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Chapter Test: Aldehydes, Ketones & Carboxylic Acids

Chapter Test: Aldehydes, Ketones & Carboxylic Acids | ChemCA.in

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CBSE Class 12 Chapter Test: ALDEHYDES, KETONES & CARBOXYLIC ACIDS

Time Allowed: 1.5 Hours Maximum Marks: 35

General Instructions:

  1. This question paper consists of 16 questions divided into 5 sections: A, B, C, D, and E.
  2. Section A comprises 5 MCQs and 2 Assertion-Reasoning questions of 1 mark each.
  3. Section B comprises 2 Very Short Answer questions of 2 marks each.
  4. Section C comprises 3 Short Answer questions of 3 marks each.
  5. Section D comprises 1 Case-Based question of 4 marks.
  6. Section E comprises 2 Long Answer questions of 5 marks each.
Section A (1 Mark Each)
[1] 1. Which of the following compounds is most reactive towards nucleophilic addition reactions?
  • Ethanal
  • Propanone
  • Benzaldehyde
  • Methanal (Formaldehyde)
[1] 2. The reagent used in the Rosenmund reduction is:
  • SnCl2 / HCl
  • H2 / Pd-BaSO4
  • Zn-Hg / Conc. HCl
  • LiAlH4
[1] 3. Which of the following compounds will give a positive Iodoform test?
  • Pentan-3-one
  • Benzaldehyde
  • Acetophenone
  • Propanal
[1] 4. Which of the following carboxylic acids is the most acidic?
  • CH3COOH
  • F-CH2COOH
  • Cl-CH2COOH
  • Br-CH2COOH
[1] 5. When formaldehyde (HCHO) is treated with 50% NaOH, it undergoes a disproportionation reaction to yield:
  • Methanol and sodium formate
  • Ethanol and sodium ethanoate
  • Methanol and carbon dioxide
  • Only sodium formate

Directions for Q6 & Q7: These questions consist of two statements, each printed as Assertion (A) and Reason (R). Select the correct answer from the codes (a), (b), (c), and (d) as given below:

  • Both A and R are true and R is the correct explanation of A.
  • Both A and R are true but R is not the correct explanation of A.
  • A is true but R is false.
  • A is false but R is true.
[1] 6. Assertion (A): Carboxylic acids do not give the characteristic nucleophilic addition reactions of aldehydes and ketones.
Reason (R): The lone pair of electrons on the oxygen atom of the -OH group is in resonance with the carbonyl carbon, drastically reducing its electrophilicity.
[1] 7. Assertion (A): Benzaldehyde gives a positive Fehling's test.
Reason (R): Benzaldehyde lacks an alpha-hydrogen atom.
Section B (2 Marks Each)
[2] 8. Give a simple chemical test to distinguish between the following pairs of compounds:
(i) Propanal and Propanone
(ii) Phenol and Benzoic acid
[2] 9. Write the chemical equations for the following Name Reactions:
(i) Hell-Volhard-Zelinsky (HVZ) reaction
(ii) Etard reaction
Section C (3 Marks Each)
[3] 10. Arrange the following in the stated order:
(i) Increasing reactivity towards HCN: Acetophenone, p-Tolualdehyde, Benzaldehyde, p-Nitrobenzaldehyde.
(ii) Increasing acidic strength: Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic acid, 4-Methoxybenzoic acid.
(iii) Increasing boiling point: CH3CHO, CH3CH2OH, CH3OCH3, CH3CH2CH3.
[3] 11. How will you carry out the following conversions? (Write chemical equations only)
(i) Benzene to Benzaldehyde
(ii) Acetaldehyde to But-2-enal
(iii) Ethylbenzene to Benzoic acid
[3] 12. Give reasons for the following:
(i) Oxidation of ketones requires extreme conditions and strong oxidizing agents compared to aldehydes.
(ii) Alpha-hydrogens of aldehydes and ketones are acidic in nature.
(iii) Benzoic acid does not undergo Friedel-Crafts alkylation reaction.
Section D: Case-Based (4 Marks)
[4] 13. Read the passage carefully and answer the questions that follow:

Nucleophiles, such as ammonia and its derivatives (H2N-Z), readily add to the carbonyl group of aldehydes and ketones. The reaction is reversible and is catalyzed by an acid. The equilibrium favors the product formation due to rapid dehydration of the intermediate to form a >C=N-Z compound. The addition of ammonia derivatives is strictly pH dependent. If the medium is too acidic, the reaction halts. If it is too basic, the reaction also slows down.

