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

Answer: (d) II and IV only

Explanation:
1. 4-Vinyltoluene, CH₃–C₆H₄–CH=CH₂ (i) O₃/CH₂Cl₂
⟶
(ii) Zn/H₂O CH₃–C₆H₄–CHO + HCHO; the aldehyde (M.F. C₈H₈O) gives Cannizzaro reaction but not haloform.
2. 2-Phenylpropene, C₆H₅–C(CH₃)=CH₂ (i) O₃/CH₂Cl₂
⟶
(ii) Zn/H₂O C₆H₅–CO–CH₃ + HCHO; acetophenone (M.F. C₈H₈O) gives haloform reaction but not Cannizzaro.
3. Hence statements II and IV can not be correct.

Teacher's Note:
1. An aldehyde with no α-H gives Cannizzaro; a methyl ketone gives haloform – no single C₈H₈O isomer here does both.

Answer: (b) B is a ketone but not methyl ketone.

Explanation:
1. Since (B) on reaction with 2,4-DNP forms a derivative, it implies that (B) has >C=O group.
2. (B) gives –ve Tollens’ test, hence it is not an aldehyde, but it is a ketone. (B) gives –ve haloform test, thus it is not a methyl ketone.
3. (B) is formed from the oxidation of (A), thus (A) is a 2° alcohol: (A) is CH₃–CH₂–CH(OH)–CH₂–CH₃ and (B) is CH₃–CH₂–CO–CH₂–CH₃.
4. >C=O Zn-Hg/HCl
⟶ –CH₂– (alkanes), so B is not converted to A by Clemmensen reduction.

Teacher's Note:
1. Read each test as a clue: 2,4-DNP = carbonyl, no Tollens’ = ketone, no haloform = not CH₃CO–.

Answer: (c) Order III, ortho effect is not considered.

Explanation:
1. Ortho-effect says that all the o-substituted benzoic acids are stronger acids than benzoic acid, so the correct order is
2. p-O₂NC₆H₄COOH (–R effect of –NO₂ group) > o-HOC₆H₄COOH (ortho-effect pronounced due to H-bonding in carboxylate ion) > o-CH₃C₆H₄COOH (ortho-effect) > C₆H₅COOH
3. Thus, Order III given is incorrect as in this ortho-effect is not considered. Order I and II are correct.

Teacher's Note:
1. Any ortho substituent, even an electron donor like –CH₃, makes benzoic acid stronger – this is the ortho effect.

Answer: (c) carboxyl group is deactivating and gets bonded to Friedel Craft’s catalyst

Explanation:
1. Carboxylic acids do not undergo Friedel Craft’s reaction because carboxyl group is deactivating and gets bonded to the catalyst in Friedel Craft’s reaction.

Teacher's Note:
1. AlCl₃ is a Lewis acid and forms a complex with the lone pairs of the –COOH oxygen, so it is no longer available as a catalyst.

Answer: (a) A is butan-2-one; B is butan-2-ol.

Explanation:
1. A undergoes Iodoform reaction hence contains a methyl ketone. So the structure of A is CH₃–CH₂–CO–CH₃ (Butan-2-one).
2. A (butan-2-one) on reduction gives butan-2-ol (B). B on heating with H₂SO₄ gives an alkene named but-2-ene (C), CH₃CH=CHCH₃.
3. C forms an ozonide D which on hydrolysis in presence of Zn dust forms acetaldehyde (E), CH₃CHO (2 moles).
4. CH₃CH₂COCH₃ (A) LiAlH₄
⟶ CH₃CH(OH)CH₂CH₃ (B) H₂SO₄
⟶
Δ CH₃CH=CHCH₃ (C) O₃
⟶ ozonide (D) Zn/H₂O
⟶ 2CH₃CHO (E)

Teacher's Note:
1. A symmetrical alkene gives a single carbonyl product on ozonolysis – that is the clue to but-2-ene here.

