Chapter 25 — Quiz

Twenty-five questions on nucleophilic addition to aldehydes and ketones, including mechanism, products, equilibria, stereochemistry, and biological context. ∗ marks questions answered in the answer key at the end.

Multiple choice

1.∗ Aldehyde + NaBH₄ in MeOH gives: (a) primary alcohol (b) secondary alcohol (c) tertiary alcohol (d) carboxylic acid

2.∗ Ketone + R'MgBr, then aqueous workup gives: (a) primary alcohol (b) secondary alcohol (c) tertiary alcohol (d) ether

3.∗ Aldehyde + 2 equivalents of R'OH + acid catalyst gives: (a) hemiacetal (b) acetal (c) alkene (d) ester

4.∗ Acetals are stable in: (a) acidic aqueous conditions (b) basic conditions (c) neither (d) both

5.∗ Acetal hydrolysis requires: (a) acid + water (b) base (c) heat alone (d) light

6.∗ Imine forms from a primary amine + carbonyl. The byproduct is: (a) ammonia (b) water (c) methanol (d) hydrogen

7.∗ Enamine forms from a secondary amine + carbonyl. The C=C is at the: (a) carbonyl C (b) α-C (c) β-C (d) none — enamines are linear

8.∗ Cyanohydrin extends the carbon chain by: (a) 1 carbon (b) 2 carbons (c) 3 carbons (d) 0 carbons

9.∗ Wittig reaction makes: (a) alcohol (b) alkene (c) ester (d) amine

10.∗ LiAlH₄ reduces: (a) aldehydes only (b) aldehydes and ketones only (c) aldehydes, ketones, esters, amides, carboxylic acids (d) ketones only

11.∗ The Bürgi-Dunitz angle is: (a) 90° (b) 107° (c) 120° (d) 180°

12.∗ A bulky hydride (L-Selectride) on 4-tert-butylcyclohexanone gives: (a) axial alcohol (cis to t-Bu) (b) equatorial alcohol (trans to t-Bu) (c) racemic mixture (d) no reaction

13.∗ Reductive amination is best done with: (a) NaBH₄ (b) LiAlH₄ (c) NaBH₃CN (d) Zn dust

14.∗ Hydration of formaldehyde at equilibrium gives: (a) >99% formaldehyde (b) >99% gem-diol (c) ~50:50 (d) none of the above

15.∗ A stabilized Wittig ylide (e.g., Ph₃P=CHCO₂Me) gives predominantly: (a) cis alkene (b) trans alkene (c) unstable mixture (d) ester

16.∗ Glucose's pyranose form is a: (a) gem-diol (b) hemiacetal (c) acetal (d) imine

17.∗ The retinal in rhodopsin is bonded to opsin via: (a) a covalent ester (b) a Schiff base imine (c) a hydrogen bond (d) a non-covalent association

18.∗ Why is glucose's open-chain aldehyde a tiny fraction at equilibrium? (a) It's actually the major form (b) The cyclic hemiacetal is more stable due to chair geometry and intramolecular C-O bond (c) It oxidizes immediately to gluconic acid (d) It evaporates

19.∗ A Grignard reagent has: (a) a partially positive C (b) a partially negative C (c) a triple bond (d) no carbon

20.∗ Which can NOT be made directly with a Grignard + aldehyde reaction? (a) primary alcohol (b) secondary alcohol (c) tertiary alcohol (from aldehyde alone) (d) all of the above can

Short answer

21. Draw the full mechanism for: cyclohexanone + EtOH + H⁺ → ethyl cyclohexanyl acetal (a hemiketal first, then ketal). Show all 7 elementary steps.

22. A student does the reaction: butanal + 2 PhMgBr + H₃O⁺. Predict the product and explain why two Grignard equivalents react.

23. Sketch the Felkin-Anh transition state for the addition of a Grignard to an α-substituted aldehyde. Explain which diastereomer is preferred.

24. Outline the mechanism of imine formation from acetaldehyde + methylamine. Note where in the mechanism the loss of water occurs and why pH ~5 is optimal.

