Summary of Alkene Reactions Mechanism Cheat Sheet, Cheat Sheet of Organic Chemistry

Download Summary of Alkene Reactions Mechanism Cheat Sheet and more Cheat Sheet Organic Chemistry in PDF only on Docsity! Chem 350 Jasperse Ch. 8 Handouts 1 Summary of Alkene Reactions, Ch. 8. Memorize Reaction, Orientation where Appropriate, Stereochemistry where Appropriate, and Mechanism where Appropriate. -all are drawn using 1-methylcyclohexene as a prototype alkene, because both orientation and stereochemistry effects are readily apparent. Orientation Stereo Mechanism 1 BrHBr (no peroxides) Markovnikov None Be able to draw completely 2 H CH3 Br both cis and trans HBr peroxides Anti-Markovnikov Nonselective. Both cis and trans Be able to draw propagation steps. 3 OH CH3 H2O, H + Markovnikov None Be able to draw completely 4 OH CH31. Hg(OAc)2, H2O 2. NaBH4 Markovnikov None Not responsible 5 H CH3 OH 1. BH3•THF 2. H2O2, NaOH Anti-Markovnikov Cis Not responsible 6 OR CH31. Hg(OAc)2, ROH 2. NaBH4 Markovnikov None Not responsible 7 H CH3 HD D H2, Pt None Cis Not responsible Chem 350 Jasperse Ch. 8 Handouts 2 Orientation Stereo Mechanism 8 Br CH3 H Br Br2 (or Cl2) None Trans Be able to draw completely 9 OH CH3 H Br Br2, H2O (or Cl2) Markovnikov Trans Be able to draw completely 10 O CH3 H PhCO3H None Cis Not responsible 11 OH CH3 H OH CH3CO3H H2O None Trans Be able to draw acid- catalyzed epoxide hydrolysis 12 OH CH3 OH H OsO4, H2O2 None Cis Not responsible 13 O H H O 1. O3 2. Me2S Note: H-bearing alkene carbon ends up as aldehyde. None None Not responsible 14 O H OH O KMnO4 H-bearing alkene carbon ends as carboxylic acid None None Not responsible Chem 350 Jasperse Ch. 8 Handouts 5 8 Br CH3 H Br Br2 (or Cl2) H HH Cation Capture Br Br BrBr Br Br 3 Notes 1. Cation intermediate is cyclic bromonium (or chloronium) ion 2. The nucleophile captures the bromonium ion via backside attack -this leads to the trans stereochemistry 3. The nucleophile attacks the bromonium ion at the *more* substituted carbon 9 OH CH3 H Br Br2, H2O (or Cl2) H HH Cation Capture Br Br Br Br OOH2 H H H Br OH-H 4 Notes 1. Cation intermediate is cyclic bromonium (or chloronium) ion 2. The nucleophile captures the bromonium ion via backside attack (ala SN2) -this leads to the trans stereochemistry 3. The nucleophile attacks the bromonium ion at the *more* substituted carbon -this explains the orientation (Markovnikov) a. There is more + charge at the more substituted carbon b. The Br-C bond to the more substituted carbon is a lot weaker HCH3 H Br Br O H H H Br OH-HMore Substituted End H Br O H H H Br OH-H Less Substituted End 4. Alcohols can function in the same way that water does, resulting in an ether OR rather than alcohol OH. Chem 350 Jasperse Ch. 8 Handouts 6 10 O CH3 H PhCO3H Ph O O O H ! CH3 " Ph ! ! ! " Ph ! ! " + ONE STEP! No ions #+ #$ Carbonyl-hydrogen Hydrogen-bonded reactant Notes 1. Complex arrow pushing 2. No ions required 3. The carbonyl oxygen picks up the hydrogen, leading directly to a neutral carboxylic acid -The peracid is already pre-organized for this' via internal H-bonding between carbonyl and H 11 OH CH3 H OH CH3CO3H H2O O CH3 H CH3 O O O H ONE STEP! No ions H OH CH3 H H Cation Capture OH OOH2 H H H OH OH-H Notes: a. The nucleophile (water) attacks from the more substituted end of the protonated epoxide  More δ+ charge there  The C-O bond to the more substituted end is much weaker b. The nucleophile adds via SN2-like backside attack. Inversion at the top stereocenter, but not the bottom, explains the trans stereochemistry. 