Aldehydes, Ketones and Carboxylic Acids

What to remember

• The carbonyl carbon (C=O) is sp² hybridised and electron-poor, so aldehydes and ketones mainly undergo nucleophilic addition; reactivity is HCHO > other aldehydes > ketones.
• Only aldehydes reduce Tollens' reagent; only aliphatic aldehydes reduce Fehling's solution; methyl ketones (CH₃CO–) and CH₃CH(OH)– compounds give the iodoform test.
• Aldehydes and ketones with α-hydrogen give aldol condensation (dilute alkali); those without α-hydrogen (HCHO, C₆H₅CHO) give the Cannizzaro reaction (concentrated alkali).
• Carboxylic acids are acidic because the carboxylate ion is stabilised by resonance; electron-withdrawing groups increase acidity and electron-donating groups decrease it. Acids are stronger than phenols and alcohols, and their NaHCO₃ test gives CO₂.

Nomenclature and structure of the carbonyl group

The carbonyl group is C=O. In an aldehyde it is attached to at least one hydrogen (R–CHO). In a ketone it is attached to two carbon groups (R–CO–R′). In a carboxylic acid it is joined to an –OH group (R–COOH).

The carbonyl carbon is sp² hybridised. It forms three σ bonds in one plane with bond angles of about 120°, and one π bond with oxygen. Oxygen is more electronegative than carbon, so the C=O bond is polar: carbon carries a partial positive charge and oxygen a partial negative charge. Because of this, the carbonyl carbon is attacked by nucleophiles.

IUPAC rules: for aldehydes, replace the final –e of the alkane name by –al (the chain is numbered from the CHO carbon). For ketones, replace –e by –one and give the lowest number to the C=O carbon. For acids, replace –e by –oic acid. A benzene ring carrying –CHO is called benzenecarbaldehyde (benzaldehyde).

Preparation of aldehydes and ketones

Ozonolysis example: CH₃CH=CH₂ with O₃ then Zn/H₂O gives CH₃CHO and HCHO. A nitrile gives an aldehyde with the same number of carbon atoms: CH₃CN → CH₃CHO.

Physical properties

Methanal is a gas at room temperature. Ethanal is a volatile liquid. Other aldehydes and ketones up to about C₁₁ are liquids. Aldehydes and ketones cannot form hydrogen bonds with each other, because they have no H on oxygen. Their boiling points are therefore higher than those of hydrocarbons and ethers of similar mass (dipole–dipole attraction), but lower than those of alcohols of similar mass.

Lower aldehydes and ketones are miscible with water because they form hydrogen bonds with water. Solubility falls as the alkyl chain grows. Carboxylic acids have even higher boiling points than alcohols of similar mass, because they form two hydrogen bonds between two molecules and exist as dimers (cyclic dimers in the pure state and in non-polar solvents). Lower acids (up to C₄) are miscible with water.

Order of boiling point (similar molar mass): carboxylic acid > alcohol > aldehyde or ketone > ether > alkane.

Nucleophilic addition reactions

Mechanism: the nucleophile (Nu⁻) attacks the carbonyl carbon from outside the plane. The carbon changes from sp² to sp³ and a tetrahedral alkoxide ion forms. This alkoxide takes H⁺ from the medium to give the neutral product.

Why aldehydes are more reactive than ketones:

• Steric effect: the carbonyl carbon of an aldehyde has one small H and one alkyl group, so the nucleophile approaches easily. A ketone has two bulky alkyl groups.
• Electronic effect: two alkyl groups (+I effect) reduce the positive charge on the carbonyl carbon of a ketone more than one alkyl group does in an aldehyde.

Reactivity order: HCHO > CH₃CHO > CH₃COCH₃ > CH₃COCH₂CH₃. Aromatic aldehydes are less reactive than aliphatic aldehydes because the benzene ring donates electrons by resonance to the carbonyl carbon.

Example: CH₃CHO + CH₃MgBr, then H₃O⁺, gives CH₃CH(OH)CH₃ (propan-2-ol, a 2° alcohol).

Addition–elimination with ammonia derivatives (Z–NH₂), in weakly acidic medium, gives C=N–Z compounds with loss of water:

The 2,4-DNP test detects any aldehyde or ketone (not acids, esters, amides).