Answer: (b) Butan-2-ol

Explanation:
1. Butan-2-ol is secondary alcohol which on oxidation with alkaline KMnO₄ solution gives butanone (ketone).
2. CH₃CH₂CH(OH)CH₃ + [O] \( \xrightarrow{\text{alk. KMnO}_{4}} \) CH₃CH₂COCH₃ + H₂O

Teacher's Note:
1. Secondary alcohols oxidise to ketones and primary alcohols to aldehydes/acids; butan-1-ol would give butanal or butanoic acid, never butanone.

Answer: (b) 2-Aminopropanal

Explanation:
1. Numbering starts from the CHO carbon (C-1): ³CH₃–²CH(NH₂)–¹CHO, i.e., 2-Aminopropanal.

Teacher's Note:
1. The –CHO carbon is always C-1 and its locant is not written; the amino group is only a prefix here.

Answer: (d) Aldehydes with no α-hydrogen

Explanation:
1. Aldehydes with no α-hydrogen undergo Cannizzaro reaction.

Teacher's Note:
1. Aldehydes that do have α-hydrogen undergo aldol condensation with dilute alkali instead; HCHO, C₆H₅CHO and (CH₃)₃CCHO are the usual Cannizzaro examples.

Answer: (b) 1-Hydroxycyclohexane-1-carbonitrile (cyclohexanone cyanohydrin)

Explanation:
1. Cyclohexanone + HCN → 1-hydroxycyclohexane-1-carbonitrile; the OH and CN groups are both attached to the former carbonyl carbon.
2. It is a nucleophilic addition reaction.

Teacher's Note:
1. In cyanohydrin formation the ring size and carbon skeleton stay the same; one new carbon (from CN) is added to the carbonyl carbon.

Answer: (b) positional isomers

Explanation:
1. CH₃CHO (ethanal) \( \xrightarrow[\text{(ii) H}_{2}\text{O}]{\text{(i) CH}_{3}\text{MgBr}} \) CH₃–CH(OH)–CH₃ (propan-2-ol) (A)
2. A \( \xrightarrow[\text{Dehydration}]{\text{H}_{2}\text{SO}_{4}\text{, }\Delta} \) CH₃–CH=CH₂ (propene) (B)
3. CH₃–CH=CH₂ \( \xrightarrow[\text{Hydroboration oxidation}]{\text{(i) BH}_{3}\text{, (ii) H}_{2}\text{O}_{2}\text{/OH}^{-}} \) CH₃–CH₂–CH₂OH (propan-1-ol) (C)
4. In compound A and C, position of –OH group is changed. So, these are positional isomers.

Teacher's Note:
1. Hydroboration–oxidation adds water in the anti-Markovnikov way, so the OH goes to the terminal carbon, unlike acid-catalysed hydration.

Answer: (a) zinc amalgam + HCl

Explanation:
1. Clemmensen reduction is used to convert carbonyl group to CH₂ group as follows:
2. >C=O \( \xrightarrow{\text{Zn(Hg) + HCl}} \) >CH₂

Teacher's Note:
1. Remember: Clemmensen = Zn-Hg/conc. HCl (acidic); Wolff-Kishner = NH₂NH₂/KOH in ethylene glycol (basic).

Answer: (b) C₆H₅COONa + CHI₃

Explanation:
1. C₆H₅COCH₃ (Acetophenone) \( \xrightarrow{\text{NaOH/I}_{2}} \) C₆H₅COONa + CHI₃ (Iodoform)

Teacher's Note:
1. Any compound with a CH₃CO– group gives the iodoform test; the acid salt formed has one carbon less than the ketone.

Answer: (a) CH₃CH₂CH₃

Explanation:
1. CH₃COCH₃ (acetone) \( \xrightarrow[\text{(ii) KOH/Glycol, }\Delta]{\text{(i) H}_{2}\text{N–NH}_{2}} \) CH₃–CH₂–CH₃ (propane)
2. It is Wolff-Kishner reduction which converts >C=O group into –CH₂– group.

Teacher's Note:
1. Wolff-Kishner reduction removes the oxygen completely; it does not stop at the alcohol stage.

