Board Previous Year Questions

How to Score Full Marks in the Board Exam

A complete bank of board-style questions with model answers for Aldehydes, Ketones and Carboxylic Acids. In the exam: write the IUPAC name carefully, draw clear structures with reagents over the arrow, name the mechanism (nucleophilic addition, aldol, Cannizzaro), and give the NCERT-canonical reason (carboxylate resonance for acidity; H-bonded dimer for boiling point; no alpha-H for Cannizzaro). For distinguishing tests, always state the reagent and the observation for each compound.

1-Mark Questions (Definitions & Direct)

Q1. Write the IUPAC name of CH$_3$COCH$_2$CH$_3$. Answer: Butan-2-one.

Q2. What is Tollens' reagent? Answer: Ammoniacal silver nitrate, $[Ag(NH3)2]^+$, used to detect aldehydes (gives a silver mirror).

Q3. Name the reaction: $2\,C6H5CHO + \text{conc. NaOH} \rightarrow C6H5CH2OH + C6H5COONa$. Answer: Cannizzaro reaction.

Q4. Which gives a positive iodoform test, propanal or propanone? Answer: Propanone (it has a $CH3CO-$ group).

Q5. Name the product of Rosenmund reduction of $CH3COCl$. Answer: Ethanal (acetaldehyde).

Q6. Why does benzaldehyde not respond to Fehling's test? Answer: Fehling's solution oxidises aliphatic aldehydes readily, but aromatic aldehydes such as benzaldehyde generally do not give this test under standard conditions.

2-Mark Questions

Q7. Why are the boiling points of aldehydes and ketones higher than those of hydrocarbons but lower than those of alcohols of comparable mass? Answer: Aldehydes and ketones are polar and attract one another by dipole-dipole forces, which are stronger than the weak forces in hydrocarbons. However, they have no O-H bond and so cannot form intermolecular hydrogen bonds (unlike alcohols), so their boiling points are lower than those of alcohols.

Q8. Why are aldehydes more reactive than ketones towards nucleophilic addition? Answer: In ketones the two alkyl groups cause steric hindrance to the approaching nucleophile and donate electron density by the $+I$ effect, reducing the positive charge on the carbonyl carbon. Aldehydes have only one such group, so their carbonyl carbon is more electrophilic and less hindered.

Q9. How will you distinguish between acetaldehyde and acetone? Answer: Add Tollens' reagent: acetaldehyde (an aldehyde) gives a silver mirror; acetone (a ketone) gives no reaction. Fehling's solution can also distinguish them.

Q10. Why is the alpha-hydrogen of an aldehyde acidic? Answer: Removing the alpha-hydrogen gives an enolate ion whose negative charge is delocalised onto the electronegative carbonyl oxygen by resonance, which stabilises it; hence the alpha-hydrogen is weakly acidic.

2-Mark Questions (continued)

Q11. Why is a carboxylic acid more acidic than a phenol? Answer: The carboxylate ion formed from a carboxylic acid is stabilised by resonance in which the negative charge is shared equally between two oxygen atoms. In the phenoxide ion the charge is delocalised over oxygen and the ring carbons, which is less effective. The more stable carboxylate makes the acid stronger.

Q12. Arrange in increasing acidity: acetic acid, chloroacetic acid, trichloroacetic acid. Answer: Acetic acid < chloroacetic acid < trichloroacetic acid. Each electron-withdrawing chlorine stabilises the carboxylate ion, so more chlorines make a stronger acid.

Q13. What happens when ethanal is treated with dilute NaOH? Name the reaction. Answer: Two molecules of ethanal undergo aldol condensation to give 3-hydroxybutanal (aldol), which on heating loses water to give but-2-enal.

Q14. How is benzaldehyde converted to cinnamaldehyde? Answer: By a crossed aldol condensation with acetaldehyde in the presence of dilute alkali; since benzaldehyde has no alpha-hydrogen it acts as the carbonyl component, giving $C6H5CH=CHCHO$.

3-Mark Questions (Conversions)

Q15. How will you convert (i) ethanol to ethanal, (ii) ethanal to ethanoic acid? Answer: (i) Oxidise ethanol with PCC (a mild oxidant) to stop at ethanal. (ii) Oxidise ethanal with acidified $KMnO4$ or $K2Cr2O7$ to ethanoic acid.