(a) What is Brady's reagent and what is its specific use in organic analysis? [1 Mark]
(b) Why is a strict optimum pH (around 3.5) required for the addition of ammonia derivatives to carbonyl compounds? [1 Mark]
(c) Write the structure of the products formed when propanone reacts with:
    (i) Hydroxylamine (H2N-OH)
    (ii) Semicarbazide (H2N-NH-CO-NH2) [2 Marks]
Section E: Long Answer (5 Marks Each)
[5] 14. An organic compound (A) with molecular formula C8H8O forms an orange-red precipitate with 2,4-DNP reagent and gives a yellow precipitate on heating with iodine and aqueous NaOH. It neither reduces Tollens' reagent nor Fehling's solution. On drastic oxidation with chromic acid, it gives a carboxylic acid (B) having molecular formula C7H6O2.
(i) Identify the compounds (A) and (B).
(ii) Write the chemical equations for the reactions of compound (A) with:
    a) 2,4-DNP reagent
    b) I2 / NaOH (Iodoform test)
    c) Chromic acid oxidation
[5] 15. Write the chemical equations for the following transformations:
(i) Clemmensen reduction of Propanone.
(ii) Wolff-Kishner reduction of Acetaldehyde.
(iii) Decarboxylation of Sodium ethanoate.
(iv) Cannizzaro reaction of Benzaldehyde.
(v) Stephen reaction to prepare Propanal.

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SOLUTIONS & MARKING SCHEME

1. Answer: (d) Methanal (Formaldehyde)

Reasoning: Reactivity in nucleophilic addition decreases with an increase in steric hindrance and the +I effect of alkyl groups. Methanal (HCHO) has only hydrogen atoms attached to the carbonyl carbon, offering zero steric hindrance and no +I effect, making it the most reactive.

2. Answer: (b) H2 / Pd-BaSO4

Reasoning: Rosenmund reduction uses Hydrogen gas over a Palladium catalyst poisoned with Barium Sulphate (BaSO4) to reduce acid chlorides to aldehydes without further reducing them to alcohols.

3. Answer: (c) Acetophenone

Reasoning: The Iodoform test is given by compounds containing the methyl ketone group (CH3-CO-). Acetophenone is C6H5-CO-CH3, which contains this group.

4. Answer: (b) F-CH2COOH

Reasoning: Acidity increases with the electron-withdrawing (-I) power of the substituent, which stabilizes the resulting carboxylate ion. Fluorine is the most electronegative halogen and exerts the strongest -I effect.

5. Answer: (a) Methanol and sodium formate

Reasoning: Formaldehyde lacks α-hydrogens. Therefore, on treatment with concentrated alkali, it undergoes the Cannizzaro reaction (disproportionation), where one molecule is oxidized to formate and another is reduced to methanol.

6. Answer: (a) Both A and R are true and R is the correct explanation of A.

Reasoning: The lone pair on the hydroxyl oxygen of the -COOH group delocalizes into the carbonyl π-system. This internal resonance heavily satisfies the electron deficiency of the carbonyl carbon, making it unresponsive to external nucleophilic attack.

7. Answer: (d) A is false but R is true.

Reasoning: Assertion is strictly false. Fehling's solution is a mild oxidizing agent that can only oxidize aliphatic aldehydes. Benzaldehyde is an aromatic aldehyde and does NOT give a positive Fehling's test. The Reason is true, benzaldehyde lacks alpha-hydrogens, but that is related to the Cannizzaro reaction, not Fehling's test.

8. Solution:

(i) Propanal and Propanone: Tollens' Test. Propanal (an aldehyde) forms a bright silver mirror on warming with Tollens' reagent. Propanone (a ketone) does not.
(ii) Phenol and Benzoic acid: Sodium Bicarbonate (NaHCO3) test. Benzoic acid reacts with aq. NaHCO3 to give brisk effervescence of CO2 gas. Phenol is a weaker acid and does not give effervescence.

9. Solution:

(i) HVZ Reaction:
CH3COOH + Cl2 →(i. Red P, ii. H2O) Cl-CH2-COOH (Chloroacetic acid) + HCl
(ii) Etard Reaction:
C6H5CH3 (Toluene) + CrO2Cl2 (in CS2) → Chromium Complex →(H3O+) C6H5CHO (Benzaldehyde)

10. Solution:

(i) Reactivity towards HCN (Nucleophilic Addition): Reactivity increases with EWG (-NO2) and decreases with EDG (-CH3). Ketones are less reactive than aldehydes.
Order: Acetophenone < p-Tolualdehyde < Benzaldehyde < p-Nitrobenzaldehyde.
(ii) Acidic strength: EWGs increase acidity, EDGs decrease acidity.
Order: 4-Methoxybenzoic acid < Benzoic acid < 4-Nitrobenzoic acid < 3,4-Dinitrobenzoic acid.
(iii) Boiling point: Alcohols form H-bonds (highest), aldehydes have dipole-dipole, ethers are weak polar, alkanes are non-polar.
Order: CH3CH2CH3 < CH3OCH3 < CH3CHO < CH3CH2OH.