Reduction, oxidation and tests

Reduction:

• To alcohols: NaBH₄ or LiAlH₄, or H₂/Ni or Pd. Aldehydes give 1° alcohols; ketones give 2° alcohols.
• To alkanes: Clemmensen reduction uses Zn–Hg and concentrated HCl. Wolff–Kishner reduction uses NH₂NH₂ and then KOH in ethylene glycol on heating. Example: CH₃COCH₃ → CH₃CH₂CH₃. The first is done in acid; the second in base.

Oxidation: aldehydes are easily oxidised to acids, even by mild oxidants. Ketones are oxidised only by strong oxidants and with C–C bond breaking, to a mixture of acids with fewer carbons.

Benzaldehyde does not reduce Fehling's solution. Methanal has no CH₃CO– group, so it does not give the iodoform test. Ethanal is the only aldehyde that gives it.

Iodoform reaction example: CH₃COCH₃ + 3 I₂ + 4 NaOH → CHI₃ + CH₃COONa + 3 NaI + 3 H₂O. The product acid has one carbon fewer than the ketone.

Reactions due to α-hydrogen, Cannizzaro and ring substitution

α-Hydrogens (on the carbon next to C=O) are acidic. The reason is that the carbanion formed after loss of H⁺ is stabilised by resonance with the carbonyl group.

Aldol condensation: two molecules of an aldehyde or ketone with α-H react in dilute alkali. The α-carbon of one adds to the carbonyl carbon of the other.

• 2 CH₃CHO → CH₃CH(OH)CH₂CHO (3-hydroxybutanal, the aldol). On heating this loses water to give CH₃CH=CHCHO (but-2-enal).
• 2 CH₃COCH₃ → (CH₃)₂C(OH)CH₂COCH₃ (4-hydroxy-4-methylpentan-2-one), which on heating gives (CH₃)₂C=CHCOCH₃ (4-methylpent-3-en-2-one).

Cross aldol: two different carbonyl compounds, both with α-H, give a mixture of four products. If one of them has no α-H (for example benzaldehyde), a single main product is formed.

Cannizzaro reaction: aldehydes with no α-H, heated with concentrated NaOH, undergo self oxidation and reduction (disproportionation). One molecule is oxidised to the acid salt and another is reduced to the alcohol.

• 2 HCHO + NaOH (conc.) → CH₃OH + HCOONa.
• 2 C₆H₅CHO + NaOH (conc.) → C₆H₅CH₂OH + C₆H₅COONa.

Electrophilic substitution: the –CHO and –COR groups are electron-withdrawing. They deactivate the benzene ring and are meta-directing. Example: benzaldehyde with HNO₃/H₂SO₄ gives 3-nitrobenzaldehyde.

Carboxylic acids: preparation and acidity

Preparation:

• Oxidation of 1° alcohols or aldehydes: KMnO₄ in acid or alkaline medium, or K₂Cr₂O₇. Example: CH₃CH₂OH → CH₃COOH.
• Oxidation of alkylbenzenes: vigorous oxidation with KMnO₄ turns any alkyl side chain that has a benzylic H into –COOH. Toluene and ethylbenzene both give benzoic acid. Tertiary butylbenzene has no benzylic H and is not oxidised.
• Hydrolysis of nitriles and amides (acid or base): CH₃CN + 2 H₂O → CH₃COOH + NH₃ (with H⁺, as NH₄⁺).
• Grignard reagent with solid CO₂ (dry ice), then H₃O⁺: RMgX + CO₂ → RCOOMgX → RCOOH. The acid has one more carbon than RMgX.
• Hydrolysis of acyl halides, anhydrides and esters.

Physical properties: simple acids are colourless liquids with sharp smell (higher members are solids). Because of dimer formation, their molar masses measured in non-polar solvents come out double.

Acidity: RCOOH ⇌ RCOO⁻ + H⁺. Both C–O bonds of the carboxylate ion are equal in length because of resonance, so the negative charge is shared between two oxygen atoms. In phenoxide ion the charge is spread over carbon atoms too, which is less favourable, so acids are stronger than phenols. The ordering is carboxylic acids > phenols > water > alcohols.

