part 4 elimination reactions – learning objectives part 4 – elimination reactions after...
TRANSCRIPT
Part 4
Elimination Reactions
Cl
H R2
R1
R3
R4
R2R1
R4 R3
R2R1
R4 R3
Loss ofStereochemistry
R3R1
R4 R2Retension
of Steroechemistry
Rate = k[R-Hal][Nu] Rate = k[R-Hal]
B
– Learning Objectives Part 4 –
Elimination Reactions
After completing PART 4 of this course you should have an understanding of, and be able to demonstrate, the following
terms, ideas and methods.
(i) Understand E2 and E1 reaction mechanisms
(ii) Understand how experimental evidence from rate equations and stereochemical outcomes in the product
lead to the proposal of reaction mechanisms
(iii) Understand the experimental factors which favour E2 or E1 reaction mechanisms
(vi) Understand the term antiperiplanar in the context of E2 reaction mechanisms
(v) Understand that in assessing the reaction outcome in an elimination reaction, the stereoelectronic of the
alkylhalide needs to be considered carefully, ie. the constitution and conformation
CHM1C3– Introduction to Chemical Reactivity of Organic
Compounds–
Elimination Reactions
Descriptor Rate Equation Stereochemical Outcome
E2 rate = k[R-Hal][Nu] Retension
E1 rate = k[R-Hal] Loss of Stereochemistry
Clearly, two different reaction mechanisms must be in operation.
It is the job of the chemists to fit the experimental data to any proposed mechanism
C
Cl
C Cl
R4
R3H
R1
R2
CCR4
R3
R1
R2
NuHB BH
The E2 Reaction Mechanism
Cl
H R2
R1
R3
R4
B
Rate = k[R-Hal][B]
Bimolecular Process
H and Cl must be antiperiplanar
Transition State – Energy Maxima
Cl
H R2
R1
R3
R4
B
HBR2R1
R4 R3
Cl
Retension of
Steroechemistry
Compare to SN2
Reaction Coordinate
Cl
H R2
R1
R3
R4
B
Cl
H R2
R1
R3
R4
B
Transition StateHB
R2R1
R4 R3
Cl
The E1 Reaction Mechanism
Cl
H R2
R1
R3
R4
UnimolecularProcess
Rate = k[R-Hal]
B
Rotation about C-C Bond
HR2
R1 R3
R4Cl
HR3
R1 R2
R4Cl
Reactive Intermediate – Energy Minima
R2R1
R4 R3
R3R1
R4 R2
Loss ofStereochemistry
Compare to SN1
Energy
Reaction Coordinate
Cl
H R2
R1
R3
R4
HR2
R1 R3R4
R2R1
R4 R3
Reactive Intermediate
Cl
H R2
R1
R3
R4
B
Stereochemistry Compared
HR2
R1 R3
R4
HR3
R1 R2
R4H, C, C and Cl are
antiperiplanar
E2 E1
R4 R1
H
R3 R2
R4 R1
H
R2 R3R4 R1
HR3 R2
Cl
B
Alkene Stability
Stability Increases
Constitutionally Different Eliminations
Br
Minor Alkene Product
Major Alkene Product
BrH 2 Equivalent
Hydrogen atomsBr
H6 Equivalent
Hydrogen atoms
Statistically favoured!
Base
Constitutional Isomers
Cl
HPh
DPh
H
Cl
Ph D
H
HPh
Rotation about C-C
bond
Cl
HPh
PhH
D
Cl
H Ph
D
HPh
B
Cl
Ph D
H
HPh
B
B
Conformational Equilibria
DiastereoisomersConformers Low Energy
Transition State
High Energy Transition State
Cl
PhH
D
HPh
B
Ph
Ph
Minor Product
D
H
Ph
PhMajor
Product
H
H
Cl
Ph D
H
HPh
B
Steric clash of thetwo phenyl groupsraises the energy ofthe transition state,as the two carboncentres become sp2 hybridised
Cl
PhH
D
HPh
B
High Energy Transition State
Low Energy Transition State
Cl
CH3
H3C CH3
NaOEt
EtOH
Rate = k[R-Cl][NaOEt] CH3
H3C CH3
CH3
H3C CH3
25% 75%
Cl
H H
Cl
H H
EtO EtO
Cyclohexane Rings – E2
Cl
H H
Two C-H bonds are antiperiplanar to the
C-Cl bond
Cyclohexane Rings – E1
No C-H bonds are antiperiplanar to the
C-Cl bond
ClH H
EtO
H HH
H H
EtO
H
Cl
CH3
H3C CH3
CH3
H3C CH3
CH3
H3C CH3
H2O
EtOH
Rate = k[R-Cl]32% 68%
Polar SolventSupports Carbocation
Formation
ClH H
– Summary Sheet Part 4 –
Elimination Reactions
The difference in electronegativity between the carbon and chlorine atoms in the C-Cl sigma () bond result in a polarised bond,
such that there is a partial positive charge (+) on the -carbon atom and a slight negative charge (-) on the halogen atom, which
in turn is transmitted to the -carbon atom and the protons associated with it. Thus, the hydrogen atoms on the -carbon atom are
slightly acidic. Thus, if we react haloalkanes with bases (chemical species which react with acids), the base will abstract the
proton atom, leading to carbon-carbon double bond being formed with cleavage of the C-Cl bond.
