JEE Chemistry Aldehydes Ketones and Carboxylic Acids MCQs Set 24

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Multiple Choice Questions

Question 1: Which one of the following pairs is not correctly matched?
(a) Clemmensen reduction: >C=O to >CH2
(b) Wolff-Kishner reduction: >C=O to >CHOH
(c) Rosenmund reduction: -COCl to -CHO
(d) Stephen reduction: -C≡N to -CHO
Show Answer & Explanation

Answer: (b) Wolff-Kishner reduction: >C=O to >CHOH

Explanation:
1. Wolff-Kishner reduction gives >CH2, not an alcohol.

Teacher's Note:
1. Clemmensen (Zn-Hg/HCl, acid) and Wolff-Kishner (N2H4/KOH, base) both convert C=O to CH2; choose by the acid or base sensitivity of the rest of the molecule.
2. Use the tests: Tollens' and Fehling's for aldehydes, iodoform for CH3CO- and CH3CH(OH)-, 2,4-DNP for any carbonyl.

Question 2: Which of the following gives an aldol condensation reaction?
(a) \( \ce{C6H5OH} \)
(b) \( \ce{C6H5COC6H5} \)
(c) \( \ce{CH3CH2COCH3} \)
(d) \( \ce{(CH3)3C-CO-C(CH3)3} \)
Show Answer & Explanation

Answer: (c) \( \ce{CH3CH2COCH3} \)

Explanation:
1. Only butanone has alpha-hydrogens.
2. Phenol, benzophenone and di-tert-butyl ketone have none.

Teacher's Note:
1. Aldol addition needs at least one alpha-hydrogen; dilute alkali gives the beta-hydroxy carbonyl compound, and heating dehydrates it to an enone.
2. Aldol needs alpha-hydrogen; Cannizzaro needs none.

Question 3: Benzaldehyde is treated with \( \ce{CH3MgBr} \) and the addition product is subjected to acid hydrolysis. The product formed is
(a) Secondary alcohol
(b) A primary alcohol
(c) Phenol
(d) Tertiary butyl alcohol
Show Answer & Explanation

Answer: (a) Secondary alcohol

Explanation:
1. Methyl adds to the carbonyl carbon to give 1-phenylethanol, a secondary alcohol.

Teacher's Note:
1. Compare reactivity with steric hindrance and electron donation: HCHO > RCHO > ArCHO > R2CO > Ar2CO.
2. Use the tests: Tollens' and Fehling's for aldehydes, iodoform for CH3CO- and CH3CH(OH)-, 2,4-DNP for any carbonyl.

Question 4: Aldol condensation will not be observed in
(a) Chloral
(b) Phenylacetaldehyde
(c) Hexanal
(d) Propanal
Show Answer & Explanation

Answer: (a) Chloral

Explanation:
1. Chloral has no alpha-hydrogen, whereas the others do.

Teacher's Note:
1. Aldol addition needs at least one alpha-hydrogen; dilute alkali gives the beta-hydroxy carbonyl compound, and heating dehydrates it to an enone.
2. Use the tests: Tollens' and Fehling's for aldehydes, iodoform for CH3CO- and CH3CH(OH)-, 2,4-DNP for any carbonyl.

Question 5: Which of the following carbonyl compounds will give a coloured crystalline compound with 2,4-dinitrophenylhydrazine?
(a) \( \ce{CH3COCl} \)
(b) \( \ce{CH3COCH3} \)
(c) \( \ce{CH3CO(OC2H5)} \)
(d) \( \ce{CH3CONH2} \)
Show Answer & Explanation

Answer: (b) \( \ce{CH3COCH3} \)

Explanation:
1. Only aldehydes and ketones form 2,4-DNP hydrazones; acid derivatives (acyl chloride, ester, amide) do not.

Teacher's Note:
1. Ammonia derivatives (NH2OH, NH2NH2, PhNHNH2, semicarbazide, 2,4-DNP) condense with C=O to give C=N products by addition then loss of water.
2. Aldol needs alpha-hydrogen; Cannizzaro needs none.

Question 6: Which of the following compounds will undergo self aldol condensation in cold dilute alkali?
(a) \( \ce{C6H5CHO} \)
(b) \( \ce{CH3CH2CHO} \)
(c) \( \ce{CH#C-CHO} \)
(d) \( \ce{CH2=CH-CHO} \)
Show Answer & Explanation

Answer: (b) \( \ce{CH3CH2CHO} \)

Explanation:
1. Only propanal has alpha-hydrogens.