A smaller pKa means a stronger acid. Rules:

• Electron-withdrawing groups (–F, –Cl, –NO₂, –CN) stabilise the anion, so they increase acidity. More such groups and a nearer position increase the effect. Strength falls as the group moves further from –COOH: CH₃CH₂CHClCOOH > CH₃CHClCH₂COOH > ClCH₂CH₂CH₂COOH.
• Halogen order: FCH₂COOH > ClCH₂COOH > BrCH₂COOH > ICH₂COOH (by –I effect of the halogen).
• Electron-donating groups (alkyl, –OCH₃) destabilise the anion, so they decrease acidity. Methanoic acid is stronger than ethanoic acid.
• Benzoic acid is stronger than ethanoic acid. In substituted benzoic acids, electron-withdrawing groups raise and donating groups lower acidity.

Reactions of carboxylic acids

Decarboxylation example: CH₃COONa + NaOH → CH₄ + Na₂CO₃ (CaO, heat).

Kolbe electrolysis example: 2 CH₃COONa + 2 H₂O → CH₃CH₃ + 2 CO₂ + H₂ + 2 NaOH. Ethane and CO₂ form at the anode; H₂ forms at the cathode. The alkane has twice the number of carbon atoms in R.

Hell–Volhard–Zelinsky (HVZ) reaction: carboxylic acids that have α-H react with Cl₂ or Br₂ in the presence of a little red phosphorus to give α-halo acids. Example: CH₃CH₂COOH + Br₂, red P → CH₃CHBrCOOH.

Ring substitution: the –COOH group is electron-withdrawing, so it deactivates the benzene ring and directs the incoming group to the meta position. Nitration of benzoic acid gives 3-nitrobenzoic acid. Benzoic acid does not undergo Friedel–Crafts reaction, because the ring is deactivated and AlCl₃ binds to –COOH.

NEET traps

• 1. Rosenmund reduction starts from an acyl chloride and uses a poisoned catalyst (Pd/BaSO₄); an unpoisoned catalyst would reduce the aldehyde further to an alcohol.
• 2. Stephen reaction uses a nitrile with SnCl₂/HCl; Rosenmund uses an acyl chloride with H₂. Do not swap them.
• 3. Fehling's solution is not reduced by benzaldehyde, but Tollens' reagent is.
• 4. Clemmensen is acidic (Zn–Hg/HCl); Wolff–Kishner is basic (KOH). An acid-sensitive compound needs Wolff–Kishner.
• 5. Ethanal gives the iodoform test; methanal does not. Ethanol gives it; methanol and propan-1-ol do not.
• 6. Cannizzaro needs no α-H; aldol needs α-H. Methanal and benzaldehyde never give aldol.
• 7. NaBH₄ reduces aldehydes and ketones but not –COOH; LiAlH₄ reduces –COOH.
• 8. FCH₂COOH is stronger than ClCH₂COOH, but Cl₃CCOOH is stronger than FCH₂COOH, because three halogens outweigh the single, stronger one.
• 9. Methanoic acid is more acidic than ethanoic acid, but benzoic acid (4.19) is weaker than methanoic acid (3.75).
• 10. Grignard with CO₂ adds one carbon; Kolbe doubles the alkyl group; soda lime loses one carbon.

One-liners

• 1. The carbonyl carbon is sp² hybridised with bond angle about 120°.
• 2. The reactivity order for nucleophilic addition is HCHO > CH₃CHO > CH₃COCH₃.
• 3. HCN adds to aldehydes and ketones to give cyanohydrins.
• 4. Acetals and ketals are formed in dry HCl with an alcohol or ethylene glycol.
• 5. 2,4-DNP gives a yellow, orange or red precipitate with aldehydes and ketones.
• 6. Semicarbazide gives semicarbazones; hydroxylamine gives oximes.
• 7. Etard and Gattermann–Koch reactions both give benzaldehyde from hydrocarbons.
• 8. Friedel–Crafts acylation of benzene with CH₃COCl/AlCl₃ gives acetophenone.
• 9. Dialkylcadmium with an acyl chloride gives a ketone.
• 10. Tollens' test gives a silver mirror with aldehydes.
• 11. The iodoform precipitate is yellow CHI₃.
• 12. Aldol condensation of ethanal gives 3-hydroxybutanal, then but-2-enal on heating.
• 13. Cannizzaro reaction of HCHO gives CH₃OH and HCOONa.
• 14. Carboxylic acids exist as hydrogen-bonded dimers.
• 15. Soda lime decarboxylation removes one carbon as CO₂ (as carbonate).