12 OH CH3 OH H OsO4, H2O2 Os OO O O O CH3 O H Os O O Os (VIII) Os (VI) Concerted cis addition H2O Osmate Ester Hydrolysis OH CH3 OH H HO HO Os O O + Os (VI) H2O2 Os OO O O Os (VIII) + H2O Osmium Reoxidation Chem 350 Jasperse Ch. 8 Handouts 7 Chapter 7 Reactions and Mechanisms, Review E2 On R-X, Normal Base CH3 Br OCH3 H H H OCH3 Br NaOCH3 H OCH3 + Mech: + + Br (Normal base) Notes 1. Trans hydrogen required for E2 2. Zaytsev elimination with normal bases 3. For 3º R-X, E2 only. But with 2º R-X, SN2 competes (and usually prevails) 4. Lots of “normal base” anions. E2, On R-X, Bulky Base Br NEt3 or KOC(CH3)3 (Bulky bases) H2 C BrMech: H NEt3 + Et3NH Br Notes: 1. Hoffman elimination with Bulky Bases 2. E2 dominates over SN2 for not only 3º R-X but also 2º R-X 3. Memorize NEt3 and KOC(CH3)3 as bulky bases. Acid- Catalyzed E1- Elimination Of Alcohols OH H2SO4 H OH+ H2SO4 + HSO4 + OH2 -H2O HSO4 + H2SO4 Protonation Elimination Deprotonation OH OH2 H H H Mech Notes: 1. Zaytsev elimination 2. Cationic intermediate means 3º > 2º > 1º 3. 3-Step mechanism Chem 350 Jasperse Ch. 8 Handouts 10 Mechanism Br H H H Br H H + Br Br Protonate Cation Capture H o Protonate first o Capture cation second o Cation formaton (step 1) is the slow step Rank the Reactivity of the following toward HBr addition. 3 (2º) 2 (3º) 1 (3º allylic) Issue: Cation stability Why Does Markovnikov’s Rule Apply? Product/Stability Reactivity Rule. o Formation of the most stable carbocation results in Markovnikov orientation H Br For unsymmetrical alkenes, protonation occurs at the less substituted alkene carbon so that the more stable cation forms (3º > 2º > 1º), in keeping with the product stability-reactivity principle or H H 2º Br Br H Br H Br 2º 1º1º Markovnikov Product anti-Markovnikov Product Slow Step o This same logic applies anytime something adds to an alkene. o You want to make the best possible intermediate in the rate-determining step. Draw the mechanis for the following reaction: HBr Br H2C H Br H3C + Br H3C Br Chem 350 Jasperse Ch. 8 Handouts 11 8.3B Free Radical Addition of HBr with Peroxide Initiator: Anti-Markovnikov Addition (Rxn 2) 2 H CH3 Br both cis and trans HBr peroxides Anti-Markovnikov Nonselective. Both cis and trans Be able to draw propagation steps. • Peroxides are radical initiators, and cause the mechanism to shift to a radical mechanism • With peroxides, the orientation is reversed to anti-Markovnikov: now the Br adds to the less substituted end and the H adds to the more substituted end of an unsymmetrical alkene o No peroxides: Br goes to more substituted end o With peroxides: Br goes to less substituted end • The anti-Markovnikov radical process works only with HBr, not HCl or HI • The radical process is faster, and wins when peroxides make it possible. In the absence of peroxides, the slower cationic process happens. Mechanism, and Reason for AntiMarkovnikov Orientation Br For unsymmetrical alkenes, bromination occurs at the less substituted alkene carbon so that the more stable radical forms (3º > 2º > 1º), in keeping with the product stability-reactivity principle or Br Br 2º radical Br H Br H 1º radical Markovnikov Product anti-Markovnikov Product Slow Step H Br H Br Examples, Predict the Products. Does Markovnikov’s Rule matter? 1 HBr, peroxides HBr, no peroxides Br Br Yes 2 HBr, peroxides HBr, no peroxides Br Br Yes 3 HBr, peroxides HBr, no peroxides Br Br No Chem 350 Jasperse Ch. 8 Handouts 12 8.4 Addition of H-OH. Direct acid-catalyzed addition. (Reaction 3) General: C C H OH C C H OH H 3 OH CH3 H2O, H + Markovnikov None Be able to draw completely Markovnikov: Hδ+OHδ-  H adds to the less substituted end of the alkene, OH adds to the more substituted end • OH ends up on more substituted end of the alkene Mechanism: 3 Steps. 