Q16. How will you convert (i) toluene to benzaldehyde, (ii) benzene to acetophenone? Answer: (i) Treat toluene with chromyl chloride ($CrO2Cl2$) and hydrolyse the product (Etard reaction) to give benzaldehyde. (ii) React benzene with acetyl chloride and anhydrous $AlCl3$ (Friedel-Crafts acylation) to give acetophenone.

Q17. How will you convert (i) propan-1-ol to propanoic acid, (ii) ethyl bromide to propanoic acid? Answer: (i) Oxidise propan-1-ol with acidified $KMnO4$ to propanoic acid. (ii) React ethyl bromide with KCN to give propanenitrile, then hydrolyse it to propanoic acid (this adds one carbon).

3-Mark Questions (Reasoning & Mechanism)

Q18. Write the mechanism of the acid-catalysed esterification of ethanoic acid with ethanol. Answer: The carbonyl oxygen of the acid is protonated, making the carbon more electrophilic. The alcohol oxygen attacks this carbon to give a tetrahedral intermediate. Proton transfers and loss of a water molecule, followed by loss of the catalyst proton, give the ester (ethyl ethanoate). The reaction is reversible.

Q19. Explain why formaldehyde undergoes the Cannizzaro reaction but acetaldehyde does not. Answer: The Cannizzaro reaction needs an aldehyde with no alpha-hydrogen. Formaldehyde (HCHO) has no alpha-hydrogen, so in concentrated alkali it disproportionates to methanol and sodium formate. Acetaldehyde has alpha-hydrogens and instead undergoes aldol condensation.

Q20. Give reasons: (a) Aldehydes are more reactive than ketones; (b) Carboxylic acids do not give the characteristic reactions of the carbonyl group. Answer: (a) Ketones have two electron-donating, bulky alkyl groups that lower the carbonyl carbon's positive charge and hinder nucleophilic attack, so aldehydes are more reactive. (b) In $-COOH$ the lone pair of the $-OH$ oxygen is delocalised into the carbonyl, reducing the carbonyl carbon's electrophilic character, so carboxylic acids do not undergo the typical nucleophilic-addition reactions of aldehydes and ketones.

3-Mark Questions (Tests & Named Reactions)

Q21. Give simple chemical tests to distinguish between: (i) pentan-2-one and pentan-3-one; (ii) benzaldehyde and acetophenone. Answer: (i) Iodoform test: pentan-2-one ($CH3CO-$) gives a yellow precipitate of iodoform; pentan-3-one does not. (ii) Tollens' test: benzaldehyde (an aldehyde) gives a silver mirror; acetophenone (a ketone) does not.

Q22. What are the Clemmensen and Wolff-Kishner reductions? Give the reagents. Answer: Both reduce a carbonyl group ($C=O$) all the way to a $CH2$ group. Clemmensen reduction uses zinc amalgam ($Zn-Hg$) and concentrated HCl (acidic conditions). Wolff-Kishner reduction uses hydrazine ($NH2NH2$) followed by KOH on heating (basic conditions).

Q23. Describe the Hell-Volhard-Zelinsky reaction with an example. Answer: A carboxylic acid having an alpha-hydrogen reacts with chlorine or bromine in the presence of a small amount of red phosphorus to give an alpha-halocarboxylic acid. For example, acetic acid gives bromoacetic acid or chloroacetic acid depending on whether bromine or chlorine is used.

5-Mark Questions (Long Answer)

Q24. (a) Write the mechanism of nucleophilic addition to a carbonyl group. (b) Give the order of reactivity of HCHO, CH$_3$CHO and CH$_3$COCH$_3$ and justify it. Answer: (a) The nucleophile attacks the electrophilic ($\delta^+$) carbonyl carbon; the pi electrons shift to oxygen, giving a tetrahedral alkoxide intermediate; protonation of the alkoxide oxygen gives the neutral addition product. The carbon changes from sp$^2$ to sp$^3$. (b) Reactivity order: HCHO > CH$_3$CHO > CH$_3$COCH$_3$. Increasing number of alkyl groups increases steric hindrance and electron donation by the $+I$ effect, reducing the positive charge on the carbonyl carbon, so reactivity decreases.

