CBSE Class 12 Chemistry Aldehydes Ketones and Carboxylic Acids MCQs Set 13

Answer: (a) C₃H₇COOH

Explanation:
1. C₃H₇MgBr + CO₂ → C₃H₇COOMgBr \( \xrightarrow{H_3O^+} \) C₃H₇COOH (butanoic acid)

Teacher's Note:
1. Carboxylation of a Grignard reagent adds one carbon: the acid has one carbon more than the alkyl group.

Answer: (d) Both sodium phenoxide and sodium benzoate

Explanation:
1. Phenyl benzoate is an ester; alkaline hydrolysis gives sodium benzoate and phenol, and phenol reacts with excess NaOH to form sodium phenoxide.
2. C₆H₅COOC₆H₅ + 2NaOH → C₆H₅COONa + C₆H₅ONa + H₂O

Teacher's Note:
1. Phenol is acidic enough to be converted into its sodium salt by NaOH, so in alkaline hydrolysis of a phenyl ester it ends up as phenoxide.

Answer: (b) Acetic acid

Explanation:
1. Carboxylic acids are stronger acids than carbonic acid and liberate CO₂ from NaHCO₃: CH₃COOH + NaHCO₃ → CH₃COONa + H₂O + CO₂↑
2. Phenol is a weaker acid than carbonic acid and does not react with NaHCO₃.

Teacher's Note:
1. The NaHCO₃ test is the standard way to tell a carboxylic acid from a phenol.

Answer: (b) CH₃COOH

Explanation:
1. Each Cl atom increases acidity by its −I effect, while the CH₃ group (+I effect) decreases it.
2. So CH₃COOH is the weakest acid; order: CH₃COOH < CH₂ClCOOH < CHCl₂COOH < CCl₃COOH.

Teacher's Note:
1. More electron-withdrawing atoms on the α-carbon means a more stable carboxylate ion and a stronger acid.

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

Explanation:
1. –OH at the para position donates electrons by resonance (+R) and decreases acidity, so p-hydroxybenzoic acid is weaker than benzoic acid.
2. –NO₂ withdraws electrons and increases acidity; in the ortho position its effect is strongest (ortho effect).
3. Hence: C < D < B < A.

Teacher's Note:
1. Almost any ortho substituent makes benzoic acid stronger (ortho effect), but a para electron donor like OH or OCH₃ makes it weaker.

Answer: (a) P₂O₅, Δ

Explanation:
1. P₂O₅ is a strong dehydrating agent: 2CH₃COOH \( \xrightarrow{P_2O_5,\,\Delta} \) (CH₃CO)₂O + H₂O
2. SOCl₂ and PCl₃ convert the acid into ethanoyl chloride, not the anhydride.

Teacher's Note:
1. An anhydride is formed by removing one H₂O from two acid molecules – think “dehydrating agent”.

Answer: (b) positional isomers

Explanation:
1. CH₃CHO + CH₃MgBr, then H₂O → CH₃–CH(OH)–CH₃, propan-2-ol (A)
2. (A) → [H₂SO₄, Δ] (dehydration) → CH₃–CH=CH₂, propene (B)
3. (B) → (i) BH₃ (ii) H₂O₂/OH⁻ (hydroboration oxidation) → CH₃–CH₂–CH₂OH, propan-1-ol (C)
4. A and C differ only in the position of –OH, so they are positional isomers.

Teacher's Note:
1. Hydroboration–oxidation adds water in an anti-Markovnikov way, so the –OH ends up on the terminal carbon.

Answer: (b) C₆H₅COO⁻K⁺ + C₆H₅CH₂OH

Explanation:
1. Benzaldehyde has no α-hydrogen, so with concentrated KOH it undergoes Cannizzaro reaction.
2. One molecule is oxidised to potassium benzoate and the other is reduced to benzyl alcohol.

Teacher's Note:
1. Cannizzaro reaction is a self redox reaction: one aldehyde molecule is oxidised, the other is reduced – look for an acid salt plus an alcohol.

Answer: (d) Prop-1-en-2-ol, tautomerism

Explanation:
1. Hydration of the alkyne (Markovnikov addition of water) first gives the enol CH₂=C(OH)–CH₃, i.e. prop-1-en-2-ol.
2. The enol changes into the more stable keto form, propanone; this keto-enol interconversion is tautomerism.

Teacher's Note:
1. Enols formed in alkyne hydration are unstable and rearrange at once to the carbonyl compound (keto-enol tautomerism).

