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 HBr Br Markovnikov None Be able to (no peroxides) draw completely 2 CH3 HBr H Anti-Markovnikov Nonselective. Be able to Both cis draw peroxides Br both cis and trans and trans propagation steps. 3 + CH3 H2O, H OH Markovnikov None Be able to draw completely 4 CH3 1. Hg(OAc)2, H2O OH Markovnikov None Not 2. NaBH4 responsible 5 CH3 1. BH3•THF H Anti-Markovnikov Cis Not 2. H2O2, NaOH responsible OH 6 CH3 1. Hg(OAc)2, ROH OR Markovnikov None Not 2. NaBH4 responsible 7 CH3 H , Pt 2 H None Cis Not D H responsible D Chem 350 Jasperse Ch. 8 Handouts 2 Orientation Stereo Mechanism CH 8 Br 3 2 Br None Trans Be able to (or Cl2) H draw Br completely CH 9 Br , H O 3 2 2 OH Markovnikov Trans Be able to (or Cl2) H draw Br completely 10 CH3 PhCO3H O None Cis Not responsible H 11 CH3 Be able to CH3CO3H OH None Trans draw H2O H acid- OH catalyzed epoxide hydrolysis 12 CH OsO , H O 3 4 2 2 OH None Cis Not OH responsible H 13 1. O3 O None None Not H 2. Me2S responsible H O Note: H-bearing alkene carbon ends up as aldehyde. 14 KMnO4 O None None Not H responsible OH O H-bearing alkene carbon ends as carboxylic acid Chem 350 Jasperse Ch. 8 Handouts 3 Summary of Mechanisms, Ch. 7 + 8. Alkene Synthesis and Reactions. 1 HBr Br (no peroxides) Note: For unsymmetrical alkenes, H Br Br Br protonation occurs at the less H substituted alkene carbon so that H Protonate Cation H H the more stable cation forms Capture H (3º > 2º > 1º), in keeping with the + Br product stability-reactivity principle CH3 CH3 vs. H H H H 3º 2º 2 CH3 HBr H peroxides Br both cis and trans Br H Br H Note 2: Hydrogenation of the radical comes from either Br + Br H Brominate Hydrogen Br face, thus cis/trans H mixture results Transfer H CH3 H Note 1: For unsymmetrical alkenes, top Br bromination occurs at the less CH3 CH3 substituted alkene carbon so that cis H vs. Br the more stable radical forms bottom Br Br H (3º > 2º > 1º), in keeping with the H H CH product stability-reactivity principle 3º H 3 2º Br trans H 3 + CH3 H2O, H OH H H OH2 O OH H -H H H Protonate Cation H H Deprotonate H Capture H H Note: For unsymmetrical alkenes, protonation again occurs at the less substituted end of the alkene, in order to produce the more stable radical intermediate (3º > 2º > 1º) Chem 350 Jasperse Ch. 8 Handouts 4 4 1. Hg(OAc) , H O CH3 2 2 OH 2. NaBH4 H HgOAc OH2 O OH HgOAc H -H H HgOAc HgOAc Cation Deprotonate - OAc H Capture H H NaBH4 Hg(OAc)2 OH H H 5 CH3 1. BH3•THF H 2. H O , NaOH 2 2 OH CH Notes 3 H2O2, NaOH CH3 H H H a. concerted addition of B-H across C=C -explains the cis stereochemistry BH2 BH2 OH b. the B-H addition is Markovnikov; the B is !+, the H is !- c. The H2O2, NaOH process is complex, but replaces the B with OH with complete retention of stereochem -the explains why the cis stereochemistry established in step one is preserved in step 2. 