11. Solution:

(i) Benzene to Benzaldehyde (Gattermann-Koch):
C6H6 + CO + HCl →(Anhydrous AlCl3 / CuCl) C6H5CHO.
(ii) Acetaldehyde to But-2-enal (Aldol Condensation):
2 CH3CHO →(dil. NaOH) CH3CH(OH)CH2CHO →(Δ, -H2O) CH3CH=CHCHO.
(iii) Ethylbenzene to Benzoic acid (Side-chain oxidation):
C6H5CH2CH3 →(Alk. KMnO4, Δ) C6H5COOK →(H3O+) C6H5COOH.

12. Solution:

(i) Ketones lack a hydrogen atom attached to the carbonyl carbon. Their oxidation involves the breaking of a strong carbon-carbon (C-C) bond, which requires drastic conditions (high heat, strong oxidizing agents).
(ii) The α-hydrogens are acidic because the resulting conjugate base (enolate anion) is highly stabilized by resonance with the electron-withdrawing carbonyl group.
(iii) Benzoic acid has a strongly deactivating (-R, -I) carboxyl group (-COOH). Additionally, the Lewis acid catalyst (AlCl3) permanently bonds to the electron-rich oxygen of the -COOH group, completely halting the reaction.

13. Solution (Case-Based):

(a) Brady's reagent is 2,4-Dinitrophenylhydrazine (2,4-DNP). It is used to chemically test for the presence of a carbonyl group (aldehydes/ketones) by forming a characteristic yellow, orange, or red crystalline precipitate.
(b) A slightly acidic medium (pH 3.5) is required to protonate the carbonyl oxygen, making the carbonyl carbon more electrophilic. If the pH drops too low (too acidic), the ammonia derivative itself gets protonated to form a salt (R-NH3+), leaving no lone pair to act as a nucleophile.
(c) (i) Propanone + Hydroxylamine:
CH3-CO-CH3 + H2N-OH → CH3-C(=N-OH)-CH3 (Propanone oxime) + H2O.
(c) (ii) Propanone + Semicarbazide:
CH3-CO-CH3 + H2N-NH-CO-NH2 → CH3-C(=N-NH-CO-NH2)-CH3 (Propanone semicarbazone) + H2O. (Note: Reaction occurs at the terminal -NH2, not the amide -NH2).

14. Solution:

(i) Identification:
- Forms 2,4-DNP derivative: It is a carbonyl compound (aldehyde or ketone).
- Does not reduce Tollens' or Fehling's: It is a Ketone.
- Gives positive Iodoform test (yellow ppt): It must be a Methyl Ketone (R-CO-CH3).
- Drastic oxidation gives C7H6O2 (Benzoic acid): The R- group must be a phenyl ring (C6H5-).
Therefore, Compound A (C8H8O) is Acetophenone (C6H5-CO-CH3).
Compound B (C7H6O2) is Benzoic Acid (C6H5COOH).

(ii) Chemical Equations:
a) With 2,4-DNP:
C6H5-CO-CH3 + H2N-NH-C6H3(NO2)2 →(H+) C6H5-C(CH3)=N-NH-C6H3(NO2)2 + H2O
b) Iodoform Test:
C6H5-CO-CH3 + 3NaOI → C6H5COONa (Sodium benzoate) + CHI3↓ (Iodoform) + 2NaOH
c) Oxidation:
C6H5-CO-CH3 →(CrO3 / H2SO4) C6H5COOH (Benzoic acid)

15. Solution:

(i) Clemmensen reduction of Propanone:
CH3-CO-CH3 + 4[H] →(Zn-Hg / Conc. HCl) CH3-CH2-CH3 (Propane) + H2O

(ii) Wolff-Kishner reduction of Acetaldehyde:
CH3-CHO + H2N-NH2 → CH3-CH=N-NH2 →(KOH / Ethylene glycol, Δ) CH3-CH3 (Ethane) + N2

(iii) Decarboxylation of Sodium ethanoate:
CH3COONa + NaOH →(CaO, Δ) CH4 (Methane) + Na2CO3

(iv) Cannizzaro reaction of Benzaldehyde:
2 C6H5CHO + Conc. NaOH → C6H5CH2OH (Benzyl alcohol) + C6H5COONa (Sodium benzoate)

(v) Stephen reaction to prepare Propanal:
CH3CH2C≡N (Propanenitrile) + SnCl2 + HCl → CH3CH2CH=NH (Imine) →(H3O+) CH3CH2CHO (Propanal)

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