JEE Chemistry Aldehydes Ketones and Carboxylic Acids MCQs Set 25

Chemistry Objective Questions and Answers: Aldehydes Ketones and Carboxylic Acids

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Multiple Choice Questions

Question 1: Silver mirror is a test for
(a) Aldehydes
(b) Thio alcohols
(c) Amines
(d) Ethers
Show Answer & Explanation

Answer: (a) Aldehydes

Explanation:
1. Tollens' reagent is reduced by aldehydes to metallic silver.

Teacher's Note:
1. Tollens' reagent, [Ag(NH3)2]+, is reduced to silver by aldehydes and not by ketones.
2. Use the tests: Tollens' and Fehling's for aldehydes, iodoform for CH3CO- and CH3CH(OH)-, 2,4-DNP for any carbonyl.

Question 2: Crotonaldehyde, \( \ce{CH3CH=CHCHO} \), is oxidised to crotonic acid without affecting the C=C bond using
(a) Alkaline \( \ce{KMnO4} \)
(b) Selenium dioxide
(c) Ammoniacal \( \ce{AgNO3} \)
(d) All of these
Show Answer & Explanation

Answer: (c) Ammoniacal \( \ce{AgNO3} \)

Explanation:
1. Tollens' reagent is a mild oxidant that attacks only the CHO group; alkaline KMnO4 would also cleave the double bond.

Teacher's Note:
1. Tollens' reagent, [Ag(NH3)2]+, is reduced to silver by aldehydes and not by ketones.
2. Aldol needs alpha-hydrogen; Cannizzaro needs none.

Question 3: Which of the following does not turn Schiff's reagent pink?
(a) Formaldehyde
(b) Benzaldehyde
(c) Acetone
(d) Acetaldehyde
Show Answer & Explanation

Answer: (c) Acetone

Explanation:
1. Schiff's reagent is restored to magenta by aldehydes, not by ketones.

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: Fehling's test is positive for
(a) Acetaldehyde
(b) Benzaldehyde
(c) Ether
(d) Ethyl alcohol
Show Answer & Explanation

Answer: (a) Acetaldehyde

Explanation:
1. Aliphatic aldehydes reduce Fehling's solution to red \( \ce{Cu2O} \).

Teacher's Note:
1. Fehling's solution is reduced by aliphatic aldehydes (red Cu2O) but not by ketones or aromatic aldehydes.
2. Use the tests: Tollens' and Fehling's for aldehydes, iodoform for CH3CO- and CH3CH(OH)-, 2,4-DNP for any carbonyl.

Question 5: The final product formed when acetaldehyde is reduced with sodium and alcohol is
(a) Ethylene
(b) Ethyl alcohol
(c) Ethene and ethanol
(d) Ethyl chloride
Show Answer & Explanation

Answer: (b) Ethyl alcohol

Explanation:
1. Nascent hydrogen from Na/alcohol reduces the carbonyl to give ethanol.

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 6: The compound obtained by the reduction of propionaldehyde by amalgamated zinc and concentrated HCl is
(a) Propanol
(b) Propane
(c) Propene
(d) Propanal
Show Answer & Explanation

Answer: (b) Propane

Explanation:
1. Clemmensen reduction converts the CHO group to CH3, giving propane.

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 7: Formaldehyde when treated with KOH gives methanol and potassium formate. The reaction is known as
(a) Perkin reaction
(b) Claisen reaction
(c) Cannizzaro reaction
(d) Knoevenagel reaction
Show Answer & Explanation

Answer: (c) Cannizzaro reaction

Explanation:
1. This is a disproportionation of an aldehyde without alpha-hydrogen, 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. Aldol needs alpha-hydrogen; Cannizzaro needs none.

Question 8: Aldehydes and ketones give addition reactions with
(a) Hydrazine
(b) Phenylhydrazine and semicarbazide only
(c) HCN only
(d) All of these
Show Answer & Explanation

Answer: (d) All of these

Explanation:
1. The carbonyl group undergoes nucleophilic addition-elimination with hydrazine, phenylhydrazine, semicarbazide and addition with HCN.

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 9: The typical reactions of aldehydes are
(a) Electrophilic addition
(b) Nucleophilic substitution
(c) Nucleophilic addition
(d) Nucleophilic elimination
Show Answer & Explanation

Answer: (c) Nucleophilic addition

Explanation:
1. The electrophilic carbonyl carbon is attacked by nucleophiles, giving nucleophilic addition.

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: Which of the following will not give acetamide on reaction with ammonia?
(a) Acetic acid
(b) Acetyl chloride
(c) Acetic anhydride
(d) Methyl formate
Show Answer & Explanation

Answer: (a) Acetic acid

Explanation:
1. Acetic acid only forms ammonium acetate with ammonia in the cold.
2. Acetyl chloride and acetic anhydride give acetamide, and methyl formate gives formamide.

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 11: The addition of HCN to carbonyl compounds is an example of
(a) Nucleophilic substitution
(b) Electrophilic addition
(c) Nucleophilic addition
(d) Electrophilic substitution
Show Answer & Explanation

Answer: (c) Nucleophilic addition

Explanation:
1. The cyanide ion attacks the carbonyl carbon first, so the reaction is a nucleophilic addition.

