Chapter6_Part_I_of_II10_17&19_2011 - Which product...

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Unformatted text preview: Which product is correct for this reaction? 1. A 2. B 3. C H+ 56% H2O 37% OH OH OH A B Answer: 3. C 8C % 1 2 3 Whic8h product is correct for this reaction? 1. A 2. B 3. C 4. A & D 5. B & E 6. C & F BD3 H2O2/NaOH H D CH3 OH D CH3 HO H A HO D Answer: 5. B&E H OH CH3 D B CH3 D H 5 3% H C CH3 D OH E 4% 18OH % D 7% 2 CH3 H F 0% 1 18% 3 4 5 6 Which product is correct for this reaction? 1. A 2. B 3. C BH3 76% H2O2/NaOH OH OH A Answer: 2. B OH B C 9% 1 2 15% 3 Which product is correct for this reaction? H+ 1. A 2. B 3. C 50% H2O 34% OH OH A Answer: 3. C OH 16% C B 1 2 3 Which product is correct for this reaction? BH3 1. A 2. B 3. C H2O2/NaOH 51% OH 28% OH 20% C B A Answer: 2. B OH 1 2 3 Organic Chemistry CHE 275 Chapter 6 Part I of II Reactions of Alkenes: Addition Reactions General Equation for Electrophilic Addition C C δ+ δ– + E—Y E C C Y When E+ is a Hydrogen Halide C C δ+ δ– + H—X H C C X Example CH2CH3 CH3CH2 CH C H C H HBr CHCl3, -30°C CH3CH2CH2CHCH2CH3 Br (76%) Mechanism • Electrophilic addition of hydrogen halides to alkenes proceeds by rate-determining formation of a carbocation intermediate. Mechanism • Electrons flow from the π system of the alkene (electron rich) toward the positively polarized proton of the hydrogen halide. Mechanism +C C H .. – .. : X: .. H C C .. X .. : .. :X .. C C H Markovnikov's Rule • When an unsymmetrically substituted alkene reacts with a hydrogen halide, the hydrogen adds to the carbon that has the greater number of hydrogen substituents, and the halogen adds to the carbon that has the fewer hydrogen substituents. Markovnikov's Rule CH3CH2CH CH2 HBr acetic acid acetic CH3CH2CHCH3 Br (80%) Example 1 Markovnikov's Rule CH3 H C CH3 C H CH3 HBr acetic acid CH3 C CH3 CH Br (90%) Example 2 Markovnikov's Rule CH3 HCl CH3 CH 0°C Cl (100%) (100%) Example 3 Mechanistic Basis for Markovnikov's Rule Protonation of a double bond occurs in direction that gives more stable of two possible carbocations. Mechanistic Basis for Markovnikov's Rule CH3CH2CH CH2 HBr acetic acid CH3CH2CHCH3 Br Mechanistic Basis for Markovnikov's Rule + CH3CH2CH2—CH2 primary carbocation is less stable: not formed + CH3CH2CH—CH3 + Br – Br HBr CH3CH2CH CH2 CH3CH2CHCH3 Br Br Mechanistic Basis for Markovnikov's Rule H HCl CH3 CH3 CH 0°C Cl H secondary secondary carbocation is less stable: not formed not + CH3 CH H HH + CH3 CH Cl– HCl H CH3 CH3 CH Cl Rearrangements H2C CHCH(CH3)2 HCl, 0°C H + CH3CHCH(CH3)2 + CH3CHC(CH3)2 CH3CHCH(CH3)2 CH3CH2C(CH3)2 Cll C (40%) (60%) Cll C Markovnikov's Rule CH3CH2CH CH2 HBr acetic acid acetic CH3CH2CHCH3 Br (80%) Application: Conversion of Alkenes to Alcohols OH OH 1. H2SO4 1. 2. H2O, heat (75%) Acid-Catalyzed Hydration of Alkenes C C + H—OH H—OH H C C OH •reaction is acid catalyzed; typical hydration medium is 50% H2SO4-50% H2O Follows Markovnikov's Rule H3C H C H3C C CH3 50% H2SO4 50% H2O CH3 CH3 C CH2CH3 OH (90%) (90%) Follows Markovnikov's Rule CH2 CH 50% H2SO4 50% CH3 50% H2O OH (80%) Mechanism •involves a carbocation intermediate •is the reverse of acid-catalyzed dehydration of alcohols to alkenes H3C C H3C CH2 + H2O H+ CH3 CH3 C OH CH3 Mechanism Step (1) Protonation of double bond Step H H3C C CH2 + + O: H H3C H slow H3C H3C H + C CH3 + :O: H Mechanism Step (2) Capture