Practice questions

1. The carbonyl carbon in an aldehyde or ketone is hybridised as

   1. sp³d
   2. sp³
   3. sp
   4. sp²
   Answer

   D. sp²

   The carbonyl carbon forms three σ bonds and one π bond, so it is sp² with trigonal planar geometry.

2. What is the IUPAC name of CH₃COCH₂CH₃?

   1. butanone (butan-2-one)
   2. butanal
   3. propanone
   4. butan-2-ol
   Answer

   A. butanone (butan-2-one)

   A four-carbon chain with C=O at carbon 2 is a ketone: butan-2-one.

3. Which reagent converts benzoyl chloride into benzaldehyde?

   1. LiAlH₄ in ether
   2. H₂ with Pd on BaSO₄
   3. Zn–Hg and conc. HCl
   4. NaBH₄ in ethanol
   Answer

   B. H₂ with Pd on BaSO₄

   Rosenmund reduction: the poisoned catalyst stops the reduction at the aldehyde stage.

4. Toluene is treated with CrO₂Cl₂ in CS₂ and the complex is then hydrolysed. The product is

   1. benzoic acid
   2. benzyl chloride
   3. benzaldehyde
   4. benzyl alcohol
   Answer

   C. benzaldehyde

   Etard reaction: the chromium complex of the –CH₃ group hydrolyses to –CHO.

5. Which compound is the most reactive towards nucleophilic addition?

   1. CH₃COCH₃
   2. CH₃CHO
   3. C₆H₅CHO
   4. HCHO
   Answer

   D. HCHO

   Methanal has two H atoms on the carbonyl carbon: least steric hindrance and no +I groups.

6. Acetophenone is converted into ethylbenzene in strongly basic medium by

   1. Zn–Hg and conc. HCl
   2. NH₂NH₂, then KOH in ethylene glycol, heat
   3. NaBH₄ in ethanol
   4. LiAlH₄ in dry ether
   Answer

   B. NH₂NH₂, then KOH in ethylene glycol, heat

   Wolff–Kishner reduction: hydrazone formation, then KOH in ethylene glycol on heating removes N₂ to give –CH₂–.

7. Which compound reduces Fehling's solution?

   1. benzaldehyde
   2. propanone
   3. propanal
   4. acetophenone
   Answer

   C. propanal

   Aliphatic aldehydes reduce Fehling's solution to red-brown Cu₂O; aromatic aldehydes and ketones do not.

8. Which compound gives a yellow precipitate with I₂ and NaOH?

   1. pentan-2-one
   2. pentan-3-one
   3. 2-methylpropanal
   4. pentanal
   Answer

   A. pentan-2-one

   Pentan-2-one has the CH₃CO– group, so it gives CHI₃. Pentan-3-one has no methyl ketone unit.

9. Which compound undergoes the Cannizzaro reaction with concentrated NaOH?

   1. ethanal
   2. propanal
   3. benzaldehyde
   4. propanone
   Answer

   C. benzaldehyde

   Benzaldehyde has no α-hydrogen. The others have α-H and give aldol-type reactions.

10. On strong heating, the aldol obtained from two molecules of ethanal gives

    1. butanal
    2. but-2-enal
    3. butanoic acid
    4. butane-1,3-diol
    Answer

    B. but-2-enal

    3-Hydroxybutanal loses water to give the α,β-unsaturated aldehyde CH₃CH=CHCHO.

11. Benzene reacts with CO and HCl in the presence of anhydrous AlCl₃/CuCl to give

    1. benzaldehyde
    2. benzoic acid
    3. acetophenone
    4. benzophenone
    Answer

    A. benzaldehyde

    This is the Gattermann–Koch reaction, which introduces –CHO.

12. Ethanenitrile is treated with SnCl₂ and HCl and then hydrolysed. The product is

    1. CH₃CH₂NH₂
    2. CH₃COOH
    3. CH₃CH₂OH
    4. CH₃CHO
    Answer

    D. CH₃CHO

    Stephen reaction: RCN → RCH=NH → RCHO on hydrolysis.

13. Which is the strongest acid?

    1. Cl₃CCOOH
    2. CH₃COOH
    3. Cl₂CHCOOH
    4. ClCH₂COOH
    Answer

    A. Cl₃CCOOH

    Three chlorine atoms give the greatest –I effect, so the anion is most stabilised (pKa about 0.65).