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 12: The product formed by the reaction of chlorine with benzaldehyde in the absence of a catalyst is
(a) Chlorobenzene
(b) Benzyl chloride
(c) Benzoyl chloride
(d) o-Chlorobenzaldehyde
Show Answer & Explanation

Answer: (c) Benzoyl chloride

Explanation:
1. Without a Lewis acid, chlorine replaces the aldehydic hydrogen, giving benzoyl chloride; with a catalyst ring substitution at the meta position occurs.

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 13: Which of the following compounds is resistant to nucleophilic attack by hydroxyl ions?
(a) Methyl acetate
(b) Acetonitrile
(c) Dimethyl ether
(d) Acetamide
Show Answer & Explanation

Answer: (c) Dimethyl ether

Explanation:
1. Ethers have no electrophilic carbon and are inert to hydroxide, whereas the others are hydrolysed.

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 14: Glucose reacts with X molecules of phenylhydrazine to yield osazone. The value of X is
(a) One
(b) Two
(c) Three
(d) Four
Show Answer & Explanation

Answer: (c) Three

Explanation:
1. Three molecules are used: one forms the hydrazone, one oxidises the adjacent CHOH to C=O and is lost as aniline and ammonia, and one forms the second hydrazone.

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. Use the tests: Tollens' and Fehling's for aldehydes, iodoform for CH3CO- and CH3CH(OH)-, 2,4-DNP for any carbonyl.

Question 15: In which reaction is an aromatic aldehyde treated with the acid anhydride in the presence of the corresponding salt of the acid to give an unsaturated aromatic acid?
(a) Friedel-Crafts reaction
(b) Perkin reaction
(c) Wurtz reaction
(d) None of these
Show Answer & Explanation

Answer: (b) Perkin reaction

Explanation:
1. Condensation of benzaldehyde with acetic anhydride and sodium acetate (Perkin reaction) gives cinnamic acid.

Teacher's Note:
1. Perkin reaction: aromatic aldehyde, acid anhydride and the sodium salt of the acid give a cinnamic acid.
2. Aldol needs alpha-hydrogen; Cannizzaro needs none.

Question 16: \( \ce{2 CH3COCH3} \) with Mg-Hg then \( \ce{H^+} \) gives the product
(a) \( \ce{(CH3)2C(OH)-C(OH)(CH3)2} \)
(b) \( \ce{CH3COOCOCH3} \)
(c) \( \ce{CH3CH(OH)CH(OH)CH3} \)
(d) None of these
Show Answer & Explanation

Answer: (a) \( \ce{(CH3)2C(OH)-C(OH)(CH3)2} \)

Explanation:
1. Pinacol coupling: two ketone molecules are reduced and joined by a C-C bond, giving 2,3-dimethylbutane-2,3-diol (pinacol).

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 17: Cinnamic acid is formed when \( \ce{C6H5-CHO} \) condenses with \( \ce{(CH3CO)2O} \) in presence of
(a) Conc. \( \ce{H2SO4} \)
(b) Sodium acetate
(c) Sodium metal
(d) Anhydrous \( \ce{ZnCl2} \)
Show Answer & Explanation

Answer: (b) Sodium acetate

Explanation:
1. Perkin reaction uses the sodium salt of the anhydride's acid as base catalyst.

Teacher's Note:
1. Perkin reaction: aromatic aldehyde, acid anhydride and the sodium salt of the acid give a cinnamic acid.
2. Use the tests: Tollens' and Fehling's for aldehydes, iodoform for CH3CO- and CH3CH(OH)-, 2,4-DNP for any carbonyl.

Question 18: A mixture of benzaldehyde and formaldehyde on heating with aqueous NaOH gives
(a) Benzyl alcohol and sodium formate
(b) Sodium benzoate and methyl alcohol
(c) Sodium benzoate and sodium formate
(d) Benzyl alcohol and methyl alcohol
Show Answer & Explanation

Answer: (a) Benzyl alcohol and sodium formate

Explanation:
1. In crossed Cannizzaro reaction the more easily oxidised formaldehyde becomes formate, and benzaldehyde is reduced to benzyl alcohol.

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 19: Formaldehyde reacts with ammonia to give urotropine. The formula of urotropine is
(a) \( \ce{(CH2)6N4} \)
(b) \( \ce{(CH2)4N3} \)
(c) \( \ce{(CH2)6N6} \)
(d) \( \ce{(CH2)3N3} \)
Show Answer & Explanation

Answer: (a) \( \ce{(CH2)6N4} \)

Explanation:
1. Urotropine (hexamethylenetetramine) is \( \ce{C6H12N4} \).

Teacher's Note:
1. Urotropine, (CH2)6N4, comes from methanal and ammonia; it is a urinary antiseptic.
2. Use the tests: Tollens' and Fehling's for aldehydes, iodoform for CH3CO- and CH3CH(OH)-, 2,4-DNP for any carbonyl.

