Chapter 24 — Quiz

Twenty-five questions on the carbonyl group, the three reactivity families, the reactivity ordering, and the diagnostic spectroscopy. Mix of multiple choice and short answer. ∗ marks questions with answers in the answer key at the end.

Multiple choice

1.∗ The carbonyl carbon (C=O) carries a partial: (a) negative charge (b) positive charge (c) zero charge (d) full +1 charge

2.∗ The carbonyl oxygen is: (a) δ⁺ (b) δ⁻ (c) sp³ hybridized (d) usually protonated

3.∗ The geometry around the carbonyl carbon is: (a) tetrahedral (b) trigonal pyramidal (c) trigonal planar (d) linear

4.∗ The C=O bond is shorter than the C=C bond. The reason is: (a) carbon is bigger than oxygen (b) oxygen is smaller than carbon and the polar bond is partially ionic (c) the carbonyl is more strained (d) C=C is rotated

5.∗ Rank reactivity toward a generic nucleophile: (a) ester > acid halide > amide (b) amide > ester > acid halide (c) acid halide > anhydride > ester > amide (d) ester > amide > acid halide

6.∗ Amides are the LEAST reactive carbonyl class because: (a) the nitrogen donates its lone pair into the carbonyl, lowering the C δ⁺ (b) amides are too crowded sterically (c) amides have a stronger C=O bond (d) amides are paired with their conjugate acid

7.∗ A typical IR C=O stretch for a saturated ketone is at approximately: (a) 1200 cm⁻¹ (b) 1450 cm⁻¹ (c) 1715 cm⁻¹ (d) 3300 cm⁻¹

8.∗ ¹³C NMR carbonyl C resonates in the range: (a) 0–50 ppm (b) 50–100 ppm (c) 100–140 ppm (d) 165–220 ppm

9.∗ Cyclopropanone has an IR C=O stretch at ~1815 cm⁻¹ (much higher than acetone). The reason is: (a) ring strain forces sp character into the C=O (b) cyclopropanone is electron-poor (c) the ring is bigger than acetone (d) the carbon is sp³

10.∗ Glucose in solution exists primarily as: (a) the open-chain aldehyde (b) a cyclic hemiacetal (c) a free ketone (d) a carboxylic acid

11.∗ A peptide bond is which functional class? (a) ester (b) amide (c) acid halide (d) anhydride

12.∗ The aldehyde C-H exhibits a characteristic IR doublet at approximately: (a) 1500 and 1580 cm⁻¹ (b) 1700 and 1715 cm⁻¹ (c) 2820 and 2720 cm⁻¹ (d) 3100 and 3300 cm⁻¹

13.∗ A nucleophile attacks an aldehyde at the: (a) carbonyl oxygen (b) carbonyl carbon (c) α-carbon (d) any carbon

14.∗ The "addition" family of carbonyl reactivity (Ch 25) applies to: (a) aldehydes and ketones (b) carboxylic acids and esters (c) amides only (d) all carbonyls equally

15.∗ "Acyl substitution" applies to: (a) aldehydes and ketones (b) carbonyls with a leaving group on the carbonyl C (esters, amides, acid halides, anhydrides, COOH) (c) only ketones (d) carbonyls in biology

16.∗ "α-Carbon chemistry" requires: (a) an aldehyde (b) an α-hydrogen on the carbon adjacent to C=O (c) a leaving group (d) a strong nucleophile

17.∗ Which of the following has the highest IR C=O stretch wavenumber? (a) acetone (1715) (b) acetic acid (1720) (c) propanoyl chloride (1800) (d) acetamide (1660)

18.∗ The amide C-N bond has: (a) freely rotating single-bond character (b) restricted rotation due to resonance (c) no double-bond character at all (d) a triple bond

19.∗ The carboxylate anion ($RCOO^-$) is even less reactive than an amide because: (a) it has a negative charge that repels nucleophiles (b) it is non-aromatic (c) it is sterically large (d) the negative oxygen donates strongly into the carbonyl

20.∗ Acetyl-CoA is a thioester. Compared to an oxoester, it is: (a) less reactive (~10⁵ times slower) (b) more reactive (~10⁵ times faster) (c) about the same (d) only reactive in water

Short answer

21. Draw the two main resonance structures of a generic carbonyl C=O. Indicate which atom is δ⁺ and which is δ⁻.

22. Why is the C=O dipole moment so much larger than the C=C dipole? Use the electronegativity difference (3.44 - 2.55 = 0.89).