Teacher's Note:
1. Aldol addition needs at least one alpha-hydrogen; dilute alkali gives the beta-hydroxy carbonyl compound, and heating dehydrates it to an enone.
2. Use the tests: Tollens' and Fehling's for aldehydes, iodoform for CH3CO- and CH3CH(OH)-, 2,4-DNP for any carbonyl.

Question 7: Acetaldehyde treated with dilute NaOH gives
(a) \( \ce{CH3CH2OH} \)
(b) \( \ce{CH3COOH} \)
(c) \( \ce{CH3CH(OH)CH2CHO} \)
(d) \( \ce{CH3-CH3} \)
Show Answer & Explanation

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

Explanation:
1. Dilute alkali gives aldol addition: 3-hydroxybutanal.

Teacher's Note:
1. Aldol addition needs at least one alpha-hydrogen; dilute alkali gives the beta-hydroxy carbonyl compound, and heating dehydrates it to an enone.
2. Aldol needs alpha-hydrogen; Cannizzaro needs none.

Question 8: \( \ce{C2H5CHO} \) and \( \ce{(CH3)2CO} \) can be distinguished by testing with
(a) Phenylhydrazine
(b) Hydroxylamine
(c) Fehling's solution
(d) Sodium bisulphite
Show Answer & Explanation

Answer: (c) Fehling's solution

Explanation:
1. Propanal reduces Fehling's solution; acetone does not.
2. The other reagents react with both.

Teacher's Note:
1. Fehling's solution is reduced by aliphatic aldehydes (red Cu2O) but not by ketones or aromatic aldehydes.
2. Compare reactivity with steric hindrance and electron donation: HCHO > RCHO > ArCHO > R2CO > Ar2CO.

Question 9: Acetaldehyde cannot show
(a) Iodoform test
(b) Lucas test
(c) Benedict's test
(d) Tollens' test
Show Answer & Explanation

Answer: (b) Lucas test

Explanation:
1. Lucas test is for alcohols.
2. Acetaldehyde gives the other three tests.

Teacher's Note:
1. Use the tests: Tollens' and Fehling's for aldehydes, iodoform for CH3CO- and CH3CH(OH)-, 2,4-DNP for any carbonyl.
2. Aldol needs alpha-hydrogen; Cannizzaro needs none.

Question 10: Benzaldehyde with NaOH gives
(a) Benzyl alcohol and sodium benzoate
(b) Hydrobenzamide
(c) Cinnamic acid
(d) Sodium cinnamate
Show Answer & Explanation

Answer: (a) Benzyl alcohol and sodium benzoate

Explanation:
1. Benzaldehyde has no alpha-hydrogen, so it gives Cannizzaro reaction, i.e. benzyl alcohol and sodium benzoate.

Teacher's Note:
1. Cannizzaro reaction: an aldehyde with no alpha-hydrogen, in strong alkali, gives equal amounts of alcohol and carboxylate through hydride transfer.
2. Aldol needs alpha-hydrogen; Cannizzaro needs none.

Question 11: The reagent that converts \( \ce{C6H5COCHO} \) to \( \ce{C6H5CHOHCOONa} \) is
(a) Aq. NaOH
(b) Acidic \( \ce{Na2S2O3} \)
(c) \( \ce{Na2Cr2O4/H2SO4} \)
(d) \( \ce{NaNO2/HCl} \)
Show Answer & Explanation

Answer: (a) Aq. NaOH

Explanation:
1. Phenylglyoxal has no alpha-hydrogen, so aqueous NaOH causes intramolecular Cannizzaro reaction, giving sodium mandelate.

Teacher's Note:
1. Cannizzaro reaction: an aldehyde with no alpha-hydrogen, in strong alkali, gives equal amounts of alcohol and carboxylate through hydride transfer.
2. Compare reactivity with steric hindrance and electron donation: HCHO > RCHO > ArCHO > R2CO > Ar2CO.

Question 12: Benzyl alcohol and sodium benzoate are obtained by the action of sodium hydroxide on benzaldehyde. This reaction is known as
(a) Perkin's reaction
(b) Cannizzaro's reaction
(c) Sandmeyer's reaction
(d) Claisen condensation
Show Answer & Explanation

Answer: (b) Cannizzaro's reaction

Explanation:
1. Disproportionation of an aldehyde lacking alpha-hydrogen in strong alkali is the Cannizzaro reaction.

