Alkene Reactions

Addition Reactions• Addition is

the opposite of elimination

• A pi bond is converted to a sigma bond

Addition Reactions• A pi bond will often act as a Lewis base (as a nucleophile

or as a Brønsted-Lowry base)

Addition / Elimination Equilibria• Because an addition is the reverse of an elimination,

often the processes are at equilibrium

• An equilibrium is a thermodynamic expression• We assess ΔG (the free energy) to determine which side

the equilibrium will favor

Addition / Elimination Equilibria

• To determine which side the equilibrium will favor, we must consider both enthalpy and entropy

Addition / Elimination Equilibria

Bonds broken – bonds formed = 166 kcal/mol – 185 kcal/mol = –19 kcal/mol

Addition / Elimination Equilibria

• Having a –ΔH (or a +ΔSsurr) favors the addition reaction rather than the elimination reaction

• To get ΔG (or ΔStot) and make a complete assessment, we must also consider the entropy of the system (ΔSsys)

Addition / Elimination Equilibria• Plugging into the formula gives…

• To favor addition, a –ΔG (or a +ΔStot) is needed• How can the temperature be adjusted to favor addition?• To favor elimination (the reverse reaction in this

example), a +ΔG (or a –ΔStot) is needed• How can the temperature be adjusted to favor

elimination?

Hydrohalogenation• Regiochemistry becomes important for asymmetrical

alkenes

• In 1869, Markovnikov showed that in general, H atoms tend to add to the carbon already bearing more H atoms

Hydrohalogenation• Markovnikov’s rule could also be stated by saying that in

general, halogen atoms tend to add to the carbon that is more substituted with other carbon groups

• This is a regioselective reaction, because one constitutional isomer is formed in greater quantity than another

• Draw the structure of the minor product

Hydrohalogenation• Anti-Markovnikov products are observed when reactions

are performed in the presence of peroxides such as H2O2

• Why would some reactions follow Markovnikov’s rule, while other reactions give Anti-Markovnikov products?

• The answer must be found in the mechanism

Hydrohalogenation Mechanism

• The mechanism is a two step process• Which step do you think is rate determining?

Hydrohalogenation Mechanism

Hydrohalogenation Mechanism• Recall that there are two possible products,

Markovnikov and anti-Markovnikiv

• Which process looks more favorable? WHY?

Hydrohalogenation Mechanism• Practice with SkillBuilder 9.1

Stereochemical Aspects• In many addition reactions, chirality centers are formed

• There are two possible Markovnikov products

• Which step in the mechanism determines the stereochemistry of the product?

Stereochemical Aspects• Recall the geometry of the carbocation

Rearrangements• Rearrangements (hydride or methyl shifts) occur for the

carbocation if the shift makes it more stable

Rearrangements

• A mixture of products limits synthetic utility

Hydration• The components of water (-H and –OH) are added

across a C=C double bond

• The acid catalyst is often shown over the arrow, because it is regenerated rather than being a reactant

Hydration Mechanism

• Why does the hydrogen add to this carbon of the alkene?

Hydration Mechanism

Hydration Equilibrium• Similar to Hydrohalogenation, hydration reactions are

also at equilibrium

Hydration Stereochemistry• Similar to Hydrohalogenation, the stereochemistry of

hydration reactions is controlled by the geometry of the carbocation

Hydrations

• Ethanol (think beer!) is mostly produced from fermentation of sugar using yeast, but industrial synthesis is also used to produce ethanol through a hydration reaction

• Predict the major product(s) for the reaction below

H3O+

H2O

Oxymercuration-Demercuration• Because rearrangements often produce a mixture of

products, the synthetic utility of Markovnikov hydration reactions is somewhat limited

• Oxymercuration-demercuration is an alternative process that can yeild Markovnikov products without the possibility of rearrangement

Oxymercuration• Oxymercuration begins with mercuric acetate

Acetate anion

• Alkene pi bond electrons can also attack the mercuric cation• Resonance stabilizes the mercurinium ion and the carbocation.

