The document discusses stereoselective and regioselective E1 and E2 elimination reactions in organic chemistry, emphasizing how factors like sterics and substrate substitution affect the stability and formation of alkenes. It details how different conformations and orientations influence the product of the reactions, with examples such as tamoxifen highlighting the importance of stereochemistry in drug design. Additionally, it contrasts stereospecific reactions, which yield a single isomer, with stereoselective reactions, where one product is favored due to lower activation energy or greater stability.

1. Mehak Shabbir
BS-chemistry

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

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

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

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

8. E1 reactions can be regioselective
q E1 eliminations that can give more than one regio isomeric alkene
q Major product is the alkene that has the more substituents, more stable of two possible products
q More 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. q More substituted alkene is more stable, does not explain why it one forms faster
Ø Transition states leading to the two alkenes
q Both form from the same carbocation, which one depends on which proton is lost
q Removal of the proton on the right leads to a transition state in which there is a
monosubstituted double bond partly formed
q Removal of the proton on the left leads to a partial double bond that is trisubstituted
q More stable—the transition state is lower in energy, more substituted alkene forms faster

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

13. q E2 eliminations take place from the anti-periplanar conformation
q E2 elimination gives mainly one of two possible stereoisomers
q 2-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. q There 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
q Next example, one proton take part in the elimination
q No choice of anti-periplanar transition states
q Whether the product is E or Z, E2 reaction has only one course to follow
q Outcome depends on which diastereoisomer of starting material is used
q When 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
q As 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. q Second 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
q First reaction is about ten times as fast as the second
• Anti-periplanar conformation only reactive one, not necessarily the most stable
q Newman 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. q Reactions in which the stereochemistry of the product is determined by the stereochemistry
of the starting material are called stereospecific
q Stereoselective reactions give one predominant product because the reaction pathway has a
choice.
q Either pathway of lower activation energy is preferred (kinetic control) or more stable
product (thermodynamic control)
q Stereospecific reactions lead to the production of a single isomer as a direct result of
mechanism of reaction and the stereochemistry of the starting material
q There is no choice
q Reaction gives a different dia-stereoisomer of the product from each stereoisomer of the
starting material