14. Soda lime is heated with sodium propanoate. The organic product is

    1. butane
    2. ethane
    3. methane
    4. propane
    Answer

    B. ethane

    Decarboxylation removes one carbon as carbonate: CH₃CH₂COONa → CH₃CH₃.

15. Propanoic acid reacts with Br₂ in the presence of red phosphorus to give mainly

    1. bromoethane
    2. 3-bromopropanoic acid
    3. propanoyl bromide
    4. 2-bromopropanoic acid
    Answer

    D. 2-bromopropanoic acid

    Hell–Volhard–Zelinsky reaction replaces an α-hydrogen by halogen.

16. Electrolysis of an aqueous solution of sodium propanoate gives at the anode a hydrocarbon. It is

    1. hexane
    2. propane
    3. butane
    4. ethane
    Answer

    C. butane

    Kolbe electrolysis joins two ethyl groups: 2 CH₃CH₂COO⁻ → CH₃CH₂CH₂CH₃ + 2 CO₂.

17. Given below are two statements: Statement I: Aromatic aldehydes are less reactive than aliphatic aldehydes towards nucleophilic addition. Statement II: The benzene ring donates electrons to the carbonyl group by resonance and lowers its positive charge. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    Both are correct; resonance donation reduces the electrophilicity of the carbonyl carbon.

18. Given below are two statements: Statement I: Benzaldehyde reduces Tollens' reagent. Statement II: Benzaldehyde reduces Fehling's solution. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    Benzaldehyde gives a silver mirror but does not reduce Fehling's solution.

19. Given below are two statements: Statement I: Ethanal undergoes the Cannizzaro reaction with concentrated NaOH. Statement II: Ethanal undergoes aldol condensation with dilute NaOH. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    Ethanal has α-hydrogens, so it gives aldol condensation, not Cannizzaro.

20. Which of the following statements are correct? 1. NaBH₄ reduces a –COOH group to –CH₂OH. 2. LiAlH₄ reduces a –COOH group to –CH₂OH. 3. Clemmensen reduction uses Zn–Hg and conc. HCl.

    1. 1 and 2 only
    2. 2 and 3 only
    3. 1 and 3 only
    4. 1, 2 and 3
    Answer

    B. 2 and 3 only

    NaBH₄ does not reduce carboxylic acids; LiAlH₄ does. Clemmensen uses Zn–Hg/HCl. So 2 and 3 are correct.

21. Given below are two statements: Statement I: Carboxylic acids have higher boiling points than alcohols of comparable molar mass. Statement II: Carboxylic acids form hydrogen-bonded dimers. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    Dimer formation gives stronger intermolecular association, so the boiling point rises. Both are correct.

22. Given below are two statements: Statement I: Methanal gives a yellow precipitate with I₂ and NaOH. Statement II: Ethanal gives a yellow precipitate with I₂ and NaOH. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    Ethanal has the CH₃CO– group; methanal does not, so it gives no iodoform.

23. Given below are two statements: Statement I: Benzoic acid undergoes Friedel–Crafts acylation easily. Statement II: The –COOH group is electron-withdrawing and directs the incoming group to the meta position. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    Benzoic acid does not undergo Friedel–Crafts reactions because the ring is deactivated. Statement II is correct.

24. Which of the following statements are correct? 1. Gattermann–Koch reaction uses CO and HCl. 2. Etard reaction uses CrO₂Cl₂. 3. Stephen reaction uses H₂ and Pd on BaSO₄.

    1. 1 and 2 only
    2. 2 and 3 only
    3. 1 and 3 only
    4. 1, 2 and 3
    Answer

    A. 1 and 2 only

    Stephen reaction uses SnCl₂/HCl on a nitrile; H₂/Pd–BaSO₄ is the Rosenmund reduction. So 1 and 2 only.

25. Given below are two statements: Statement I: Acetic acid is a stronger acid than methanoic acid. Statement II: Phenol is a stronger acid than ethanoic acid. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    B. Both Statement I and Statement II are incorrect

    Methanoic acid (pKa 3.75) is stronger than ethanoic acid (4.76). Carboxylic acids are stronger than phenols. Both are incorrect.

26. Assertion (A): Aldehydes are more reactive than ketones in nucleophilic addition. Reason (R): The carbonyl carbon of a ketone carries two electron-releasing alkyl groups and is more crowded.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    A. Both A and R are true and R is the correct explanation of A

    Both +I effect and steric crowding lower ketone reactivity, so R explains A.

