JEE Chemistry Aldehydes Ketones and Carboxylic Acids MCQs Set 10

Answer: (d) \(\ce{C6H5CHO}\)

Explanation:
1. The aldol reaction needs an alpha hydrogen.
2. Benzaldehyde has none. The alpha carbon of \(\ce{CD3CHO}\) has deuterium, which behaves just like hydrogen here, so it still undergoes the reaction.

Teacher's Note:
1. Isotopic substitution does not remove acidity of the alpha position.
2. Benzaldehyde can only act as the electrophile in a crossed aldol reaction.

Answer: (c) \(\ce{CH3CH=CHCH2CH2CHO}\)

Explanation:
1. DIBAL-H reduces a nitrile to an imine-aluminium complex that hydrolyses to an aldehyde.
2. It does not reduce an isolated \(\ce{C=C}\) bond, so the double bond is untouched.

Teacher's Note:
1. DIBAL-H is a bulky, mild hydride that stops at the aldehyde stage for nitriles and esters at low temperature.
2. It leaves isolated alkenes alone.

Answer: (b) 4-(1,3-Dioxolan-2-yl)cyclohexanone

Explanation:
1. Aldehydes form acetals much faster than ketones because they are less hindered and more electrophilic.
2. With one mole of the diol only the aldehyde is protected as a cyclic acetal, leaving the ring ketone free.

Teacher's Note:
1. Selective protection of an aldehyde in the presence of a ketone is possible with one equivalent of ethylene glycol.
2. The cyclic acetal is stable to base and nucleophiles.

Answer: (b) \(\ce{(CH3)2C=N-NH-CO-NH2}\)

Explanation:
1. The terminal \(\ce{NH2}\) on the hydrazine side of semicarbazide is the nucleophilic nitrogen, because the nitrogen next to the carbonyl is amide-like and not nucleophilic.
2. Condensation with loss of water gives acetone semicarbazone, \(\ce{(CH3)2C=N-NH-CO-NH2}\).

Teacher's Note:
1. The amide-type nitrogen is delocalised into the carbonyl, so only the hydrazine-type terminal nitrogen attacks.
2. The semicarbazone is a crystalline solid used to identify carbonyl compounds.

Answer: (b) 3-(4-Methylphenyl)-1-phenylprop-2-en-1-one

Explanation:
1. This is a Claisen-Schmidt (crossed aldol) condensation. The enolate of acetophenone attacks the aldehyde group of 4-methylbenzaldehyde.
2. Dehydration gives \(\ce{PhCO-CH=CH-C6H4-CH3}\), in which the carbonyl is attached to the unsubstituted phenyl ring and the tolyl ring sits on the double bond.

Teacher's Note:
1. The aldehyde provides the beta aryl group and the ketone provides the carbonyl and its aryl group.
2. This is the general way to make chalcones.

Answer: (b) III, I, II, IV

Explanation:
1. The nitro group withdraws electrons, so 4-nitrobenzaldehyde is most electrophilic.
2. The methyl group donates electrons and lowers reactivity compared to benzaldehyde, and the ketone is the least electrophilic because it has two electron-donating groups and more steric hindrance.

Teacher's Note:
1. Electron-withdrawing groups on the ring increase the electrophilicity of the carbonyl carbon.
2. Ketones are always less reactive than aldehydes.

Answer: (b) \(\ce{RCOCl}\)

Explanation:
1. Chloride is the best leaving group of the four.
2. So the acid chloride is hydrolysed fastest, followed by the anhydride, ester and amide.

Teacher's Note:
1. Acid chlorides react violently even with cold water.
2. Amides need prolonged heating with strong acid or base.

Answer: (b) Acetophenone

Explanation:
1. \(\ce{P2O5}\) dehydrates ethanoic acid to ethanoic anhydride.
2. The anhydride and \(\ce{PCl5}\) give ethanoyl chloride, and Friedel-Crafts acylation of benzene with it gives acetophenone.