Answer: (b) C₆H₅CHO

Explanation:
1. A = C₆H₅COOH (benzoic acid); B = C₆H₅COCl (benzoyl chloride).
2. Rosenmund reduction of B with H₂/Pd–BaSO₄ gives C = C₆H₅CHO.

Teacher's Note:
1. BaSO₄ poisons the Pd catalyst so that the reduction stops at the aldehyde stage.

Answer: (b) iv < iii < ii < i

Explanation:
1. Reactivity towards nucleophilic addition falls as electron-releasing and bulky groups on the carbonyl carbon increase.
2. Phenyl groups reduce the electrophilicity (by resonance) and add steric hindrance, so: PhCOPh < PhCOCH₃ < CH₃COCH₃ < HCHO.

Teacher's Note:
1. Two factors decide the order: electronic (+I/+R groups lower the positive charge on C) and steric (bulky groups block the nucleophile).

Answer: (b) HCOOH

Explanation:
1. Lowest pK_a means strongest acid. HCOOH has no electron-releasing alkyl group, so its anion is most stable.
2. Alkyl groups (+I) decrease acid strength, and ethanol is a much weaker acid than any carboxylic acid.

Teacher's Note:
1. Lower pK_a = stronger acid; read the question carefully, since examiners often ask for the highest pK_a instead.

Answer: (d) iii > i > ii > iv

Explanation:
1. Electron-withdrawing groups stabilise the carboxylate ion and increase acidity; –NO₂ > –F > –Cl in −I effect.
2. –CH₃ is electron releasing, so CH₃COOH is the weakest: iii > i > ii > iv.

Teacher's Note:
1. Remember the −I order NO₂ > CN > F > Cl > Br > I; the nitro group is the strongest here.

Answer: (a) CH₂=CH–CH(OH)–COOH

Explanation:
1. Cross aldol condensation on heating gives A = CH₂=CH–CHO (acrolein).
2. HCN adds to the C=O group to form the cyanohydrin; hydrolysis converts –CN to –COOH: B = CH₂=CH–CH(OH)–COOH.

Teacher's Note:
1. Cyanohydrin followed by acid hydrolysis is the standard route to α-hydroxy acids – one extra carbon is added as –COOH.

Answer: (d) 2-methyl-3-oxohex-4-enal

Explanation:
1. The aldehyde group has priority, so its carbon is C-1 and the name ends in ‘-al’; the ketone is shown by the prefix ‘oxo’.
2. Numbering: C-1 CHO, C-2 bears CH₃, C-3 is C=O, double bond at C-4: 2-methyl-3-oxohex-4-enal.

Teacher's Note:
1. When an aldehyde and ketone are both present, the aldehyde is the principal group and the ketone becomes ‘oxo’ (not ‘keto’) in IUPAC names.

Answer: (b) X-ethanol, A-ethanal, Y-but-2-enal, Z-semicarbazone

Explanation:
1. X = CH₃CH₂OH is dehydrogenated by Cu at 573 K to A = CH₃CHO (it reduces Tollens' reagent).
2. Aldol condensation of A gives Y = CH₃CH=CHCHO (but-2-enal); with semicarbazide A gives Z = CH₃CH=N–NHCONH₂ (semicarbazone).

Teacher's Note:
1. Cu at 573 K converts 1° alcohols to aldehydes, 2° alcohols to ketones and 3° alcohols to alkenes – a frequent identification clue.

Answer: (a) CH₃–CBr₂–COOH

Explanation:
1. Hell-Volhard-Zelinsky reaction replaces only the α-hydrogen atoms of the acid by halogen.
2. CH₃CH₂COOH has two α-H atoms, so the dibromo product is CH₃CBr₂COOH (2,2-dibromopropanoic acid).

Teacher's Note:
1. HVZ halogenation is selective for the α-carbon; β-hydrogens are not replaced.

Answer: (d) Clemmensen reduction

Explanation:
1. Zn-Hg/conc. HCl reduces >C=O to >CH₂; this is Clemmensen reduction.
2. Wolff-Kishner reduction does the same conversion with NH₂NH₂ and KOH/ethylene glycol.

Teacher's Note:
1. Clemmensen = acidic conditions, Wolff-Kishner = basic conditions; choose according to which groups in the molecule are acid or base sensitive.

Answer: (d) C₆H₅–C(CH₃)=CH–CO–C₆H₅

Explanation:
1. Acetophenone has α-hydrogens, so in base it undergoes aldol condensation: the aldol C₆H₅C(OH)(CH₃)CH₂COC₆H₅ is formed first.
2. It loses water to give the stable conjugated α,β-unsaturated ketone C₆H₅C(CH₃)=CHCOC₆H₅.

