CBSE Class 12 Chemistry Aldehydes Ketones And Carboxylic Acids Question Bank Set 06

Short and Long Answer Questions

Question 1. Write the IUPAC names of: (i) CH₃CH(CH₃)CH₂CHO, (ii) CH₃CH(CH₃)COCH₂CH₃, (iii) C₆H₅CH₂CHO, (iv) CH₃CH=CHCOCH₃.
Answer:
(i) 3-Methylbutanal: the chain from CHO has four carbons and the methyl group is on C-3.
(ii) 2-Methylpentan-3-one: five-carbon chain with C=O on C-3 and a methyl group on C-2.
(iii) 2-Phenylethanal: CHO is C-1 and the phenyl group is on C-2.
(iv) Pent-3-en-2-one: C=O on C-2 (lowest locant) and the C=C between C-3 and C-4.
In simple words: Number from the carbonyl end, then name the substituents.

Teacher's Note:
1. Aldehydes always start at C-1.
2. In ketones, give C=O the lowest number first.

Question 2. Write the structural formulae of: (i) 3-methylbutanal, (ii) 4-chloropentan-2-one, (iii) 4-methylbenzaldehyde.
Answer:
(i) (CH₃)₂CHCH₂CHO
(ii) CH₃COCH₂CH(Cl)CH₃
(iii) CH₃–C₆H₄–CHO with the CH₃ and CHO groups para to each other (positions 1 and 4 of the ring).
In simple words: Build the chain from the name, then place the substituents on the numbered carbons.

Teacher's Note:
1. Count the parent chain first.
2. Check that each carbon has four bonds.

Question 3. Why are the boiling points of aldehydes and ketones higher than those of alkanes of comparable molar mass but lower than those of alcohols?
Answer:
1. The C=O bond is polar, so carbonyl molecules attract each other by dipole–dipole forces. Alkanes are non-polar and have only weak van der Waals forces, so aldehydes and ketones boil higher.
2. Alcohols have O–H bonds and form strong intermolecular hydrogen bonds. Aldehydes and ketones have no O–H or N–H bond and cannot hydrogen-bond with each other.
3. Example: butane 273 K, propanal 322 K, propanone 329 K, propan-1-ol 370 K.
In simple words: Dipole attraction is stronger than van der Waals attraction but weaker than hydrogen bonding.

Teacher's Note:
1. Compare the type of intermolecular force.
2. Quote the order alcohol > ketone > aldehyde > alkane.

Question 4. Explain why ethanal is more reactive than propanone towards nucleophilic addition.
Answer:
1. In propanone the carbonyl carbon carries two electron-releasing methyl groups (+I effect), which reduce its positive charge. In ethanal there is only one methyl group, so the carbon is more positive.
2. Steric factor: the tetrahedral intermediate from propanone has two bulky groups on the carbon and is more crowded than that from ethanal, so attack by the nucleophile is slower.
In simple words: Ethanal has less crowding and a more positive carbon.

Teacher's Note:
1. Mention both electronic and steric reasons.
2. The same logic puts methanal at the top.

Question 5. Write chemical equations for the preparation of (i) ethanal from ethanoyl chloride, (ii) benzaldehyde from toluene by the Etard reaction, (iii) acetophenone from benzene.
Answer:
(i) CH₃COCl + H₂ —(Pd/BaSO₄)→ CH₃CHO + HCl (Rosenmund reduction)
(ii) C₆H₅CH₃ + CrO₂Cl₂ —(CS₂)→ chromium complex —(H₃O⁺)→ C₆H₅CHO
(iii) C₆H₆ + CH₃COCl —(anhyd. AlCl₃)→ C₆H₅COCH₃ + HCl (Friedel–Crafts acylation)
In simple words: Each named reaction uses its own reagent: Pd/BaSO4, CrO2Cl2 or AlCl3.

Teacher's Note:
1. Write the catalyst over the arrow.
2. Check that the equation is balanced.

Question 6. How will you convert ethanol into 3-hydroxybutanal? Write the reactions.
Answer:
Step 1: oxidise ethanol to ethanal with PCC: CH₃CH₂OH —(PCC)→ CH₃CHO.
Step 2: aldol addition with dilute alkali: 2CH₃CHO —(dil. NaOH)→ CH₃CH(OH)CH₂CHO (3-hydroxybutanal).
A mild oxidant is needed in step 1 so that the aldehyde is not oxidised on to ethanoic acid.
In simple words: First make the aldehyde, then let two molecules join by aldol addition.

Teacher's Note:
1. Aldol needs an α-hydrogen.
2. Do not heat the aldol if you want to keep the –OH.

Question 7. Give one chemical test to distinguish between each pair: (i) ethanal and propanone, (ii) benzaldehyde and acetophenone, (iii) pentan-2-one and pentan-3-one, (iv) benzoic acid and phenol.
Answer:
(i) Tollens' test: ethanal gives a silver mirror, propanone does not.
(ii) Iodoform test (I₂/NaOH): acetophenone, a methyl ketone, gives yellow CHI₃; benzaldehyde does not.
(iii) Iodoform test: pentan-2-one (CH₃CO–) gives yellow CHI₃; pentan-3-one does not.
(iv) NaHCO₃ test: benzoic acid gives effervescence of CO₂; phenol does not.
In simple words: Look for a methyl ketone, an aldehyde group or a carboxyl group.

Teacher's Note:
1. Name the reagent and the visible change.
2. Write the answer for both compounds of the pair.

