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Gibbs free energy of alkene reactions

Enthalpy favors alkene to alkane reactions

At higher temperatures, entropy favors the reverse (alkene to alkane) reactions


Electrophilic addition to alkenes overall mechanism

H of acid attacks less substituted carbon

Carbocation forms

Initial leaving group attacks the carbocation


regioselectivity of electrophilic addition to alkenes

H goes to less substituted carbon

Oftentimes, a halogen goes to the more substituted carbon


stereoselectivity of electrophilic addition to alkenes

there isn't any

the halogen can attack the carbocation from either top or bottom, resulting in 2 enantiomers being formed


What does hydration of an alkene require?

Acid catalyst


Why does hydration require an acid catalyst?

Protonate H2O, so that OH- is not the initial leaving group


Is hydration reversible?

Yes through E1 reactions


E1 reaction

leaving group leaves and carbocation is formed

then nucleophile deprotonates H from other carbon and alkene is formed


Hydration overall mechanism

Acid protonates H2O

H3O+ deprotonates alkene and forms carbocation

H2O attacks carbocation

H2O deprotonates alkane to make neutral

Neutral alkane and H3O+ are formed


Regioselectivity of hydration

The -OH group will be added to the more substituted carbon due to more substituted carbon having the cation that H2O attacks


Markovinkov alcohol

-OH group is added to more substituted carbon


When are markovinkov alcohols seen?


Halogenation with a nucleophilic solvent


Stereoselectivity of hydration

there isn't any

H2O can attack the carbocation from either top or bottom


How does an acid catalyst lower the LUMO of H2O?

Makes the LUMO sigma*O+ H

The + charge on oxygen means that there is a greater Zeff and the orbital's energy will be lowered, making a lower LUMO


Solvent of halogenation

Nucleophilic or non-nucleophilic solvent matters


Regioselectivity of halogenation?

None seen with a non-nucleophilic solvent

Nucleophilic solvent will only attack the more substituted carbon since halonium cation intermediate creates a slight positive charge on the more substituted carbon


Halonium cation

intermediate in halogenation where there is a positively charged halogen intermediate


Stereoselectivity of halogenation?

Nucleophile will only attack anti- in second step since this is where the sigma* lobe is exposed

Similar to a SN2 reaction


What does anti- attack in halogenation produce?

Only the trans- configuation

Syn would result in the cis- configuration (Meso) which we don't see


LUMO of halogenation

Sigma* Br-Br or Sigma* Cl-Cl

Great LUMOs because of:
1) low starting position on MO diagram
2) poor overlap of bulky orbitals


Peroxycarboxlyic acid LUMO

O-O is a low lumo due to initial low starting energy


Is expoxidation concerted? What indicates this?

Yes expoxidation is concerted

This is indicated by only seeing cis- (meso) configuration or only seeing trans- configurations, not both, in products

(* cis or trans depends on E or Z starting configuration)

Essentially the concerted mechanism "locks" the original configuration in place


E configuration alkene products of expoxidation

2 Trans enantiomers


Z configuration alkene products of expoxidation

1 meso cis- configuration


Regiochemistry of expoxidation in base

Base will attack the less substituted side of the intermediate since it attacks in a SN2-like nature


Regiochemistry of expoxidation in acid

Acid will create a cation intermediate

This positive charge will allow the nucleophile to attack the more substituted (and more positive) carbon


Configurational diastereomers

can only happen with 2 or more asymmetric carbons

configurations are not the same or mirror images

ex: RS and RR or E and Z


Configurational enantiomers

can happen with only 1 or more asymmetric carbons

carbons that are completely symmetric

ex: RRR to SSS


Example of Conformational enantiomers

rotating around a sigma bond to make gauche on the other-side


Meso compound checklist

1. Has at least 2 stereocenters
2. There is an internal mirror plane
3. Ensure tht each half is infact a mirror image by doing R/S
4. Each stereocenter must have the same connectivity