1. Organic Chemistry, 8th Edition L. G. Wade, Jr.
Chapter 8 Lecture
Reactions of Alkenes
Rizalia Klausmeyer
Baylor University
Waco, TX
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2. Review of Substitution/Elimination
SN1 Reactions
Works best when carbocation intermediate is stabilized by substitution
Works best with strong leaving group (leaving group is stable in solvent)
Frequently occur under neutral or acidic conditions
Polar, protic solvents that stabilize nucleophile are best
Works best with non-basic nucleophile
SN2 Reactions
Work faster in polar aprotic solvents (acetonitrile, acetone, DMF, DMSO)
Work best with strong nucleophile that can reach carbon (e.g., primary halide)
Negatively-charged nucleophiles generally stronger than neutral nucleophiles
Work best with good leaving group I < Br < Cl < F

3. Types of Alkene Reactions
C=C
The double-bond in alkenes make them chemically reactive (pi bond electrons held more loosely)
Four major types of alkene reactions
Addition
Elimination
Substitution
Rearrangement

4. A + B  C Addition Reactions
Two reactants add together to form a single new product that includes all original atoms.
Double bond becomes 2 single bonds.

5. Bonding in Alkenes
Electrons in pi bond are loosely held.  More reactive
The double bond acts as a nucleophile attacking electrophilic species.
Carbocations are intermediates in some of these reactions.
These reactions are called electrophilic additions.
sp2
sp3
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6. Electrophilic Addition
Step 1: Pi electrons attack the electrophile, forming new bond between E+ and one of the carbons from the alkene.
Step 2: A nucleophile attacks the carbocation.
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7. Types of Additions
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8. Mechanism of Addition of HX (HBr, HCl, HI)
Step 1: Protonation of the double bond (Nu- attacks H+).
Stability of carbocations: 3° > 2° > 1° > +CH3
An electrophile adds to a double bond to give most stable intermediate carbocation.
Step 2: Nucleophilic attack of the halide on the carbocation.
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9. Selective Addition: Markovnikov’s Rule
Markovnikov’s rule: Proton added to double bond of an alkene goes to the carbon atom that already holds the greater number of hydrogens.
Markovnikov’s rule (extended): In an electrophilic addition to the alkene, the electrophile adds in such a way that it generates the most stable intermediate.
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10. Markovnikov’s Rule: Examples
Less stable
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11. Example Problems Copyright © 2006 Pearson Prentice Hall, Inc.
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12. Free-Radical Addition of HBr
In the presence of peroxides, HBr adds to an alkene to form the “anti-Markovnikov” product.
Peroxides produce free radicals.
Only HBr has just the right reactivity for each step of the free-radical chain reaction to take place.
The peroxide effect is not seen with HCl or HI because the reaction of an alkyl radical with HCl or HI is strongly endothermic.
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13. Free-Radical Addition of HBr
The peroxide bond breaks homolytically to form the first radical
The alkoxy radical attacks proton, forming alcohol and bromide radical
Bromine adds to the double bond, forming the most stable radical possible
Carbon radical attack proton, regenerates bromide radical
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14. Anti-Markovnikov Stereochemistry
The intermediate tertiary radical forms faster because it is more stable.
Stability of radicals: 3° > 2° > 1° > •CH3
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15. Chapter 8, Key Mechanism 3a

16. Example Problems Copyright © 2006 Pearson Prentice Hall, Inc.
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17. Hydration of Alkenes
Addition of water to the double bond forms an alcohol.
The addition follows Markovnikov’s rule.
This is the reverse of the dehydration of alcohol.
Uses dilute solutions of H2SO4 or H3PO4 to drive equilibrium toward hydration.
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18. Mechanism for Hydration
Step 1: Protonation of the double bond forms a carbocation.
Step 2: Nucleophilic attack of water.
Step 3: Deprotonation of the alcohol.
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19. Orientation of Hydration
The protonation follows Markovnikov’s rule: The proton adds to the less substituted end of the double bond, so the positive charge appears at the more substituted end (most stable carbocation).
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20. Rearrangements Possible in Electrophilic Addition Reactions
Methyl Shift:
A methyl shift after protonation will produce the more stable tertiary carbocation.
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21. Explain the Product of This Reaction

22. Oxymercuration–Demercuration Reaction
Reagent is mercury(II) acetate, which dissociates slightly to form +Hg(OAc).
+Hg(OAc) is the electrophile that adds to the pi bond.
The intermediate is a three-membered ring called the mercurinium ion.
Markovnikov addition of water to the double bond.
Milder conditions than direct hydration.
No rearrangements or polymerization.
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23. Mechanism of Oxymercuration
Anti addition of water on more substituted carbon
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24. Demercuration Reaction
In the demercuration reaction, a hydride furnished by the sodium borohydride (NaBH4) replaces the mercuric acetate.
The overall reaction gives the Markovnikov product with the hydroxy group on the most substituted carbon.
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25. Oxymercuration–Demercuration of 3,3,-Dimethylbut-1-ene
The reaction does not suffer from rearrangements because there is no carbocation intermediate.
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26. Alkoxymercuration–Demercuration
If the nucleophile is an alcohol, ROH, instead of water an ether (R-O-R) is produced.
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27. Solved Problem Solution
Show the intermediates and products that result from alkoxymercuration–demercuration of
1-methylcyclopentene, using methanol as the solvent.
Solution
Mercuric acetate adds to 1-methylcyclopentene to give the cyclic mercurinium ion. This ion has a considerable amount of positive charge on the more substituted tertiary carbon atom. Methanol attacks this carbon. Note movement of methyl group when methoxy group added!
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Reduction of the intermediate gives the Markovnikov product, 1-methoxy-1-methylcyclopentane.
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28. Mercuration–Demercuration Reactions

