09 alkereac st

© 2013 Pearson Education, Inc.

Chapter 8Lecture

Reactions of Alkenes



Overview

1. Electrophilic additions HX, H2O(3 reactions) X2, Br2/H2O, H2, Carbenes,

2. Free Radical addition of HBr

Chapter 8 2


Alkenes undergo electrophilic additions.

• Electrons in pi bond are loosely held.

• The double bond acts as a nucleophile attacking electrophilic species.

Chapter 8 3


Types of Additions

Chapter 8 4


Addition of HX to Alkenes

• ___________ intermediate: rearrangements• HBr, HCl, and HI can be added through this

reaction.• ______________: From two possible products

one is formed preferentially. • _________________: the electrophile adds to

the double bond to give the more stable carbocation (H goes where there are more Hs; the rich gets richer)

Chapter 8 5


Addition of HX to double bonds

Chapter 8 6

© 2013 Pearson Education, Inc. Chapter 8 7


Write the product and curved arrow mechanism for each one of the following

Chapter 8 8

+ HBr

+ HBr

a)

b)


Acid catalyzed hydration of Alkenes

• Addition of water to the double bond forms an ______.• The addition follows ______________ regioselectivity.• This is the reverse of the dehydration of alcohol.• Uses dilute solutions of H2SO4 or H3PO4 to drive equilibrium

toward hydration. Concentrated acid for dehydration.

Chapter 8 9


Mechanism for HydrationStep 1: Protonation of the double bond forms a carbocation Markovnikov’s regioselective.

Step 2: Nucleophilic attack of water.

Step 3: Deprotonation of the alcohol.

Chapter 8 10


Rearrangements Are Possible

• A ____________ will produce the more stable tertiary carbocation.

Chapter 8 11

+ H2O H+


Free-Radical Addition of HBr• In the presence of peroxides, HBr adds to an

alkene to form the “________________” product.

• Peroxides produce _____________ 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.

Chapter 8 12


Free-Radical mechanismThe peroxide bond breaks homolytically to form the first radical

Hydrogen is abstracted from HBr.Bromine adds to the double bond, forming the most stable

radical

Chapter 8 13

Hydrogen is abstracted from HBr


Synthesize 1-bromo-2-methylcyclohexane starting with 1-methylcyclohexanol: This synthesis requires the conversion of an alcohol to an alkyl

bromide with the bromine atom at the neighboring carbon atom.

1-Methylcyclohexene is easily synthesized by the dehydration of 1-methylcyclohexanol. The most substituted alkene is the desired product.

Chapter 8 14


Oxymercuration–Reduction Reaction

• ________________ regioselectivity.• Milder conditions than acid catalyzed hydration.• The intermediate is a three-membered ring called the mercurinium ion; No

rearrangements or polymerization.• Mercury(II) acetate, dissociates slightly to form the electrophile +Hg(OAc)2.

Chapter 8 15


Mechanism of Oxymercuration

Chapter 8 16

Regioselective: Water adds to the more substituted carbon to form the Markovnikov product. Stereoselective: Water approaches the mercurinium ion from the side opposite the ring (anti addition).


Demercuration Reaction

• In the demercuration reaction, a hydride furnished by the sodium borohydride ___________replaces the mercuric acetate.

• The overall reaction gives the Markovnikov product with the hydroxy group on the ______________substituted carbon.

Chapter 8 17


Write the product and curved arrow mechanism of the Oxymercuration–

Demercuration of 3,3,-Dimethylbut-1-ene

Compare to product of acid catalyzed hydration

Chapter 8 18


Alkoxymercuration–Demercuration

Chapter 8 19

If the nucleophile is an alcohol, ROH, instead of water an ________is produced.

Show the intermediates and products that result from alkoxymercuration–demercuration of 1-methylcyclopentene, using methanol as the solvent.


Addition of Halogens

Cl2, Br2, and sometimes I2 add to a double bond to form a vicinal dihalide.The intermediate is a three-membered ring called ___________________.Stereoselectivity is an _____________addition of halides.

Chapter 8 20


Stereochemistry of Halogen Addition: Anti stereoselective

• Anti stereochemistry results from the back-side attack of the nucleophile on the bromonium ion.

• This back-side attack assures anti stereochemistry of addition.