Teacher's Note:
1. The ‘stable’ aldol product is the dehydrated one, because the new C=C is conjugated with both the C=O group and the benzene ring.

Answer: (d) Both sodium phenoxide and sodium benzoate

Explanation:
1. NaOH hydrolyses the ester (saponification) to benzoic acid and phenol.
2. In excess NaOH both are present as salts: sodium benzoate and sodium phenoxide.

Teacher's Note:
1. Phenol is acidic enough to be converted to sodium phenoxide by NaOH, so it does not stay as free phenol.

Answer: (b) Cyclohexanone to cyclohexane

Explanation:
1. Clemmensen reduction (Zn-Hg/conc. HCl) converts >C=O into >CH₂.
2. Cyclohexanone → cyclohexane is such a change; the other options are reductions to an alcohol or an aldehyde.

Teacher's Note:
1. Clemmensen gives a hydrocarbon, never an alcohol – if the product still has oxygen, it is not Clemmensen.

Answer: (d) Both (a) and (c)

Explanation:
1. Grignard reagents add a new alkyl group to a carbonyl carbon, and aldol condensation joins two carbonyl molecules – both form new C–C bonds.
2. Cannizzaro reaction only transfers hydride, so the chain length does not change.

Teacher's Note:
1. C–C bond-forming reactions to remember: Grignard addition, aldol, cyanohydrin formation, Friedel-Crafts, Wurtz.

Answer: (d) Both (a) and (b)

Explanation:
1. Friedel-Crafts acylation of benzene with C₆H₅COCl/anhyd. AlCl₃ gives C₆H₅COC₆H₅.
2. 2C₆H₅COCl + (C₆H₅)₂Cd → 2C₆H₅COC₆H₅ + CdCl₂. A Grignard reagent reacts further with the ketone, giving a 3° alcohol.

Teacher's Note:
1. Dialkyl/diaryl cadmium stops at the ketone stage, while Grignard reagents go on to form tertiary alcohols with acid chlorides.

Answer: (b) cyclopentanecarbaldehyde

Explanation:
1. When a –CHO group is attached to a ring, the suffix ‘carbaldehyde’ is added to the name of the ring: cyclopentanecarbaldehyde.

Teacher's Note:
1. The ring carbon cannot be the aldehyde carbon, so ‘-al’ is wrong here; use ‘-carbaldehyde’ (e.g. benzenecarbaldehyde, cyclohexanecarbaldehyde).

Answer: (c) carboxylic acids

Explanation:
1. Saturated monocarboxylic acids C_nH_2n+1COOH have the formula C_nH_2nO₂ (e.g. CH₃COOH = C₂H₄O₂).
2. Open-chain dialdehydes and diketones are C_nH_2n−2O₂ and diols are C_nH_2n+2O₂.

Teacher's Note:
1. Count degrees of unsaturation: C_nH_2nO₂ has one, which fits a single C=O as in –COOH (esters also fit).

Answer: (a) C₆H₅COOH + Zn/Hg – conc. HCl

Explanation:
1. Zn/Hg and conc. HCl (Clemmensen reagent) does not reduce –COOH to –CHO, so benzoic acid cannot give benzaldehyde this way.
2. (b) is Etard reaction, (c) is Rosenmund reduction and (d) is Gattermann-Koch reaction – all give benzaldehyde.

Teacher's Note:
1. Remember the three named routes to benzaldehyde: Etard (toluene + CrO₂Cl₂), Rosenmund (C₆H₅COCl + H₂/Pd-BaSO₄) and Gattermann-Koch (C₆H₆ + CO + HCl).

Answer: (c) Dimethylcadmium

Explanation:
1. 2R–COCl + (CH₃)₂Cd →_{(dry ether)} 2R–CO–CH₃ (ketone) + CdCl₂

Teacher's Note:
1. Dialkylcadmium stops at the ketone stage; Grignard reagent reacts further with the ketone to give a 3° alcohol.

Answer: (b) PCC

Explanation:
1. PCC (pyridinium chlorochromate) is a mild oxidising agent that oxidises 1° alcohols to aldehydes and stops there.
2. The other reagents are strong oxidants that carry the oxidation further to the carboxylic acid.

Teacher's Note:
1. PCC works in the absence of water, so the aldehyde is not converted to its hydrate and is not oxidised further.

Answer: (a) acetophenone

Explanation:
1. (C₆H₅COO)₂Ca + (CH₃COO)₂Ca →_{(dry distil)} 2C₆H₅COCH₃ + 2CaCO₃
2. The mixed salt combines one C₆H₅CO– part with one CH₃ group to give acetophenone.