of carbocation by water Step H3C H + C :O: CH3 + H3C fast CH3 CH3 C CH3 H + O: H H Mechanism Step (3) Deprotonation of oxonium ion Step CH3 CH3 C H : +O CH3 H + :O: H H fast CH3 CH3 C CH3 H .. O: H + H + O: H Mechanism H3CCH CH2 + H O OSO2OH slow - O OSO2OH + H3CCH CH3 fast O OSO2OH H3CCH CH3 Mechanism CH3CHCH3 O + H—OH H—OH SO2OH heat CH3CHCH3 O + HO—SO2OH H The sulfate then undergoes hydrolysis in hot water Relative Rates Acid-catalyzed hydration • ethylene CH2=CH2 1.0 • propene CH3CH=CH2 1.6 x 106 • 2-methylpropene (CH3)2C=CH2 2.5 x 1011 • The more stable the carbocation, the faster it is formed, and the faster the reaction rate. Principle of Microscopic Reversibility H3C C CH2 + H2O H+ CH3 CH3 H3C C OH •In an equilibrium process, the same intermediates and transition states are encountered in the forward direction and the reverse, but in the opposite order. CH3 Hydration-Dehydration Equilibrium H H C H C H H+ + H2O H H H C C H H How do we control the position of the How equilibrium and maximize the product? equilibrium OH Le Chatelier’s Principle A system at equilibrium adjusts so to system minimize any stress applies to it. minimize For the hydration-dehydration equilibria, the For key stress is water. key Adding water pushes the equilibrium toward Adding more product (alcohol). more Removing water pushes the equilibrium Removing toward more reactant (alkene). toward Le Chatelier’s Principle At constant temperature and pressure a reaction proceeds in a direction which is spontaneous or decreases free energy (G). spontaneous The sign of G is always positive, but ∆ G can The can be positive or negative. be ∆ G = Gproducts – Greactants Spontaneous when ∆ G < 0 Spontaneous Le Chatelier’s Principle For a reversible reaction: For aA + bB ↔ cC + dD The relationship between ∆ G and ∆ G o is: The and is: ∆ G = DG o + RT l n [C] c[D ]d a [A ] [B ] b R = 8.314 J/(mol.K) and T is the temperature in K Le Chatelier’s Principle At equilibrium ∆ G = 0 and the following and becomes true: becomes Keq = [C]c[D]d [A]a[B]b Substituting Keq into the previous equation gives: ∆ Go = - RT lnKeq Reactions for ∆ Go positive are endergonic and for endergonic ∆ Go negative are exergonic. exergonic Synthesis Suppose you wanted to prepare 1-decanol Suppose from 1-decene? from OH OH •Needed: a method for hydration of alkenes with a regioselectivity opposite to Markovnikov's rule. Synthesis Two-step reaction sequence called hydroborationTwo-step oxidation converts alkenes to alcohols with a oxidation regiochemistry opposite to Markovnikov's rule. regiochemistry 1. hydroboration 2. oxidation OH OH Hydroboration Step C C + H—BH2 H—BH H C C BH2 Hydroboration can be viewed as the addition of borane (BH3) to the double bond. But BH3 is not the reagent actually used. not Hydroboration Step C C + H—BH2 H—BH H C C BH2 Hydroboration reagents: H H2B BH2 H Diborane (B2H6) normally used in an normally ether- like solvent ethercalled "diglyme" Hydroboration Step C C + H—BH2 H—BH H C C Hydroboration reagents: •• +O – BH3 Borane-tetrahydrofuran complex (H3B-THF) BH2 Oxidation Step H2O2, HO– H C C BH2 H C C Organoborane formed in the hydroboration step is oxidized with hydrogen peroxide. OH Example 1. B2H6, diglyme 1. 