CBSE Class 12 Chemistry Aldehydes Ketones and Carboxylic Acids MCQs Set 11

Answer: (b) CH₃COCH₃

Explanation:
1. Aldehydes give silver mirror test with Tollen’s reagent while ketones form oximes with hydroxylamine. Hence the compound is a ketone.
2. Alcohol and ethers do not give this test.

Teacher's Note:
1. Oxime formation proves a C=O group; a negative Tollens’ test then rules out the aldehyde.

Answer: (a) Iodoform test

Explanation:
1. Pentan-2-one and pentan-3-one can be differentiated by iodoform test.
2. CH₃CH₂–CO–CH₂CH₃ + I₂ + NaOH → No reaction; CH₃CH₂CH₂–CO–CH₃ + I₂ + NaOH → CHI₃ (Yellow ppt.)

Teacher's Note:
1. Benedict’s and Fehling’s tests are for aldehydes, so they cannot separate two ketones.

Answer: (d) H–CO–H

Explanation:
1. HCHO will be most polar due to lowest electron density on carbon of carbonyl group.

Teacher's Note:
1. Every alkyl group pushes electrons into the carbonyl carbon (+I), reducing its partial positive charge.

Answer: (a) HCHO > RCHO > ArCHO > R₂CO > Ar₂CO

Explanation:
1. Aldehydes are more reactive than ketones towards nucleophilic addition reactions.
2. Aromatic aldehydes and ketones are less reactive than corresponding aliphatic aldehydes and ketones.

Teacher's Note:
1. The aryl ring lowers reactivity by +R (resonance) donation into the carbonyl carbon.

Answer: (d) Butanal and propanal can be distinguished by iodoform test.

Explanation:
1. Both butanal and propanal do not give iodoform test, hence cannot be distinguished from each other.

Teacher's Note:
1. Among aldehydes only ethanal gives the iodoform test, since it alone has the CH₃CO– unit.

Answer: (a) It is used for purification of aldehydes and ketones.

Explanation:
1. Aldehydes and ketones form insoluble crystalline compounds with NaHSO₃ which can be filtered. These on distillation with saturated solution of Na₂CO₃ again give the aldehydes and ketones.

Teacher's Note:
1. The bisulphite addition is reversible, which is exactly what makes it useful for separating and purifying carbonyl compounds.

Answer: (d) All of these

Explanation:
1. All the given aldehydes will give Cannizzaro reaction, as none of them has an α-hydrogen.

Teacher's Note:
1. (CH₃)₃CCHO looks aliphatic, but its α-carbon is quaternary – no α-H, so it undergoes Cannizzaro reaction.

Answer: (a) Treat with HCN followed by acidic hydrolysis.

Explanation:
1. CH₃CHO (Ethanal) HCN
⟶ CH₃CH(OH)CN H₂O
⟶
H⁺ CH₃CH(OH)COOH (α-Hydroxypropanoic acid)

Teacher's Note:
1. Cyanohydrin formation adds one carbon; hydrolysis of –CN gives –COOH, so α-hydroxy acids have one carbon more than the aldehyde.

Answer: (b) CH₃COCH₃ I₂
⟶
NaOH CH₃COONa H⁺
⟶
NH₃ CH₃COONH₄ Δ
⟶ CH₃CONH₂

Explanation:
1. CH₃COCH₃ (Acetone) I₂
⟶
NaOH CH₃COONa H⁺
⟶ CH₃COOH NH₃
⟶ CH₃COONH₄ Δ
⟶ CH₃CONH₂ (Acetamide)

Teacher's Note:
1. Acid chlorides are made with SOCl₂ or PCl₅, not HCl – that alone rules out option (d).

Answer: (a) intermolecular hydrogen bonding in butan-1-ol

Explanation:
1. Butan-1-ol has higher boiling point due to intermolecular hydrogen bonding.

Teacher's Note:
1. Aldehydes have no O–H bond, so they cannot hydrogen bond with each other; their molar masses are nearly the same as the alcohols.