Chapter 8 21


Halogenetaion of 2-butene is stereospecificcis -> enantiomers

trans -> meso

Chapter 8 22


Bromine Test for Unsaturation• Add Br2 in CCl4 (dark, red-

brown color) to an alkene.• The color quickly disappears

as the bromine adds to the double bond (left-side test tube).

• If there is no double bond present, the brown color will remain (right side).

• “Decolorizing bromine” is the chemical test for the presence of a double bond.

Chapter 8 23


Formation of Halohydrins

• If a halogen is added in the presence of _______ as solvent, a ____________is formed. __________ is the nucleophile that opens the halonium ion.

• This is a ______________ addition: The bromide (electrophile) will add to the __________ substituted carbon.

• Because the mechanism involves a halonium ion, the stereochemistry of addition is ________, as in halogenation.

• Attack by water occurs on the more substituted carbon with ________________ regioselectivity.

Chapter 8 24


Mechanism of Halohydrin Formation (compare to halogenation)

Chapter 8 25

© 2013 Pearson Education, Inc. Chapter 8 26

The opening of a halonium ion isdriven by its electrophilic nature. The weak nucleophile

attacks the carbon bearing more positive charge

Br2

H2O


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.

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.

Chapter 8 27


Hydroboration of Alkenes

H. C. 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.

Chapter 8 28


The complex of borane with tetrahydrofuran (BH3•THF) is the most commonly used form of borane.

Chapter 8 29

Borane adds to the double bond in a_______________, with boron adding to the __________ substituted carbon and hydrogen adding to the __________substituted carbon. This orientation places the partial positive charge in the transition state on the more highly substituted carbon atom.


Stoichiometry of Hydroboration: Three moles of alkene can react with each mole of BH3.

Chapter 8 30

Oxidation of the alkyl borane with basic hydrogen peroxide produces the anti-Markovnikov alcohol.


Working backward, use hydroboration–oxidation to form 2-methyl cyclopentanol from 1-methylcyclopentene. The use of (1) and (2) above and below the reaction arrow indicates individual steps in a two-step sequence.

1-Methylcyclopentene is the most substituted alkene that results from dehydration of 1-methylcyclopentanol. Dehydration of the alcohol would give the correct alkene that upon hydroboration oxidation produces the trans 2-methylcyclopentanol

Synthesize 2-methylcyclopentanol from 1-methylcyclopentanol

Chapter 8 31


The syn addition of BH3 across the double bond of norbornene takes place mostly from the more accessible outside (exo) face of the double bond. Oxidation gives a product with both the hydrogen atom and the hydroxyl group in exo positions. (The less accessible inner face of the double bond is called the endo face.)

A norbornene molecule labeled with deuterium is subjected to hydroboration–oxidation. Give the structures of the intermediates and products.

Chapter 8 32

BH3•THF H2O2, OH-

exo

endo face

racemic


Catalytic Hydrogenation of Alkenes

• Hydrogen (H2) can be added across the double bond in a process known as ________________________.

• The reaction only takes place if a catalyst is used.

Chapter 8 33


Mechanism of catalytic hydrogenation

34

Interaction of H2 with the _______________ Rupture of the HH bond and formation of two _________bondsA pi bond interacts with this activated HAdsorption probably by metal H sigma bonds and metal C bondsTransfer of one H to the alkeneTransfer of another HDesorption and regeneration of catalyst


The reaction has a syn stereochemistry since both hydrogens will add to the same

side of the double bond.

•

Chapter 8 35


Chiral Hydrogenation Catalysts

• 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.

Chapter 8 36


The Need for Chiral Catalysts

• Only the (-)-enantiomer of dopa can cross the blood-brain barrier and be transformed into dopamine; the other enantiomer is toxic to the patient.

Chapter 8 37


Addition of Carbenes

• The insertion of the —CH2 group into a double bond produces a __________________ ring.

• Three methods: Diazomethane (CH3N2, UV light or heat). Simmons–Smith (CH2I2 and Zn(Cu)). Alpha elimination of a haloform (CHX3, NaOH, H2O).

Chapter 8 38


Carbenes: Diazomethane Method

N N CH2 N N CH2

diazomethane

N N CH2

heat or UV lightN2 +

carbene

C

H

H

Problems with diazomethane: 1. Extremely toxic and explosive. 2. The carbene can insert into C—H bonds, too.

Chapter 8 39


Simmons–Smith Reaction

Best method for preparing cyclopropanes.