29. Hydroboration of Alkenes
Herbert Brown, of Purdue University, discovered that diborane (B2H6) adds to alkenes with anti-Markovnikov orientation to form alkylboranes, which after oxidation give anti-Markovnikov alcohols.
Brown received the Nobel Prize in Chemistry in 1979 for his work in the field of borane chemistry.
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30. Diborane and the borane/THF complex
Diborane (B2H6) is a dimer of borane and is in equilibrium with a small amount of BH3.
The complex of borane with tetrahydrofuran (BH3•THF) is the most commonly used form of borane.
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31. Mechanism of Hydroboration
Borane adds to double bond in a single step, with boron adding to the less substituted carbon (anti-Markovnikov).
Places the partial positive charge in the transition state on the more highly substituted carbon atom.
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32. Stoichiometry of Hydroboration
1 mole of BH3 can react with 3 moles of alkene.
Boron adds to less substituted carbon.
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33. Oxidation to Alcohol
Oxidation of alkyl borane with basic hydrogen peroxide produces the anti-Markovnikov alcohol (syn addition).
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34. Oxidation of a Trialkylborane
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35. Solved Problem Solution
Show how you would convert 1-methylcyclopentanol to 2-methylcyclopentanol.
Solution
Working backward, use hydroboration–oxidation to form 2-methyl cyclopentanol from 1-methylcyclopentene.
The 2-methylcyclopentanol that results from this synthesis is the pure trans isomer. 1-Methylcyclopentene is the most substituted alkene that results from dehydration of 1-methylcyclopentanol. Dehydration of the alcohol would give the correct alkene.
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36. Addition of Halogens (halogenation)
Cl2, Br2, and sometimes I2 add to a double bond to form a vicinal dihalide.
This is an anti addition of halides.
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37. Mechanism of Halogenation
The intermediate is a three-membered ring called the halonium ion.
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38. Stereochemistry of Halogen Addition
Anti stereochemistry results from the back-side attack of the nucleophile on the bromonium ion.
This back-side attack assures anti stereochemistry of addition.
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39. Examples of Stereospecificity
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40. In the Lab: Bromine Test for Unsaturation
Add Br2 in CCl4 (dark, red-brown color) to an alkene.
The brown color of Br2 remains apparent in the absence of a double bond (right test tube).
The color quickly disappears in the presence of a double bond (left-side test tube).
Double bond No
Double bond
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41. Formation of Halohydrins
If a halogen is added in the presence of water as solvent, a halohydrin is formed.
Water is the nucleophile.
This is a Markovnikov addition: The bromide (electrophile) will add to the less substituted carbon.
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42. Mechanism of Halohydrin Formation
X goes to less substituted carbon.
OH goes to more highly-substituted carbon.
Anti stereochemistry.
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43. Orientation of Halohydrin Formation
Attack by water (weak nucleophile) occurs on the more substituted carbon (bearing more positive charge) to give the Markovnikov product.
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44. Solved Problem Solution
Propose a mechanism for the reaction of 1-methylcyclopentene with bromine water.
Solution
1-Methylcyclopentene reacts with bromine to give a bromonium ion. Attack by water could occur at either the secondary carbon or the tertiary carbon of the bromonium ion. Attack actually occurs at the more substituted carbon, which bears more of the positive charge. The product is formed as a racemic mixture.
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45. Solved Problem Solution
When cyclohexene is treated with bromine in saturated aqueous sodium chloride, a mixture of trans-2-bromocyclohexanol and trans-1-bromo-2-chlorocyclohexane results. Propose a mechanism to account for these two products.
Solution
Cyclohexene reacts with bromine to give a bromonium ion, which will react with any available nucleophile. The most abundant nucleophiles in saturated aqueous sodium chloride solution are water and chloride ions. Attack by water gives the bromohydrin, and attack by chloride gives the dihalide. Either of these attacks gives anti stereochemistry.
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46. Catalytic Hydrogenation of Alkenes
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47. Effects of Steric Hindrance on Catalytic Hydrogenation
Both sides of alkene can interact with metal surface H H
Steric hindrance allows only 1 side of alkene to interact with metal surface