Answer: (b) (b)

Explanation:
1. CH₃COOC₂H₅ (Ethyl ethanoate, X) H₂O/H⁺
⟶ CH₃COOH (Y) + C₂H₅OH (Z)
2. C₂H₅OH (Z) CrO₃
⟶ CH₃COOH (Y)

Teacher's Note:
1. If the alcohol oxidises to the same acid, both halves of the ester must have the same number of carbons.

Answer: (a) in carboxylic acid –OH group is attached to electron withdrawing carbonyl group

Explanation:
1. In alcohols –OH group is attached to an electron releasing group while in carboxylic acids –OH group is attached to an electron withdrawing group making it more acidic.

Teacher's Note:
1. The carbonyl group also stabilises the carboxylate ion by resonance, which is the main reason for the acidity.

Answer: (b) ICH₂COOH > BrCH₂COOH > ClCH₂COOH > FCH₂COOH

Explanation:
1. The correct order is ICH₂COOH < BrCH₂COOH < ClCH₂COOH < FCH₂COOH, so the order given in (b) is reversed.

Teacher's Note:
1. The –I effect fades with distance and grows with electronegativity and with the number of halogens – use these three checks on each order.

Answer: (b)

Explanation:
1. R–COOH –H⁺
⟶ R–COO⁻
2. The carboxylate ion is a resonance hybrid of the structures R–C(=O)–O⁻ <→ R–C(–O⁻)=O, so both C–O bonds are equivalent and the negative charge is spread equally over the two oxygen atoms.

Teacher's Note:
1. In a carboxylate ion both C–O bond lengths are equal, intermediate between C=O and C–O.

Answer: (d) C₆H₅–C(CH₃)₃ KMnO₄/KOH
⟶
Δ C₆H₅–COOH

Explanation:
1. Side-chain oxidation by alkaline KMnO₄ needs at least one benzylic hydrogen.
2. In tert-butylbenzene the carbon attached to the ring has no hydrogen, so it is not oxidised to benzoic acid.

Teacher's Note:
1. Whatever the length of the side chain, it is oxidised to –COOH as long as the benzylic carbon carries an H.

Answer: (b) – 5-Methylcyclohexanecarbaldehyde

Explanation:
1. The correct name of the compound in (b) is 3-methylcyclohexanecarbaldehyde (ring carbons are numbered from the carbon bearing –CHO so that the methyl gets the lowest number).

Teacher's Note:
1. Always choose the numbering direction that gives the substituent the lowest locant: 3, not 5.

Answer: (c) propanone

Explanation:
1. CH≡CH Hg²⁺
⟶
H₂SO₄ CH₃CHO (X) CH₃MgX
⟶ CH₃–CH(CH₃)–OMgX H₂O
⟶ CH₃–CH(OH)–CH₃ (Y) [O]
⟶ CH₃COCH₃ (Z)

Teacher's Note:
1. Ethanal + Grignard gives a secondary alcohol, and oxidation of a 2° alcohol stops at the ketone.

Answer: (a) X = Acetophenone, Y = m-Nitroacetophenone

Explanation:
1. C₆H₆ (CH₃CO)₂O
⟶
AlCl₃ C₆H₅COCH₃ (Acetophenone, X) (Friedel-Craft’s acylation)
2. C₆H₅COCH₃ conc. HNO₃
⟶
conc. H₂SO₄ m-O₂NC₆H₄COCH₃ (m-Nitroacetophenone, Y)

Teacher's Note:
1. The acyl group is electron-withdrawing and therefore meta-directing in nitration.

Answer: (b) CO, HCl, Gattermann–Koch reaction

Explanation:
1. Benzaldehyde can be prepared from benzene by passing vapours of CO and hydrochloric acid in its solution in presence of catalyst mixture of AlCl₃/CuCl. The reaction is known as Gattermann–Koch reaction.
2. C₆H₆ + CO + HCl AlCl₃/CuCl
⟶ C₆H₅CHO

Teacher's Note:
1. Gattermann–Koch introduces –CHO directly onto the benzene ring; Etard reaction makes benzaldehyde from toluene instead.

Answer: (b) Tollen’s reagent

Explanation:
1. Tollen’s reagent oxidises only –CHO to –COOH group.
2. CH₃CH=CHCHO Ag(NH₃)₂⁺
⟶ CH₃CH=CHCOOH

Teacher's Note:
1. Baeyer’s reagent and acidified dichromate also attack the C=C bond; Tollens’ reagent is a mild, selective oxidant.