27. Assertion (A): Benzaldehyde does not reduce Fehling's solution. Reason (R): Benzaldehyde has no α-hydrogen.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    B. Both A and R are true but R is not the correct explanation of A

    Both are true, but absence of α-H is not the reason for the Fehling's result; it is the resonance with the ring.

28. Assertion (A): Rosenmund reduction is done with Pd that is poisoned by BaSO₄. Reason (R): Poisoning prevents further reduction of the aldehyde to an alcohol.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    A. Both A and R are true and R is the correct explanation of A

    The poisoned catalyst is less active, so the reaction stops at RCHO.

29. Assertion (A): Propanone undergoes aldol condensation in dilute alkali. Reason (R): Propanone has no α-hydrogen atoms.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    C. A is true but R is false

    Propanone has six α-hydrogens, so A is true and R is false.

30. Assertion (A): Methanoic acid is a weaker acid than ethanoic acid. Reason (R): The alkyl group in ethanoic acid has a +I effect and destabilises the carboxylate ion.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    D. A is false but R is true

    A is false: methanoic acid is stronger. R is true but explains why ethanoic acid is weaker.

31. Assertion (A): Carboxylic acids have higher boiling points than alcohols of similar molar mass. Reason (R): Carboxylic acids are more acidic than alcohols.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    B. Both A and R are true but R is not the correct explanation of A

    Both true; the reason for the high boiling point is dimer formation, not acidity.

32. Match List I with List II. List I: a. Rosenmund reduction b. Stephen reaction c. Etard reaction d. Gattermann–Koch reaction List II: i. CrO₂Cl₂ in CS₂ ii. CO, HCl, AlCl₃/CuCl iii. SnCl₂, HCl iv. H₂, Pd–BaSO₄ Choose the correct answer.

    1. a-i, b-iii, c-iv, d-ii
    2. a-iii, b-iv, c-i, d-ii
    3. a-iv, b-iii, c-i, d-ii
    4. a-iv, b-iii, c-ii, d-i
    Answer

    C. a-iv, b-iii, c-i, d-ii

    Rosenmund: H₂/Pd–BaSO₄; Stephen: SnCl₂/HCl; Etard: CrO₂Cl₂; Gattermann–Koch: CO/HCl/AlCl₃.

33. Match List I with List II. List I: a. Hydroxylamine b. Hydrazine c. Semicarbazide d. 2,4-Dinitrophenylhydrazine List II: i. hydrazone ii. oxime iii. semicarbazone iv. 2,4-dinitrophenylhydrazone Choose the correct answer.

    1. a-i, b-ii, c-iii, d-iv
    2. a-iii, b-i, c-ii, d-iv
    3. a-ii, b-i, c-iii, d-iv
    4. a-ii, b-i, c-iv, d-iii
    Answer

    C. a-ii, b-i, c-iii, d-iv

    NH₂OH gives oxime, NH₂NH₂ gives hydrazone, semicarbazide gives semicarbazone, 2,4-DNP gives its own hydrazone.

34. Match List I with List II. List I: a. Clemmensen reduction b. Wolff–Kishner reduction c. Kolbe electrolysis d. Hell–Volhard–Zelinsky reaction List II: i. electrolysis of sodium salt ii. Zn–Hg, conc. HCl iii. NH₂NH₂, KOH, glycol iv. X₂, red phosphorus Choose the correct answer.

    1. a-iii, b-ii, c-i, d-iv
    2. a-ii, b-iii, c-i, d-iv
    3. a-ii, b-iii, c-iv, d-i
    4. a-i, b-iii, c-ii, d-iv
    Answer

    B. a-ii, b-iii, c-i, d-iv

    Clemmensen: Zn–Hg/HCl; Wolff–Kishner: NH₂NH₂/KOH; Kolbe: electrolysis; HVZ: X₂/red P.

35. Match List I with List II. List I: a. Ethanal b. Propanone c. Benzaldehyde d. Methanal List II: i. gives iodoform test and reduces Fehling's solution ii. gives iodoform test but does not reduce Fehling's solution iii. reduces Tollens' reagent but not Fehling's solution iv. reduces Fehling's solution but gives no iodoform test Choose the correct answer.

