CBSE Class 12 Chemistry Aldehydes Ketones and Carboxylic Acids MCQs Set 16

Answer: (c) Ethanal

Explanation:
1. Aldehydes are more reactive than ketones (less steric hindrance, larger δ+ on carbon).
2. Benzaldehyde is less reactive than ethanal because the ring donates electrons by resonance.
3. Benzophenone is the least reactive.

Teacher's Note:
1. Order: aliphatic aldehyde > aromatic aldehyde > ketone.

Answer: (a) Of less steric hindrance and a more positive carbonyl carbon

Explanation:
1. A ketone has two alkyl groups, which hinder the nucleophile and release electrons (+I).
2. An aldehyde has only one alkyl group, so the carbonyl carbon is more positive and more accessible.

Teacher's Note:
1. Both steric and electronic effects favour aldehydes.

Answer: (d) 2-Hydroxy-2-methylpropanoic acid

Explanation:
1. CH₃COCH₃ + HCN gives (CH₃)₂C(OH)CN.
2. Hydrolysis changes –CN to –COOH, giving (CH₃)₂C(OH)COOH.

Teacher's Note:
1. The carbon skeleton of the ketone stays the same; one carbon is added.

Answer: (b) Generates the stronger nucleophile CN⁻

Explanation:
1. HCN is a weak acid, so very little CN⁻ is present.
2. Base removes H⁺ from HCN to give CN⁻, which attacks the carbonyl carbon much faster.

Teacher's Note:
1. The nucleophile is CN⁻ and not HCN.

Answer: (c) Benzophenone

Explanation:
1. Bisulphite addition works for aldehydes, methyl ketones and small cyclic ketones.
2. Benzophenone has two bulky phenyl groups that block the attack.

Teacher's Note:
1. Steric hindrance prevents bulky ketones from reacting.

Answer: (d) 1,1-Diethoxyethane

Explanation:
1. The aldehyde first gives the hemiacetal CH₃CH(OH)OC₂H₅.
2. With excess alcohol and dry HCl the acetal CH₃CH(OC₂H₅)₂ forms.

Teacher's Note:
1. An acetal has two –OR groups on the same carbon.

Answer: (a) (CH₃)₂C=NOH, propanone oxime

Explanation:
1. The NH₂ of NH₂OH attacks the carbonyl carbon.
2. Water is lost to give C=N–OH, an oxime.

Teacher's Note:
1. Addition followed by elimination of H₂O gives C=N–Z.

Answer: (b) Test for the carbonyl group in aldehydes and ketones

Explanation:
1. Both aldehydes and ketones give a yellow, orange or red 2,4-DNP derivative.
2. So it detects C=O but cannot tell the two apart.

Teacher's Note:
1. Tollens' or Fehling's tests separate aldehydes from ketones.

Answer: (c) CH₃CH=N–NH–CO–NH₂

Explanation:
1. The NH₂ attached to NH (not the amide NH₂) attacks C=O.
2. Loss of water gives CH₃CH=NNHCONH₂, the semicarbazone.

Teacher's Note:
1. Only the hydrazine-type NH₂ is nucleophilic in semicarbazide.

Answer: (a) Zn–Hg and conc. HCl

Explanation:
1. Zn–Hg/conc. HCl (Clemmensen reduction) converts C=O into CH₂.
2. NaBH₄, LiAlH₄ and H₂/Ni reduce C=O to CHOH, giving cyclohexanol.

Teacher's Note:
1. Metal hydrides give alcohols; Clemmensen and Wolff–Kishner give alkanes.

Answer: (d) NH₂NH₂ followed by KOH in ethylene glycol at high temperature

Explanation:
1. The ketone forms a hydrazone with NH₂NH₂.
2. Heating with KOH in glycol removes N₂ and gives the CH₂ group.

Teacher's Note:
1. Wolff–Kishner is carried out in basic medium.

Answer: (b) Ammoniacal silver nitrate solution

Explanation:
1. Tollens' reagent is AgNO₃ in aqueous ammonia, containing [Ag(NH₃)₂]⁺.
2. The alkaline copper tartrate solution is Fehling's solution.

Teacher's Note:
1. Silver mirror means Tollens' test.

Answer: (a) Propanal

Explanation:
1. Aliphatic aldehydes reduce Cu²⁺ to red-brown Cu₂O.
2. Aromatic aldehydes and ketones do not respond to Fehling's solution.

Teacher's Note:
1. Only aliphatic aldehydes give a positive Fehling's test.

Answer: (c) Tollens' reagent

Explanation:
1. Ethanal gives a silver mirror with Tollens' reagent.
2. Propanone gives no reaction.
3. 2,4-DNP, NaHSO₃ and NH₂OH react with both.

Teacher's Note:
1. Use a reagent that oxidises aldehydes.

Answer: (d) Pentan-3-one

Explanation:
1. The test needs a CH₃CO– group or CH₃CH(OH)– group.
2. Pentan-3-one, CH₃CH₂COCH₂CH₃, has no methyl ketone unit.

Teacher's Note:
1. Look for CH₃–C=O or CH₃–CH(OH)–.

Answer: (b) Benzaldehyde

Explanation:
1. The Cannizzaro reaction needs an aldehyde with no α-hydrogen.
2. Benzaldehyde has none; ethanal and propanal have α-hydrogens and propanone is a ketone.

Teacher's Note:
1. No α-hydrogen means no enolate can form, so disproportionation occurs.

Answer: (c) Methanol and sodium methanoate

Explanation:
1. 2HCHO + NaOH → CH₃OH + HCOONa.
2. One molecule is oxidised to the salt and the other is reduced to the alcohol.

Teacher's Note:
1. Cannizzaro always gives one alcohol and one carboxylate salt.

Answer: (a) 3-Hydroxybutanal

Explanation:
1. This is aldol addition.
2. The enolate of one ethanal adds to the carbonyl carbon of another, giving CH₃CH(OH)CH₂CHO.

Teacher's Note:
1. Aldol = aldehyde + alcohol: a β-hydroxy aldehyde.

Answer: (d) But-2-enal

Explanation:
1. The aldol loses water between the α-carbon and the β-carbon bearing –OH.
2. This gives the conjugated product CH₃CH=CHCHO.

Teacher's Note:
1. Aldol condensation ends in an α,β-unsaturated carbonyl compound.

Answer: (b) 3-Nitrobenzaldehyde

Explanation:
1. The –CHO group is electron withdrawing, so it deactivates the ring and directs the incoming group to the meta position.

Teacher's Note:
1. Carbonyl groups are meta directing.