Answer: (a) CH₃CHO

Explanation:
1. Acetophenone < Benzaldehyde < Propanone < Ethanal – reactivity towards nucleophilic addition.
2. Aldehydes are more reactive than aliphatic ketones. Aliphatic ketones are more reactive than aromatic ketones.
3. The +I effect is more in ketone than in aldehyde. Thus, ketone will be least reactive in nucleophilic addition reactions. The presence of electron withdrawing group increases the reactivity towards the addition while the presence of electron donating group decreases the reactivity of compound towards nucleophilic addition.
4. Benzaldehyde does not favour nucleophilic addition reaction due to resonance stabilisation.

Teacher's Note:
1. Two factors decide this order: steric crowding at the carbonyl carbon and electron release (+I or resonance) that lowers its positive charge.

Answer: (b) Ethanol

Explanation:
1. HCHO + CH₃MgBr → H–CH(CH₃)–OMgBr (adduct)
2. H–CH(CH₃)–OMgBr \( \xrightarrow{\text{H}_{2}\text{O/H}^{+}} \) CH₃CH₂OH (Ethanol) + Mg(Br)OH

Teacher's Note:
1. Formaldehyde + any Grignard reagent gives a primary alcohol with one carbon more than the alkyl group; other aldehydes give 2° and ketones give 3° alcohols.

Answer: (a) Oxalic acid

Explanation:
1. Structural formula of Ethane-1,2-dioic acid is HOOC–COOH.
2. ∴ It is oxalic acid.

Teacher's Note:
1. Learn the common names of dicarboxylic acids in order: oxalic (C₂), malonic (C₃), succinic (C₄), glutaric (C₅), adipic (C₆).

Answer: (d) (CH₃)₃CCOOH

Explanation:
1. The carboxylic acids having α-hydrogen atom undergo HVZ reaction. Since (CH₃)₃C–COOH does not contain α-H-atom; so, it does not undergo HVZ reaction.

Teacher's Note:
1. Look at the carbon next to COOH: in (CH₃)₃C–COOH it is a quaternary carbon with no hydrogen to replace.

Answer: (a) Formic acid

Explanation:
1. Formic acid (HCOOH) does not form anhydride because it does not contain α-C-atom.

Teacher's Note:
1. On heating with dehydrating agents, formic acid decomposes into CO and H₂O instead of forming an anhydride.

Answer: (a) Acetic acid

Explanation:
1. Acetic acid is the strongest acid because it loses H⁺ ion to form carboxylate ion (CH₃COO⁻) which gets stabilised by resonance.

Teacher's Note:
1. Order of acidity: carboxylic acid > phenol > water > alcohol; the stability of the conjugate base decides it.

Answer: (b) Kolbe’s electrolysis decarboxylation

Explanation:
1. It is Kolbe’s electrolytic decarboxylation.
2. RCOONa(aq) → RCOO⁻ + Na⁺
3. At anode, 2RCOO⁻ → R–R (Alkane) + 2CO₂ + 2e⁻
4. At cathode, 2H₂O + 2e⁻ → H₂ + 2OH⁻

Teacher's Note:
1. Kolbe electrolysis joins two alkyl groups (R–R), while soda-lime decarboxylation gives R–H; compare the carbon counts.

Answer: (c) (iii) < (i) < (ii)

Explanation:
1. The electron withdrawing group (–NO₂) increases the acid strength of aromatic acids while electron releasing group (–CH₃) decreases the acid strength of aromatic acids.
2. Hence, the increasing order of acid strength is: p-methylbenzoic acid (least acidic) < benzoic acid < p-nitrobenzoic acid (most acidic).

Teacher's Note:
1. Electron-withdrawing groups stabilise the carboxylate anion and raise acidity; electron-donating groups do the opposite.

Answer: (c) CH₃COONa + NaOH \( \xrightarrow{\text{CaO}} \) CH₄ + Na₂CO₃

Explanation:
1. CH₃COONa + NaOH \( \xrightarrow{\text{CaO}} \) CH₄ (Methane) + Na₂CO₃

Teacher's Note:
1. In soda-lime decarboxylation the COONa group leaves as Na₂CO₃ and is replaced by H, so the alkane has one carbon less.