Question 20: Aldol condensation will not take place in
(a) \( \ce{HCHO} \)
(b) \( \ce{CH3CH2CHO} \)
(c) \( \ce{CH3CHO} \)
(d) \( \ce{CH3COCH3} \)
Show Answer & Explanation

Answer: (a) \( \ce{HCHO} \)

Explanation:
1. Methanal has no alpha-hydrogen.

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 21: Contents of three bottles react as follows: (i) neither Fehling's solution nor Tollens' reagent, (ii) Tollens' reagent only, (iii) both. If they contain ethanal, propanone or benzaldehyde, the bottles contain
(a) (i) benzal, (ii) ethanal, (iii) propanone
(b) (i) benzal, (ii) propanone, (iii) ethanal
(c) (i) propanone, (ii) benzal, (iii) ethanal
(d) (i) propanone, (ii) ethanal, (iii) benzal
Show Answer & Explanation

Answer: (c) (i) propanone, (ii) benzal, (iii) ethanal

Explanation:
1. Propanone reacts with neither.
2. Benzaldehyde reduces Tollens' reagent but not Fehling's.
3. Ethanal reduces both.

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.

Question 22: Action of hydrazine on aldehydes and ketones gives a compound of the general structure
(a) \( \ce{>C=N-NH2} \)
(b) \( \ce{>C=N-OH} \)
(c) \( \ce{>C=N-NH-CONH2} \)
(d) \( \ce{>C=N-NH-C6H5} \)
Show Answer & Explanation

Answer: (a) \( \ce{>C=N-NH2} \)

Explanation:
1. Hydrazine, \( \ce{NH2NH2} \), condenses with the carbonyl group and loses water to form a hydrazone, \( \ce{>C=N-NH2} \).

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. Use the tests: Tollens' and Fehling's for aldehydes, iodoform for CH3CO- and CH3CH(OH)-, 2,4-DNP for any carbonyl.

Question 23: The reaction in which sodium cyanide is used is
(a) Perkin reaction
(b) Reimer-Tiemann reaction
(c) Benzoin condensation
(d) Rosenmund reaction
Show Answer & Explanation

Answer: (c) Benzoin condensation

Explanation:
1. Benzoin condensation of benzaldehyde uses cyanide ion as catalyst.

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: Which of the following reactions is a method for converting a ketone into a hydrocarbon?
(a) Aldol condensation
(b) Reimer-Tiemann reaction
(c) Cannizzaro reaction
(d) Wolff-Kishner reduction
Show Answer & Explanation

Answer: (d) Wolff-Kishner reduction

Explanation:
1. Hydrazine and a strong base in glycol reduce >C=O to >CH2.
2. The others do not remove the oxygen.

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 25: Bakelite is a polymer of
(a) \( \ce{HCHO} \) and phenol
(b) \( \ce{HCHO} \) and acetaldehyde
(c) Phenol and \( \ce{H2SO4} \)
(d) \( \ce{HCHO} \) and acetone
Show Answer & Explanation

Answer: (a) \( \ce{HCHO} \) and phenol

Explanation:
1. Phenol-formaldehyde condensation gives the thermosetting resin Bakelite.

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 26: Enol content is highest in
(a) Acetone
(b) Acetophenone
(c) Acetic acid
(d) Acetyl acetone
Show Answer & Explanation

Answer: (d) Acetyl acetone

Explanation:
1. In pentane-2,4-dione the enol is stabilised by conjugation and intramolecular hydrogen bonding, so its enol content is highest.

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. Compare reactivity with steric hindrance and electron donation: HCHO > RCHO > ArCHO > R2CO > Ar2CO.

Question 27: Which of the following reacts with HCN and Tollens' reagent, but is not oxidised by Fehling's solution?
(a) Methanal
(b) Ethanal
(c) Benzaldehyde
(d) Acetone
Show Answer & Explanation

Answer: (c) Benzaldehyde

Explanation:
1. Aromatic aldehydes react with Tollens' but not Fehling's.
2. Methanal and ethanal reduce both, and acetone reduces neither.

Teacher's Note:
1. Tollens' reagent, [Ag(NH3)2]+, is reduced to silver by aldehydes and not by ketones.
2. Use the tests: Tollens' and Fehling's for aldehydes, iodoform for CH3CO- and CH3CH(OH)-, 2,4-DNP for any carbonyl.

Question 28: During reaction of benzaldehyde with alkali one of the products is
(a) Phenol
(b) Benzyl alcohol
(c) Benzene
(d) Benzophenone
Show Answer & Explanation

Answer: (b) Benzyl alcohol

Explanation:
1. Cannizzaro reaction gives benzyl alcohol and 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 29: Cannizzaro reaction is given by
(a) \( \ce{HCHO} \)
(b) \( \ce{CH3COCH3} \)
(c) \( \ce{CH3CHO} \)
(d) \( \ce{CH3CH2OH} \)
Show Answer & Explanation

Answer: (a) \( \ce{HCHO} \)

Explanation:
1. Only methanal lacks alpha-hydrogens.
2. Acetaldehyde undergoes aldol reaction and the others do not have the required group.

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.

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Aldehydes Ketones and Carboxylic Acids Objective Questions & Solutions for JEE Chemistry

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