Stereochemistry of Elimination Reactions

Stereochemistry
Elimination reactions
E1 eliminations
E2 eliminations
Sterospecific
Stereoselective
Regioselective
Regiospecific
E or Z Isomers
Minor and Major products

1. Mehak Shabbir
BS-chemistry

2. E1 reactions can be stereoselective
qFor some eliminations only one product is possible
qFor others may be a choice of two (or more) alkene products
§Differ either in the location or stereochemistry of the double bond
qFactors that control the stereochemistry and regiochemistry of the alkenes
qStarting with E1 reactions

4. vFor steric reasons
q E-alkenes are lower in energy than Z-alkenes
§Substituents can get farther apart from one another
qReaction that can choose which it forms is likely to favour the formation of E-alkenes
qFor alkenes formed by E1 elimination, less hindered E-alkene is favoured

5. qGeometry of the product is determined when the proton is lost from intermediate carbocation
qNew p bond only form if vacant p orbital of carbocation and breaking C–H bond are aligned
parallel
qTwo possible conformations of the carbocation with parallel orientations, one is more stable
than the other
§Suffers less steric hindrance
qTransition states on the route to the alkenes
§E-alkene is lower in energy and more E-alkene than Z-alkene is formed
qStereoselective, reaction chooses to form predominantly one of two possible stereoisomeric
products

7. qTamoxifen, important drug in the fight against breast cancer, one of the most common forms
of cancer
qWorks by blocking the action of the female sex hormone estrogen
qTetrasubstituted double bond can be introduced by an E1 elimination

8. E1 reactions can be regioselective
qE1 eliminations that can give more than one regio isomeric alkene
qMajor product is the alkene that has the more substituents, more stable of two possible products
qMore substituted alkenes are more stable
§Stabilized when empty p* antibonding orbital can interact with filled orbitals of parallel C–H
and C–C bonds
§More C–C or C–H bonds there are, more stable the alkene

10. qMore substituted alkene is more stable, does not explain why it one forms faster
ØTransition states leading to the two alkenes
qBoth form from the same carbocation, which one depends on which proton is lost
qRemoval of the proton on the right leads to a transition state in which there is a
monosubstituted double bond partly formed
qRemoval of the proton on the left leads to a partial double bond that is trisubstituted
qMore stable—the transition state is lower in energy, more substituted alkene forms faster

12. E2 eliminations have anti-periplanar transition states
qNew p bond is formed by the overlap of C–H s bond with C–X s* antibonding orbital
qTwo orbitals have to lie in same plane for best overlap
qTwo conformations that allow this
§One has H and X syn-periplanar
§Other anti-periplanar
qAnti-periplanar conformation more stable, staggered
qSyn-periplanar conformation is eclipsed but,
qOnly in the anti-periplanar conformation are the bonds (and therefore the orbitals) truly
parallel

13. qE2 eliminations take place from the anti-periplanar conformation
qE2 elimination gives mainly one of two possible stereoisomers
q2-Bromobutane has two conformations with H and Br anti-periplanar
§One that is less hindered leads to more of the product, and the E-alkene
predominates

14. qThere is a choice of protons to be eliminated
•Stereochemistry of the product results from which proton is anti-periplanar to the
leaving group
•when the reaction takes place, and the reaction is stereoselective as a result

15. E2 eliminations can be stereospecific
qNext example, one proton take part in the elimination
qNo choice of anti-periplanar transition states
qWhether the product is E or Z, E2 reaction has only one course to follow
qOutcome depends on which diastereoisomer of starting material is used
qWhen first diastereoisomer is drawn with the proton
•Bromine anti-periplanar, as required, in the plane of the page,
•Two phenyl groups have to lie one in front and one behind the plane of the paper
qAs hydroxide attacks the C–H bond and eliminates Br
•this arrangement is preserved and the two phenyl groups end up trans (the alkene is E)

17. qSecond dia-stereoisomer forms Z-alkene for the same reasons:
•Two phenyl groups are on the same side of H–C–C–Br plane in reactive anti-
periplanar conformation, end up cis in the product
•Each diastereoisomer gives a different alkene geometry at different rates
qFirst reaction is about ten times as fast as the second
•Anti-periplanar conformation only reactive one, not necessarily the most stable
qNewman projection for second reaction shows that two phenyl groups have to lie synclinal
(gauche) to one another:
•Steric interaction between these large groups at any time
•Relatively small proportion of molecules adopt the right conformation for elimination,
slowing the process down

18. qReactions in which the stereochemistry of the product is determined by the stereochemistry
of the starting material are called stereospecific
qStereoselective reactions give one predominant product because the reaction pathway has a
choice.
qEither pathway of lower activation energy is preferred (kinetic control) or more stable
product (thermodynamic control)
qStereospecific reactions lead to the production of a single isomer as a direct result of
mechanism of reaction and the stereochemistry of the starting material
qThere is no choice
qReaction gives a different dia-stereoisomer of the product from each stereoisomer of the
starting material