9.1 Addition Reactions - WordPress.com · will be added across the C=C double bond. 2. Determine...

• Addition is the opposite of elimination.

• A pi bond is converted to a sigma bond.

9.1 Addition Reactions

Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -1

• 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?

9.1 Addition Reactions


• 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?



• 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?



• 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?



• 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.



• 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.



• Anti-Markovnikov products are observed when reactions are performed in the presence of peroxides such as H2O2.

• Why would some reactions yield Markovnikov products, while other reactions give anti-Markovnikov products? – The answer must be found in the mechanism.

• Practice with CONCEPTUAL CHECKPOINT 9.1.



• 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.



• Recall that there are two possible products, Markovnikov and anti-Markovnikov products.

• Which process looks more favorable? WHY?



• Practice with SKILLBUILDER 9.1.



• 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 Hydrohalogenation – Stereochemical Aspects


• Recall the geometry of the carbocation.


9.3 Hydrohalogenation – Stereochemical Aspects


• Rearrangements (hydride or methyl shifts) occur for a carbocation if a shift makes it more stable.

9.3 Hydrohalogenation – 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 – Rearrangements


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

9.3 Hydrohalogenation – Example


HCl

• 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.

9.4 Hydration


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

9.4 Hydration


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

9.4 Hydration – Mechanism


• Could a stronger base help promote the last step? • Practice with CONCEPTUAL CHECKPOINT 9.10.

9.4 Hydration – Mechanism


• 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.

9.4 Hydration – Thermodynamics


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




• 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.




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


9.4 Hydration – Examples


H3O+

H2O

• 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 yield Markovnikov products more cleanly.

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?



• Similar to how we saw the alkene attack a proton previously, it can also attack the mercuric cation.

• Resonance stabilizes the mercurinium ion. Can you draw a reasonable resonance hybrid?



• 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.



• 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.



• 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.



• 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.



• 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?



• Let’s examine the first step of the hydroboration.

9.6 Hydroboration-Oxidation – The Hydroboration Step


• 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.


9.6 Hydroboration-Oxidation – The Hydroboration Step


9.6 Hydroboration-Oxidation – The Oxidation Step


9.6 Hydroboration-Oxidation – The Oxidation Step


• 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.



• When TWO chirality centers are formed, a racemic mixture results.

– WHY aren’t the other stereoisomers formed?




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



1) BH3

THF

2) H2O2

/ NaOH

1) BH3

THF

2) H2O2

/ NaOH

• The addition of H2 across a C=C double bond:

• If a chirality center is formed, SYN addition is observed. – Draw the other stereoisomers

that are NOT produced.

9.7 Hydrogenation – Catalytic


• Analyze the energy diagram. – Why is a catalyst

necessary? – Does the catalyst affect

the spontaneity of the process?

• Typical catalysts include Pt, Pd, and Ni.



• 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.



• 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?




• 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?



• 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 Hydrogenation – Asymmetric


• A chiral catalyst allows one enantiomer to be formed more frequently in the reaction mixture.

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



• BINAP is a chiral ligand that gives very pronounced enantioselectivity.

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






H2

Pt

H2

Pt

• 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’s reactivity 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?

9.8 Halogenation


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

formation of a carbocation.

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

9.8 Halogenation


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



9.8 Halogenation


Br2

CCl4

Cl2CCl4

• 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 Halogenation – 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 Halogenation – Halohydrin Formation


• 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 Halogenation – 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. • Practice with SKILLBUILDER 9.6.






Br2

H2O

Cl2H2O

• 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)?



• Water is a poor nucleophile, so the epoxide is activated. with an acid



• 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 in an SN2 fashion.




• 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.



• MnO41- 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 MnO4

1- is limited. • Practice with CONCEPTUAL CHECKPOINT 9.33.



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

• Ozonolysis is one such process.

• Ozone exists as a resonance hybrid of two contributors.

9.11 Oxidative Cleavage with O3


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





• Predict the reactant used to form the product below.



1) O3

SS2)

O

H O

O

H

O

H

1) O3

SS2)

1. Analyze the reagents used to determine what groups will be added across the C=C double bond.

2. 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



9.12 Predicting Addition Products


• 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 Application – 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.


9.13 Application – One-step Syntheses


+ En

BrOH

• 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 Application – Multi-step Syntheses


• For the strategy to work, the regioselectivity must be correct.

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



• For the strategy to work, the regioselectivity must be correct.

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

9.13 Syntheses – 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.



• 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?



• Use reagents that give anti-Markovnikov products.

• Is stereochemistry an issue in this specific reaction?




• Solve the multistep syntheses below.

– This is not a simple substitution, addition or elimination, so

two processes must be combined.

– What reagents should be used?




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