Teacher's Note:
1. Cannizzaro reaction: an aldehyde with no alpha-hydrogen, in strong alkali, gives equal amounts of alcohol and carboxylate through hydride transfer.
2. Use the tests: Tollens' and Fehling's for aldehydes, iodoform for CH3CO- and CH3CH(OH)-, 2,4-DNP for any carbonyl.

Question 13: Which of the following does not give the iodoform test?
(a) \( \ce{CH3CH2OH} \)
(b) \( \ce{CH3OH} \)
(c) \( \ce{CH3CHO} \)
(d) \( \ce{PhCOCH3} \)
Show Answer & Explanation

Answer: (b) \( \ce{CH3OH} \)

Explanation:
1. Methanol has no CH3CH(OH)- or CH3CO- unit.

Teacher's Note:
1. The iodoform test is positive for CH3CO- groups and CH3CH(OH)- groups (ethanol, propan-2-ol, ethanal, methyl ketones); methanal, benzophenone and 3-pentanone do not respond.
2. Compare reactivity with steric hindrance and electron donation: HCHO > RCHO > ArCHO > R2CO > Ar2CO.

Question 14: Which of the following will not give the iodoform test?
(a) Ethanal
(b) Ethanol
(c) 2-Propanone
(d) 3-Pentanone
Show Answer & Explanation

Answer: (d) 3-Pentanone

Explanation:
1. 3-Pentanone has no methyl ketone group and no CH3CH(OH)- group.

Teacher's Note:
1. The iodoform test is positive for CH3CO- groups and CH3CH(OH)- groups (ethanol, propan-2-ol, ethanal, methyl ketones); methanal, benzophenone and 3-pentanone do not respond.
2. Use the tests: Tollens' and Fehling's for aldehydes, iodoform for CH3CO- and CH3CH(OH)-, 2,4-DNP for any carbonyl.

Question 15: Which of the following will not give the iodoform test?
(a) Acetophenone
(b) Ethanal
(c) Benzophenone
(d) Ethanol
Show Answer & Explanation

Answer: (c) Benzophenone

Explanation:
1. Benzophenone has no CH3CO- group.

Teacher's Note:
1. The iodoform test is positive for CH3CO- groups and CH3CH(OH)- groups (ethanol, propan-2-ol, ethanal, methyl ketones); methanal, benzophenone and 3-pentanone do not respond.
2. Aldol needs alpha-hydrogen; Cannizzaro needs none.

Question 16: The haloform test is given by
(a) \( \ce{HCHO} \)
(b) \( \ce{(CH3)2CO} \)
(c) \( \ce{CH3OCH3} \)
(d) \( \ce{CH3CH2Cl} \)
Show Answer & Explanation

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

Explanation:
1. Acetone has the CH3CO- group.

Teacher's Note:
1. The iodoform test is positive for CH3CO- groups and CH3CH(OH)- groups (ethanol, propan-2-ol, ethanal, methyl ketones); methanal, benzophenone and 3-pentanone do not respond.
2. Compare reactivity with steric hindrance and electron donation: HCHO > RCHO > ArCHO > R2CO > Ar2CO.

Question 17: Dimethyl ketones are usually characterised through
(a) Tollens' reagent
(b) Schiff's test
(c) Iodoform test
(d) Benedict's reagent
Show Answer & Explanation

Answer: (c) Iodoform test

Explanation:
1. The CH3CO- group of methyl ketones gives yellow iodoform.

Teacher's Note:
1. The iodoform test is positive for CH3CO- groups and CH3CH(OH)- groups (ethanol, propan-2-ol, ethanal, methyl ketones); methanal, benzophenone and 3-pentanone do not respond.
2. Use the tests: Tollens' and Fehling's for aldehydes, iodoform for CH3CO- and CH3CH(OH)-, 2,4-DNP for any carbonyl.

Question 18: The light yellow compound produced when acetone reacts with iodine and alkali is
(a) \( \ce{CH3COCH2I} \)
(b) \( \ce{CH3I} \)
(c) \( \ce{CHI3} \)
(d) None of these
Show Answer & Explanation

Answer: (c) \( \ce{CHI3} \)

Explanation:
1. Triiodination of the methyl group followed by hydroxide cleavage gives iodoform and a carboxylate.

