JEE Chemistry Aldehydes Ketones and Carboxylic Acids MCQs Set 30

Answer: (a) Acetaldehyde

Explanation:
1. Aldehydes are more reactive than ketones because they have less steric hindrance and the carbonyl carbon carries a larger positive charge (only one electron-donating alkyl group).
2. Benzaldehyde is less reactive than acetaldehyde because the ring donates electrons by resonance.
3. So acetaldehyde is the most reactive.

Teacher's Note:
1. Reactivity order: HCHO > aliphatic aldehydes > aromatic aldehydes > aliphatic ketones > aromatic ketones.

Answer: (c) Fehling's solution

Explanation:
1. Both carbonyl compounds add \(\ce{NaHSO3}\) and \(\ce{RMgX}\) and form hydrazones with phenylhydrazine.
2. Fehling's solution oxidises only aliphatic aldehydes. Acetone is a ketone and benzaldehyde is aromatic, so neither reacts.

Teacher's Note:
1. Fehling's test is given by aliphatic aldehydes only; aromatic aldehydes and ketones fail it.

Answer: (d) ethanal

Explanation:
1. The Cannizzaro reaction needs an aldehyde without alpha hydrogen atoms.
2. Ethanal has alpha hydrogens, so it undergoes aldol condensation instead.
3. The other three have no alpha hydrogen.

Teacher's Note:
1. No alpha hydrogen, then Cannizzaro; with alpha hydrogen, then aldol.

Answer: (b) positional isomers

Explanation:
1. A = propan-2-ol (secondary alcohol from an aldehyde and Grignard reagent).
2. B = propene by dehydration.
3. Hydroboration-oxidation follows anti-Markovnikov addition, so C = propan-1-ol.
4. Propan-2-ol and propan-1-ol differ in the position of the -OH group, hence positional isomers.

Teacher's Note:
1. Hydroboration-oxidation gives the anti-Markovnikov alcohol, acid-catalysed hydration the Markovnikov alcohol.

Answer: (a) (ii) and (iv) only

Explanation:
1. Aldol condensation requires at least one alpha hydrogen.
2. Acetaldehyde and acetone have alpha hydrogens.
3. Benzaldehyde and 2,2-dimethylpropanal (pivaldehyde) have none.

Teacher's Note:
1. Check the carbon next to the carbonyl for hydrogen atoms.

Answer: (a) (ii) and (iii) only

Explanation:
1. Alkaline hydrolysis of the ester gives the carboxylate and the phenol.
2. In NaOH solution the phenol is converted to sodium phenoxide because phenol is acidic.
3. Hence sodium phenoxide and sodium benzoate are obtained.

Teacher's Note:
1. Alkali hydrolysis of esters is called saponification, and it is irreversible.

Answer: (a) (ii) and (iv) only

Explanation:
1. Clemmensen reduction (Zn-Hg/HCl) converts a carbonyl group into a methylene group.
2. Cyclohexanone gives cyclohexane and benzophenone gives diphenylmethane.
3. Benzaldehyde would give toluene (not benzyl alcohol), and benzoyl chloride to benzaldehyde is the Rosenmund reduction.

Teacher's Note:
1. Clemmensen and Wolff-Kishner both reduce C=O to CH2, one in acid and the other in base.

Answer: (a) (i) and (iii) only

Explanation:
1. A Grignard reagent adds an alkyl group to a carbonyl carbon, and aldol condensation joins two carbonyl molecules by a new C-C bond.
2. The Cannizzaro reaction is a redox disproportionation, and the HVZ reaction introduces only a halogen at the alpha position, so neither increases the chain.

Teacher's Note:
1. A new C-C bond is needed to lengthen the chain.

Answer: (a) (i) and (ii) only

Explanation:
1. Friedel-Crafts acylation of benzene with benzoyl chloride gives benzophenone, and organocadmium compounds react with acid chlorides to stop at the ketone.
2. A Grignard reagent reacts further with the ketone and gives the tertiary alcohol triphenylmethanol.
3. Benzene, CO and HCl (not \(\ce{ZnCl2}\) alone) are the Gattermann-Koch reagents for benzaldehyde.

Teacher's Note:
1. Dialkylcadmium reagents are less reactive than Grignard reagents, so the reaction stops at the ketone.

Answer: (a) Both Assertion and Reason are true and the Reason is a correct explanation of the Assertion.

Explanation:
1. The carbonyl carbon of formaldehyde is sp2 hybridised, with bond angles of about 120 degrees, so the molecule is planar.
2. Statement 2 is true and it explains Statement 1.

Teacher's Note:
1. Trigonal planar geometry follows from sp2 hybridisation.

Answer: (c) Assertion is true but the Reason is false.

Explanation:
1. Both lack alpha hydrogens, so the Assertion is true.
2. Aromatic aldehydes are much less reactive than formaldehyde towards nucleophilic addition because of resonance with the ring, so the Reason is false.

Teacher's Note:
1. Formaldehyde is the most reactive carbonyl compound.

Answer: (c) ethylene

Explanation:
1. \(\ce{LiAlH4}\) reduces acetic acid to ethanol (A).
2. \(\ce{PCl5}\) converts ethanol to chloroethane (B).
3. Alcoholic KOH eliminates HCl and gives ethylene (C).

Teacher's Note:
1. Alcoholic KOH gives elimination, aqueous KOH gives substitution.

