JEE Chemistry Aldehydes Ketones and Carboxylic Acids MCQs Set 35

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

Explanation:
1. Aldehydes are more reactive than ketones, so acetophenone is last.
2. The nitro group increases the electrophilicity of the carbonyl carbon, and the methyl group decreases it.
3. The order is p-nitrobenzaldehyde, benzaldehyde, p-methylbenzaldehyde, acetophenone.

Teacher's Note:
1. Electron-withdrawing groups raise reactivity.
2. Ketones are the least reactive.

Answer: (d) phenylacetic acid

Explanation:
1. Bromine in light brominates the methyl group to give benzyl bromide.
2. KCN replaces Br by CN to give benzyl cyanide, and hydrolysis gives phenylacetic acid.

Teacher's Note:
1. Light means side-chain substitution.
2. The nitrile adds one carbon.

Answer: (c) \( \ce{CH3CH2CH(OD)CH2CH2CH3} \)

Explanation:
1. The hydride from \( \ce{NaBH4} \) (H, not D) adds to the carbonyl carbon.
2. The oxygen anion is then protonated by \( \ce{D2O} \), so the OH becomes OD.

Teacher's Note:
1. Carbon gets H from the hydride.
2. Oxygen gets D from D2O.

Answer: (b) It reacts with Tollens' reagent to form a silver mirror

Explanation:
1. Acetophenone is a methyl ketone and does not reduce Tollens' reagent.
2. It gives the iodoform test and a 2,4-DNP derivative.
3. Strong oxidation cleaves it to benzoic acid.

Teacher's Note:
1. Ketones do not give silver mirror.
2. The CH3CO group gives iodoform.

Answer: (b) formamide

Explanation:
1. A compound that gives ammonia with soda lime is an amide.
2. Formamide is a liquid at room temperature and gives sodium formate and ammonia, while propanamide is a solid.

Teacher's Note:
1. Amides with soda lime give salt plus ammonia.
2. Check the physical state.

Answer: (d) resonance stabilisation of their conjugate base

Explanation:
1. The carboxylate ion is stabilised by delocalisation of the negative charge over two oxygen atoms.
2. A more stable conjugate base means a stronger acid.

Teacher's Note:
1. Acidity depends on conjugate base stability.
2. Phenoxide has less delocalisation of charge on the oxygen.

Answer: (a) acetic acid

Explanation:
1. Only carboxylic acids are strong enough acids to decompose bicarbonate with evolution of \( \ce{CO2} \).
2. Phenols and alcohols are weaker acids than carbonic acid.

Teacher's Note:
1. This is the test for carboxylic acids.
2. Phenol is less acidic than carbonic acid.

Answer: (d) ethyl benzoate

Explanation:
1. Alkaline hydrolysis gives the potassium salt of the acid, and acidification liberates the free acid.
2. Only benzoic acid is a white crystalline solid, while the other acids are liquids soluble in water.

Teacher's Note:
1. Benzoic acid is sparingly soluble in cold water.
2. Acetic and formic acids are liquids.

Answer: (b) methyl ethanoate

Explanation:
1. Kolbe electrolysis of acetate gives ethane.
2. The acid must be acetic acid, so the ester is of ethanoic acid, i.e. methyl ethanoate.

Teacher's Note:
1. Two acetate radicals couple to give ethane.
2. Formic acid would give hydrogen.

Answer: (c) 1,4-dibromobutane

Explanation:
1. Both active hydrogens of malonic ester are alkylated by the dibromide to form a ring.
2. 1,4-Dibromobutane gives a five-membered ring including the malonate carbon.
3. Hydrolysis and decarboxylation give cyclopentanecarboxylic acid.

Teacher's Note:
1. Ring size = number of carbons in the dihalide plus one.
2. Heating removes one COOH as \( \ce{CO2} \).

Answer: (b) \( \ce{(CH3)2CHCH2COOH} \)

Explanation:
1. \( \ce{LiAlH4} \) reduces the acid to 2-methylpropan-1-ol, and \( \ce{PBr3} \) converts it to 1-bromo-2-methylpropane.
2. KCN gives the nitrile, which on hydrolysis gives 3-methylbutanoic acid (one carbon longer).

