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C HAPTER 21 HW: A LDEHYDES + K ETONES NOMENCLATURE 1. Give the name for each compound (IUPAC or common name). CHO OH 6 O 5 Structure 1 H 4 NO2 O 3,3-dimethyl-2-pentanone Name (or 1,1-dimethylpropyl methyl ketone) 3 5 Structure 5-hydroxy-4-methylhexanal O 1 H 1 1-penten-3-one or ethyl vinyl ketone 2 4 O O 6 5 4 7 O Name m-nitrobenzaldehyde 2,4-dioxohexanal 3 1 O 2-methyl-5-heptyn-3-one 2. Draw each compound. O O zame zide F Structure F dicyclohexyl ketone Name CH3 H F O 1,1,1-trifluoro-3-pentanone (Z)-3,6-dimethyl-3-heptenal SPECTROSCOPY 3. In the 1H NMR spectrum of butanal, the signal at 2.40 ppm is a triplet of doublets (approximate J’s are 2 Hz and 7 Hz). Explain the splitting of this signal, including a sketch of a “tree diagram”. O H C H H C C C H H H Signal at 2.40 ppm (next to carbonyl) H H The CH2 next to the carbonyl is the signal at 2.40 ppm. It is split in a large way by its two CH2 neighbors (7 Hz, splitting first level into a triplet), and in a smaller way by the aldehyde H (2 Hz, splitting second level into a doublet). This results in a triplet of doublets. Page 1 4. Of compounds A-D, O O OCH3 O O OCH3 Br Br A Br Br B C D a. Which compound would have the following 13C NMR spectra (briefly explain)? δ = 165.9, 132.1, 131.5, 129.1, 127.4, 51.5 ppm Compound A 165.9 ppm is the C=O, 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 13C signals, so it is A (B would have 8 13C signals). b. Which compound would have the following 13C NMR spectra (briefly explain)? δ = 199.8, 140.3, 135.4, 131.5, 131.0, 127.7, 121.2, 28.6 ppm Compound D 199.8 ppm is the C=O, 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 13C signals, so it is D (C would have 6 13C signals). 5. Below are 1H and 13C NMR spectra of a compound with a formula of C5H8O. Determine the structure, then assign the peaks in the 1H NMR spectrum. 1H a 10 220 b 1H 9 200 8 180 7 O aH 5 PPM 140 120 c f b PPM 4 100 3 80 2H f 2 60 1 40 20 0 0 e/f assignment could be switched d e c d 1H 1H 6 160 2H e c is a wider ~doublet so must involve a trans coupling (to b). Page 2 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=O bonds. 2-cyclohexenone has a lower wavenumber absorbance, as it has a different C=O 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=O is weaker, and thus absorbs at a lower wavenumber (since v α f). O O O O O 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? Briefly explain. Both compounds have very conjugated C=O bonds, and conjugation lowers the O O IR wavenumber (more single bond character). The first compound has twice as much conjugation / resonance, so it’s C=O signal is the lowest. C=O Stretch 1666 cm-1 C=O: 1692 cm-1 WITTIG REACTION 8. Give the curved arrow mechanism for each reaction. H a. a. PPh3 CH3CH2Br PPh3 H3C b. nBuLi CH3CH2Br S N2 H H3C C O H PPh3 H3C C PPh3 Ph P H b. C Ph Ph Ph H3C C (nBuLi) H P Ph Ph H3C CH PPh3 O C H3C CH PPh3 O O H3C PPh3 H C H3C O H PPh3 H C CH3 Ph P Ph Ph Page 3 8 continued O c. PPh3 H PPh3 O Ph Ph H Ph H PPh3 O PPh3 O Ph H H 9. Draw both stereoisomers that can be formed in this Wittig reaction. Ph3P O H H cis trans H 10. Give the major product of each reaction (one stereoisomer is sufficient). H O CH3CH2CH=PPh3 a. O C O CH2CH3 Ph3P c. OCH3 CHO PPh3 OCH3 H2C PPh3 d. b. O H2C O 11. Use a Wittig reaction to produce each alkene, starting