Oxymercuration• The mercurinium ion is also a good electrophile, and it

can easily be attacked by a nucleophile, even a weak nucleophile such as water

• NaBH4 is generally used to replace the –HgOAc group with a –H group via a free radical mechanism

• For laboratory-scale hydration of an alkene, use mercuric acetate in THF followed by sodium borohydride

• Produces Markovnikov orientation via mercurinium ion• Works with water and alcohols

Oxymercuration Intermediates

1. +Hg(OAc), H2O/THF

2. NaBH4

OH

1. +Hg(OAc), THF

2. NaBH4

OH

+ OO

Hydroboration-Oxidation• To achieve anti-Markovnikov hydration, Hydroboration-

Oxidation (2 steps) is often used

Hydroboration-Oxidation• Hydroboration-Oxidation reactions achieve syn addition

• Anti addition is NOT observed

• To answer WHY, we must investigate the mechanism

Hydroboration-Oxidation• Let’s examine how this new set of reagents might react• The BH3 molecule is similar to a carbocation but not as

reactive, because it does not carry a formal charge

Hydroboration-Oxidation• Because of their broken octet, BH3 molecules undergo

intermolecular resonance to help fulfill their octets

• The hybrid that results from the resonance (diborane) involves a new type of bonding called banana bonds

Hydroboration-Oxidation• In the hydroboration reaction, BH3•THF is used. BH3•THF

is formed when borane is stabilized with THF (tetrahydrofuran)

Copyright © 2015 John Wiley & Sons, Inc. All rights reserved. 9-34 Klein, Organic Chemistry 2e

9.6 Hydroboration-Oxidation

• Let’s examine the first step of the Hydroboration mechanism on the next slide

Hydroboration

• Addition of H-BH2 (from BH3-THF complex) to three

alkenes gives a trialkylborane• Oxidation with alkaline hydrogen peroxide in water

produces the alcohol derived from the alkene

Hydroboration-Oxidation Forms an Alcohol from an Alkene

• Regiochemistry is opposite to Markovnikov orientation– OH is added to carbon with most H’s

• H and OH add with syn stereochemistry, to the same face of the alkene (opposite of anti addition)

Orientation in Hydration via Hydroboration

• Borane is a Lewis acid• Alkene is Lewis base• Transition state involves anionic development on B• The components of BH3 are added across C=C• More stable carbocation is also consistent with steric preferences

Mechanism of Hydroboration

Hydroboration-Oxidation

Oxidation

Hydroboration-Oxidation

StartHere

Oxidation

Hydroboration-Oxidation• When ONE chirality center is formed, a racemic mixture

results

• The squiggle bond above shows two products, a 50/50 mixture of the R and the S enantiomer

Hydroboration-Oxidation• When TWO chirality centers are formed, a racemic

mixture results

Hydroboration-Oxidation• Predict the major product(s) for the reactions below

1) BH3-THF

2) H2O2,NaOH

OH

1) BH3-THF

2) H2O2,NaOH OH

1) BH3-THF

2) H2O2,NaOH

OH

1) BH3-THF

2) H2O2,NaOH OH

• Addition of H-H across C=C• Reduction in general is addition of H2 or its equivalent • Requires Pt or Pd as powders on carbon and H2

• Hydrogen is first adsorbed on catalyst• Reaction is heterogeneous (process is not in solution)

Reduction of Alkenes: Hydrogenation

• Selective for C=C. No reaction with C=O, C=N• Polyunsaturated liquid oils become solids• If one side is blocked, hydrogen adds to other

Hydrogen Addition - Selectivity

• Heterogeneous – reaction between phases

• Addition of H-H is syn

Mechanism of Catalytic Hydrogenation

Catalytic Hydrogenation• Draw product(s) for the reaction below. Pay close

attention to stereochemistry

Halogenation• Halogenation involves adding two halogen atoms across

a C=C double bond

• Halogenation is a key step in the production of PVC

Halogenation• Halogenation with Cl2 and Br2 is generally effective, but

halogenation with I2 is too slow and halogenation with F2 is too violent

• Halogenation occurs with anti addition

Halogenation• Let’s look at the reactivity of Br2. Cl2 is similar• It is nonpolar, but it is polarizable.• Does the Br2

molecule have a good leaving group attached to it?

• We know alkenes can act as nucleophiles• Imagine an alkene attacking Br2. You might imagine the

formation of a carbocation

Halogenation

• However, this mechanism DOES NOT match the stereospecificity of the reaction.

