CHAPTER 21 HW: ALDEHYDES + KETONES

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1 CAPTER 21 W: ALDEYDES + KETES MECLATURE 1. Give the name for each compound (IUPAC or common name). Structure C 2 ame 3,3-dimethyl-2-pentanone (or 1,1-dimethylpropyl methyl ketone) 5-hydroxy-4-methylhexanal m-nitrobenzaldehyde Structure ame 1-penten-3-one or ethyl vinyl ketone 2,4-dioxohexanal 2-methyl-5-heptyn-3-one 2. Draw each compound. Structure F F F C 3 zame zide ame dicyclohexyl ketone 1,1,1-trifluoro-3-pentanone (Z)-3,6-dimethyl-3-heptenal SPECTRSCPY 3. In the 1 MR spectrum of butanal, the signal at 2.40 ppm is a triplet of doublets (approximate J s are 2 z and 7 z). Explain the splitting of this signal, including a sketch of a tree diagram. C C C C Signal at 2.40 ppm (next to carbonyl) The C 2 next to the carbonyl is the signal at 2.40 ppm. It is split in a large way by its two C 2 neighbors (7 z, splitting first level into a triplet), and in a smaller way by the aldehyde (2 z, splitting second level into a doublet). This results in a triplet of doublets. Page 1

2 4. f compounds A-D, C 3 C 3 A B C D a. Which compound would have the following 13 C MR spectra (briefly explain)? δ = 165.9, 132.1, 131.5, 129.1, 127.4, 51.5 ppm Compound A ppm is the C=, and the low range means it could be part of an ester, carboxylic acid or amide. The 51.5 ppm signal is the carbon next to oxygen (could be A or B). There are only 6 total 13 C signals, so it is A (B would have 8 13 C signals). Which compound would have the following 13 C MR spectra (briefly explain)? δ = 199.8, 140.3, 135.4, 131.5, 131.0, 127.7, 121.2, 28.6 ppm Compound D ppm is the C=, and a high range means it could part of an aldehyde or ketone. The 28.6 ppm signal is low, and so is not next to oxygen (could be C or D). There are 8 total 13 C signals, so it is D (C would have 6 13 C signals). 5. Below are 1 and 13 C MR spectra of a compound with a formula of C 5 8. Determine the structure, then assign the peaks in the 1 MR spectrum. 1 a b c d e 2 f PPM PPM a e f b c d e/f assignment could be switched c is a wider ~doublet so must involve a trans coupling (to b). Page 2

3 6. Cyclohexanone has a strong signal in its IR spectrum at 1718 cm -1, while 2-cyclohexenone has a strong signal at 1691 cm -1. Both signals represent vibration of the same kind of bond. Explain why the absorption in 2-cyclohexenone is at a lower wavenumber, including resonance structures. Both signals represent the IR stretching of the C= bonds. 2-cyclohexenone has a lower wavenumber absorbance, as it has a different C= bond strength than cyclohexanone. All carbonyls have a +/- resonance structure (shown with cyclohexanone below), but 2-cyclohexenone has an additional resonance structure, which causes its resonance hybrid to have more single bond character. With greater single bondedness, the C= is weaker, and thus absorbs at a lower wavenumber (since v α f). vs. 7. An IR is taken of a mixture of the two compounds below. Two strong signals are noted in the mixed IR spectrum at 1666 and 1692 cm -1, representing the carbonyl stretching modes. Which signal corresponds to which compound? iefly explain. Both compounds have very conjugated C= bonds, and conjugation lowers the IR wavenumber (more single bond character). The first compound has twice as much conjugation / resonance, C= Stretch 1666 cm -1 C=: 1692 cm -1 so it s C= signal is the lowest. WITTIG REACTI 8. Give the curved arrow mechanism for each reaction. a. C 3 C 2 a. P 3 nbuli 3 C C P 3 3 C C P 3 S 2 P 3 C 3 C 2 3 C C P (nbuli) 3 C C P 3 C C P 3 3 C C P 3 3 C C P 3 C P 3 3 C C C 3 P Page 3