Teacher's Note:
1. Friedel-Crafts acylation of benzene with ethanoyl chloride is the standard route to acetophenone.
2. The ketone formed deactivates the ring, so only one acyl group enters.

Answer: (d) Iodoform, anaesthetic

Explanation:
1. Ethanoic acid is a solvent, esters of benzoic acid are used in perfumes, and sodium benzoate is a food preservative.
2. Iodoform is used as an antiseptic, not as an anaesthetic.

Teacher's Note:
1. Iodoform was used in dressings because it releases iodine slowly.
2. Chloroform and ether are the classical anaesthetics.

Answer: (d) Propane

Explanation:
1. Tollens' reagent oxidises butanal to butanoate or butanoic acid (X), and \(\ce{NaHCO3}\) gives sodium butanoate (Y).
2. Heating with soda lime decarboxylates it, giving propane (Z) which has one carbon less.

Teacher's Note:
1. Soda lime decarboxylation removes the carboxyl carbon and replaces it with hydrogen.
2. The product alkane always has one carbon fewer than the acid.

Answer: (c) Propanone

Explanation:
1. Methanal gives a primary alcohol and other aldehydes give secondary alcohols.
2. Ketones such as propanone give tertiary alcohols, here 2-methylpropan-2-ol.

Teacher's Note:
1. Rule: methanal gives primary, other aldehydes give secondary, ketones give tertiary alcohols.
2. Esters and acid chlorides give tertiary alcohols with two equivalents of Grignard reagent.

Answer: (b) 2-Cyclopentylidenecyclopentanone

Explanation:
1. Cyclopentanone has alpha hydrogens and undergoes aldol addition to give the beta-hydroxy ketone.
2. On heating it dehydrates to the conjugated 2-cyclopentylidenecyclopentanone.

Teacher's Note:
1. Dehydration of an aldol is easy when it gives a double bond conjugated with the carbonyl.
2. The new bond forms at the alpha carbon, which is carbon 2.

Answer: (d) Ethanal and methanal

Explanation:
1. Baeyer's reagent hydroxylates propene to propane-1,2-diol.
2. Periodic acid cleaves the carbon-carbon bond between the two hydroxyl-bearing carbons, giving ethanal and methanal.

Teacher's Note:
1. Periodic acid cleaves vicinal diols into carbonyl compounds.
2. A secondary carbinol carbon gives an aldehyde and a primary one gives methanal.

Answer: (b) Cyclopentanone

Explanation:
1. Hot \(\ce{KMnO4}\) cleaves the ring to hexanedioic acid (X), and calcium hydroxide gives calcium hexanedioate (Y).
2. Pyrolysis of this salt loses \(\ce{CaCO3}\) and gives cyclopentanone (Z).

Teacher's Note:
1. Hexanedioate gives a five-membered ring and heptanedioate gives a six-membered ring.
2. The ring formed always has one carbon fewer than the diacid chain.

Answer: (d) Pent-2-ene, ethanal, propanal

Explanation:
1. For pent-2-ene, \(\ce{CH3CH=CHCH2CH3}\), ozonolysis gives ethanal (Y) and propanal (Z).
2. Ethanal reduces Fehling's solution and propanal gives no iodoform test, so it fits.
3. 2-Methylbut-1-ene gives butanone and 2-methylbut-2-ene gives propanone, both of which give iodoform, and but-1-ene is not \(\ce{C5H10}\).

Teacher's Note:
1. Work backwards by joining the carbonyl carbons of the two fragments with a double bond.
2. Check the formula first, it eliminates but-1-ene at once.

Answer: (a) 2-Hydroxybut-3-enoic acid

Explanation:
1. Methanal has no alpha hydrogen, so only ethanal forms the enolate, and the crossed aldol with methanal followed by dehydration gives propenal (acrolein), \(\ce{CH2=CHCHO}\).
2. \(\ce{HCN}\) adds to the carbonyl group to give a cyanohydrin and hydrolysis changes \(\ce{CN}\) into \(\ce{COOH}\), giving \(\ce{CH2=CHCH(OH)COOH}\).

