Chem 350 Jasperse Ch. 8 Notes 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 CH 3 HBr H Anti-Markovnikov Nonselective. Be able to Both cis draw peroxides Br and trans propagation both cis and trans steps. 3 CH H O, H+ OH3 2 Markovnikov None Be able to draw completely 4 CH 1. Hg(OAc) , H O 3 2 2 OH Markovnikov None Not 2. NaBH4 responsible 5 CH 1. BH •THF 3 3 H Anti-Markovnikov Cis Not 2. H2O2, NaOH OH responsible 6 CH 1. Hg(OAc) , ROH 3 2 OR Markovnikov None Not 2. NaBH4 responsible 7 CH 3 H2, Pt H None Cis Not responsible D H D Chem 350 Jasperse Ch. 8 Notes 2 Orientation Stereo Mechanism 8 CH Br2 Br3 None Trans Be able to draw (or Cl ) H 2 Br completely 9 CH Br2, H2O OH3 Markovnikov Trans Be able to draw (or Cl ) H 2 Br completely 10 CH 3 PhCO H 3 None Cis Not O responsible H 11 CH Be able to CH CO H 3 3 3 OH None Trans draw H O H acid- 2 OH catalyzed epoxide hydrolysis 12 CH 3 OsO4, H2O2 OH None Cis Not OH responsible H 13 1. O 3 None None Not O H2. Me2S responsible H O Note: H-bearing alkene carbon ends up as aldehyde. 14 KMnO4 None None Not O H responsible OH O H-bearing alkene carbon ends as carboxylic acid Chem 350 Jasperse Ch. 8 Notes 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 the more stable cation forms H Capture H (3º > 2º > 1º), in keeping with the + Br product stability-reactivity principle CH CH 3 3 vs. H H H H 3º 2º 2 CH 3 HBr H peroxides Br both cis and trans Br H Br H Note 2: Hydrogenation of H Brominate Br Hydrogen Br + Br tfhaece r,a tdhiucas l cciso/mtraens sfrom either H Transfer H mixture results CH 3 H Note 1: For unsymmetrical alkenes, top Br bromination occurs at the less CH CH 3 3 cis H substituted alkene carbon so that vs. Br bottom the more stable radical forms Br H Br (3º > 2º > 1º), in keeping with the product stability-reactivity principle 3º H 2º H H CH3 Br trans H 3 CH H O, H+ OH3 2 H H OH2 O OH H -H H H Protonate Cation H H H Capture H Deprotonate 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 Notes 4 4 CH 1. Hg(OAc) , H O 3 2 2 OH 2. NaBH4 H HgOAc OH2 O OH HgOAc H -H H Cation HgOAc HgOAc - OAc H Capture H Deprotonate H NaBH 4 Hg(OAc) 2 OH H H 5 CH 1. BH •THF 3 3 H 2. H2O2, NaOH OH Notes H CHH3 H2O2, NaOH CHH3 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 δ+, the H is δ- c. The H O , NaOH process is complex, 2 2 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 CH 1. Hg(OAc) , ROH 3 2 OR 2. NaBH4 H HgOAc HgOAc HOCH3 O CH -H OCH3 3 H Cation HgOAc HgOAc - OAc H Capture H Deprotonate H NaBH 4 Hg(OAc) 2 OCH 3 H H Chem 350 Jasperse Ch. 8 Notes 5 8 CH Br2 Br3 (or Cl ) H 2 Br 3 Notes Br Br Br Br 1. Cation intermediate is cyclic Br H Br bromonium (or chloronium) ion H CCaatpiotunre H 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 CH Br2, H2O OH3 (or Cl ) H 2 Br H Br Br Br OH2 O H -H OH H Br Cation 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 End Br H CH3 H Br Less Substituted H End Br Br H -H OH O 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 Notes 6 10 CH 3 PhCO H 3 O H ONE CH3 Η δ+ H O Ο Η STEP! Ο + Ο Ο O O Ο Ο Nioons Η Ph δ− 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 CH CO H 3 3 3 OH H O H 2 OH H OH OSTNEEP! COH3 H COHH3 OH2 O H -H OH O O No Cation OH 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 S 2-like backside attack. Inversion at the top stereocenter, but not the N bottom, explains the trans stereochemistry. 12 CH 3 OsO4, H2O2 OH OH H O O Ccios nacdedritteiodn CHO3 OsO H2O CHO3H + HO OsO Os (VI) Os O O OH HO O O O Osmate H H Ester Osmium Os (VIII) Os (VI) Hydrolysis H2O2 Reoxidation O O + H O Os 2 O O Os (VIII) Chem 350 Jasperse Ch. 8 Notes 7 Chapter 7 Reactions and Mechanisms, Review E2 Br Br Mech: On CH NaOCH 3 3 R-X, Normal (Normal +H OCH3 H OCH 3 base) H + H OCH + Br Base 3 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, S 2 competes (and usually prevails) N 4. Lots of “normal base” anions. EO2n, Br NKOEtC3 (oCrH ) Mech: BrHC2 NEt3 + Et NH Br 3 3 H 3 R-X, Bulky (Bulky Base bases) Notes: 1. Hoffman elimination with Bulky Bases 2. E2 dominates over S 2 for not only 3º R-X but also 2º R-X N 3. Memorize NEt and KOC(CH ) as bulky bases. 3 3 3 Acid- OH Catalyzed H2SO4 +H OH E1- Elimination Of Mech Alcohols OH 2 OH H2SO4 -H2O Deprotonation H Protonation Elimination H + HSO4 H HSO 4 + H SO + OH2 2 4 Notes: 1. Zaytsev elimination 2. Cationic intermediate means 3º > 2º > 1º 3. 3-Step mechanism Chem 350 Jasperse Ch. 8 Notes 8 Ch. 8 Reactions of Alkenes 8-1,2 Introduction CH 3 CH 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 CH CH CH CH3 Cation 3 Cation 3 3 A B B vs A + B δ+ δ− Formation A Capture A B H A B H E H F H or CH3 CH CH3 A Doesn't Happen 3 B Because + B A Inferior Cation A H H C Product E H 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 Notes 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. Does Markovnikov’s Rule matter? 1 HBr 2 HCl 3 HI 4 HBr 5 HBr 6 I Cl Chem 350 Jasperse Ch. 8 Notes 10 Mechanism H Br Br Br H H Protonate Cation H H Capture H + Br 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. Issue: Why Does Markovnikov’s Rule Apply? Product/Stability Reactivity Rule. o Formation of the most stable carbocation results in Markovnikov orientation Br H Br H Br H Markovnikov Product Slow Step 2º 2º or For unsymmetrical alkenes, protonation occurs at the H less substituted alkene carbon H so that the more stable cation forms Br Br (3º > 2º > 1º), in keeping with the anti-Markovnikov Product product stability-reactivity principle 1º 1º 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. HBr Draw the mechanis for the following reaction:
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