Answer: (b) Hell-Volhard-Zelinsky reaction

Explanation:
1. Halogenation of carboxylic acids at the α-position by X₂/red phosphorus is the Hell-Volhard-Zelinsky reaction.

Teacher's Note:
1. Etard (CrO₂Cl₂) and Stephen (SnCl₂/HCl) reactions are preparations of aldehydes, not halogenation of acids.

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

Explanation:
1. Hydration of propyne gives the enol CH₃–C(OH)=CH₂ (prop-1-en-2-ol).
2. The enol changes into the keto form (propanone) by keto-enol tautomerism.

Teacher's Note:
1. Keto-enol interconversion involves shift of a proton and a double bond – that is tautomerism, not metamerism.

Answer: (d) Nucleophilic addition

Explanation:
1. CN⁻ (nucleophile) attacks the carbonyl carbon and H⁺ then adds to oxygen: (CH₃)₂C=O + HCN → (CH₃)₂C(OH)CN.

Teacher's Note:
1. The characteristic reaction of the C=O group of aldehydes and ketones is nucleophilic addition, because the carbonyl carbon is electron deficient.

Answer: (c) I₂ and NaOH solution

Explanation:
1. A methyl ketone is converted into an acid with one carbon atom less by the haloform (iodoform) reaction with I₂/NaOH; the C=C double bond is not attacked.
2. CH₃CH=CHCH₂COCH₃ + 3I₂ + 4NaOH → CH₃CH=CHCH₂COONa + CHI₃ + 3NaI + 3H₂O, followed by acidification.

Teacher's Note:
1. Notice the product has one carbon fewer – that is the signature of the haloform reaction on a –COCH₃ group.

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

Explanation:
1. Aldol condensation requires at least one α-hydrogen atom.
2. Benzaldehyde (C₆H₅CHO) and 2,2-dimethylpropanal ((CH₃)₃CCHO) have no α-hydrogen, so neither undergoes aldol condensation; ethanal has three α-hydrogens and does.

Teacher's Note:
1. Count hydrogens on the carbon next to C=O: zero α-H means no aldol, but such aldehydes give the Cannizzaro reaction instead.

Answer: (d) CH₃CHO

Explanation:
1. Cannizzaro reaction is given by aldehydes having no α-hydrogen atom.
2. CH₃CHO has three α-hydrogens, so it undergoes aldol condensation rather than Cannizzaro reaction. The other three have no α-hydrogen.

Teacher's Note:
1. In option (a) the ring carbon carrying the CHO also carries CH₃, so it has no hydrogen – check the α-carbon itself, not the whole molecule.

Answer: (b) C₆H₅COO⁻K⁺ + C₆H₅CH₂OH

Explanation:
1. Benzaldehyde has no α-hydrogen, so it undergoes Cannizzaro reaction (disproportionation) with conc. KOH:
2. 2C₆H₅CHO + KOH → C₆H₅COOK + C₆H₅CH₂OH

Teacher's Note:
1. In Cannizzaro reaction one molecule is oxidised to the acid salt and the other is reduced to the alcohol – the ring itself is untouched.

Answer: (c)

Explanation:
1. Formation of 2,4-DNP derivative shows a carbonyl group; reduction of Tollens’ reagent and Cannizzaro reaction show an aldehyde with no α-hydrogen, i.e., –CHO attached directly to the ring.
2. Vigorous oxidation gives 1,2-benzenedicarboxylic acid (phthalic acid), so the two side chains are ortho to each other. Hence X is 2-ethylbenzaldehyde (option c).

Teacher's Note:
1. Any alkyl side chain on benzene, however long, is oxidised to a single –COOH group by vigorous oxidation – so the oxidation product tells you only the positions of the side chains.

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

Explanation:
1. Carboxylic acids are more acidic than phenols as the carboxylate ion, the conjugate base of carboxylic acid is stabilized by two equivalent resonance structures. Thus, the negative charge is delocalized effectively. However, in phenols, negative charge is less effectively delocalized over oxygen atom and carbon atoms in phenoxide ion.
2. R–C(=O)–O⁻ ↔ R–C(–O⁻)=O

Teacher's Note:
1. In the carboxylate ion the charge is shared by two equally electronegative O atoms; in phenoxide it spreads onto less electronegative ring carbons.

