CBSE Class 12 Chemistry Aldehydes Ketones and Carboxylic Acids MCQs Set 14

Answer: (c) C₆H₅CHO

Explanation:
1. C₆H₅CH₃ (toluene) →^{KMnO₄, KOH, Δ} C₆H₅COOH (A) →^SOCl₂ C₆H₅COCl (B)
2. C₆H₅COCl (B) →^{H₂/Pd-BaSO₄} C₆H₅CHO (Rosenmund reaction)

Teacher's Note:
1. Pd poisoned with BaSO₄ stops the reduction of an acid chloride at the aldehyde stage.

Answer: (b) C₆H₅COOH

Explanation:
1. C₆H₅MgBr + CO₂ → C₆H₅COOMgBr →^H₃O+ C₆H₅COOH + Mg(OH)Br

Teacher's Note:
1. Grignard reagent + CO₂ (dry ice) then hydrolysis always gives an acid with one carbon more than the alkyl/aryl group.

Answer: (c) NaOH, I₂/H⁺

Explanation:
1. CH₃COCH₂CH₃ (butan-2-one) →^{(i) NaOH/I₂, (ii) H+} CHI₃ + CH₃CH₂COOH (iodoform reaction)

Teacher's Note:
1. Iodoform reaction converts a methyl ketone into an acid with one carbon less; ketones do not respond to Tollens’ or Fehling’s reagents.

Answer: (b) 3 > 1 > 2

Explanation:
1. Carboxylic acids form strong intermolecular H-bonds (dimers), so the acid boils highest; the alcohol is H-bonded but less strongly; the aldehyde has only dipole-dipole attractions.

Teacher's Note:
1. For comparable molecular masses: carboxylic acids > alcohols > aldehydes/ketones > ethers/alkanes in boiling point.

Answer: (c) D < C < B < A

Explanation:
1. Reactivity towards nucleophilic addition decreases as the +I/+R effect and steric hindrance of the groups on the carbonyl carbon increase: HCHO > CH₃COCH₃ > PhCOCH₃ > PhCOPh.
2. So the increasing order is D < C < B < A.

Teacher's Note:
1. Phenyl groups lower reactivity both by resonance (+R) and by bulk, so benzophenone is the least reactive.

Answer: (b) p-O₂N–C₆H₄–CHO

Explanation:
1. Due to −R-effect of the NO₂, the +ve charge on carbonyl carbon increases and hence p-nitrobenzaldehyde is most reactive.

Teacher's Note:
1. Electron-withdrawing groups on the ring increase, and electron-donating groups (like CH₃) decrease, the reactivity of the C=O group towards nucleophiles.

Answer: (d) NH₂NH₂ in the presence of feebly acidic solution

Explanation:
1. >C=O + H₂N–NH₂ → [>C(OH)–NH–NH₂] → >C=N–NH₂ (hydrazone) + H₂O
2. HCN, RMgX and NaHSO₃ give simple addition products without loss of water.

Teacher's Note:
1. Ammonia derivatives (NH₂–Z) react by addition-elimination; the weakly acidic medium activates the C=O without protonating the nucleophile completely.

Answer: (c) a beta hydroxyaldehyde or ketone

Explanation:
1. Aldehydes/ketones with α-H in dilute alkali give β-hydroxy aldehydes (aldols) or β-hydroxy ketones (ketols), e.g. 2CH₃CHO → CH₃CH(OH)CH₂CHO.

Teacher's Note:
1. On heating, the aldol loses water to give the α,β-unsaturated aldehyde or ketone – note it is not an ester.

Answer: (c) formaldehyde

Explanation:
1. Formaldehyde has no α-H and undergoes Cannizzaro reaction with conc. NaOH: 2HCHO → CH₃OH + HCOONa.
2. One molecule is oxidised (to formate) and another is reduced (to methanol).

Teacher's Note:
1. Self-oxidation-reduction (disproportionation) of aldehydes is the Cannizzaro reaction; it needs an aldehyde without α-hydrogen.

Answer: (b) acetaldehyde

Explanation:
1. Acetaldehyde (CH₃CHO) has α-hydrogens, so with alkali it undergoes aldol condensation instead of Cannizzaro reaction.
2. HCHO, C₆H₅CHO and (CH₃)₃CCHO have no α-H and give Cannizzaro reaction.

Teacher's Note:
1. Check the carbon next to –CHO: if it carries H, expect aldol; if it has no H, expect Cannizzaro.