2. H2O2, HO– OH OH (93%) Example H3C CH3 C H3C C H H 1. H3B-THF 2. H2O2, HO– CH3 OH C C CH3 H (98%) (98%) CH3 Features of HydroborationOxidation • hydration of alkenes • regioselectivity opposite to Markovnikov's rule • no rearrangement • stereospecific syn addition syn-Addition •H and OH become attached to same face of double bond H CH3 CH 1. B2H6 2. H2O2, NaOH H CH3 HO H only product is trans-2-methylcyclopentanol only trans (86%) yield 1-Methylcyclopentene + BH3 •syn addition of H and B to double bond •B adds to less substituted carbon Organoborane Intermediate Add Hydrogen Peroxide • OH replaces B on same side trans-2-Methylcyclopentanol General Features C C + X2 X C •electrophilic addition to double bond •forms a vicinal dihalide C X Example CH3CH CHCH(CH3)2 Br2 CHCl3 0°C CH3CHCHCH(CH3)2 CH Br Br (100%) Scope limited to Cl2 and Br2 limited F2 addition proceeds with explosive violence I2 addition is endothermic: vicinal diiodides dissociate to an alkene and I2 Example H H H Br2 Br Br Br Br H trans-1,2-Dibromocyclopentane 80% yield; only product Example H H Cl Cll2 C H H Cl trans-1,2-Dichlorocyclooctane 73% yield; only product Mechanism is Electrophilic Addition Br2 is not polar, but it is polarizable 2 steps (1) formation of bromonium ion (2) nucleophilic attack on bromonium ion by bromide Relative Rates of Bromination ethylene H2C=CH2 propene CH3CH=CH2 61 2-methylpropene (CH3)2C=CH2 5400 2,3-dimethyl-2-butene (CH3)2C=C(CH3)2 - More highly substituted double bonds react faster. - Alkyl groups on the double bond make it more “electron rich.” 1 920,000 Mechanism? H2C + CH2 Br2 BrCH2CH2Br ? C + C .. – + : Br : .. : Br : .. No obvious explanation for anti addition provided by this mechanism. Mechanism H2C CH2 + Br2 C C BrCH2CH2Br .. – + : Br : .. : Br : + • Cyclic bromonium ion Mechanism of Br2 Addition • Br+ adds to an alkene producing a cyclic ion • Bromonium ion, bromine shares charge with carbon • Gives trans addition Bromonium Ion Mechanism • Electrophilic addition of bromine to give a cation is followed by cyclization to give a bromonium ion • This bromoniun ion is a reactive electrophile and bromide ion is a good nucleophile Bromonium Ions • Bromonium were postulated more than 60 years ago to explain the stereochemical course of the addition (to give the trans-dibromide from a cyclic alkene • George Olah (Nobel Prize, 1994, Chemistry) showed that bromonium ions are stable in liquid SO2 with SbF5 and can be studied directly .. – : Br : .. Stereochemistry .. Br + .. .. : Br : .. attack of Br– from side opposite attack C—Br bond of bromonium ion gives anti addition .. Br : .. Addition Br2 to Cyclopentene • Addition is exclusively trans • Why don't we see any of the cis product? • Can you draw a mechanism for this transformation? Cyclopentene +Br2 Bromonium ion – – – Bromide ion attacks the bromonium ion from side opposite carbon-bromine bond trans-Stereochemistry in vicinal dibromide Epoxides •are examples of heterocyclic compounds •three-membered rings that contain oxygen • ethylene oxide CH2 H2C O propylene oxide CHCH3 H2C O Epoxide Nomenclature •Substitutive nomenclature: named as epoxy-substituted alkanes. •“epoxy” precedes name of alkane •1,2-epoxypropane epoxybutane 2-methyl-2,31 H3C CHCH3 H2C O 2 3 CHCH3 C H3C 4 O Problem: Give the IUPAC name, including stereochemistry, for disparlure H O H •cis-2-Methyl-7,8-epoxyoctadecane Epoxidation of Alkenes O C C + RCOOH peroxy acid peroxy O C C O + RCOH Example O + CH3COOH CH O O + CH3COH (52%) (52%) Stereochemistry