑methylpropanal.
Hence the IUPAC name is 2‑Methylpropanal.
Calcium acetate on dry distillation undergoes thermal decomposition: (CH3COO)2Ca → CaCO3 + CH3COCH3.
The acetate ions combine, losing CO2 and forming a carbonyl compound.
The product formed is the ketone propanone, not an aldehyde, acid or alkane.
Hence the correct answer is Propanone.
NaHSO₃ adds across the C=O bond of aldehydes and some ketones, giving a bisulphite adduct where the carbonyl carbon is attached to –SO₃⁻ and –OH.
The addition is reversible and the product precipitates as a solid crystalline bisulphite addition compound.
Thus the reaction forms a crystalline bisulphite addition product.
Ethanol under dilute NaOH forms its enolate ion (CH₃CH=O⁻).
The enolate attacks the carbonyl carbon of a second ethanal molecule, giving a β‑hydroxy aldehyde.
The product has a hydroxyl group on C‑3 and an aldehyde on C‑1, i.e., 3‑hydroxybutanal.
Hence the aldol product is 3‑Hydroxybutanal.
NaBH4 is a mild reducing agent that delivers hydride to the carbonyl carbon of a ketone.
The carbonyl C=O is converted to a C–OH group while the carbon skeleton remains unchanged.
Since the carbon bearing the carbonyl was attached to two other carbon atoms, the product is a secondary alcohol.
More electron‑withdrawing halogens increase the stability of the conjugate base by delocalising the negative charge.
Each –Cl substituent inductively pulls electron density away from the carboxylate, making the anion more stable.
With three –Cl groups, trichloroacetic acid’s conjugate base is the most stabilized, so it dissociates most readily.
Hence trichloroacetic acid is the strongest acid.
Ethanal is the IUPAC name for the aldehyde with two carbon atoms (CH₃CHO).
The common (trivial) name for a two‑carbon aldehyde is acetaldehyde.
Hence the correct answer is Acetaldehyde.
Primary alcohol is first oxidized to a carbonyl compound; PCC is a mild oxidant that stops oxidation at the aldehyde stage without further hydration. No excess water or strong oxidizer is present to convert the aldehyde to a carboxylic acid. Hence RCH₂OH → RCHO, giving an aldehyde.
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