Answer: (a) Ethanal

Explanation:
1. Ethanal (CH₃CHO) has a CH₃CO– group, so it gives iodoform, and being an aliphatic aldehyde it reduces Fehling’s solution.
2. Propanone and ethanol give only the iodoform test; methanol gives neither.

Teacher's Note:
1. Ethanal is the only aldehyde that gives the iodoform test.

Answer: (d) I₂/NaOH

Explanation:
1. Both CH₃CHO and CH₃COCH₃ contain the CH₃CO– group and give iodoform with I₂/NaOH; acetone does not react with Tollens’ or Fehling’s reagent.

Teacher's Note:
1. Tollens’ and Fehling’s tests distinguish aldehydes from ketones; the iodoform test detects the CH₃CO– (or CH₃CH(OH)–) group.

Answer: (b) Iodoform test

Explanation:
1. Methyl ketones (R–CO–CH₃) give a yellow precipitate of iodoform with I₂/NaOH; the other three reagents are tests for aldehydes.

Teacher's Note:
1. Yellow ppt. of CHI₃ with characteristic smell confirms a CH₃CO– group.

Answer: (b) Reacts with Tollens’ reagent to form silver mirror

Explanation:
1. Acetophenone (C₆H₅COCH₃) is a ketone and does not reduce Tollens’ reagent.
2. Being a carbonyl compound it forms a 2,4-DNP derivative, being a methyl ketone it gives iodoform, and it is oxidised to benzoic acid.

Teacher's Note:
1. Only aldehydes give the silver mirror test; ketones need strong oxidants and then break C–C bonds.

Answer: (d) I₂/NaOH

Explanation:
1. Acetophenone (C₆H₅COCH₃) is a methyl ketone and gives yellow ppt. of iodoform with I₂/NaOH; benzophenone (C₆H₅COC₆H₅) does not.
2. Both react with 2,4-DNP; neither reacts with Benedict’s reagent or (being aromatic ketones) appreciably with NaHSO₃.

Teacher's Note:
1. A distinguishing test must give a positive result with only one of the two compounds.

Answer: (d) Benzoic acid.

Explanation:
1. C₆H₅CH₃ →^{Cl₂, hν, heat} C₆H₅CCl₃ (side-chain chlorination) →^{aq. NaOH} [C₆H₅C(OH)₃] → C₆H₅COOH (as its salt; benzoic acid on acidification)

Teacher's Note:
1. Light and heat lead to side-chain substitution; ring chlorination needs a Lewis acid such as FeCl₃ in the dark.

Answer: (d) bicarbonate.

Explanation:
1. The acid transfers its proton to HCO₃⁻: CH₃CH₂COOH + NaHCO₃ → CH₃CH₂COO⁻Na⁺ + [HO–CO–OH] (carbonic acid, unstable)
2. H₂CO₃ → CO₂ + H₂O, so the carbon of CO₂ comes from the bicarbonate.

Teacher's Note:
1. This is a simple acid-base reaction; the carbon skeleton of the acid stays intact as the carboxylate ion.

Answer: (c) D < A < B < C

Explanation:
1. Electron-donating alkyl groups (+I) decrease acidity: (CH₃)₂CHCOOH < CH₃COOH.
2. The –OCH₃ group exerts −I effect and three F atoms exert a very strong −I effect, so MeOCH₂COOH < CF₃COOH. Order: D < A < B < C.

Teacher's Note:
1. Across a series like this, compare inductive effects only: more and stronger −I groups close to –COOH mean a stronger acid.

Answer: (c) B > A > D > C

Explanation:
1. F is more electronegative than Cl, so CF₃COOH > CCl₃COOH.
2. Formic acid (pK_a 3.75) is stronger than benzoic acid (pK_a 4.19). Order: B > A > D > C.

Teacher's Note:
1. Learn the benchmark pK_a values: HCOOH 3.75, C₆H₅COOH 4.19, CH₃COOH 4.76.

Answer: (c) Carboxylate ion gets stabilized by resonance

Explanation:
1. Because of resonance in the carboxyl group (and in the carboxylate ion), the carbonyl carbon is less electrophilic, so –COOH does not show the typical nucleophilic addition and addition-elimination reactions of >C=O.

Teacher's Note:
1. The –COOH group does contain C=O, but its electrophilic character is reduced by resonance with the –OH oxygen.