Teacher's Note:
1. Cyanohydrin formation and hydrolysis converts an aldehyde into a 2-hydroxy acid.
2. Hydrolysis changes the nitrile carbon into the carboxyl carbon.

Answer: (c) Butane-2,3-dione, \(\ce{CH3COCOCH3}\)

Explanation:
1. An intramolecular Cannizzaro (benzilic-type hydride shift) needs an aldehyde group and no alpha hydrogen.
2. Butane-2,3-dione has no aldehyde group and has alpha hydrogens, so it would undergo aldol-type reactions or benzilic acid rearrangement but not this reaction.

Teacher's Note:
1. The reaction needs a hydride donor, which is an aldehyde hydrate.
2. Compounds with a 1,2-dicarbonyl unit and a CHO group migrate the hydride internally.

Answer: (a) \(\ce{DCOO-}\) and \(\ce{CH2DOH}\)

Explanation:
1. Hydroxide adds to give the tetrahedral hydrate anion, which transfers a hydride ion (here the hydrogen) to a second molecule.
2. The molecule that gave up \(\ce{H-}\) becomes \(\ce{DCOO-}\), and the molecule that accepted it becomes \(\ce{CH2DOH}\).

Teacher's Note:
1. The hydrogen atom on the oxygen of the alcohol comes from the solvent, not from the aldehyde.
2. The carbon-bound hydrogens and deuterium of the products come from the aldehyde.

Answer: (a) Wolff-Kishner reduction

Explanation:
1. The carbonyl group must be reduced to \(\ce{CH2}\) and the alcohol must survive.
2. Wolff-Kishner uses hydrazine and base, which leaves the hydroxyl group alone, while Clemmensen is strongly acidic and could dehydrate the alcohol. \(\ce{LiAlH4}\) and \(\ce{NaBH4}\) would give a diol.

Teacher's Note:
1. Choose Wolff-Kishner for acid-sensitive substrates and Clemmensen for base-sensitive ones.
2. Metal hydrides reduce the carbonyl to a hydroxyl and not to a methylene.

Answer: (b) 2-Methylpropanal

Explanation:
1. Reaction with 2,4-DNP shows a carbonyl group, which rules out the two alcohols.
2. Butan-2-one is a methyl ketone and gives a haloform test, so the compound must be 2-methylpropanal, which has no \(\ce{CH3CO-}\) unit.

Teacher's Note:
1. Check the formula against the degree of unsaturation (one here) before choosing.
2. Butanal is also \(\ce{C4H8O}\) and would also fit if it were an option.

Answer: (d) Cyclohexanone

Explanation:
1. Hydrate formation is favoured by electron-withdrawing neighbouring groups, as in the vicinal tricarbonyl compounds, and by relief of angle strain, as in cyclopropanone.
2. Cyclohexanone has neither feature, so its hydrate is unstable.

Teacher's Note:
1. The equilibrium for hydration lies far to the left for simple ketones.
2. Chloral hydrate and ninhydrin are the classical stable hydrates.

Answer: (b) \(\ce{PhC(CH3)=N-NH-CO-NH2}\)

Explanation:
1. PCC oxidises the secondary alcohol to acetophenone (A).
2. Acetophenone condenses with the hydrazine-type \(\ce{NH2}\) of semicarbazide to give \(\ce{PhC(CH3)=N-NH-CO-NH2}\), acetophenone semicarbazone.

Teacher's Note:
1. Keep the methyl group on the carbon, because the carbon skeleton does not change in a condensation.
2. The semicarbazone of a ketone has two substituents on the \(\ce{C=N}\) carbon.

Answer: (a) less electrophilic

Explanation:
1. In a carboxylic acid the lone pair on the hydroxyl oxygen is delocalised into the carbonyl group.
2. This reduces the positive character of the carbon, so it is less electrophilic than the carbonyl carbon of an aldehyde or ketone.

Teacher's Note:
1. This is why carboxylic acids do not give the addition reactions typical of aldehydes and ketones.
2. The same resonance is what makes the acid group a poor site for nucleophilic addition.