Halogenation

Halogenation• Only anti addition is observed

Halogenation• Predict the major product(s) for the reactions below

Br2

CCl4

Cl2CCl4

Halohydrin Formation• Halohydrins are formed when halogens (Cl2 or Br2) are

added to an alkene with WATER as the solvent• The bromonium ion forms from Br2 + alkene, and then it

is attacked by water

Halohydrin Formation• A proton transfer completes the mechanism producing a

neutral halohydrin product

• The net reaction is the addition of –X and –OH across a C=C double bond

Halohydrin Regioselectivity• The –OH group adds to the more substituted carbon

• The key step that determines regioselectivity is the attack of water on the bromonium ion at the more substituted carbon

Halohydrin Regioselectivity• When water attacks the bromonium ion, it will attack the side

that goes through the lower energy transition state (the carbon that is more stable due to presence of R groups).

• Water is a small molecule that can easily access the more sterically hindered site

Transition state

Halohydrin Regioselectivity• Predict the major product(s) for the reactions below

Br2

H2O

Anti Dihydroxylation• Dihydroxylation occurs when two –OH groups are added

across a C=C double bond

• Anti dihydroxylation is achieved through a multi-step process

Anti Dihydroxylation• First, an epoxide is formed

• Replacing the relatively unstable O-O single bond is the thermodynamic driving force for this process

Anti Dihydroxylation• Water is a

poor nucleophile, so the epoxide is activated with an acid

Anti Dihydroxylation• Note the similarities between three key intermediates

• Ring strain and a +1 formal charge makes these structures GREAT electrophiles

• They also each yield anti products, because the nucleophile must attack from the side opposite the leaving group

Syn Dihydroxylation• Like other syn additions, syn dihydroxylation adds across

the C=C double bond in ONE step

Syn Dihydroxylation• MnO4

1- is similar to OsO4 but more reactive

• Syn dihydroxylation can be achieved with KMnO4 but only under mild conditions (cold temperatures)

• Diols are often further oxidized by MnO41-, and MnO4

1- is reactive toward many other functional groups as well

• The synthetic utility of MnO41- is limited; therefore, we

aren’t going to worry about this reaction

• Common reducing agents include dimethyl sulfide and Zn/H2O.

Oxidative Cleavage with O3

Oxidative Cleavage with O3

• Predict the major product(s) for the reaction below

• Predict a bicyclic reactant used to form the product below

1) O3

S2)

O

H O

O

H

O

H

1) O3

S2)

Predicting Addition Products1. Analyze the reagents used to determine what groups

will be added across the C=C double bond2. Determine the regioselectivity (Markovnikov or anti-

Markovnikov)3. Determine the stereospecificity (syn or anti addition)• Each step can be achieved with minor reagent

memorization and a firm grasp of the mechanistic rational

• The more familiar you are with the mechanisms, the easier predicting products will be

One Step Syntheses• To set up a synthesis, assess the reactants and products

to see what changes need to be made• Label each of the processes below

One Step Syntheses• To set up a synthesis, assess the reactants and products

to see what changes need to be made• Give reagents and conditions for the following

+ En

BrOH

Multi-Step Syntheses• Multistep syntheses are more challenging, but the

same strategy applies

• This is not a simple substitution, addition or elimination, so two processes must be combined

Multi-Step Syntheses

• For the strategy to work, the regioselectivty must be correct

• A smaller base should be used to produce the more stable Zaitsev product

Multi-Step Syntheses

• For the strategy to work, the regioselectivty must be correct

• Will the regioselectivity for the HBr reaction give the desired product?

Multistep Rxns The synthesis of most alcohols may require multiple steps (i.e., to get

product X from reactant A, a product (B, C …X) must be created). To solve these problems, work backwards from the final product.

O

OOH (O)+ H-OH+

Oxidize 2° alcohol to form ketone.

+ OHH-OHH2SO4

Use acid-catalyzed hydration (addition) to form alcohol.

+ OHH-OHH2SO4

(O)+ O H-OH+

The completed series of rxns.

Additional Practice Problems• Predict the major product for the addition reaction

below. Be aware of possible rearrangements and stereochemical concerns.

H-Br

-30 degrees

Copyright © 2015 John Wiley & Sons, Inc. All rights reserved. 9-75 Klein, Organic Chemistry 2e

Additional Practice Problems• What reagents are necessary to achieve the following

synthesis?

Br

Copyright © 2015 John Wiley & Sons, Inc. All rights reserved. 9-76 Klein, Organic Chemistry 2e