Answer: (d) ClCH₂CH₂COOH

Explanation:
1. Due to reduced −I-effect of Cl at β-position, ClCH₂CH₂COOH (pK_a 3.98) is a weaker acid than HCOOH (pK_a 3.75).

Teacher's Note:
1. The −I effect falls off quickly with distance; a Cl atom at the β-position has only a small effect on acidity.

Answer: (d) CH₃COCH₃

Explanation:
1. Acetone is a ketone and does not reduce Tollens’ reagent; HCOOH (which has an H on the carboxyl carbon), HCHO and CH₃CHO all give the silver mirror.

Teacher's Note:
1. Formic acid is the only common carboxylic acid that reduces Tollens’ reagent because it contains an aldehyde-like H–C=O unit.

Answer: (b) dil. acidified KMnO₄ solution

Explanation:
1. HCOOH is a reducing agent and decolourises dil. acidified KMnO₄ (it is oxidised to CO₂ and H₂O); acetic acid does not.
2. Both acids give CO₂ with NaHCO₃ and H₂ with Na; neither gives a 2,4-DNP derivative.

Teacher's Note:
1. Formic acid behaves both as an acid and as an aldehyde (reducing agent) – tests for reducing behaviour distinguish it from other acids.

Answer: (d) acetic acid < benzoic acid < formic acid

Explanation:
1. Acetic acid < benzoic acid < formic acid (pK_a: 4.76, 4.19, 3.75).

Teacher's Note:
1. Lower pK_a means a stronger acid; the +I effect of CH₃ makes acetic acid the weakest of the three.

Answer: (b) CH₃CH₂CF₂COOH

Explanation:
1. CH₃CH₂CF₂COOH having two strongly electronegative F atoms at the α-position is the strongest acid and hence has the smallest pK_a.

Teacher's Note:
1. Strongest acid = smallest pK_a; electronegative atoms nearest to –COOH have the largest effect.

Answer: (a) FCH₂COOH > ClCH₂COOH > BrCH₂COOH > CH₃COOH

Explanation:
1. The −I effect decreases in the order F > Cl > Br, and all halogens increase acidity relative to CH₃COOH, so FCH₂COOH > ClCH₂COOH > BrCH₂COOH > CH₃COOH.

Teacher's Note:
1. Acidity of haloacetic acids follows the electronegativity of the halogen.

Answer: (a) (III) > (IV) > (II) > (I)

Explanation:
1. Carboxylic acids are stronger acids than phenols. Further, electron-donating groups decrease while electron-withdrawing groups increase the acidity of phenols/carboxylic acids.
2. Thus, the overall acidity decreases in the order: III > IV > II > I, i.e., option (a) is correct.

Teacher's Note:
1. Even a carboxylic acid weakened by a CH₃ group is far stronger than a phenol bearing a Cl atom.

Answer: (c) IV < I < II < III

Explanation:
1. Electron-donating groups decrease while electron-withdrawing groups increase the acidity of carboxylic acids. Further, larger the number of electron-withdrawing groups, more is the acidity.
2. Thus, 3, 4-dinitrobenzoic acid (III) is more acidic than 4-nitrobenzoic acid (II) and both are more acidic than benzoic acid (I). 4-Methoxybenzoic acid (IV) is, however, the weakest acid.
3. Thus, correct increasing order of acid strength is: IV < I < II < III, i.e., option (c) is correct.

Teacher's Note:
1. At the para-position, –OCH₃ acts mainly by its +R effect and so weakens the acid.

Answer: (a) CH₃COCl

Explanation:
1. Cl has the strongest −I effect and the weakest +R effect (poor 3p–2p overlap), and Cl⁻ is the best leaving group.
2. Order of reactivity: acid chloride > anhydride > ester > amide.

Teacher's Note:
1. Reactivity of acid derivatives decreases as the leaving group becomes a stronger base: Cl⁻ < RCOO⁻ < RO⁻ < NH₂⁻.

Answer: (a) I and II

Explanation:
1. (CH₃CO)₂O reacts with H₂O to produce acetic acid (CH₃COOH). Thus, both (CH₃CO)₂O and CH₃COOH give effervescence with an aqueous solution of NaHCO₃.

Teacher's Note:
1. Phenol is too weak an acid to liberate CO₂ from NaHCO₃; this is the standard test to tell carboxylic acids from phenols.

Answer: (c) CH₃COOH

Explanation:
1. Acid (CH₃COOH) reacts with ethanol to form ester (CH₃COOCH₂CH₃) having fruity smell.