    1. a-iv, b-ii, c-iii, d-i
    2. a-i, b-iii, c-ii, d-iv
    3. a-i, b-ii, c-iii, d-iv
    4. a-ii, b-i, c-iii, d-iv
    Answer

    C. a-i, b-ii, c-iii, d-iv

    Ethanal: both; propanone: iodoform only; benzaldehyde: Tollens' only; methanal: Fehling's only.

36. Match List I with List II. List I: a. HOOC–COOH b. HOOC–CH₂–COOH c. HOOC–(CH₂)₂–COOH d. HOOC–(CH₂)₄–COOH List II: i. succinic acid ii. adipic acid iii. oxalic acid iv. malonic acid Choose the correct answer.

    1. a-iii, b-iv, c-i, d-ii
    2. a-iv, b-iii, c-i, d-ii
    3. a-i, b-iv, c-iii, d-ii
    4. a-iii, b-iv, c-ii, d-i
    Answer

    A. a-iii, b-iv, c-i, d-ii

    Oxalic (C2), malonic (C3), succinic (C4), adipic (C6).

37. Match List I with List II. List I: a. Cl₃CCOOH b. ClCH₂COOH c. CH₃COOH d. HCOOH List II: i. 4.76 ii. 0.65 iii. 3.75 iv. 2.86 Choose the correct answer.

    1. a-ii, b-iv, c-iii, d-i
    2. a-iv, b-ii, c-i, d-iii
    3. a-ii, b-iv, c-i, d-iii
    4. a-ii, b-iii, c-i, d-iv
    Answer

    C. a-ii, b-iv, c-i, d-iii

    pKa values: Cl₃CCOOH 0.65, ClCH₂COOH 2.86, CH₃COOH 4.76, HCOOH 3.75.

38. A compound C₃H₆O gives a yellow precipitate with I₂/NaOH and does not give a silver mirror with Tollens' reagent. The compound is

    1. propan-1-ol
    2. propanone
    3. propanal
    4. prop-2-en-1-ol
    Answer

    B. propanone

    A methyl ketone gives iodoform and no Tollens' reaction; propanal gives Tollens' and no iodoform.

39. Benzaldehyde is heated with concentrated NaOH. The products are

    1. benzyl alcohol only
    2. sodium benzoate only
    3. toluene and sodium benzoate
    4. benzyl alcohol and sodium benzoate
    Answer

    D. benzyl alcohol and sodium benzoate

    Cannizzaro reaction: one molecule is oxidised to the acid salt, another reduced to the alcohol.

40. Ethanal is treated with CH₃MgBr in dry ether and then with dilute acid. The product is

    1. propan-2-ol
    2. propan-1-ol
    3. propanone
    4. ethanol
    Answer

    A. propan-2-ol

    An aldehyde other than methanal gives a secondary alcohol: CH₃CH(OH)CH₃.

41. Which hydrocarbon is not oxidised to a carboxylic acid on heating with alkaline KMnO₄?

    1. ethylbenzene
    2. toluene
    3. tert-butylbenzene
    4. isopropylbenzene
    Answer

    C. tert-butylbenzene

    Oxidation needs a benzylic hydrogen. tert-Butylbenzene has none.

42. An aromatic compound C₇H₆O forms a silver mirror with Tollens' reagent, does not reduce Fehling's solution and gives an orange precipitate with 2,4-DNP. It is

    1. phenol
    2. benzoic acid
    3. acetophenone
    4. benzaldehyde
    Answer

    D. benzaldehyde

    An aromatic aldehyde: positive Tollens' and 2,4-DNP, negative Fehling's. C₇H₆O fits benzaldehyde.

43. To convert CH₃MgBr into ethanoic acid, it is treated with

    1. HCHO, then dilute acid
    2. CH₃CHO, then dilute acid
    3. O₂ only
    4. solid CO₂, then dilute acid
    Answer

    D. solid CO₂, then dilute acid

    CH₃MgBr + CO₂ gives CH₃COOMgBr, which on acidification gives CH₃COOH.

44. Ethanoic acid and ethanol are heated with a few drops of conc. H₂SO₄. The product and the role of the acid are

    1. diethyl ether; reducing agent
    2. ethyl ethanoate; catalyst and dehydrating agent
    3. ethanoyl chloride; chlorinating agent
    4. ethyl ethanoate; oxidising agent
    Answer

    B. ethyl ethanoate; catalyst and dehydrating agent

    Fischer esterification: H₂SO₄ catalyses the reversible reaction and removes water.