Chapter 9 Addition Reactions and Alkenes Organic Chemistry Second Edition David Klein Copyright ©...

Chapter 9Addition Reactions and Alkenes

Organic ChemistrySecond Edition

David Klein

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

9.1 Addition Reactions• Addition is

the opposite of elimination

• A pi bond is converted to a sigma bond

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

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

or as a Brønsted-Lowry base)

• Why are pi bonds more reactive in this sense than sigma bonds?


9.2 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


9.2 Addition / Elimination Equilibria

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



• Typical addition reactions have a –ΔH. WHY?


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


• Typical addition reactions have a –ΔH• Will heat be absorbed by or released into the

surroundings?• What will the sign (+/-) be for ΔSsurr?• Will the enthalpy term favor the reactants or products?• The heat change (ΔH) will remain roughly constant

regardless of temperature



• 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)

• What will the sign (+/-) be for ΔSsys? WHY?

• What will the sign (+/-) be for -TΔSsys? • Will the enthalpy term favor the reactants or products?


9.2 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?


9.3 Hydrohalogenation• Note the temperature used in this addition reaction

• Does it matter whether the Br adds to the right side of the C=C double bond or whether it adds to the left?


9.3 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


9.3 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


9.3 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• Practice with conceptual checkpoint 9.1


9.3 Hydrohalogenation Mechanism

• The mechanism is a two step process• Which step do you think is rate determining?• Write a rate law for the reaction


9.3 Hydrohalogenation Mechanism

• Explain the FREE energy changes in each step


9.3 Hydrohalogenation Mechanism• Recall that there are two possible products,

Markovnikov and anti-Markovnikiv

• Which process looks more favorable? WHY?


9.3 Hydrohalogenation Mechanism• Practice with SkillBuilder 9.1


9.3 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?


9.3 Stereochemical Aspects• Recall the geometry of the carbocation

• Practice with conceptual checkpoint 9.6


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

carbocation if the shift makes it more stable


9.3 Rearrangements

• A mixture of products limits synthetic utility• With an INTRAmolecular rearrangement, WHY isn’t the

rearrangement product an even greater percentage?• How might [Cl-] be used to alter the ratio of products?• Practice with SkillBuilder 9.2


9.3 Hydrohalogenation Example• Predict the major product(s) for the reaction below

HCl


9.4 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


• Given the data below, do you think the acid catalyzed hydration goes through a mechanism that involves a carbocation?

9.4 Hydration


9.4 Hydration Mechanism

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

• Mechanism continues on next slide


9.4 Hydration Mechanism

• Could a stronger base help promote the last step?• Practice with conceptual checkpoint 9.10


9.4 Hydration Thermodynamics• Similar to Hydrohalogenation, hydration reactions are

also at equilibrium

• Explain HOW and WHY temperature could be used to shift the equilibrium to the right or left


Addition

Elimination

9.4 Hydration Thermodynamics

• How could Le Châtelier’s principle be used to shift the equilibrium to the right or left?



9.4 Hydration Thermodynamics• Similar to Hydrohalogenation, the stereochemistry of

hydration reactions is controlled by the geometry of the carbocation

• Draw the complete mechanism for the reaction above to show WHY a racemic mixture is formed

• Practice with SkillBuilder 9.3


9.4 Hydrations• Ethanol 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


9.5 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


9.5 Oxymercuration-Demercuration• Oxymercuration begins with mercuric acetate

• How would you classify the mercuric cation? – As a nucleophile or an electrophile? – As a Lewis acid or Lewis base?

• How might an alkene react with the mercuric cation?


9.5 Oxymercuration-Demercuration• Similar to how we saw the alkene attack a proton

previously, it can also attack the mercuric cation

• Resonance stabilizes the mercurinium ion and the carbocation. Draw a reasonable resonance hybrid


9.5 Oxymercuration-Demercuration• 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


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

Oxidation is often used

• Note that the process occurs in two steps


9.6 Hydroboration-Oxidation• Hydroboration-Oxidation reactions achieve syn addition

• Anti addition is NOT observed

• To answer WHY, we must investigate the mechanism


9.6 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


9.6 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


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

is formed when borane is stabilized with THF (tetrahydrofuran)

• What general role do you think BH3•THF is likely to play in a reaction?


9.6 Hydroboration-Oxidation

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


Hydroboration


• What evidence is there for a concerted addition of the B-H bond across the C=C double bond?

• Use sterics and electronics to explain the regioselectivity of the reaction

• Practice with conceptual checkpoint 9.17Copyright © 2015 John Wiley & Sons, Inc. All rights reserved. 9-41 Klein, Organic Chemistry 2e



Oxidation


StartHere


Oxidation

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

results

• WHY? What is the geometry of the alkene as the borane attacks?

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


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

mixture results

• Why aren’t the other stereoisomers formed?



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

1) BH3 THF

2) H2O2 / NaOH

1) BH3 THF

2) H2O2 / NaOH


9.7 Catalytic Hydrogenation• The addition of H2 across a C=C double bond

• If a chirality center is formed, syn addition is observed

• Draw the stereoisomers that are produced


9.7 Catalytic Hydrogenation• Analyze the energy diagram below

• Why is a catalyst necessary?

• Does the catalyst affect the spontaneity of the process?