from an ylide. CH3 CH2 a. C CH3 CH2 CH2 CH3 C CH3 CH2 C Route 1 H CH2 CH3 O + Ph3P C CH3 CH2 CH3 CH2 Route 2 C CH2 CH3 PPh3 + O C CH3 CH2 O b. H H H or PPh3 PPh3 product H Route 1 Route 2 product H H H product PPh3 H product O Page 4 12. Use a Wittig reaction to produce this alkene, starting from an alkyl halide. H Br PPh3 PPh3 Route 1 PPh3 O nBuLi O PPh3 Ph3P Br Route 2 nBuLi Ph3P HYDRATES 13. Give the curved arrow mechanism that shows the formation of hydrate under each set of conditions. O trace OH- HO OH a. H2O H H HO O O H O H HO trace H+ HO OH H H H H O O OH b. H2O H H O O trace H O H+ H H H H H H H O OH O H HO OH H H 14. 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. O O O H H E F E F-hydrate E-hydrate O F OH OH OH OH H 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=O, 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. O O H H One EDG Page 5 15. Compound G has a smaller equilibrium constant (Keq) for hydrate formation than compound H. Explain this trend, including energy diagrams with your explanation. O H O O CH3 G Keq= 1.06 H CF3 O O H CF3 H H Keq= 2.9 x 104 CF3 H R H2O HO Keq H OH R EWG K = hydrate / carbonyl; so a large Keq means a higher % of hydrate. Carbonyl with CF3 has more hydrate. The reactivity differences arise from O different starting carbonyl energies (the H CF3 H hydrate energies are also nearly HO OH HO OH equivalent). CF3 is a strong EWG, so O H CH3 H CF3 destabilizes the δ+ of the carbonyl, H CH3 G G+H-hydrate making the CF3 carbonyl higher in energy than the aldehyde. This makes the hydrate reaction of the CF3 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. Briefly explain each answer. Brief Explanation: O O The aromatic group can participate in resonance with the C=O 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 A Brief Explanation: O O C=O 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 B Brief Explanation: O Pair C O H H CH3 OCH3 Brief Explanation: O Pair D O H H3N Methoxy 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. H The aldehyde is more reactive (makes more hydrate) because it has a higher energy carbonyl form (reaction is downhill). The aromatic has an EWG (-NH3+), which destabilizes the δ+ of the carbonyl. Page 6 ACETALS 17. Give the curved arrow mechanism for this acetal formation reaction. O OCH2CH3 H+ OCH2CH3 CH3CH2OH H O O H+ H CH3CH2OH Attack Protonate OCH2CH3 O OH2 Leave OCH2CH3 OCH2CH3 OH CH3CH2OH OH H+ Deprotonate O CH2CH3 OCH2CH3 Deprotonate OCH2CH3 CH2CH3 product H Attack CH3CH2OH CH3CH2OH Protonate CH3CH2OH Note: 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 Ea). 19. Give the curved arrow mechanism for this reaction. O H+ O H H Protonate HO H2O H HO HO O H Leave OH H HO HO H HO O H H Attack O O O H+ H O OH HO H Attack OH HO HO O H Deprotonate H O HO O O HO H OH Deprotonate H+ Protonate O O H Page 7 20. Give the major organic product for each reaction. O CH3CH2O OCH2CH3 H+ a. H O O d. O OCH3 CHO H+ OH H+ O OH O b. CH3CH2OH HO c. H H+ OCH3 CH3OH O 21. 