CBSE Class 12 Chemistry HOTs Aldehyde Ketones And Carboxylic Acids Set 02

Higher Order Thinking Questions

Question 1. Assertion (A): Ethanal is more reactive than propanone towards nucleophilic addition.
Reason (R): Ketones have two electron-releasing alkyl groups and more steric hindrance at the carbonyl carbon.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
(a) Both A and R are true, and R is the correct explanation of A.
Both statements are true. The two alkyl groups reduce the positive charge on carbon and crowd the transition state, which is exactly why ketones are less reactive.
In simple words: Fewer alkyl groups mean a more positive and more open carbonyl carbon.

Teacher's Note:
1. Check each statement separately first.
2. Then ask whether R explains A.

Question 2. Assertion (A): Benzaldehyde does not give Fehling's test.
Reason (R): Benzaldehyde has no α-hydrogen atom.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
(b) Both A and R are true, but R is not the correct explanation of A.
A is true: aromatic aldehydes do not reduce Fehling's solution.
R is also true, but the absence of α-hydrogen is the condition for the Cannizzaro reaction, not the reason for the Fehling's result. So R does not explain A.
In simple words: A fact can be true and still not be the reason.

Teacher's Note:
1. Fehling's test depends on easy oxidation of CHO, not on α-hydrogen.
2. Remember: aliphatic aldehydes only.

Question 3. Assertion (A): Methanal gives the Cannizzaro reaction with concentrated alkali.
Reason (R): Methanal has no α-hydrogen atom.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
(a) Both A and R are true, and R is the correct explanation of A.
Both are true. With no α-hydrogen no enolate can form, so methanal disproportionates into methanol and sodium methanoate.
In simple words: No α-hydrogen leaves redox disproportionation as the only route.

Teacher's Note:
1. Link the reason to the mechanism.
2. Write the products to be safe.

Question 4. Assertion (A): Carboxylic acids have higher boiling points than alcohols of comparable molar mass.
Reason (R): Carboxylic acids form hydrogen-bonded dimers, giving more extensive association.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
(a) Both A and R are true, and R is the correct explanation of A.
Both are true. Two O–H···O=C hydrogen bonds join two acid molecules into a dimer, so more energy is needed to separate them.
In simple words: Acids hold together in pairs, so they boil higher than alcohols.

Teacher's Note:
1. Dimer formation is the key idea.
2. The dimer persists even in the vapour phase.

Question 5. Assertion (A): Carboxylic acids do not give the typical nucleophilic addition reactions of aldehydes and ketones.
Reason (R): The lone pair on the –OH oxygen is delocalised into the C=O group, which lowers the positive charge on the carbonyl carbon.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
(a) Both A and R are true, and R is the correct explanation of A.
Both are true. Resonance of the hydroxyl lone pair with C=O reduces the electrophilicity of the carbon, so addition is much less favourable.
In simple words: The –OH group feeds electrons to the carbonyl carbon.

Teacher's Note:
1. Resonance reduces δ+ on carbon.
2. Compare with aldehydes, which have no such donor.

Question 6. Assertion (A): Methanoic acid gives a silver mirror with Tollens' reagent.
Reason (R): Methanoic acid contains an H–C(=O)– group like an aldehyde.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
(a) Both A and R are true, and R is the correct explanation of A.
Both are true. HCOOH has a hydrogen attached to the carbonyl carbon, so it is oxidised to carbonic acid (CO₂ and H₂O) by the silver(I) complex.
In simple words: It has an aldehyde-like hydrogen, so it reduces Ag⁺.

Teacher's Note:
1. HCOOH is the only simple acid that reduces Tollens' reagent.
2. Its structure is H–COOH.

Question 7. Assertion (A): Sodium hydrogensulphite is used to separate ethanal from a mixture with hydrocarbons.
Reason (R): The bisulphite adduct of ethanal is a water-soluble ionic compound that can be changed back to ethanal by dilute acid or alkali.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
(a) Both A and R are true, and R is the correct explanation of A.
Both are true. The adduct dissolves in water, leaving the hydrocarbon in the organic layer, and ethanal is regenerated on treating the aqueous solution with acid or alkali.
In simple words: Make the aldehyde water soluble, wash away the hydrocarbon, then reverse the reaction.