Teacher's Note:
1. The iodoform test is positive for CH3CO- groups and CH3CH(OH)- groups (ethanol, propan-2-ol, ethanal, methyl ketones); methanal, benzophenone and 3-pentanone do not respond.
2. Aldol needs alpha-hydrogen; Cannizzaro needs none.

Question 19: If formaldehyde and KOH are heated, then we get
(a) Acetylene
(b) Methane
(c) Methyl alcohol
(d) Ethyl formate
Show Answer & Explanation

Answer: (c) Methyl alcohol

Explanation:
1. Cannizzaro reaction gives methanol and potassium formate.

Teacher's Note:
1. Cannizzaro reaction: an aldehyde with no alpha-hydrogen, in strong alkali, gives equal amounts of alcohol and carboxylate through hydride transfer.
2. Compare reactivity with steric hindrance and electron donation: HCHO > RCHO > ArCHO > R2CO > Ar2CO.

Question 20: Acetaldehyde reacts with \( \ce{C2H5MgCl} \) and the final product is
(a) An aldehyde
(b) A ketone
(c) A primary alcohol
(d) A secondary alcohol
Show Answer & Explanation

Answer: (d) A secondary alcohol

Explanation:
1. The product is \( \ce{CH3CH(OH)C2H5} \), butan-2-ol.

Teacher's Note:
1. Aldol needs alpha-hydrogen; Cannizzaro needs none.
2. Compare reactivity with steric hindrance and electron donation: HCHO > RCHO > ArCHO > R2CO > Ar2CO.

Question 21: Treatment of propionaldehyde with dilute NaOH solution gives
(a) \( \ce{CH3CH2COOCH2CH2CH3} \)
(b) \( \ce{CH3CH2CH(OH)CH(CH3)CHO} \)
(c) \( \ce{CH3CH2CH(OH)CH2CH2CHO} \)
(d) \( \ce{CH3CH2COCH2CH2CHO} \)
Show Answer & Explanation

Answer: (b) \( \ce{CH3CH2CH(OH)CH(CH3)CHO} \)

Explanation:
1. Aldol addition of propanal: the alpha-carbon of one molecule attacks the carbonyl of another, giving 3-hydroxy-2-methylpentanal.

Teacher's Note:
1. Aldol addition needs at least one alpha-hydrogen; dilute alkali gives the beta-hydroxy carbonyl compound, and heating dehydrates it to an enone.
2. Compare reactivity with steric hindrance and electron donation: HCHO > RCHO > ArCHO > R2CO > Ar2CO.

Question 22: Aldol condensation of acetaldehyde involves the formation of which of the following intermediate?
(a) An acetate ion
(b) A carbanion (enolate)
(c) A carbonium ion
(d) A free radical
Show Answer & Explanation

Answer: (b) A carbanion (enolate)

Explanation:
1. Base removes an alpha-hydrogen to give a resonance-stabilised enolate (carbanion), which attacks another carbonyl.

Teacher's Note:
1. Aldol addition needs at least one alpha-hydrogen; dilute alkali gives the beta-hydroxy carbonyl compound, and heating dehydrates it to an enone.
2. Use the tests: Tollens' and Fehling's for aldehydes, iodoform for CH3CO- and CH3CH(OH)-, 2,4-DNP for any carbonyl.

Question 23: Three molecules of acetone with HCl give B, \( \ce{(CH3)2C=CH-CO-CH=C(CH3)2} \). B is
(a) Phorone
(b) Formose
(c) Diacetone alcohol
(d) Mesityl oxide
Show Answer & Explanation

Answer: (a) Phorone

Explanation:
1. Three acetone molecules condense with loss of two water molecules to give phorone.

Teacher's Note:
1. Aldol needs alpha-hydrogen; Cannizzaro needs none.
2. Compare reactivity with steric hindrance and electron donation: HCHO > RCHO > ArCHO > R2CO > Ar2CO.

Question 24: Aromatic aldehydes undergo disproportionation in presence of sodium or potassium hydroxide to give the corresponding alcohol and acid. The reaction is known as
(a) Wurtz reaction
(b) Cannizzaro reaction
(c) Friedel-Crafts reaction
(d) Claisen reaction
Show Answer & Explanation

Answer: (b) Cannizzaro reaction

Explanation:
1. Cannizzaro reaction is the disproportionation of aldehydes lacking alpha-hydrogen.