1. Protonation 2. Cation Capture 3. Deprotonation Cation Capture OH2 -H Deprotonate For unsymmetrical alkenes, protonation occurs at the less substituted alkene carbon so that the more stable cation forms (3º > 2º > 1º), in keeping with the product stability-reactivity principle or H H 2º H O H O 2º 1º1º Markovnikov Product anti-Markovnikov Product Slow Step OH2 H H H H -H Deprotonate H OH 2º H O 1º H H • The sequence in which key step (cation capture in this case) is sandwiched by proton on- proton off protonation-deprotonation is super common for acid-catalyzed reactions. o Whenever you see an acid-catalyzed process, expect to use H+ in first step and to deprotonate in the last step • Cation stability dictates reactivity • Cation stability explains why the Markovnikov orientation occurs. This involves the more substituted, more stable carbocation product in the rate-determining step. • The actual reaction is an equilibrium. o The reverse of alcohol dehydration to make alkenes! o A key drive is to have excess water. That pushes the equilibrium to the alcohol side. o Under alcohol  alkene conditions, the equilibrium is often driven to the alkene side by having no water, or by distilling off the lower-boiling alkene as it forms. Chem 350 Jasperse Ch. 8 Handouts 15 Mechanism (For interest sake. Not for memorization, not for test.) C C C C OH HgOAcHg(OAc)2 Overall pathway: H2O NaBH4 C C OH H "Oxymercuration" "Demercuration" Hg(OAc)2 O H HH Cation Capture HgOAc HgOAc H H -H OH H HgOAc Deprotonate HgOAc NaBH4 OH H H - OAc OH2 OAc = "acetate" = O O "Mercurinium Ion" More Details for the Oxymercuration Phase Notes: 1. “demercuration” with NaBH4 replaces the mercury with a hydrogen 2. The initial “oxymercuration” essentially adds (HgOAc)δ+(OH)δ-, and follows Markov.’s rule 3. The interesting new thing here is the “mercuronium” ion 4. This is normally drawn as a 3-ring, but can also be viewed as a resonance structure of a hybrid Mercuronium Ion H HgOAc H HgOAc H HgOAc Mercuronium Ring 3º Cation 2º Cation A B C Both participation from structures A and B are required to explain everything o A explains why you don’t get cation rearrangments, ever: you don’t have a free carbocation o A also explains structure studies, which show that the mercury is weakly bonded to the more substituted carbon o B helps to explain why water adds to the more substituted carbon, which has extensive positive charge o C doesn’t contribute, isn’t really involved o In the real thing, there is a long, very weak and super breakable bond between mercury and the more substituted carbon. The bond to the less substituted carbon is much shorter and stronger. "!-complex" C C C C Electrophilic HgOAc Adds C C HgOAc HgOAc "!-Complex" Perspective Chem 350 Jasperse Ch. 8 Handouts 16 8.7 Indirect anti-Markovnikov Addition of H-OH via Hydroboration/Oxidation. Reaction 5. C C C C H BH2 Overall pathway: 1. BH3•THF 2. H2O2, NaOH C C H OH "Hydroboration" "Oxidation" 5 H CH3 OH 1. BH3•THF 2. H2O2, NaOH plus enantiomer Anti-Markovnikov Cis Not responsible Notes: 1. Anti-Markovnikov orientation: the OH ends up on the less substituted end of an unsymmetrical alkene; the H adds to the more substituted end 2. Cis addition. Both the H and the OH add from the same side. 3. When does cis/trans addition stereochemistry matter? o Only when both alkene carbons turn into chiral centers in the product. o If one does but not both, then the relative stereochemistry doesn’t matter o For Markovnikov additions involving H-Br or H-OH, the H usually adds to a carbon that already has an H, so that in the product it is not a stereocenter. o In anti-Markovnikov additions, much more common for both carbons to become chiral carbons 4. Chiral products are Racemic (two enantiomers form) but not optically active o When only one chiral center forms (often in the Markovnikov additions), any chiral product will always be racemic o When two chiral centers form, as in the example above, of the four possible stereoisomers, you get only two of them, in racemic mixture. H CH3 OH 1. BH3•THF 2. H2O2, NaOH H CH3 OH + H CH3 OH H CH3 OH A B C D Cis Addition Enantiomers Do Form Trans Addition Enantiomers Do NOT Form Examples, Predict the Products. Does Markov. Matter? Does Stereo Matter? 