Teacher's Note:
1. This sequence is a one-carbon chain extension.
2. The nitrile is hydrolysed to COOH.

Answer: (c) Tollens' reagent is used in detection of unsaturation

Explanation:
1. Tollens' reagent detects aldehydes (silver mirror), not unsaturation.
2. Bromine water or Baeyer's reagent is used for unsaturation.

Teacher's Note:
1. Match each reagent with its test.
2. Tollens' means reducing aldehydes.

Answer: (c) \( \ce{KMnO4} \) (aq); HI/P

Explanation:
1. Alkaline permanganate oxidises the side chain of toluene to benzoic acid.
2. Strong reduction with HI and red phosphorus converts the carboxyl group back to a methyl group.

Teacher's Note:
1. Alkyl side chains oxidise to COOH.
2. HI/P is a powerful reducing agent.

Answer: (c) acetone

Explanation:
1. Hydration of ethyne gives acetaldehyde (A).
2. Methyl Grignard adds to give propan-2-ol (B).
3. Oxidation of the secondary alcohol gives acetone (C).

Teacher's Note:
1. Acetylene hydrates to acetaldehyde.
2. Secondary alcohol oxidises to a ketone.

Answer: (d) \( \ce{CH3CH(OH)CH2CHO} \)

Explanation:
1. Ethyne is hydrated to acetaldehyde (A).
2. Dilute NaOH then causes aldol addition of two molecules of acetaldehyde, giving 3-hydroxybutanal.

Teacher's Note:
1. Acetaldehyde has alpha hydrogens.
2. Aldol addition gives a beta-hydroxy aldehyde.

Answer: (b) \( \ce{CH3CH(OH)CH2CH3} \)

Explanation:
1. The oxidation product is a methyl ketone, butanone, so the original compound is the secondary alcohol butan-2-ol.
2. Dehydration with conc. \( \ce{H2SO4} \) gives butene (\( \ce{C4H8} \)).

Teacher's Note:
1. Iodoform positive means \( \ce{CH3CO-} \).
2. Tertiary alcohols do not oxidise easily.

Answer: (a) glycol, then \( \ce{LiAlH4} \), then \( \ce{H3O+} \)

Explanation:
1. The ketone must be protected as a cyclic ketal with glycol before the ester is reduced.
2. \( \ce{LiAlH4} \) reduces the ester to the primary alcohol and acid then removes the ketal and restores the ketone.

Teacher's Note:
1. Use a protecting group to differentiate two carbonyl groups.
2. \( \ce{LiAlH4} \) alone would reduce both groups.

Answer: (b) \( \ce{HCHO} \) > \( \ce{CH3CHO} \) > \( \ce{CH3COCH3} \)

Explanation:
1. Alkyl groups donate electrons and hinder the approach of the nucleophile.
2. Formaldehyde has none, acetaldehyde has one and acetone has two.

Teacher's Note:
1. Fewer alkyl groups mean higher reactivity.
2. HCHO is the most reactive carbonyl compound.

Answer: (d) Clemmensen reduction

Explanation:
1. Zinc amalgam with conc. HCl reduces the carbonyl group to CH2.
2. This is the Clemmensen reduction.

Teacher's Note:
1. Clemmensen uses Zn(Hg)/HCl.
2. Wolff-Kishner uses hydrazine and KOH.

Answer: (d) Hell-Volhard-Zelinsky reaction

Explanation:
1. The alpha-halogenation of carboxylic acids with halogen and red phosphorus is the Hell-Volhard-Zelinsky reaction.

Teacher's Note:
1. Red P forms PBr3 in situ.
2. The alpha hydrogen is replaced.

Answer: (d) fluorescein

Explanation:
1. Phthalic anhydride condenses with two molecules of resorcinol in the presence of acid.
2. The product is the dye fluorescein, while phenol gives phenolphthalein.

Teacher's Note:
1. Phenol gives phenolphthalein.
2. Resorcinol gives fluorescein.