Answer: (a) 2,2,2-trichloroethanol

Explanation:
1. Chloral has no alpha hydrogen, so it undergoes Cannizzaro disproportionation.
2. One molecule is oxidised to \(\ce{CCl3COONa}\), and the other is reduced to \(\ce{CCl3CH2OH}\), 2,2,2-trichloroethanol.

Teacher's Note:
1. In a Cannizzaro reaction one half of the aldehyde gives the acid salt and the other half the alcohol.

Answer: (a) maleic anhydride

Explanation:
1. Vapour-phase catalytic oxidation of benzene over \(\ce{V2O5}\) at about 773 K breaks the ring and gives maleic anhydride.

Teacher's Note:
1. This is the industrial route to maleic anhydride, besides oxidation of butane.

Answer: (d) 4-nitrophenyl acetate

Explanation:
1. In alkaline hydrolysis the phenoxide ion is the leaving group.
2. The nitro group stabilises the leaving phenoxide most (-I and -R), so 4-nitrophenyl acetate is hydrolysed fastest.
3. The methoxy group destabilises the leaving group and slows hydrolysis.

Teacher's Note:
1. The better the leaving group (the more stable the phenoxide), the faster the ester hydrolysis.

Answer: (a) (i), (iii) and (iv) only

Explanation:
1. (i) is the Gattermann-Koch reaction, (iii) is the Etard reaction and (iv) is the Rosenmund reduction; all give benzaldehyde.
2. (ii) Clemmensen reduction of benzoic acid does not give benzaldehyde (the carboxyl group is not reduced this way).

Teacher's Note:
1. Know the named routes to aldehydes: Rosenmund, Stephen, Etard, Gattermann-Koch and DIBAL reduction.

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

Explanation:
1. Electron-withdrawing groups strengthen acids; fluorine is more electronegative than chlorine, so \(\ce{CF3COOH}\) > \(\ce{CCl3COOH}\).
2. Formic acid (pKa 3.75) is stronger than acetic acid (4.76) because acetic acid has an electron-donating methyl group.
3. Order: B > A > D > C.

Teacher's Note:
1. Remember: CF3COOH > CCl3COOH > HCOOH > CH3COOH.

Answer: (d) (A)-(ii), (B)-(i), (C)-(iv), (D)-(iii)

Explanation:
1. Benzaldehyde undergoes the benzoin condensation.
2. Phthalic anhydride condenses with phenol to give phenolphthalein.
3. Phenyl esters such as phenyl benzoate undergo the Fries rearrangement to hydroxyketones.
4. Methyl salicylate is oil of wintergreen.

Teacher's Note:
1. Connect each compound with its best-known reaction or use.

Answer: (a) (i), (iii) and (iv) only

Explanation:
1. Reimer-Tiemann joins a -CHO carbon to the ring, Wurtz couples two alkyl groups and Friedel-Crafts acylation attaches an acyl group to the ring, all forming C-C bonds.
2. The Cannizzaro reaction is a redox reaction with no new C-C bond.

Teacher's Note:
1. Check whether the product has a new carbon-carbon linkage not present in the reactants.

Answer: (d) \(\ce{CH3COCl}\)

Explanation:
1. Reactivity depends on the leaving group ability and on how little the substituent donates electrons to the carbonyl.
2. Chloride is the best leaving group and chlorine is a poor pi-donor, so acetyl chloride is the most reactive.
3. Order: acid chloride > anhydride > ester > amide.

Teacher's Note:
1. Memorise the order of acid derivative reactivity: RCOCl > (RCO)2O > RCOOR > RCONH2.

Answer: (c) 1,1,1-trichloro-2,2-bis(4-chlorophenyl)ethane (DDT)

Explanation:
1. Chloral is protonated at the carbonyl oxygen and attacks chlorobenzene at the para position twice (electrophilic substitution).
2. Loss of water gives \(\ce{(p-ClC6H4)2CH-CCl3}\), the insecticide DDT.

Teacher's Note:
1. DDT is made from chloral and chlorobenzene with conc. sulphuric acid.

Answer: (c) 1,3-diphenylbut-2-en-1-one (dypnone)

Explanation:
1. Base removes an alpha hydrogen of the acetyl group to give an enolate which adds to a second molecule of acetophenone (aldol addition).
2. Dehydration of the beta-hydroxy ketone gives the conjugated enone \(\ce{PhC(CH3)=CH-COPh}\), called dypnone.

Teacher's Note:
1. Aldol condensation of ketones is driven by dehydration to a conjugated product.

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

Explanation:
1. Electron-withdrawing groups increase the positive charge on the carbonyl carbon and speed up nucleophilic attack.
2. The nitro group is the strongest withdrawing group, followed by the formyl group.
3. Benzoyl chloride is next and the methyl-substituted compound (electron donating) is the slowest.
4. Order: (ii) > (iv) > (i) > (iii).

Teacher's Note:
1. Electron-withdrawing substituents increase the reactivity of acid chlorides towards nucleophiles.

Answer: (a) (ii) and (iv) only

Explanation:
1. A racemic mixture requires a chiral carbon in the product.
2. Acetaldehyde gives \(\ce{CH3CH(OH)COOH}\) and benzaldehyde gives mandelic acid, both with four different groups on the alpha carbon.
3. Formaldehyde and acetone give achiral products.

Teacher's Note:
1. A cyanohydrin carbon is chiral only if the original carbonyl has two different groups (one of them may be H).