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

Explanation:
1. The carbonyl carbon in HCHO is sp² hybridised; its three σ-bonds lie in one plane at about 120°, so the molecule is planar. R explains A.

Teacher's Note:
1. All atoms directly attached to an sp² carbonyl carbon lie in one plane – this is also why a nucleophile can attack from either face.

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

Explanation:
1. Lower aldehydes and ketones dissolve because they form hydrogen bonds with water; as the alkyl (hydrophobic) part grows, solubility falls.
2. Alkyl groups are indeed electron-releasing, but that is not why solubility decreases.

Teacher's Note:
1. Solubility trends are about the size of the hydrophobic part, not about inductive effects – keep the two ideas separate.

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

Explanation:
1. α-Hydrogens of carbonyl compounds are acidic because of the strong electron-withdrawing effect of the C=O group and resonance stabilisation of the conjugate base (enolate ion). So A is false and R is true.

Teacher's Note:
1. The acidity of α-hydrogens is the reason aldehydes and ketones undergo aldol condensation.

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

Explanation:
1. Acetaldehyde has α-hydrogens and undergoes aldol condensation, so A is true.
2. Aldehydes without α-hydrogen do not undergo aldol condensation (they give the Cannizzaro reaction), so R is false.

Teacher's Note:
1. Aldol needs α-H; Cannizzaro needs no α-H – this pair is tested again and again.

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

Explanation:
1. In RCOO⁻ the negative charge is shared equally by two oxygen atoms through two equivalent resonance structures; the resonance structures of RCOOH involve charge separation and are not equivalent. R explains A.

Teacher's Note:
1. Equivalent resonance structures give the greatest resonance stabilisation – the same idea explains why both C–O bonds in a carboxylate ion are equal in length.

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

Explanation:
1. Both statements are true. The addition of CH₃⁻ (e.g., from a Grignard reagent) is irreversible because CH₃⁻ is a very strong base and an extremely poor leaving group, so it cannot be expelled back; CN⁻ is a weak base and a good leaving group, so its addition is reversible.

Teacher's Note:
1. Reversibility of a nucleophilic addition depends on whether the nucleophile can leave again – weak bases like CN⁻ can, carbanions cannot.

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

Explanation:
1. Aromatic acids do not undergo Friedel-Crafts reaction because the –COOH group deactivates the ring and the catalyst AlCl₃ (a Lewis acid) gets bonded to the carboxyl group.
2. –COOH is indeed meta-directing, but that does not explain the absence of Friedel-Crafts reaction.

Teacher's Note:
1. Being meta-directing tells you where substitution occurs, not whether it occurs – nitration and bromination of benzoic acid still happen at the meta position.

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

Explanation:
1. C₆H₅CH₃ + CrO₂Cl₂ (in CS₂) → chromium complex →^H₃O+ C₆H₅CHO
2. Chromyl chloride oxidises the methyl group only up to the aldehyde stage, which is why toluene gives benzaldehyde in Etard reaction.

Teacher's Note:
1. Etard reaction = toluene + CrO₂Cl₂; a similar partial oxidation uses CrO₃ in acetic anhydride (benzylidene diacetate).

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

Explanation:
1. Correct A. LiAlH₄ reduces esters to 1° alcohols but DIBAL-H reduces esters to aldehydes.

Teacher's Note:
1. DIBAL-H [(i-Bu)₂AlH] at low temperature stops at the aldehyde; it also reduces nitriles to aldehydes.

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

Explanation:
1. Correct R. In presence of sunlight chlorination occurs in the side chain.
2. C₆H₅CH₃ →^Cl₂/hν C₆H₅CHCl₂ →^{H₂O, 373 K} C₆H₅CHO

Teacher's Note:
1. Sunlight/heat → free-radical side-chain substitution; FeCl₃ in the dark → ring substitution.