Question 8. Write the Cannizzaro reaction of benzaldehyde and state why ethanal does not give it.
Answer:
2C₆H₅CHO + conc. NaOH → C₆H₅CH₂OH + C₆H₅COONa
One molecule is oxidised to sodium benzoate and the other is reduced to benzyl alcohol, so it is a disproportionation.
Ethanal has α-hydrogens, so with alkali it undergoes aldol condensation rather than the Cannizzaro reaction.
In simple words: No α-hydrogen means the aldehyde can only oxidise and reduce itself.

Teacher's Note:
1. Cannizzaro needs conc. alkali.
2. Ketones never give this reaction.

Question 9. What is aldol condensation? Write the reaction of ethanal with dilute NaOH and the product obtained on heating.
Answer:
Aldehydes and ketones having at least one α-hydrogen react in dilute alkali to give β-hydroxy aldehydes (aldols) or β-hydroxy ketones (ketols); on heating these lose water to give α,β-unsaturated carbonyl compounds.
2CH₃CHO —(dil. NaOH)→ CH₃CH(OH)CH₂CHO (3-hydroxybutanal) —(heat, –H₂O)→ CH₃CH=CHCHO (but-2-enal)
In simple words: Two carbonyl molecules join through the α-carbon and may then lose water.

Teacher's Note:
1. α-Hydrogen is essential.
2. Name the aldol and the final unsaturated aldehyde.

Question 10. Differentiate between Clemmensen reduction and Wolff–Kishner reduction. Write the reaction of propanone in each case.
Answer:
Both convert C=O into CH2. Clemmensen uses Zn–Hg and conc. HCl (acidic medium), and Wolff–Kishner uses NH2NH2 then KOH in ethylene glycol on heating (basic medium).
CH₃COCH₃ —(Zn–Hg, conc. HCl)→ CH₃CH₂CH₃ + H₂O
CH₃COCH₃ --(i) NH₂NH₂, (ii) KOH/glycol, heat-→ CH₃CH₂CH₃ + N₂
Use Clemmensen for acid-stable compounds and Wolff–Kishner for base-stable compounds.
In simple words: Same result, but one works in acid and the other in base.

Teacher's Note:
1. Mention the medium.
2. Both products are propane.

Question 11. Arrange in increasing order of acid strength and give reasons: ClCH₂COOH, CH₃COOH, Cl₂CHCOOH, Cl₃CCOOH.
Answer:
Order: CH3COOH < ClCH2COOH < Cl2CHCOOH < Cl3CCOOH (pKa 4.76, 2.86, 1.29, 0.65).
Chlorine withdraws electrons (–I effect). It disperses the negative charge on the carboxylate ion and stabilises it. Each extra chlorine adds to this stabilisation, so acidity increases.
In simple words: More electron-withdrawing chlorine atoms give a more stable anion and a stronger acid.

Teacher's Note:
1. Quote the pKa values to show you know the trend.
2. Methyl group does the opposite: it weakens acidity.

Question 12. 6.0 g of ethanoic acid is esterified with excess ethanol. If the yield is 70%, calculate the mass of ethyl ethanoate obtained. (C = 12.011, H = 1.008, O = 15.999)
Answer:
CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O
Molar mass of CH₃COOH = 60.05 g/mol and of CH₃COOC₂H₅ = 88.11 g/mol.
Moles of acid = 6.0 / 60.05 = 0.0999 mol. At 100% yield the ester is 8.80 g.
Actual mass = 0.70 × 8.80 = 6.16 g (about 6.2 g).
In simple words: One mole of acid gives one mole of ester, then multiply by the yield.

Teacher's Note:
1. Use the 1:1 mole ratio.
2. Apply the percentage yield at the end.

Question 13. 2.2 g of ethanal is added to excess Tollens' reagent. Calculate the mass of silver deposited. (Ag = 107.868, C = 12.011, H = 1.008, O = 15.999)
Answer:
CH₃CHO + 2[Ag(NH₃)₂]⁺ + 3OH⁻ → CH₃COO⁻ + 2Ag + 4NH₃ + 2H₂O
Moles of ethanal = 2.2 / 44.05 = 0.0499 mol.
One mole of ethanal gives 2 mol Ag, so moles of Ag = 0.0999 mol.
Mass of Ag = 0.0999 × 107.868 = 10.77 g (about 10.8 g).
In simple words: Each aldehyde molecule reduces two silver ions.

Teacher's Note:
1. Balance the equation first to get the 1:2 ratio.
2. Convert moles of Ag to grams with the atomic mass.

Question 14. Write the reactions of propanoic acid with (i) PCl₅, (ii) NaHCO₃, (iii) LiAlH₄ followed by water, (iv) Br₂ and red phosphorus followed by water.
Answer:
(i) CH₃CH₂COOH + PCl₅ → CH₃CH₂COCl + POCl₃ + HCl
(ii) CH₃CH₂COOH + NaHCO₃ → CH₃CH₂COONa + H₂O + CO₂
(iii) CH₃CH₂COOH --(i) LiAlH₄, (ii) H₂O-→ CH₃CH₂CH₂OH (propan-1-ol)
(iv) CH₃CH₂COOH --(i) Br₂/red P, (ii) H₂O-→ CH₃CHBrCOOH (2-bromopropanoic acid), the HVZ reaction
In simple words: The acid can lose H⁺, swap OH for Cl, be reduced, or be halogenated at the α-carbon.

Teacher's Note:
1. Name the product in each part.
2. Remember only LiAlH4, not NaBH4, reduces –COOH.