Q25. Describe the preparation of benzaldehyde by (i) the Etard reaction, (ii) the Gattermann-Koch reaction, and (iii) Rosenmund reduction. Give reagents for each. Answer: (i) Etard: toluene is treated with chromyl chloride ($CrO2Cl2$); the chromium complex is hydrolysed to benzaldehyde. (ii) Gattermann-Koch: benzene is treated with carbon monoxide and hydrogen chloride in the presence of anhydrous $AlCl3$ (or CuCl) to give benzaldehyde. (iii) Rosenmund: benzoyl chloride is hydrogenated over palladium on barium sulphate (poisoned catalyst), which stops the reduction at benzaldehyde.

5-Mark Questions (continued)

Q26. (a) Explain the effect of electron-withdrawing and electron-donating substituents on the acidity of substituted benzoic acids with examples. (b) Which is more acidic: benzoic acid or ethanoic acid? Answer: (a) Electron-withdrawing groups (e.g. $-NO2$) stabilise the carboxylate ion and increase acidity, so 4-nitrobenzoic acid is more acidic than benzoic acid. Electron-donating groups (e.g. $-OCH3$, $-CH3$) destabilise the carboxylate and decrease acidity, so 4-methoxybenzoic acid is less acidic than benzoic acid. (b) Benzoic acid is slightly more acidic than ethanoic acid, because the phenyl group is mildly electron-withdrawing relative to the methyl group, which is electron-donating.

Q27. How are the following carried out? (i) Aldol condensation of acetaldehyde (ii) Cannizzaro reaction of benzaldehyde (iii) Decarboxylation of sodium acetate. Answer: (i) Acetaldehyde is treated with dilute NaOH; two molecules combine to give 3-hydroxybutanal, which on heating loses water to give but-2-enal. (ii) Benzaldehyde is warmed with concentrated NaOH; it disproportionates into benzyl alcohol and sodium benzoate. (iii) Sodium acetate is heated with soda lime (NaOH + CaO); it loses $CO2$ to give methane.

Quick-Fire Board Favourites

Q28. Why do carboxylic acids have higher boiling points than alcohols of comparable molecular mass? Answer: Carboxylic acids form hydrogen-bonded cyclic dimers (two O-H…O bonds per pair), giving stronger and more extensive association than the single hydrogen bonds in alcohols, so they boil higher.

Q29. Name the reagent used to convert a carboxylic acid into an acyl chloride. Answer: $SOCl2$ (thionyl chloride); $PCl3$ or $PCl5$ can also be used.

Q30. Why is chloroacetic acid stronger than acetic acid? Answer: The electron-withdrawing chlorine stabilises the carboxylate ion by its $-I$ effect, so chloroacetic acid loses its proton more readily and is the stronger acid.

Q31. Which reagent reduces a carboxylic acid to a primary alcohol? Answer: Lithium aluminium hydride ($LiAlH4$). ($NaBH4$ does not reduce carboxylic acids.)

Q32. What is formalin? Answer: A 40% aqueous solution of methanal (formaldehyde), used as a preservative and disinfectant.

More Board Favourites

Q33. How will you bring about the conversion of acetic acid to methane? Answer: Convert acetic acid to sodium acetate (with NaOH), then heat with soda lime; decarboxylation gives methane.

Q34. Why does benzoic acid undergo electrophilic substitution at the meta position? Answer: The $-COOH$ group is electron-withdrawing (deactivating) and destabilises the intermediate for ortho/para attack, so the electrophile enters mainly at the meta position.

Q35. Give the products and name the reaction: $CH3CH2CN + SnCl2/HCl$, then $H2O$. Answer: Propanal ($CH3CH2CHO$); the reaction is the Stephen reaction (reduction of a nitrile to an aldehyde).

Q36. State two uses of acetone and one use of benzoic acid. Answer: Acetone is used as an organic solvent (e.g. for paints/varnishes and as nail-polish remover) and as a chemical feedstock. Benzoic acid (as sodium benzoate) is used as a food preservative.