1 1. BH3•THF 2. H2O2, NaOH OH Yes No Chem 350 Jasperse Ch. 8 Handouts 17 Does Markov. Matter? Does Stereo Matter? 2 1. BH3•THF 2. H2O2, NaOH HO Yes No 3 1. BH3•THF 2. H2O2, NaOH OH No No 4 1. BH3•THF 2. H2O2, NaOH OH No No 5 1. BH3-THF 2. NaOH, H2O2 1. Hg(OAc)2, H2O 2. NaBH4 H2O, H + OHHH3C OH OH H Yes Yes No No 1. Which starting alkenes would produce the following products following hydroboration- oxidation? Factor in the stereochemistry of the products in considering what starting materials would work. 1. BH3-THF 2. NaOH, H2O2 1. BH3-THF 2. NaOH, H2O2 Ph H H HO H3C Ph OH H H H3C Ph H CH3 H Ph CH3 CH3 Ph CH3 H HO H H3C 2. Fill in recipes for converting 1-butene into the three derivatives shown. 1. BH3-THF 2. NaOH, H2O2 OH OH 1. Hg(OAc)2, H2O 2. NaBH4 H2O, H + or Chem 350 Jasperse Ch. 8 Handouts 20 Ether Synthesis: Two Routes 1. From Alkene and Alcohol: By Oxymercuration/Demercuration 2. From R-Br and Alkoxide Anion: By SN2 3. Multistep Syntheses: Design Reactants for the Following Conversions • Note: It is often most strategic to think backward from product to precursor. • Then think back how you could access the precursor from the starting material. • There may sometimes be more than one suitable route. a. OCH3 1. HBr, peroxides 2. NaOCH3 b. 1. Hg(OAc)2, CH3OH 2. NaBH4 OCH3 c. OH CH3 OH CH31. H2SO4, heat 2. BH3-THF 3. NaOH, H2O2 d. 1. Br2, hv 2. NaOCH3 (or other small, normal base) e. 1. Br2, hv 2. NEt3 (or KOCMe3) f. OH CH3 Br CH31. H2SO4, heat 2. HBr g. 1. H2SO4, heat 2. HBr, peroxides OH CH3 CH3 Br Chem 350 Jasperse Ch. 8 Handouts 21 8-10. H-H addition. Catalytic Hydrogenation (Reaction 7) General: C C C C H H H2, Pt (or Pd, or Ni, etc.) 7 H CH3 HD D H2, Pt plus enantiomer Orientation: None Stereo: Cis Mech: Not responsible Notes: 1. Since both atoms adding are the same (H), Markovnikov orientation issues don’t apply • You’re adding a hydrogen to both the more and less substituted alkene carbon! 2. Stereochemistry isn’t often relevant, but when it is it’s cis • Rarely relevant because if either alkene carbon has even one hydrogen attached, addition of an additional hydrogen will result in an achiral carbon. 3. The reaction is substantially exothermic 4. But some kind of transition-metal catalyst is required to active the otherwise strong H-H bonds. Examples, Predict the Products. Does Mark’s Rule matter? Does Stereo? 1 H2, Pt No No 2 H2, Pt No No 3 H2, Pt D H3C D H H No Yes 4 H2, Pt CH3 CH3 H H No Yes 5 CH3 H H H H2, Pt No No 6 H2, Pt D CH3 D H H No Yes Chem 350 Jasperse Ch. 8 Handouts 22 8.8 X-X Dihalogen Addition: Trans Addition (Reaction 8) General: C C C C Br BrBr2 or Cl2 or C C Cl Cl Orientation Stereo Mechanism 8 Br CH3 H Br Br2 (or Cl2) plus enantiomer None Trans Be able to draw completely Notes: 1. Orientation: Non-issue, since you’re adding the same atom to each alkene carbon 2. Trans addition 3. Solvent matters: to get X-X addition, you need a solvent other than water or alcohol. • With water or alcohol, you get different products, see reaction 9 Examples, Predict the Products. Does Mark. matter? Does Stereo? Chiral? 1 Br2 Br Br No No Yes 2 Br2 Br CH3 H Br No Yes Yes 3 Cl2 Cl Cl No No Yes 4 Cl2 Cl Cl No Yes Yes 5 Br2 Br Br Br Br Br Br meso No Yes No 6 Br2 Br Br Br Br chiral No Yes Yes Notes: 1. Cis and trans reactants give different products! 2. For any product (in this and other reactions), be able to identify whether it is chiral or not Chem 350 Jasperse Ch. 8 Handouts 25 Draw the mechanism for the following reaction: Br2 H2O CH3 OH Br H Br Br Br CH3 OH2 CH3 O Br H HH Br CH3 OH Br H Examples, Predict the Products. Does Mark. matter? Does Stereo? Chiral? 