Aldehydes, Ketones and Carboxylic Acids — Questions & Answers

Sample questions from Aldehydes, Ketones and Carboxylic Acids

1. Q1. The relative rates of alkaline hydrolysis of the esters CH3COOCH3, ClCH2COOCH3 and (CH3)3CCOOCH3 are best predicted as

   • A.(CH3)3CCOOCH3 > CH3COOCH3 > ClCH2COOCH3
   • B.ClCH2COOCH3 > CH3COOCH3 > (CH3)3CCOOCH3✓
   • C.CH3COOCH3 > ClCH2COOCH3 > (CH3)3CCOOCH3
   • D.all three hydrolyse at the same rate
   Solution

   Saponification begins with hydroxide attacking the carbonyl carbon, so it is accelerated by electron withdrawal and slowed by steric crowding. The -I chlorine makes the chloro ester fastest, and the bulky tert-butyl group shields the carbonyl carbon and also donates electrons, making that ester slowest.

2. Q2. Which order correctly ranks the compounds by increasing enol content at equilibrium: propanone, pentane-2,4-dione, ethanal?

   • A.ethanal < propanone < pentane-2,4-dione✓
   • B.pentane-2,4-dione < propanone < ethanal
   • C.propanone < ethanal < pentane-2,4-dione
   • D.ethanal < pentane-2,4-dione < propanone
   Solution

   Enol content rises with the stabilisation available to the enol. Ethanal has the least stable enol of the three, propanone slightly more (an extra alkyl group on the C=C), while the enol of pentane-2,4-dione is both conjugated with the second carbonyl and locked by a strong six-membered intramolecular hydrogen bond, so its enol content is very high.

3. Q3. Which explanation accounts for why an intramolecular Cannizzaro-type disproportionation is not typically observed for simple monoaldehydes lacking a second carbonyl group nearby, while certain specially designed dialdehydes CAN undergo an intramolecular version?

   • A.An intramolecular Cannizzaro reaction requires two carbonyl groups positioned appropriately within the same molecule so that hydride transfer can occur intramolecularly between them, a simple monoaldehyde has only one carbonyl and must instead react intermolecularly (or not at all if alpha-hydrogens are present)✓
   • B.All aldehydes undergo intramolecular Cannizzaro reactions regardless of structure
   • C.Intramolecular Cannizzaro reactions are mechanistically impossible under any circumstances
   • D.The number of carbonyl groups present has no bearing on whether Cannizzaro chemistry can occur intramolecularly
   Solution

   Just as intramolecular versions of many reactions (aldol, Williamson) are favoured when geometrically feasible, an intramolecular Cannizzaro-type hydride transfer requires a second, appropriately positioned carbonyl group within the same molecule to serve as the hydride acceptor, simple monoaldehydes lack this second carbonyl and so undergo (if at all, when no alpha-hydrogens are present) only the standard intermolecular Cannizzaro reaction with a separate aldehyde molecule.

4. Q4. Which pair of compounds can be distinguished using sodium bicarbonate solution alone?

   • A.Propanal and propanone
   • B.Ethanoic acid and ethanol✓
   • C.Butan-2-one and pentan-3-one
   • D.Benzaldehyde and ethanal
   Solution

   Ethanoic acid effervesces with NaHCO3 because CO2 is liberated, while ethanol gives no reaction. The other pairs need Tollens, iodoform or Fehling tests, as neither member of those pairs is acidic enough.

5. Q5. Which of the following carbonyl compounds does NOT undergo the aldol condensation because it lacks alpha-hydrogen atoms?

   • A.Acetaldehyde
   • B.Acetone
   • C.Benzaldehyde✓
   • D.Propanal
   Solution

   Benzaldehyde has no alpha-hydrogen atoms on a carbon adjacent to the carbonyl (the adjacent carbon is part of the aromatic ring), so it cannot undergo self-aldol condensation, though it can participate in a crossed (Claisen-Schmidt) reaction with a partner that does have alpha-hydrogens.

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