Answer: (c) propanal and methanal

Explanation:
1. CH₃CH₂–CH=CH₂ O₃
⟶ ozonide Zn/H₂O
⟶ HCHO (Methanal) + CH₃CH₂CHO (Propanal)

Teacher's Note:
1. Ozonolysis splits the molecule exactly at the C=C, so the carbons on each side stay together.

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

Explanation:
1. CH₃–C≡CH 40% H₂SO₄
⟶
1% HgSO₄ CH₃–C(OH)=CH₂ (Prop-1-en-2-ol, A) Tautomerization
⟶ CH₃–CO–CH₃ (Acetone)
2. Prop-1-en-2-ol (A) and acetone show tautomerism.

Teacher's Note:
1. Markovnikov addition puts the OH on the more substituted carbon (C-2), and the enol then rearranges to the keto form.

Answer: (a) OHC–CHO + 2HCHO

Explanation:
1. CH₂=CH–CH=CH₂ O₃
⟶ diozonide Zn/H₂O
⟶ OHC–CHO + 2HCHO

Teacher's Note:
1. With two C=C bonds the molecule is cut twice: the two terminal CH₂ give HCHO and the middle two carbons give glyoxal.

Answer: (b) CH₃–CH₂–CO–CH₃

Explanation:
1. CH₃CH₂C≡CH H⁺/H₂O, Hg²⁺
⟶ CH₃CH₂COCH₃

Teacher's Note:
1. Any terminal alkyne other than ethyne gives a methyl ketone on hydration (Markovnikov’s rule).

Answer: (c) p-ethylbenzaldehyde

Explanation:
1. p-C₂H₅–C₆H₄–CHO (X) [O]
⟶ p-HOOC–C₆H₄–COOH (1,4-Benzenedicarboxylic acid)
2. C₉H₁₀O fits p-ethylbenzaldehyde; it is an aldehyde with no α-H, so it gives Tollens’ test and Cannizzaro reaction.

Teacher's Note:
1. Vigorous oxidation converts every side chain on a benzene ring to –COOH, so the positions of the acid groups reveal the positions of the side chains.

Answer: (b) R–CH=CH–CH=N–NH–CO–NH₂

Explanation:
1. R–CH=CH–CHO + H₂N–NH–CO–NH₂ H⁺
⟶ R–CH=CH–CH=N–NH–CO–NH₂ + H₂O

Teacher's Note:
1. Semicarbazide reacts through its terminal –NH₂ (the one attached to NH), and the product keeps the C=C bond untouched.

Answer: (c) Ethyl ethanoate

Explanation:
1. CH₃COOC₂H₅ will not give acetic acid on oxidation.

Teacher's Note:
1. An ester gives acetic acid on hydrolysis, not on oxidation – read whether the question asks about oxidation or hydrolysis.

Answer: (b) C₆H₅COCH₃

Explanation:
1. Aromatic ketones are less reactive than aliphatic ketones which in turn are less reactive than aldehyde. Hence, acetophenone does not react with NaHSO₃.

Teacher's Note:
1. The bulky HSO₃⁻ adds readily only to aldehydes and methyl ketones; aryl ketones are too hindered and too stabilised.

Answer: (a) (a)

Explanation:
1. (CH₃CO)₂O (A) + C₂H₅OH → CH₃COOH (B) + CH₃COOC₂H₅ (C)
2. CH₃COOC₂H₅ (C) + HOH H⁺
⟶ CH₃COOH (B) + C₂H₅OH (D)
3. C₂H₅OH (D) [O]
⟶ CH₃COOH (B)
4. 2CH₃COOH (B) + Ca(OH)₂ → (CH₃COO)₂Ca + H₂O dry distillation
⟶ CH₃COCH₃ (Acetone)

Teacher's Note:
1. Start from the end: acetone comes from calcium acetate, so B must be acetic acid; then check which option is consistent.