4 8 continued c. P 3 P 3 P 3 P 3 9. Draw both stereoisomers that can be formed in this Wittig reaction. 3 P trans cis 10. Give the major product of each reaction (one stereoisomer is sufficient). C C 3 C 2 C=P a. 3 C 2 C 3 3 P c. C C 3 C 3 P 3 d. 2 C P 3 2 C 11. Use a Wittig reaction to produce each alkene, starting from an ylide. C 3 C 2 C 2 C 3 a. C C C 3 C 2 Route 1 C 3 C 2 C 2 C 3 C + 3 P C C 3 C 2 product Route 2 C C 3 C 2 C 2 C 3 C P 3 + C 3 C 2 product Route 1 P 3 or P 3 product Route 2 P 3 product Page 4

5 12. Use a Wittig reaction to produce this alkene, starting from an alkyl halide. Route 1 P 3 P 3 nbuli P 3 Route 2 P 3 3 P nbuli 3 P YDRATES 13. Give the curved arrow mechanism that shows the formation of hydrate under each set of conditions. a. trace - 2 trace + 2 trace Compound E has a higher percentage of hydrate relative to carbonyl in aqueous solution than compound F. Explain this trend, including energy diagrams with your explanation. E F-hydrate E-hydrate E F F The difference in reactivity often arises from the relative energies of the carbonyl species (starting reactant energies). The carbonyl carbon is δ+, and EDG lower the energy. The aldehyde has one alkyl group (EDG) attached to the C=, but the ketone has 2 EDG. Therefore, the ketone stabilizes the δ+ more and starts at a lower energy than the aldehyde. This causes the hydrate reaction to be uphill for the ketone (so higher hydrate % for aldehyde). Although it s only a minor factor, we can also be complete by noting that the ketone hydrate energies should also be somewhat higher E because 2 alkyl groups are more crowded than just one. This would make the ketone reaction even more uphill. ne EDG Page 5

6 15. Compound G has a smaller equilibrium constant (K eq ) for hydrate formation than compound. Explain this trend, including energy diagrams with your explanation. 2 C 3 CF 3 G K eq = 1.06 K eq = 2.9 x 10 4 CF 3 EWG CF 3 K eq R R K = hydrate / carbonyl; so a large K eq means a higher % of hydrate. Carbonyl with CF 3 has more hydrate. The reactivity differences arise from different starting carbonyl energies (the hydrate energies are also nearly CF 3 equivalent). CF 3 is a strong EWG, so destabilizes the δ+ of the carbonyl, C 3 CF 3 C making the CF 3 carbonyl higher in 3 G G+-hydrate energy than the aldehyde. This makes the hydrate reaction of the CF 3 carbonyl downhill, resulting in a higher amount of hydrate. 16. In each pair, predict which would have a greater percentage of hydrate relative to carbonyl when the two forms are at equilibrium in water. iefly explain each answer. Pair A ief Explanation: The aromatic group can participate in resonance with the C= which greatly stabilizes the δ+ of the carbonyl. Therefore, the aromatic carbonyl starts at a lower energy, and reacts less (reaction is more uphill). Pair B ief Explanation: C= energies are probably similar. But hydrate energies are more sensitive to steric issues. The dicyclohexyl compound s hydrate will be higher energy, so reaction is more uphill. Pair C C 3 C 3 ief Explanation: thoxy is a stronger EDG than methyl (resonance, not hyperconjugation), so best stabilizes the δ+ of the carbonyl. The methoxy carbonyl starts at a lower energy, so is more uphill. Pair D 3 ief Explanation: The aldehyde is more reactive (makes more hydrate) because it has a higher energy carbonyl form (reaction is downhill). The aromatic has an EWG (- + 3 ), which destabilizes the δ+ of the carbonyl. Page 6

7 ACETALS 17. Give the curved arrow mechanism for this acetal formation reaction. C 2 C 3 + C 2 C 3 C 3 C 2 C 2 C 3 + C 3 C 2 C 3 C 2 Protonate Attack Deprotonate C 2 C 3 + Protonate C 2 C 3 2 C2 C 3 C2 C 3 C 2 C 3 C 2 C 3 Deprotonate Leave C 3 C 2 C 3 C 2 Attack C 3 C 2 product ote: attack of either resonance structure is acceptable; you don t need to show both. 18. Explain why acid is catalytic in the formation of an acetal. (Use the mechanism in the previous problem.) Acid is a catalyst because it satisfies both criteria: Acid is unconsumed. For every protonate step where it is used, there is a deprotonate step where it is regenerated. Acid lowers the activation barrier of the reaction. It makes the carbonyl more reactive to attack, as it puts charge on the carbonyl, making the carbon of the carbonyl more δ+ (starts at a higher energy, thus lowering E a ). 19. Give the curved arrow mechanism for this reaction Protonate Attack Deprotonate Protonate 2 Leave Attack Deprotonate Page 7