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. H+ OCH3 H2O a. b. O OCH3 O d. H+ H3CO O H2O O c. O CH3OH O OH f. OH + CH3CHO H3CO + CH3CH2OH OH H+ H3CO OCH3 H+ H2O e. HO H2O O + 2 CH3OH O H H+ OCH3 H2O O + 2 CH OH 3 O H+ OCH3 H2O H3CO H + 2 CH3OH Page 8 23. Give the curved arrow mechanism for each reaction. OCH3 OCH3 + CH3OH H2O OCH3 H OCH3 H+ OCH3 OCH3 Protonate H O CH3 O H+ OH OH Leave O + H2O OH O O OH OH OH O H2O acetone O + CH3CH2OH HO H2O H+ H CH2CH3 OCH2CH3 O H O O O H H2O O H2O O (Mech could be begun by protonating either oxygen) OH H O H+ H OH O O HO H d. O H H OH H+ O O H H O H OCH2CH3 c. O OH H+ H2O H2O O O OH OH OH O H+ O H2O H Deprotonate H O H O product OH Protonate H+ O H2O Attack H OCH3 H HO 2 b. OCH3 OCH3 Leave H Deprotonate OCH3 O O H+ a. O H HO O H+ OH H2O O OH O HO H+ H O OH O 1 O 4 3 OH O O 4 OH O 3 O H product 1 OH HOR Page 9 24. Milder conditions can be used to hydrolyze acetal J than to hydrolyze acetal K. Explain their difference in reactivity. OCH3 OCH3 OCH3 OCH3 J K 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). PROTECTING GROUPS 25. Design a synthesis that uses cyclopentanone and 4-bromobutanal to efficiently produce the aldehyde shown. + O Br O OH H H O OH HO H+, H2O, heat (hydrolysis) H+, heat (protect aldehyde) H Br O a) Mg H O b) OH O O O c) H+ workup 26. The following multi-step synthesis converts benzene into a dicarbonyl species. a. Give the reagents needed to complete each step in the sequence. O Br Br2 Br Br Cl AlCl3 FeBr3 OH HO H+, heat O O O Mg OH O O H+, H2O H heat b) H+ workup O O O MgBr O a) CrO3, H+ or PCC O O O O b. Briefly explain why the synthesis below does not work well. O O Cl AlCl3 O O Cl AlCl3 O 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 IMINES + ENAMINES 27. Give the curved arrow mechanism for each reaction. CH3NH2 a. + H2O N trace H+ O CH3 H+ CH3NH2 O N H2O O H+ CH3 N H H N HO H H N CH3NH2 N HO H CH3 H CH3NH2 CH3 H CH3 CH3 N + H2O CH3 CH3 b. H N CH3 trace O + H2O H+ N H3C CH3 CH3 CH3 O H O H+ HO H N CH3 H3C N H3C O c. H CH3 H H O O H N NH2 H O NH2 NH2 NH2 NH2 H O NH2 H OH O H N O NH2 H O (Half-way point) H N N N H H H OH RNH2 N H OH N N H H+ O N OH2 H+ H N H O N CH3 NH2 H RNH2 H+ N H OH RNH 2 N N NH2 H N N CH3 N H OH2 H3C + H2O H H H+ N N O H N H3C trace H+ O HO CH3 CH3 NH2 O CH3 H H+ NH2 H N H3C CH3 N N H CH3 H H H2O N N RNH2 Page 11 28. Draw both stereoisomers that can be formed in this reaction. NH2 E N mild acid O Z N 29. Give the major organic product for each reaction (one stereoisomer is sufficient). CH3NH2 CHO a. CH3 N trace H+ NH2 d. N trace H+ O CH3 O b. N (CH3)2NH O N N H e. CH3 trace H+ pH 5 O c. N H O f. N H NH NH3 trace H+ trace H+ H 30. Give the curved arrow mechanism for each reaction. H+ N a. CH2CH3 H+ N CH2CH3 O H2O H N + CH3CH2NH2 H H2O H N CH2CH3 O H H N CH2CH3 H2O H+ CH2CH3 OH H N OH H CH2CH3 O H2O O H O H Page 12 30 continued O H+ b. N + CH3CH2NH2 H2O H H H H+ OH OH N H O H H H O H2O H CH3 H2O O CH3 N H H H O Me Me N CH3 H H H O Me Me N Me Me N Me Me H+ O O H O H2O H2O N H H + H+ Me N H H+ H3C N + CH3CH2NH2 c. N H+ OH H2O H2O N N H O N H Me O H H2O 31. Give the major organic product for each reaction. a. H+ N CH3 N b. O H2O d. + CH3NH2 H+ O N Ph NH2 H N CH3 O H+ + CH3NH-NH2 H2O H N Ph H NH2 H+ e. H2O H2O O H CH3 c. N CH3 H+ H2O H O N CH3 CH3 H3C f. N Ph H+ H2O H3C N H Ph O Page 13 REACTION SUMMARY 32. Give the major organic product for each reaction. O O a. OH KOH O HO HO O a. NaNH2 b. b. cyclopentanone c. H+ workup O OCH2CH3 OCH2CH3 H+ c. CH3CH2OH H H OH O a. Li (s) Br d. Li H b. pentanal c. H+ workup H3C O Ph3P=CHCH3 e. O f. (The Z isomer is also OK) OD NaBD4 D CH3CH2OD N O CH2CH3 cat. H+ g. CH3CH2NH2 O O H+ h. H2O O i. OH O + 2 CH3CH2OH H O a. (CH3CH2)2CuLi H b. H+ workup H Page 14