• Typical catalysts include Pt, Pd, and Ni


9.7 Catalytic Hydrogenation• The metal catalyst is believed to both adsorb the H

atoms and coordinate the alkene

• The H atoms add to the same side of the alkene pi system


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

attention to stereochemistry

• How many chirality centers are there in the alkene reactant above?

• How does the term, mesocompound, describe the product(s) of the reaction?



9.7 Catalytic Hydrogenation• If catalysis takes place on the surface of a solid

surrounded by solution, the catalyst is heterogeneous. WHY?

• Homogeneous catalysts also exist

• What advantage might a homogeneous catalyst have?


9.7 Asymmetric Hydrogenation• In 1968, Knowles modified Wilkinson’s

catalyst by using a chiral phosphine ligand

• A chiral catalyst can produce one desired enantiomer over another. HOW?

• Why would someone want to synthesize one enantiomer rather than a racemic mixture?


9.7 Asymmetric Hydrogenation• A chiral catalyst allows

one enantiomer to be formed more frequently in the reaction mixture

• Some chiral catalysts give better enantioselectivity than others. WHY?


9.7 Asymmetric Hydrogenation• BINAP is a chiral ligand that gives pronounced

enantioselectivity

• For any reaction, stereoselectivity can only be achieved if at least one reagent (reactant or catalyst) is chiral


9.7 Asymmetric Hydrogenation• Predict the major product(s) for the reactions below

H2

Pt

H2

Pt


9.8 Halogenation• Halogenation involves adding two halogen atoms across

a C=C double bond

• Halogenation is a key step in the production of PVC


9.8 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

• Given the stereospecificity, is it likely to be a concerted or a multi-step process?


9.8 Halogenation• Let’s look at the reactivity of Br2. Cl2 is similar• It is nonpolar, but it is polarizable. WHY?• What type of

attraction exists between the Nuc:1- and Br2?

• 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

9.8 Halogenation

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


9.8 Halogenation

• Mechanism continued on next slide


9.8 Halogenation

• Only anti addition is observed. WHY?• Prove to yourself that the products are enantiomers

rather than identical


9.8 Halogenation• Only anti addition is observed

• Can you design a synthesis for ?



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

Br2

CCl4

Cl2CCl4


9.8 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

• Why is the bromonium attacked by water rather than a Br1- ion? Is water a better nucleophile?


9.8 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


9.8 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


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

the side that goes through the lower energy transition state

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



Transition state

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

Br2

H2O

Cl2H2O


9.9 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


9.9 Anti Dihydroxylation• First, an epoxide is formed

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

• Is there anything unstable about an epoxide?• Is an epoxide likely to react as a nucleophile (Lewis base)

or as an electrophile (Lewis acid)?


9.9 Anti Dihydroxylation• Water is a

poor nucleophile, so the epoxide is activated with an acid


9.9 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

• Practice with SkillBuilder 9.7Copyright © 2015 John Wiley & Sons, Inc. All rights reserved. 9-72 Klein, Organic Chemistry 2e

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

the C=C double bond in ONE step


9.10 Syn Dihydroxylation• Because OsO4 is expensive and toxic, conditions have

been developed where the OsO4 is regenerated after reacting, so only catalytic amounts are needed


9.10 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

• Practice with conceptual checkpoint 9.33Copyright © 2015 John Wiley & Sons, Inc. All rights reserved. 9-75 Klein, Organic Chemistry 2e

9.11 Oxidative Cleavage with O3

• C=C double bonds are also reactive toward oxidative cleavage

• Ozonolysis is one such process

• Ozone exists as a resonance hybrid of two contributors


• Common reducing agents include dimethyl sulfide and Zn/H2O. Practice with SkillBuilder 9.8




• 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)


9.12 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



9.12 Predicting Addition Products• Predict the major product(s) for the reaction below


9.13 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


9.13 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


9.13 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


9.13 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



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

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


9.13 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


9.13 Multi-Step Syntheses• How can the alcohol be eliminated to give the less

stable Hoffmann product?

• H3O+ will give the Zaitsev product• OH- is too poor of a leaving group to use the bulky

base, t-BuOK • The OH must first be converted to a better leaving

group, and then t-BuOK can be used



• In the last step, –H and –OH must be added across the C=C double bond

• Is the desired addition Markovnikov or anti-Markovnikov?


9.13 Multi-Step Syntheses• Use reagents that give anti-Markovnikov products

• Is stereochemistry an issue in this specific reaction?



9.13 Multi-Step Syntheses• Solve the multistep syntheses below

• Again, two processes must be combined

• What reagents should be used?• Practice with SkillBuilder 9.12


Additional Practice Problems• If you want to favor addition rather than elimination,

do you generally want a high or low temperature, and why?


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

below. Be aware of possible rearrangements and stereochemical concerns.

H-Br

-30 degrees


Additional Practice Problems• How and why will the concentration of acid affect

whether an acid catalyzed hydration will favor products or reactants at equilibrium?


Additional Practice Problems• Give an example reaction for Markovnikov hydration

without the possibility of rearrangement.

• Give an example reaction for syn antiMarkovnikov hydration.


Additional Practice Problems• Should a halogenation reaction be overall first or

second order kinetics? Also, Explain why it gives anti addition rather than syn.


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

synthesis?

Br


Chapter 9 Addition Reactions and Alkenes Organic Chemistry Second Edition David Klein Copyright ©...

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