Preparation of the Simmons-Smith reagent:

CH2I2 + Zn(Cu) ICH2ZnISimmons–Smith

reagent

Chapter 8 40


Alpha Elimination Reaction

• In the presence of a base, chloroform or bromoform can be dehydrohalogenated to form a _____________.

Chapter 8 41


Stereospecificity

• The cylopropanes will _____________ the cis or trans stereochemistry of the alkene.

Chapter 8 42


Carbene Examples

CH2I2Zn, CuCl

CHBr3KOH/H2O

Br

Br

Simmons–Smith Reaction

Alpha Elimination Reaction

Chapter 8 43


Epoxidation

• Alkene reacts with a peroxyacid to form an epoxide (also called ______________).

• The most common peroxyacid used is meta-chloroperoxybenzoic acid (MCPBA).

Chapter 8 44


Mechanism

• The peroxyacid and the alkene react with each other in a one-step reaction to produce the _____________and a molecule of ____________.

Chapter 8 45


Stereochemistry of EpoxidationThis reaction is stereospecific

Chapter 8 46

O

O

OHCl

O

O

OHCl


Opening the Epoxide Ring

• This process is ____________catalyzed.• Water attacks the protonated epoxide on the

____________ side of the ring (anti-addition).• _____________product is formed.

Chapter 8 47


Syn Hydroxylation of Alkenes

Two reagents undergo syn addition to double bonds. Osmium tetroxide, OsO4, followed by

hydrogen peroxide and Cold, dilute solution of KMnO4 in base

Chapter 8 48


Mechanism with OsO4

• The OsO4 adds to the double bond of an alkene in a ______________mechanism forming an _________ ester.

• The osmate ester can be hydrolyzed to produce a _____-glycol and regenerate the OsO4.

Chapter 8 49


Permanganate Dihydroxylation

• A cold, dilute solution of KMnO4 also hydroxylates alkenes with ______________ stereochemistry.

• The basic solution hydrolyzes the manganate ester, liberating the glycol and producing a brown precipitate of manganese dioxide__________

Chapter 8 50


Oxidative Cleavage with KMnO4

• If the solution is warm or acidic or too concentrated, oxidative cleavage of the glycol may occur.

• Disubstituted carbons become ketones.• Monosubstituted carbons become carboxylic acids.

Chapter 8 51


Ozonolysis

• Ozone will oxidatively cleave the double bond to produce _____________________.

• Ozonolysis is milder than KMnO4 and will not oxidize aldehydes further.

• A second step of the ozonolysis is the reduction of the intermediate by zinc or dimethyl sulfide.

Chapter 8 52


Mechanism of Ozonolysis

• The ozone adds to the double bond, forming a five-membered ring intermediate called ________________, which rearranges to form the ________________.

Chapter 8 53


Reduction of the Ozonide

• The ozonide is not isolated, but is immediately reduced by a mild reducing agent, such as zinc or dimethyl sulfide, to give the aldehydes and ketones as the main products.

• When dimethyl sulfide is used, the sulfur atom gets oxidized, forming dimethyl sulfoxide (DMSO).

Chapter 8 54


Comparison of Permanganate Cleavage and Ozonolysis

Chapter 8 55


Osmium tetroxide, cold, dilute KMnO4, and epoxidation oxidize the pi bond of an alkene but leave the

sigma bond intact. Ozone and warm, concentrated KMnO4 break the double bond

entirely to give carbonyl compounds.

Chapter 8 56


Ozonolysis–reduction of an unknown alkene gives an equimolar mixture of cyclohexanecarbaldehyde and butan-2-one. Determine the structure of the original alkene.

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.

Chapter 8 57


Chapter 8 58

H2OBr2

DMSO


Chapter 8 59

+ Br2

CH2Cl2

+ H2OH2SO4

+ HBr


Polymerization

• An alkene (____________) can add to another molecule like itself to form a chain (______________).

• Three methods: Cationic, a carbocation intermediate. Free radical. Anionic, a carbanion intermediate (rare).

Chapter 8 60


Cationic Polymerization

Chapter 8 61


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.

Chapter 8 62


Cationic Polymerization Using BF3 as Catalyst

Chapter 8 63


Radical Polymerization

• In the presence of an initiator such as peroxide, free-radical polymerization occurs.

Chapter 8 64


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.

Chapter 8 65


Chapter 8 66

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