8 20. Give the major organic product for each reaction. a. + C 3 C 2 C 3 C 2 C 2 C 3 c. + + d. C + C 3 C 3 C In order to achieve good yields for most acetal formations, they need to be driven by Le Châtelier s Principle. Explain why good yields are easier to achieve when reacting aldehydes than when reacting ketones. Acetal formation reactions have nearly the same energetics as hydrate formation. For aldehydes, K=1 for acetal formation, while ketones K<1. Ketones have a more stabilized carbonyl (2 EDG stabilizing the δ+ of the carbonyl), so their acetal reaction is uphill. This means it is easier to get good yields with aldehydes (K=1) compared to ketones (uphill). 22. Show all organic products of these reactions. a. C 3 C C 3 d C 3 C 2 3 C + 2 C 3 e. 3 C C C3 c C 3 C f. 3 C C 3 C C + 2 C 3 Page 8

9 23. Give the curved arrow mechanism for each reaction. a. C 3 C C 3 C 3 + C 3 C 3 C3 2 C 3 Protonate C 3 Leave Attack C 3 C 3 + C Deprotonate 3 2 product 2 Protonate Leave Deprotonate acetone c. C 2 C C 3 C 2 + C 2 C 3 (ch could be begun by protonating either oxygen) C 2 C d product R Page 9

10 24. Milder conditions can be used to hydrolyze acetal J than to hydrolyze acetal K. Explain their difference in reactivity. C 3 C 3 J C 3 K C3 Acetal hydrolysis reactions have similar energetics to hydrate reactions. Since J produces a ketone instead of an aldehyde (where the carbonyl is more stabilized by 2 EDGs), it s a more favorable hydrolysis reaction (reaction is easier and can use milder conditions). PRTECTIG GRUPS 25. Design a synthesis that uses cyclopentanone and 4-bromobutanal to efficiently produce the aldehyde shown. + +, heat (protect aldehyde) a) Mg b) c) + workup +, 2, heat (hydrolysis) 26. The following multi-step synthesis converts benzene into a dicarbonyl species. a. Give the reagents needed to complete each step in the sequence. 2 Cl Fe 3 AlCl 3 +, heat Mg +, 2 Cr 3, + or PCC a) Mg heat b) + workup iefly explain why the synthesis below does not work well. Cl AlCl 3 Cl AlCl 3 Step 2 should have problems. The carbonyl is a meta director, and also Friedel Crafts reactions don t work well on deactivated rings. Page 10

11 IMIES + EAMIES 27. Give the curved arrow mechanism for each reaction. a. C 3 2 trace + C C C 3 C 3 2 C 3 2 C 3 C 3 C3 C 3 2 C3 + 2 C 3 C 3 trace + 3 C C C 3 C 3 C 3 C C C 3 3 C C3 C 3 C 3 3 C C 3 3 C C3 3 C C 3 + c. 2 2 trace R 2 2 R R 2 (alf-way point) + 2 R 2 Page 11

12 28. Draw both stereoisomers that can be formed in this reaction. 2 mild acid E Z 29. Give the major organic product for each reaction (one stereoisomer is sufficient). a. C C 3 2 trace + C 3 d. trace + 2 (C3 ) 2 p 5 C 3 C 3 e. trace + c. trace + f. 3 trace Give the curved arrow mechanism for each reaction. a. + C 2 2 C 3 + C 3 C C 2 C 3 C 2 C 3 C 2 C 3 2 C 2 C 3 C 2 C 3 2 Page 12

13 30 continued C 3 C C 3 C 2 2 c. 3 C C C 3 C Give the major organic product for each reaction. + C 3 a. 2 + C 3 2 C3 d C e c. C 3 C C 3 C 3 3 C + 3 C f. 2 Page 13

14 REACTI SUMMARY 32. Give the major organic product for each reaction. a. K a. a 2 cyclopentanone c. + workup c. + C 3 C 2 C 2 C 3 C 2 C 3 d. a. Li (s) pentanal c. + workup Li 3 C e. 3 P=CC 3 (The Z isomer is also K) f. abd 4 C 3 C 2 D D D g. C 2 C 3 cat. + C 3 C 2 2 h C 3 C 2 i. a. (C 3 C 2 ) 2 CuLi + workup Page 14