of Epoxidation O C C + RCOOH syn addition O C C O + RCOH Problem: Give the structure of the alkene, including stereochemistry, that you would choose as the starting material in a preparation of synthetic disparlure. H H peroxy acid peroxy H O H Relative Rates of Epoxidation ethylene H2C=CH2 propene CH3CH=CH2 22 2-methylpropene (CH3)2C=CH2 484 2-methyl-2-butene (CH3)2C=CHCH3 More highly substituted double bonds react faster. Alkyl groups on the double bond make it more “electron rich.” 1 6526 Mechanism of Epoxidation Addition Reactions of Alkenes •The characteristic reaction of alkenes is addition to the double bond. C C + A—B A—B A C CB Hydrogenation of Ethylene H C H π H + C σ H—H H—H H H σ H •exothermic ∆ H° = –136 kJ/mol •catalyzed by finely divided Pt, Pd, Rh, Ni H C C H H σ H Example CH2 H3C H3C H2, Pt CH3 H3C H3C H (73%) Problem • What three alkenes yield 2-methylbutane on catalytic hydrogenation? H2, Pt Mechanism of Catalytic Hydrogenation B HH A HH Y C C X Mechanism of Catalytic Hydrogenation B A H Y C C X H H H Mechanism of Catalytic Hydrogenation H A H B Y C X C H H Mechanism of Catalytic Hydrogenation H A H B Y C X C H H Mechanism of Catalytic Hydrogenation H A B Y C H X C H H Mechanism of Catalytic Hydrogenation A B Y C H H X C H H Heats of Hydrogenation • can be used to measure relative stability of isomeric alkenes • correlation with structure is same as when heats of combustion are measured Heats of Hydrogenation of Isomers 126 126 119 119 CH3CH2CH2CH3 CH 115 115 Heats of Hydrogenation (kJ/mol) •Ethylene 136 •Monosubstituted 125-126 •cis-Disubstituted 117-119 •trans-Disubstituted 114-115 •Terminally disubstituted 116-117 •Trisubstituted 112 •Tetrasubstituted 110 Problem Match each alkene of Problem 6.1 with its correct Match heat of hydrogenation. 126 kJ/mol highest heat of hydrogenation; least stable isomer 118 kJ/mol 112 kJ/mol lowest heat of hydrogenation; most stable isomer Stereochemical Aspects of Alkene Hydrogenation: (1) syn addition of both H atoms to double bond (2) hydrogenation is stereoselective, corresponding to addition to less crowded face of double bond syn-Additon Vs. anti-Addition syn addition syn anti addition Example of Syn Addition CO2CH3 H H2, Pt CO2CH3 CO2CH3 CO CO2CH3 H (100%) Stereochemical Aspects of Alkene Hydrogenation (2) hydrogenation is stereoselective, corresponding to addition to less crowded face of double bond A reaction in which a single starting material can give two or more stereoisomeric products but yields one of them in greater amounts than the other (or even to the exclusion of the other) is said to be stereoselective. H3C CH3 Example H H3C H3C H H H3C H H2, cat CH3 H3C H Both products H correspond to H syn addition of H2. H3C CH3 CH H3C CH3 Example H H3C H3C H H2, cat CH3 H H3C H Only this product is formed Addition of H2 prefers less hindered approach H3C CH3 H H3C H2, cat Top face of double Top face of double bond blocked by bond blocked by this methyl group H3C this methyl group CH3 CH H H H3C H 3 Example H3C CH3 H H3C H3C H H H3C H H2, cat CH3 H22 adds to H adds to bottom face of bottom face of double bond. double bond. Example ...
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This note was uploaded on 02/15/2012 for the course ORGO 101 taught by Professor Staff during the Fall '11 term at Syracuse.

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