6 CH3 1. Hg(OAc)2, ROH OR 2. NaBH4 H HgOAc HOCH3 O OCH3 HgOAc CH3 -H H HgOAc HgOAc Cation Deprotonate - OAc H Capture H H NaBH4 Hg(OAc)2 OCH3 H H Chem 350 Jasperse Ch. 8 Handouts 5 8 CH3 Br2 Br (or Cl2) H Br 3 Notes Br Br Br Br Br 1. Cation intermediate is cyclic H Br bromonium (or chloronium) ion Cation 2. The nucleophile captures the H H Capture bromonium ion via backside attack -this leads to the trans stereochemistry 3. The nucleophile attacks the bromonium ion at the *more* substituted carbon 9 CH3 Br2, H2O OH (or Cl2) H Br H Br Br OH2 O Br H -H OH H Cation Br Br H Capture H 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 H O More H -H OH Substituted Br Br End H CH3 H Br Less H Substituted End Br Br H -H O OH H H H 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 CH3 PhCO3H O H #+ ONE CH3 " H O " STEP! ! O ! ! + ! O ! ! No Ph #$ ions " Ph Ph 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 CH CO H CH3 3 3 OH H2O H OH H ONE CH3 CH3 H OH2 O H STEP! -H OH O O OH H O OH O No Cation OH ions H H Capture H H CH3 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 CH3 OsO4, H2O2 OH OH H Concerted CH H2O CH 3 O 3 O O O cis addition O OH HO Os + Os Os (VI) Os O O OH HO O O O Osmate H Ester H H O Osmium Os (VIII) Os (VI) Hydrolysis 2 2 Reoxidation O O + H O Os 2 O O Os (VIII) Chem 350 Jasperse Ch. 8 Handouts 7 Chapter 7 Reactions and Mechanisms, Review E2 Br Br Mech: CH On NaOCH 3 R-X, 3 (Normal + H OCH3 H Normal OCH3 base) H + H OCH3 + Br 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, Br NEt or Mech: Br H2 3 NEt3 On KOC(CH ) C + Et NH Br 3 3 H 3 R-X, Bulky (Bulky Base bases) 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- OH Catalyzed H2SO4 + H OH E1- Elimination Of Mech Alcohols OH2 OH H2SO4 -H O 2 Deprotonation Protonation H + HSO Elimination H H 4 HSO4 + H2SO4 + OH2 Notes: 1. Zaytsev elimination 2. Cationic intermediate means 3º > 2º > 1º 3. 3-Step mechanism Chem 350 Jasperse Ch. 8 Handouts 8 Ch. 8 Reactions of Alkenes 8-1,2 Introduction CH CH 3 3 B + A B Addition Reaction A H H 1. Thermodynamics: Usually exothermic . 1 π + 1 σ 2 σ bonds 2. Kinetics: π bond is exposed and accessible Generic Electrophilic Addition Mechanism CH3 CH3 CH3 CH3 Cation Cation A B B A + B vs !+ !" Formation A Capture A B H H H H A B E F or CH3 CH3 CH3 A Doesn't Happen B Because + B A Inferior Cation A H C Product H H E Forms D 2 Steps: Cation formation and cation capture • Cation formation is the slow step o Cation stability will routinely determine the orientation in the first step . Which is preferred, A B or A C? • Often the cation is a normal cation B. Sometimes 3-membered ring cations D will be involved. • In some cases, the cation will be captured by a neutral species (like water), in which case an extra deprotonation step will be involved 4 Aspects to Watch For 1. Orientation • Matters only if both of two things are true: a. The alkene is unsymmetrical, and b. The electrophile is unsymmetrical 2. Relative Stereochemistry o Matters only if both the first and the second alkene carbons are transformed into chiral centers 3. Mechanism 4. Relative Reactivity of Different Alkenes o Stability of cation formed is key Chem 350 Jasperse Ch. 8 Handouts 9 8.3 H-X Hydrogen Halide Addition: Ionic/Cationic Addition in the Absence of Peroxides (Reaction 1) H X H X General: C C C C Orientation Stereo Mechanism 1 HBr Br Markovnikov None Be able to (no peroxides) draw completely Markovnikov’s Rule (For Predicting Products): When H-X (or any unsymmetrical species Aδ+Bδ-) adds to an unsymmetrical alkene: o the H+ (or Aδ+) adds to the less substituted carbon (the one with more H’s) o the X- (or Bδ-) adds to the more substituted carbon (the one with more non-H’s). o Note: Markovnikov’s rule does not apply if either the alkene or the atoms that are adding are symmetrical Examples, Predict the Products.