Teacher's Note:
1. Fruity smell on warming with an alcohol and conc. H₂SO₄ (esterification) is a test for carboxylic acids.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Correct explanation. Reduction of C=O group is easier than –COOH group.
2. R correctly states the reagent but does not explain why –COOH is not reduced.

Teacher's Note:
1. Clemmensen (Zn-Hg/conc. HCl) and Wolff-Kishner (NH₂NH₂/KOH, glycol) both convert >C=O into >CH₂.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. CH₃CHO + SeO₂ → OHC–CHO (glyoxal) + Se + H₂O
2. SeO₂ oxidises the CH₃/CH₂ group next to the carbonyl group into a carbonyl group.

Teacher's Note:
1. SeO₂ gives 1,2-dicarbonyl compounds: ethanal → glyoxal, acetone → methylglyoxal (CH₃COCHO).

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Due to resonance in the carboxyl group, R–C(=O)–OH ↔ R–C(–O⁻)=O⁺H, the lone pair of the –OH oxygen reduces the positive charge on the carboxyl carbon. Hence it does not show the nucleophilic addition reactions of aldehydes and ketones.

Teacher's Note:
1. Carboxylic acids do not form oximes, hydrazones or cyanohydrins – a good one-line distinction from aldehydes and ketones.

Answer: (c) A is true, but R is false.

Explanation:
1. Correct R. C₆H₅ group is less electron-donating than CH₃ group.
2. Hence benzoic acid (pK_a 4.19) is a stronger acid than acetic acid (pK_a 4.76).

Teacher's Note:
1. The sp² carbon of the phenyl ring is more electronegative than the sp³ carbon of CH₃, so phenyl is electron-withdrawing relative to methyl.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Correct explanation. Chlorine has both +R-effect and −I-effect but −I-effect outweighs +R-effect. As a result, p-chlorobenzoic acid is a stronger acid than benzoic acid.

Teacher's Note:
1. For halogens on a benzene ring, −I dominates +R; so halogen-substituted benzoic acids are stronger than benzoic acid.

Answer: (d) A is false, but R is true.

Explanation:
1. Correct assertion. Phenol and benzoic acid cannot be distinguished by NaOH since both of them dissolve in NaOH solution. They can, however, be distinguished by NaHCO₃ solution.

Teacher's Note:
1. Benzoic acid gives brisk effervescence (CO₂) with NaHCO₃; phenol does not because it is a much weaker acid.

Answer: (d) A is false, but R is true.

Explanation:
1. Correct A. Propionic acid reacts with Br₂ in presence of red P to give α-bromopropionic acid.
2. CH₃CH₂COOH →^{Br₂/red P} CH₃CHBrCOOH (HVZ reaction)

Teacher's Note:
1. HVZ reaction always halogenates the α-carbon, never the β-carbon.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Correct explanation. Phthalic acid first reacts with NH₃ to form ammonium phthalate which upon strong heating loses a molecule of NH₃ and H₂O to form phthalimide.
2. C₆H₄(COOH)₂ (phthalic acid) →^2NH₃ C₆H₄(COONH₄)₂ (amm. phthalate) →^{strong heating, −NH₃, −2H₂O} C₆H₄(CO)₂NH (phthalimide)

Teacher's Note:
1. The acid-base step only forms the ammonium salt; it is the dehydration on strong heating that gives the imide.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. HVZ reaction substitutes an α-hydrogen by halogen. In (CH₃)₃C–COOH the α-carbon is quaternary and carries no hydrogen, so the reaction is not possible.

Teacher's Note:
1. Always check the carbon attached to –COOH for hydrogen before writing an HVZ product.

Answer: (c) A is true, but R is false.

Explanation:
1. Correct R. RO⁻ is a better leaving group than NH₂⁻.

Teacher's Note:
1. Weaker base = better leaving group; NH₂⁻ is a stronger base than RO⁻, so amides are the least reactive acid derivatives.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. HCOOH contains an H atom on the carboxyl carbon (H–COOH) and is easily oxidised: HCOOH + 2[Ag(NH₃)₂]⁺ + 2OH⁻ → 2Ag + CO₂ + 2H₂O + 4NH₃
2. Acetic acid has no such H and is not oxidised by Tollens’ reagent.

Teacher's Note:
1. Formic acid also reduces Fehling’s solution and decolourises acidified KMnO₄.