Teacher's Note:
1. The reaction must be reversible.
2. Adduct formation is slow for bulky ketones.

Question 8. Which of the following give a yellow precipitate with I₂ and NaOH?
(a) Ethanol
(b) Propan-2-ol
(c) Propan-1-ol
(d) Acetophenone
Select all that apply.
Answer:
(a), (b) and (d)
Ethanol is oxidised by NaOI to ethanal (CH3CHO), propan-2-ol to propanone and acetophenone already has CH3CO–, so all three give CHI3.
Propan-1-ol is oxidised to propanal, which has no CH3CO– group, so it does not respond.
In simple words: Look for CH3CO– or CH3CH(OH)– in the starting material.

Teacher's Note:
1. Alcohols must first oxidise to a methyl carbonyl.
2. Ethanol is the only primary alcohol that gives the test.

Question 9. Explain why p-nitrobenzoic acid is a stronger acid than benzoic acid, whereas p-methylbenzoic acid is weaker.
Answer:
The –NO2 group withdraws electrons by –I and –R effects. It disperses the negative charge on the carboxylate ion and stabilises it, so the acid is stronger (the pKa falls).
The –CH3 group releases electrons (+I effect), which intensifies the negative charge on the carboxylate ion and destabilises it, so p-methylbenzoic acid is weaker than benzoic acid.
In simple words: Electron-withdrawing groups strengthen an acid and electron-donating groups weaken it.

Teacher's Note:
1. Think about the stability of the conjugate base.
2. Name the effect (–I, –R or +I).

Question 10. Semicarbazide has two –NH₂ groups, but only one of them takes part in the formation of a semicarbazone. Explain.
Answer:
Semicarbazide is NH2–NH–CO–NH2. The NH2 attached to C=O is an amide-type nitrogen: its lone pair is delocalised over the carbonyl group by resonance, so it is not nucleophilic.
The other NH2 (attached to NH) has a free lone pair and attacks the carbonyl carbon of the aldehyde or ketone, giving C=N–NH–CO–NH2.
In simple words: One nitrogen is tied up by resonance with C=O; the other is free to attack.

Teacher's Note:
1. Compare the lone pairs on the two nitrogens.
2. Resonance reduces nucleophilicity.

Question 11. Case study. A compound A (C₃H₆O) does not reduce Tollens' reagent but gives a yellow precipitate with I₂ and NaOH. A reacts with HCN to give B, and B on hydrolysis gives C. A is also converted into D by Zn–Hg and conc. HCl. Identify A, B, C and D.
Answer:
A is propanone, CH3COCH3: it is a ketone (no silver mirror) with a CH3CO– group (iodoform test).
B: (CH₃)₂C(OH)CN, 2-hydroxy-2-methylpropanenitrile (cyanohydrin).
C: (CH₃)₂C(OH)COOH, 2-hydroxy-2-methylpropanoic acid.
D: CH₃CH₂CH₃, propane, by Clemmensen reduction.
In simple words: A methyl ketone with three carbons can only be propanone.

Teacher's Note:
1. Use each clue in turn.
2. Hydrolysis changes –CN to –COOH.

Question 12. Case study. Toluene is treated with CrO₂Cl₂ in CS₂ and then hydrolysed to give A. A, on heating with concentrated NaOH, gives B (a sodium salt) and C (an alcohol). B on acidification gives D. D and C react in the presence of conc. H₂SO₄ to give E. Identify A to E.
Answer:
A = benzaldehyde, C6H5CHO (Etard reaction).
B = sodium benzoate, C₆H₅COONa, and C = benzyl alcohol, C₆H₅CH₂OH (Cannizzaro reaction).
D = benzoic acid, C₆H₅COOH.
E = benzyl benzoate, C₆H₅COOCH₂C₆H₅ (esterification: C₆H₅COOH + C₆H₅CH₂OH ⇌ C₆H₅COOCH₂C₆H₅ + H₂O).
In simple words: Etard gives the aldehyde, Cannizzaro splits it into salt and alcohol, and the acid and alcohol then make the ester.

Teacher's Note:
1. Follow the reactions one step at a time.
2. Benzaldehyde has no α-hydrogen, so the Cannizzaro reaction is possible.