The mechanism of this -elimination (or 1,2 elimination) can take two limiting forms described as Bimolecular Elimination (E2)
and Unimolecular Elimination (E1).
The E2 mechanism fits with a rate equation which is dependent on both the base and haloalkane, and that the product retains
the stereochemical information about the C-C bond. This retension of stereochemical integrity requires an antiperiplanar
relationship of the eliminated atoms.
In contrast, The E1 mechanism fits with a rate equation which is dependent on only the haloalkane, and that the product
undergoes a loss of the stereochemical information about the C-C bond. Thus, with appropriately substituted haloalkane a
pair of diastereomeric alkenes are formed, as result of rotation around the C-C bond upon formation of the carbocationic
intermediate.
CHM1C3– Introduction to Chemical Reactivity of Organic
Compounds–
Exercise 1: Substitution/Elimination ReactionsRationalise the experimental results that when 1 is reacted with NaOEt in EtOH, two alkenes are formed, whereas 2 under the same conditions affords an inverted substitution product.
H
HCl
1
NaOEtEtOH
H
H
H
H
H
HCl
2
H
HOEt
NaOEtEtOH
Answer 1: Substitution/Elimination ReactionsRationalise the experimental results that when 1 is reacted with NaOEt in EtOH, two alkenes are formed, whereas 2 under the same conditions affords an inverted substitution product.
H
HCl
1
NaOEtEtOH
H
H
H
H
H
HCl
2
H
HOEt
NaOEtEtOH
As 2 undergoes an inversion of stereochemistry one must assume SN2 mechanism.As 1 is subject to the same reaction conditions as 2 one must assume that elimination of HCl does not involve the formation of a carbocation, and thus E2 mechanism must operate.
Cl
H
H
H
EtO
H
H
H
Cl
H
H
H
EtO
H
H
Cl
H
H
EtO
OEt
H
H
Antiperiplanar Conformational Relationships
E2 E2
SN2
Cl Cl
NaOEt
EtOH
MODERATE Temperature
rate = k[R-Cl][NaOEt]
NaOEt
EtOH
HIGHTemperature
rate = k[R-Cl][NaOEt]
Exercise 2: Elimination ReactionsRationalise the following
Cl Cl
NaOEt
EtOH
MODERATE Temperature
rate = k[R-Cl][NaOEt]
NaOEt
EtOH
HIGHTemperature
rate = k[R-Cl][NaOEt]
Cl Cl
NaOEt
EtOH
MODERATE Temperature
rate = k[R-Cl][NaOEt]
NaOEt
EtOH
HIGHTemperature
rate = k[R-Cl][NaOEt]
Answer2: Elimination ReactionsRationalise the following
Cl Cl
NaOEt
EtOH
MODERATE Temperature
rate = k[R-Cl][NaOEt]
NaOEt
EtOH
HIGHTemperature
rate = k[R-Cl][NaOEt]
HCl
Cl
H
Cl
HH
Cl
H
Cl
H
Cl
HHEtO
EtO
TS2
TS1
OEt
OEt
The energy to attain this transition state TS2 geometry is much higher that TS1, because the largest substituent (t-Bu) and the Cl are both in the axial positions, which leads to large steric clashes. Thus, more energy, i.e. higher reaction temperatures, are required to attain TS2 relative to TS1.