23. A compound has IR peaks at 1660 and 3300 cm⁻¹. What functional class is this most likely? Give two reasons.

24. Identify all carbonyl groups in aspirin (acetylsalicylic acid) and classify each by family. Then predict which one is targeted in the body's mechanism (acyl transfer to COX serine).

25. Why is the cyclic hemiacetal form of glucose so much more abundant (>99%) than the open-chain aldehyde form? Give two thermodynamic and one entropic reason.

Answer key

  1. b — C is partially positive because O is more electronegative.
  2. b — O is partially negative.
  3. c — sp² hybridized carbon → trigonal planar.
  4. b — Polar bonds are stronger and shorter than nonpolar covalent bonds with similar atoms; oxygen is smaller.
  5. c — Acid halide is most reactive; amide least.
  6. a — Resonance donation from N's lone pair stabilizes the C and lowers electrophilicity.
  7. c — 1715 cm⁻¹ is the canonical saturated ketone position.
  8. d — Carbonyl carbons are very deshielded.
  9. a — Strain forces more s-character into the C=O bonds, making the bond stronger and more polar.
  10. b — Glucose's C5 OH attacks the C1 aldehyde, forming a 6-membered hemiacetal.
  11. b — A peptide bond is an amide.
  12. c — The two C-H stretching modes of the aldehyde C-H.
  13. b — Carbonyl C is the electrophile.
  14. a — Aldehydes/ketones add nucleophiles because their substituents (H, R) are bad leaving groups.
  15. b — Acyl substitution requires a leaving group on the C=O.
  16. b — α-H is required to form the enolate.
  17. c — Acid halides have the highest C=O stretch due to the inductive effect of the chlorine.
  18. b — N's lone pair donation gives partial double-bond character; rotation is restricted at room temperature.
  19. d — A negative charge on the substituent oxygen donates very strongly, lowering the C electrophilicity.
  20. b — S is a poor π donor (so C is more electrophilic); S is a better leaving group; net rate is ~10⁵ faster than oxoester.

21. Structure 1: covalent C=O with no charges. Structure 2: C-O single bond with a positive on C and negative on O. The hybrid is closer to structure 1 but has measurable contribution from structure 2 — explaining the polarization.

22. The C=O electronegativity difference (~0.89) puts about 30% partial-ionic character on the bond. Compare to C=C (Δχ = 0). The dipole adds vector contributions from each polarized bond, giving ~2.7 D for a typical carbonyl.

23. Most likely an amide. Reasons: (1) C=O at 1660 is shifted to lower wavenumber compared to a regular ketone (1715), consistent with amide N-H lone-pair donation lengthening the C=O bond. (2) The 3300 cm⁻¹ N-H stretch is diagnostic of an amide N-H.

24. Two carbonyls: (a) the acetyl C=O (an ester linked to the salicylate oxygen) and (b) the salicylate's COOH. The body targets the ester C=O for acyl substitution onto the COX serine OH; this transfers the acetyl group to COX, irreversibly inactivating the enzyme.

25. Thermodynamics: (a) The cyclic hemiacetal C-O σ bond is stronger than the C=O π bond it replaces in the equilibrium. (b) Forming a 6-membered ring places all substituents in equatorial positions, the stable chair conformation. Entropic: (c) Cyclization brings two atoms of the same molecule close together (intramolecular), making this entropically more favorable than attacking with a separate water molecule.