Teacher's Note:
1. Cannizzaro reaction: an aldehyde with no alpha-hydrogen, in strong alkali, gives equal amounts of alcohol and carboxylate through hydride transfer.
2. Compare reactivity with steric hindrance and electron donation: HCHO > RCHO > ArCHO > R2CO > Ar2CO.

Question 25: m-Chlorobenzaldehyde on reaction with conc. KOH at room temperature gives
(a) Potassium m-chlorobenzoate and m-hydroxybenzaldehyde
(b) m-Hydroxybenzaldehyde and m-chlorobenzyl alcohol
(c) m-Chlorobenzyl alcohol and m-hydroxybenzyl alcohol
(d) Potassium m-chlorobenzoate and m-chlorobenzyl alcohol
Show Answer & Explanation

Answer: (d) Potassium m-chlorobenzoate and m-chlorobenzyl alcohol

Explanation:
1. Cannizzaro reaction takes place and the chlorine stays on the ring.

Teacher's Note:
1. Cannizzaro reaction: an aldehyde with no alpha-hydrogen, in strong alkali, gives equal amounts of alcohol and carboxylate through hydride transfer.
2. Use the tests: Tollens' and Fehling's for aldehydes, iodoform for CH3CO- and CH3CH(OH)-, 2,4-DNP for any carbonyl.

Question 26: Which of the following does not give a yellow precipitate with \( \ce{NaOH + I2} \)?
(a) Acetone
(b) Acetaldehyde
(c) Benzaldehyde
(d) Acetophenone
Show Answer & Explanation

Answer: (c) Benzaldehyde

Explanation:
1. Benzaldehyde has no CH3CO- group, so it gives no iodoform.

Teacher's Note:
1. The iodoform test is positive for CH3CO- groups and CH3CH(OH)- groups (ethanol, propan-2-ol, ethanal, methyl ketones); methanal, benzophenone and 3-pentanone do not respond.
2. Aldol needs alpha-hydrogen; Cannizzaro needs none.

Question 27: The correct order of reactivity of \( \ce{CH3CHO} \), \( \ce{C2H5COCH3} \) and \( \ce{CH3COCH3} \) towards nucleophilic addition is
(a) \( \ce{CH3CHO > CH3COCH3 > CH3COC2H5} \)
(b) \( \ce{C2H5COCH3 > CH3COCH3 > CH3CHO} \)
(c) \( \ce{CH3COCH3 > CH3CHO > C2H5COCH3} \)
(d) \( \ce{CH3COCH3 > C2H5COCH3 > CH3CHO} \)
Show Answer & Explanation

Answer: (a) \( \ce{CH3CHO > CH3COCH3 > CH3COC2H5} \)

Explanation:
1. Aldehydes are more reactive than ketones, and larger alkyl groups reduce reactivity by steric and inductive effects.

Teacher's Note:
1. Use the tests: Tollens' and Fehling's for aldehydes, iodoform for CH3CO- and CH3CH(OH)-, 2,4-DNP for any carbonyl.
2. Aldol needs alpha-hydrogen; Cannizzaro needs none.

Question 28: One mole of an organic compound requires 0.5 mole of oxygen to produce an acid. The compound may be
(a) Alcohol
(b) Ether
(c) Ketone
(d) Aldehyde
Show Answer & Explanation

Answer: (d) Aldehyde

Explanation:
1. RCHO + 1/2 O2 gives RCOOH; the other classes need more oxygen or are not converted to an acid with this stoichiometry.

Teacher's Note:
1. Aldol needs alpha-hydrogen; Cannizzaro needs none.
2. Compare reactivity with steric hindrance and electron donation: HCHO > RCHO > ArCHO > R2CO > Ar2CO.

Question 29: Aldehydes can be oxidised by
(a) Tollens' reagent
(b) Fehling solution
(c) Benedict solution
(d) All of these
Show Answer & Explanation

Answer: (d) All of these

Explanation:
1. Tollens', Fehling and Benedict reagents are all mild oxidants that oxidise aldehydes.

Teacher's Note:
1. Tollens' reagent, [Ag(NH3)2]+, is reduced to silver by aldehydes and not by ketones.
2. Compare reactivity with steric hindrance and electron donation: HCHO > RCHO > ArCHO > R2CO > Ar2CO.

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