48. Chiral Hydrogenation Catalysts (Advanced Organic)
Rhodium and ruthenium phosphines are effective homogeneous catalysts for hydrogenation.
Chiral ligands can be attached to accomplish asymmetric induction, the creation of a new asymmetric carbon as mostly one enantiomer.
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49. The Need for Chiral Catalysts (Advanced Organic)
Only the (-)-enantiomer of dopa can cross the blood-brain barrier and be transformed into dopamine; the other enantiomer is toxic to the patient.
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50. Addition of Carbenes
The sp2 hybridized carbene is uncharged, reactive intermediate.
The insertion of the —CH2 group into a double bond produces a cyclopropane ring.
Three methods (we will look at last 2):
Diazomethane (CH3N2, UV light or heat).  Poor method.
Simmons–Smith (CH2I2 and Zn(Cu)).
α elimination of a haloform (CHX3, NaOH, H2O).
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51. Simmons–Smith Reaction
Best method for preparing cyclopropanes.
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52. α Elimination Reaction
In the presence of a base, chloroform or bromoform can be dehydrohalogenated (i.e., loss of H and X) by α elimination (i.e., from the same carbon) to form a carbene.
The dibromocarbene adds to alkene to form cyclopropane ring
Start numbering at bridgehead, number largest ring to smallest 2 1 3 7 4 6 5
7,7-dibromobicyclo[4.1.0]heptane
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53. Stereospecificity of α elimination reaction
The cylopropanes will retain the cis or trans stereochemistry of the alkene.
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54. Epoxidation
Alkene reacts with a peroxyacid to form an epoxide (also called oxirane).
An epoxide is an ether that is more reactive due to ring strain.
One step rxn forming epoxide, H transferred to carbonyl oxygen.
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55. Stereochemistry of Epoxidation
Most common peroxyacid used is meta-chloroperoxybenzoic acid (MCPBA).
Stereochemistry preserved due to syn addition.
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56. Acid-Catalyzed Opening of the Epoxide Ring
H30+ donates proton to epoxide oxygen.
Back-side attack of water on protonated epoxide.
Anti-diol is formed.
Q: Would acidity of solvent increase, decrease or remain the same?
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57. Base-Catalyzed Opening of the Epoxide Ring
Strain of the three-membered ring is relieved on ring opening
Hydroxide cleaves epoxides at elevated temperatures to give trans 1,2-diols
This is an SN2 rxn mechanism

58. Syn Hydroxylation of Alkenes
Alkene is converted to a syn-1,2-diol
Two reagents:
Osmium tetroxide, OsO4, followed by hydrogen peroxide or
Cold, dilute solution of KMnO4 in base.
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59. Mechanism with OsO4
The OsO4 adds to the double bond of an alkene in a concerted mechanism forming an osmate ester.
The osmate ester can be hydrolyzed to produce a cis-glycol and regenerate the OsO4.
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60. Permanganate Dihydroxylation
A cold, dilute solution of KMnO4 also hydroxylates alkenes with syn stereochemistry.
The basic solution hydrolyzes the manganate ester, liberating the glycol and producing a brown precipitate of manganese dioxide, MnO2.
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61. Oxidative Cleavage with KMnO4
Following initial dihydroxylation, oxidative cleavage of glycol can occur if solution is warm or acidic or too concentrated.
Disubstituted carbons become ketones.
Monosubstituted carbons become aldehydes, which are further oxidized to produce carboxylic acids.
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62. Examples: Oxidative Cleavage with KMnO4
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63. Ozonolysis
Ozone will oxidatively cleave (break) double bond  aldehydes and ketones.
Ozonolysis milder than KMnO4, won’t oxidize aldehydes.
In second step, intermediate reduced by zinc or dimethyl sulfide.
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64. Mechanism of Ozonolysis
Ozone adds to double bond, forms 5-membered ring intermediate (molozonide), which rearranges to form the ozonide.
Ozonide is immediately reduced by zinc or dimethyl sulfide, to give aldehydes and ketones.
When dimethyl sulfide is used, sulfur atom is oxidized, forming dimethyl sulfoxide (DMSO).
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65. Comparison of Permanganate Cleavage and Ozonolysis
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66. Solved Problem Solution
Ozonolysis–reduction of an unknown alkene gives an equimolar mixture of cyclohexanecarbaldehyde and butan-2-one. Determine the structure of the original alkene.
Solution
We can reconstruct the alkene by removing the two oxygen atoms of the carbonyl groups (C=O) and connecting the remaining carbon atoms with a double bond. One uncertainty remains, however: The original alkene might be either of two possible geometric isomers.
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67. Polymerization
An alkene (monomer) can add to another molecule like itself to form a chain (polymer).
Three methods:
Cationic, a carbocation intermediate.
Free radical.
Anionic, a carbanion intermediate (rare).
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68. Cationic Polymerization
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69. Termination Step of Cationic Polymerization
The chain growth ends when a proton is abstracted by the weak base of the acid used to initiate the reaction.
The loss of a hydrogen forms an alkene and ends the chain growth, so this is a termination step.
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70. Cationic Polymerization Using BF3 as Catalyst
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71. Radical Polymerization
In the presence of an initiator such as peroxide, free-radical polymerization occurs.
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72. Anionic Polymerization
For an alkene to gain electrons, strong electron-withdrawing groups such as nitro, cyano, or carbonyl must be attached to the carbons in the double bond.
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