1.
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2.
E1 reactions can be stereoselective
 For some eliminations only one product is possible
 For others may be a choice of two (or more) alkene products
 Differ either in the location or stereochemistry of the double bond
 Factors that control the stereochemistry and regiochemistry of the alkenes
 Starting with E1 reactions
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3.
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4.
 For steric reasons
 E-alkenes are lower in energy than Z-alkenes
 Substituents can get farther apart from one another
 Reaction that can choose which it forms is likely to favour the formation of E-alkenes
 For alkenes formed by E1 elimination, less hindered E-alkene is favoured
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5.
 Geometry of the product is determined when the proton is lost from intermediate
carbocation
 New pi bond only form if vacant p orbital of carbocation and breaking C–H bond are
aligned parallel
 Two possible conformations of the carbocation with parallel orientations, one is more
stable than the other
 Suffers less steric hindrance
 Transition states on the route to the alkenes
 E-alkene is lower in energy and more E-alkene than Z-alkene is formed
 Stereoselective, reaction chooses to form predominantly one of two possible
stereoisomeric products
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6.
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7.
 Tamoxifen, important drug in the fight against breast cancer, one of the most
common forms of cancer
 Works by blocking the action of the female sex hormone estrogen
 Tetrasubstituted double bond can be introduced by an E1 elimination
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8.
E1 reactions can be regioselective
 E1 eliminations that can give more than one regio isomeric
alkene
 Major product is the alkene that has the more substituents, more stable of two
possible products
 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 8

9.
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10.
 More substituted alkene is more stable, does not explain why it one forms faster
 Transition states leading to the two alkenes
 Both form from the same carbocation, which one depends on which proton is lost
 Removal of the proton on the right leads to a transition state in which there is a
monosubstituted double bond partly formed
 Removal of the proton on the left leads to a partial double bond that is trisubstituted
 More stable—the transition state is lower in energy, more substituted alkene forms
faster
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11.
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12.
E2 eliminations have anti-periplanar transition states
 New pi bond is formed by the overlap of C–H s bond with C–X s* antibonding orbital
 Two orbitals have to lie in same plane for best overlap
 Two conformations that allow this
 One has H and X syn-periplanar
 Other anti-periplanar
 Anti-periplanar conformation more stable, staggered
 Syn-periplanar conformation is eclipsed but,
 Only in the anti-periplanar conformation are the bonds (and therefore the orbitals)
truly parallel 12

13.
 E2 eliminations take place from the anti-periplanar conformation
 E2 elimination gives mainly one of two possible stereoisomers
 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
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14.
 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
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15.
E2 eliminations can be stereospecific
 Next example, one proton take part in the elimination
 No choice of anti-periplanar transition states
 Whether the product is E or Z, E2 reaction has only one course to follow
 Outcome depends on which diastereoisomer of starting material is used
 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
 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) 15

16.
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17.
 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
 First reaction is about ten times as fast as the second
• Anti-periplanar conformation only reactive one, not necessarily the most
stable
 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
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18.
 Reactions in which the stereochemistry of the product is determined by the
stereochemistry of the starting material are called stereospecific
 Stereoselective reactions give one predominant product because the reaction
pathway has a choice.
 Either pathway of lower activation energy is preferred (kinetic control) or more stable
product (thermodynamic control)
 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
 There is no choice
 Reaction gives a different dia-stereoisomer of the product from each stereoisomer of
the starting material
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