25. Why does NaBH₄ not reduce esters but LiAlH₄ does? Connect to the Ch 24 reactivity ranking.

Answer key

  1. a — Aldehyde reduces to a primary alcohol.
  2. c — Ketone + R-MgX → tertiary alcohol (after both R groups are now on the C).
  3. b — Acid catalysis takes hemiacetal forward to acetal by losing water and adding a second alcohol.
  4. b — Acetals are stable in basic conditions.
  5. a — Acid + water hydrolyzes acetal to aldehyde + alcohol.
  6. b — Water is lost from the hemiaminal in imine formation.
  7. b — Enamine has C=C-N functional group; the α-H is what leaves.
  8. a — One CN attached at the carbonyl carbon adds 1 carbon.
  9. b — Wittig gives alkene + Ph₃P=O.
  10. c — LiAlH₄ is the universal hydride reducer.
  11. b — Bürgi-Dunitz angle is approximately 107°.
  12. a — Bulky hydride attacks from the less-hindered axial direction (the opposite of what NaBH₄ does).
  13. c — NaBH₃CN reduces iminium selectively but doesn't reduce the parent C=O.
  14. b — Formaldehyde is essentially fully hydrated in water.
  15. b — Stabilized ylides give trans alkenes (E).
  16. b — Glucose's pyranose is a cyclic hemiacetal.
  17. b — Schiff base imine.
  18. b — The cyclic hemiacetal has a strong intramolecular C-O σ bond, all-equatorial chair, and is entropically favored. Open-chain has the weaker C=O π bond and no chair conformation.
  19. b — The C-Mg bond is highly polar with the C as the negative end (carbanion-like).
  20. c — A tertiary alcohol cannot be made from an aldehyde alone (you need a ketone). Aldehyde + 1 equiv R-MgX = secondary alcohol; ketone + 1 equiv = tertiary alcohol.

21. Step 1: H⁺ protonates the carbonyl O, making C more electrophilic. Step 2: EtOH attacks C; alkoxide forms (still protonated). Step 3: deprotonation gives hemiketal. Step 4: H⁺ reprotonates the OH, making it a leaving group. Step 5: water leaves, generating an oxocarbenium ion (C⁺ stabilized by remaining OEt lone pairs). Step 6: second EtOH attacks C. Step 7: deprotonation gives the acetal. Equilibrium favors acetal under conditions of dryness (water removal); favors hemiketal/ketone with water present.

22. Two equivalents of PhMgBr can react if the substrate is, e.g., an ester. With butanal (an aldehyde), the first PhMgBr addition gives an alkoxide that does not undergo a second addition because the aldehyde is gone. So the answer depends on the actual substrate. If butanal: first addition gives 1-phenylpentan-1-ol; the alkoxide is not electrophilic enough for a second attack. With an ester (e.g., methyl pentanoate), the first addition gives a tetrahedral intermediate that collapses to a ketone (and methoxide); the ketone is then attacked by a second PhMgBr → tertiary alcohol.

23. Felkin-Anh: the largest α-substituent (L) is anti-periplanar to the incoming Nu. The medium substituent (M) is gauche to the carbonyl; the smallest (S) is also gauche but closer to the C=O. Nu attacks at the Bürgi-Dunitz angle (107° from C=O). The result is the diastereomer with Nu and L anti (Felkin product), opposite to the Cram/anti-Felkin product.

24. Step 1: methylamine attacks the C=O carbon; alkoxide forms. Step 2: protonation of alkoxide by acid to give hemiaminal. Step 3: protonation of OH by acid → -OH₂⁺ leaving group. Step 4: water leaves; iminium ion formed. Step 5: deprotonation of N-H (which is now on the same N as the C=N) gives the neutral imine. The pH ~5 optimum balances: needing acid (to protonate the OH and drive water out) with not protonating the amine itself (the amine pKaH is ~9–10; at pH 5, ~99% of amine is protonated, but ~1% is still free, enough to attack).

25. NaBH₄ is a relatively mild hydride. Its reactivity matches the electrophilicity of aldehydes and ketones (high) but is insufficient to reduce esters, amides, or COOH (lower electrophilicity due to resonance donation by OR, NR₂, OH groups; reactivity ordering Ch 24). LiAlH₄, with its much higher charge density and higher nucleophilicity, overcomes the resonance-dampened electrophilicity and reduces all the carbonyl families.