Answer: (b) (CH₃)₂C=NNHCONH₂

Explanation:
1. CH₃–CH(OH)–CH₃ (X) K₂Cr₂O₇
⟶
H⁺ CH₃–CO–CH₃ (Y)
2. (CH₃)₂C=O (Y) NH₂CONHNH₂·HCl + CH₃COONa
⟶ (CH₃)₂C=NNHCONH₂ (Z)

Teacher's Note:
1. A negative Tollens’ test tells you Y is a ketone, so X must be the secondary alcohol propan-2-ol.

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

Explanation:
1. Due to overall electron-donating effect of the phenyl group, benzoate ion is less stable than formate ion.

Teacher's Note:
1. Lower pK_a means stronger acid; HCOOH (pK_a 3.75) is stronger than C₆H₅COOH (pK_a 4.19).

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

Explanation:
1. Carbonyl compounds form solid additive products with NaHSO₃ which are separated out. The solid bisulphites of carbonyl compounds on hydrolysis with dilute acid regenerate original carbonyl compounds and thus, this property is used for the purification of carbonyl compounds as well as for their separation.

Teacher's Note:
1. Purification uses the reversibility of bisulphite addition; uses in perfumery are unrelated to it.

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

Explanation:
1. Boiling points of carboxylic acids are higher due to their tendency to associate and form dimers to a greater extent by hydrogen bonding.

Teacher's Note:
1. High boiling points of acids come from strong intermolecular hydrogen bonding (dimers), not from resonance.

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

Explanation:
1. o-Substituted benzoic acids are generally stronger acids than benzoic acid. This is regardless of the nature (+I or –I) of the substituent. This is called ortho-effect and is probably due to a combination of steric and electronic factors.

Teacher's Note:
1. Even o-toluic acid, with an electron-donating –CH₃, is stronger than benzoic acid – a classic exam example of the ortho effect.

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

Explanation:
1. Oxidation of aldehydes are easier than ketones, so (A) is false; (R) is a true statement.

Teacher's Note:
1. Aldehydes are oxidised even by mild agents (Tollens’, Fehling’s) because only a C–H bond breaks; ketones need strong agents that break C–C bonds.

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

Explanation:
1. The carbon atom of the carbonyl group of benzaldehyde is less electrophilic than carbon atom of carbonyl group present in ethanal. The polarity of the carbonyl group is reduced in benzaldehyde due to resonance hence it is less reactive than ethanal towards nucleophilic addition reaction.
2. Ethanal (small CH₃ group) is less sterically hindered than benzaldehyde (bulky phenyl group), so (R) is false.

Teacher's Note:
1. When judging a reason, check the comparison itself: the bulkier group (C₆H₅) gives more hindrance, not less.

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

Explanation:
1. Both aldehydes and ketones have carbonyl group but only aldehydes react with Tollens’ reagent to give silver mirror.

Teacher's Note:
1. Tollens’ test distinguishes aldehydes from ketones; it depends on the oxidisable C–H of –CHO, not on the C=O group itself.

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

Explanation:
1. The carboxyl group (–COOH) is deactivating group because it is electron withdrawing group. It decreases the electron density at benzene ring, hence deactivates it towards electrophilic substitution reactions.

Teacher's Note:
1. –COOH is a deactivating, meta-directing group; in addition the AlCl₃ catalyst gets bonded to it.

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

Explanation:
1. Compounds containing –CHO group are easily oxidised to corresponding carboxylic acids because due to electron withdrawing nature of C=O group, C–H bond in aldehydes is weak and easily oxidised to the corresponding carboxylic acids even with mild oxidising agents like Fehling’s solution and Tollen’s reagent.
2. The reduction of carboxylic acids by LiAlH₄ is a true statement but it does not explain (A).

Teacher's Note:
1. Both statements can be true yet unrelated: one is about oxidation of aldehydes, the other about reduction of acids.

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

Explanation:
1. Aromatic carboxylic acids do not undergo Friedel-Crafts reaction because carboxyl group is deactivating and the catalyst aluminium chloride gets bonded to the carboxyl group.

Teacher's Note:
1. This is a favourite 1-mark reason; quote both parts: the ring is deactivated and the Lewis acid AlCl₃ is tied up by the –COOH group.