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ChemistryMediumClass 12

Aldol Condensation + Cannizzaro

Aldehydes Ketones Carboxylic Acids

50

JEE Qs

8%

Hard

90

min

Always check for the presence or absence of alpha-hydrogens on the carbonyl compound(s) first, as this immediately determines whether Aldol or Cannizzaro is possible, then apply the correct mechanism and conditions.

🧮 Key Formulas

Aldol Addition: R-CH2-CHO + R'-CH2-CHO --(Dil. base/acid)--> R-CH2-CH(OH)-CHR'-CHO (beta-hydroxy aldehyde/ketone)
Aldol Condensation: R-CH2-CH(OH)-CHR'-CHO --(Heat, acid/base)--> R-CH2-CH=CR'-CHO + H2O (alpha,beta-unsaturated aldehyde/ketone)
Cannizzaro Reaction: 2 R-CHO (no alpha-H) --(Conc. base, e.g., NaOH/KOH)--> R-CH2OH + R-COO(-)Na(+) (Alcohol + Carboxylate salt)

✅ Key Points for JEE

  • 1Aldol Condensation requires at least one alpha-hydrogen on the carbonyl compound. It proceeds via enolate formation (in base) or enol formation (in acid).
  • 2Cannizzaro reaction occurs with aldehydes lacking alpha-hydrogens. It is a disproportionation reaction (self-oxidation and self-reduction).
  • 3In crossed aldol condensation, if one reactant has no alpha-H (e.g., formaldehyde, benzaldehyde) and the other has alpha-H, the compound without alpha-H typically acts as the electrophile, while the compound with alpha-H forms the enolate (nucleophile).
  • 4Intramolecular aldol condensation occurs in dicarbonyl compounds to form cyclic products, usually 5- or 6-membered rings are favored.
  • 5Crossed Cannizzaro involves two different aldehydes lacking alpha-hydrogens. The more reactive aldehyde (usually the less sterically hindered one, e.g., formaldehyde) is oxidized to the carboxylic acid, and the other is reduced to the alcohol.

⚠️ Common Mistakes

  • ✕Confusing the conditions and requirements (presence/absence of alpha-H) for Aldol vs. Cannizzaro reactions.
  • ✕Incorrectly identifying alpha-hydrogens, especially in complex or cyclic structures.
  • ✕Failing to perform the dehydration step for aldol condensation when heat is applied or implied, leading to beta-hydroxy carbonyl products instead of alpha,beta-unsaturated ones.
  • ✕In crossed reactions, incorrectly predicting which carbonyl compound acts as the nucleophile (enolate former) or electrophile (attacked carbonyl).
  • ✕Missing intramolecular reaction possibilities for dicarbonyl compounds.

📝 Practice Questions

See all

Q65.Given below are two statements : Statement (I) : On nitration of m-xylene with HNO3, H2SO4 followed by oxidation, 4-nitrobenzene-1,3-dicarboxylic acid is obtained as the major product. Statement (II) : −CH3 group is o/p-directing while −NO2 group is m-directing group. In the light of the above statements, choose the correct answer from the options given below : 2025 (29 Jan Shift 2) JEE Main Previous Year Paper (1) Both Statement I and Statement II are false (2) Statement I is false but Statement II is true (3) Statement I is true but Statement II is false (4) Both Statement I and Statement II are true

2025·Assertion ReasoningMedium

Q57. The compounds which give positive Fehling's test are : (A) (B) (C) HOCH2 −CO −(CHOH)3 −CH2 −OH (D) (E) Choose the correct answer from the options given below : (1) (A), (D) and (E) Only (2) (C), (D) and (E) Only (3) (A), (C) and (D) Only (4) (A), (B) and (C) Only

2025·MCQMedium

Q67. The IUPAC name of the following compound is : (1) Methyl-6-carboxy-2,5-dimethylhexanoate. (2) 2-Carboxy-5-methoxycarbonylhexane. (3) 6-Methoxycarbonyl-2,5-dimethylhexanoic acid. (4) Methyl-5-carboxy-2-methylhexanoate.

2025·MCQMedium

Q73. Consider the following sequence of reactions : Molar mass of the product formed (A) is _______ gmol−1 .

2025·NumericalMedium

Q64. Residue (A) + HCl (dil) → Compound (B) Structure of residue (A) and compound (B) formed respectively is : (1) (2) (3) (4)

2025·MCQMedium

Q63.The major product of the following reaction is: CH3CH2CH = O → ? reflux (1) (2) (3) (4)

2025·MCQMedium

NCERT Chapters

  • Class 12 Chemistry Part 2 Ch 12: Aldehydes, Ketones and Carboxylic Acids