Teacher's Note:
1. In dry distillation of calcium salts the ketone carbon comes from one carboxylate and the other alkyl/aryl group from the second carboxylate; calcium formate is needed to get an aldehyde.

Answer: (a) Ph–CO–CH₂–CH₃

Explanation:
1. Hg²⁺ forms a complex with the triple bond; H₂O attacks the carbon next to Ph (the carbon better able to carry the positive charge), and loss of H⁺ and Hg²⁺ gives the enol Ph–C(OH)=CH–CH₃.
2. The enol tautomerises to the ketone Ph–CO–CH₂CH₃ (propiophenone).

Teacher's Note:
1. The enol (option c) is only an intermediate; it rapidly tautomerises to the more stable keto form.

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

Explanation:
1. PCC is used in anhydrous CH₂Cl₂. Without water the aldehyde cannot form its hydrate, which is the species that gets oxidised to the acid, so oxidation stops at RCHO.

Teacher's Note:
1. Aqueous acidic K₂Cr₂O₇ or KMnO₄ take a 1° alcohol all the way to the carboxylic acid.

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

Explanation:
1. Correct R. Ketals are hydrolysed back to ketones by aqueous acid.

Teacher's Note:
1. Ketals (acetals) are stable to bases, which is why they are used to protect a C=O group during reactions in alkaline medium.

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

Explanation:
1. Correct explanation. The −I-effect of the three Cl-atoms pulls the electrons of OH bonds thereby preventing the loss of H₂O.
2. R is a true fact (CCl₃CHO + NaOH → CHCl₃ + HCOONa) but it does not explain the stability of the hydrate.

Teacher's Note:
1. Chloral hydrate, CCl₃CH(OH)₂, is an exception to the rule that a carbon cannot hold two OH groups stably.

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

Explanation:
1. The benzene ring donates electrons to the carbonyl group by resonance (+R-effect), so the carbonyl carbon of acetophenone carries less positive charge and is attacked less readily by nucleophiles.

Teacher's Note:
1. Aromatic aldehydes and ketones are less reactive than aliphatic ones because of this resonance with the ring.

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

Explanation:
1. Correct explanation. Due to smaller +I-effect of one CH₃ group as compared to larger +I-effect of two CH₃ groups, the +ve charge on the carbonyl group of acetaldehyde is more than in acetone and hence nucleophilic addition reaction of HCN to acetaldehyde occurs more readily than in acetone.

Teacher's Note:
1. Reducing behaviour (oxidation) and nucleophilic addition are different properties – one cannot explain the other.

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

Explanation:
1. Due to strong −I-effect of ⁺NH₃ group, the lone pair of electrons on NH₂ group of protonated hydrazine (H₂N–NH₂ + H⁺ → H₃N⁺–NH₂) is not available for nucleophilic attack on the C=O group and hence hydrazine formation does not occur.

Teacher's Note:
1. Ammonia derivatives react best at about pH 3.5: enough acid to activate the C=O but not so much that the nucleophile is fully protonated.

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

Explanation:
1. HCHO + C₆H₅CHO + OH⁻ → HCOO⁻ + C₆H₅CH₂OH
2. Being the better hydride donor, HCHO is oxidised to formate and transfers H⁻ to benzaldehyde, which is reduced to benzyl alcohol.

Teacher's Note:
1. In crossed Cannizzaro reactions involving HCHO, the formaldehyde always ends up as formate.

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

Explanation:
1. Correct A. Cross Cannizzaro reaction between C₆H₅CHO and HCHO gives benzyl alcohol and formate ion.

Teacher's Note:
1. Ask which aldehyde gives the hydride: the donor is oxidised (to its carboxylate), the acceptor is reduced (to its alcohol).

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

Explanation:
1. 2-Methylpropanal, (CH₃)₂CH–CHO, does not undergo Cannizzaro reaction at room temperature.
2. It has one α-hydrogen atom, so under ordinary conditions it gives aldol-type reactions instead; R is true.

Teacher's Note:
1. Aldehydes with α-H normally undergo aldol condensation, not Cannizzaro reaction, with alkali.

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

Explanation:
1. Correct reason. Anions left after loss of an α-hydrogen are stabilized by resonance.
2. –CH₂⁻–CH=O ↔ –CH₂=CH–O⁻ (enolate ion)

Teacher's Note:
1. The acidity of α-H is the basis of aldol condensation and haloform reaction.