1 H2O Br2 Br2 Br Br OH Br No Yes No No Yes Yes 2 H2O Br2 Br2 Br CH3 Br Br CH3 HO No Yes Yes Yes Yes Yes 3 Cl OH H2O Cl2 No Yes Yes 5 Br2 H2O Br OH Br OH Br OH No Yes Yes 6 Br OH Br OH Br2 H2O No Yes Yes Chem 350 Jasperse Ch. 8 Handouts 26 8-12 Epoxidation. Addition of one Oxygen (Reaction General: C C C C OPhCO3H "peracid" "epoxide" 10 O CH3 H PhCO3H plus enantiomer None Cis Not responsible Notes: 1. No orientation issues, since the same oxygen atom connects to both bonds 2. Cis addition: both oxygen bonds come from the same direction Mechanism: No test Responsibility Ph O O O H ! CH3 " Ph ! ! ! " Ph ! ! " + ONE STEP! No ions #+ #$ Carbonyl-hydrogen Hydrogen-bonded reactant • Any peracid with formula RCO3H has an extra oxygen relative to a carboxylic acid. • Any peracid can deposit the extra oxygen onto the p-bond to make the epoxide • No ions are actually involved, because the leaving group is the neutral carboxylic acid Examples, Predict the Products. Does Mark. matter? Does Stereo? Chiral? 1 PhCO3H O O No Yes Yes 2 PhCO3H O No Yes No 3 PhCO3H O O No Yes No 4 PhCO3H O O No Yes Yes Chem 350 Jasperse Ch. 8 Handouts 27 8-13 Trans OH-OH addition. Epoxidation in water. The initially formed epoxide undergoes Acid- Catalyzed Ring Opening. Reaction 11. General: C C C C CH3CO3H H2O OH OH 11 OH CH3 H OH CH3CO3H H2O plus enantiomer None Trans Be able to draw acid- catalyzed epoxide hydrolysis Examples, Predict the Products. Does Mark. matter? Does Stereo? Chiral? 1 CH3CO3H H2O OH OH No No Yes 2 CH3CO3H H2O OH OH No Yes Yes 3 CH3CO3H H2O HO OH No Yes Yes 4 CH3CO3H H2O OH OH HO OH No Yes No Mech 11 O CH3 H CH3 O O O H ONE STEP! No ions H OH CH3 H H Cation Capture OH OOH2 H H H OH OH-H Notes: a. The nucleophile (water) attacks from the more substituted end of the protonated epoxide  More δ+ charge there  The C-O bond to the more substituted end is much weaker c. The nucleophile adds via SN2-like backside attack. Inversion at the top stereocenter, but not the bottom, explains the trans stereochemistry. Chem 350 Jasperse Ch. 8 Handouts 30 8.15-B Ozonolysis. Cleavage of Alkenes. Reaction 13 General: C C 1. O3 2. Me2S C CO O 13 O H H O 1. O3 2. Me2S Note: H-bearing alkene carbon ends up as aldehyde. None None Not responsible Notes 1. Double bond gets sliced in half 2. Get two corresponding carbonyls 3. Alkene bonds and nothing else are oxidized. 4. Get ketones and/or aldehydes and/or formaldehyde 8.15-A Oxidative Cleavage of Alkenes by Permanganate. Reaction 14 General: C C H C C OH O O KMnO4 heat 14 O H OH O KMnO4 H-bearing alkene carbon ends as carboxylic acid None None Not responsible Notes 1. Double bond gets sliced in half 2. Get two corresponding carbonyls 3. Alkene C-H bonds are also oxidized to C-OH bonds. 4. Get ketones and/or carboxylic acids and/or carbonic acid. 1 KMnO41. O3 2. Me2S O O H O O OH 2 KMnO41. O3 2. Me2S OO OH H H O OH OH OH 3 KMnO4 1. O3 2. Me2SO O OO H OH Chem 350 Jasperse Ch. 8 Handouts 31 4. KMnO4 1. O3 2. Me2S O O H O H H O H O O H O OH HO O OH 5. Identify reactants. 1. O3 2. Me2S H O O O + H2C=O H H H or KMnO4 OH O O H 6. Identify A, B, and C. 1. O3 2. Me2S H O O Unknown A C9H16 Br2, H2O CH3CO3H, H2O Product B Product C H A HO H3C Br H HO H3C HO H Review Problems. 7. “Roadmap format”. Identify products A-D. Br2, hv A NaOH B Br2, CH3OH C NEt3 D Br OCH3 Br OCH3 Chem 350 Jasperse Ch. 8 Handouts 32 8. Design a synthetic plan for the following conversions. (Several on test) a. OH OHOH 1. H2SO4, heat 2. OsO4, H2O2 b. OH1. Br2, H2O 2. NEt3 c. Br Cl Cl 1. NEt3 2. Cl2 d. O O 1. Br2, hv 2. NaOH (or other normal base) 3. O3 4. Me2S e. OH OH 1. Br2, hv 2. NaOH (or other normal base) 3. CH3CO3H, H2O f. OH O H3C CH3 1. H2SO4, heat 2. PhCO3H