![Part 4
Elimination Reactions
Cl
H R2
R1
R3
R4
R2
R1
R4
R3
R2
R1
R4
R3
Loss of
Stereochemistry
R3
R1
R4
R2
Retension
of
Steroechemistry
Rate =k[R-Hal][Nu] Rate =k[R-Hal]
B](https://image.slidesharecdn.com/elimnations-220608164629-66ce7a4c/85/elimnationsppt-1-320.jpg)
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Elimnations.ppt
- 1. Part 4 Elimination Reactions Cl H R2 R1 R3 R4 R2 R1 R4 R3 R2 R1 R4 R3 Loss of Stereochemistry R3 R1 R4 R2 Retension of Steroechemistry Rate =k[R-Hal][Nu] Rate =k[R-Hal] B
- 2. – 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–
- 3. 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 R3 H R1 R2 C C R4 R3 R1 R2 NuH B BH
- 5. 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 H B R2 R1 R4 R3 Cl Retension of Steroechemistry Compare to SN2
- 6. Reaction Coordinate Cl H R2 R1 R3 R4 B Cl H R2 R1 R3 R4 B Transition State H B R2 R1 R4 R3 Cl
- 7. The E1 Reaction Mechanism Cl H R2 R1 R3 R4 Unimolecular Process Rate = k[R-Hal] B Rotation about C-C Bond H R2 R1 R3 R4 Cl H R3 R1 R2 R4 Cl Reactive Intermediate – Energy Minima R2 R1 R4 R3 R3 R1 R4 R2 Loss of Stereochemistry Compare to SN1
- 9. Cl H R2 R1 R3 R4 B Stereochemistry Compared H R2 R1 R3 R4 H R3 R1 R2 R4 H, C, C and Cl are antiperiplanar E2 E1 R4 R1 H R3 R2 R4 R1 H R2 R3 R4 R1 H R3 R2 Cl B
- 12. Constitutionally Different Eliminations Br Minor Alkene Product Major Alkene Product Br H 2 Equivalent Hydrogen atoms Br H 6 Equivalent Hydrogen atoms Statistically favoured! Base Constitutional Isomers
- 13. Cl H Ph D Ph H Cl Ph D H H Ph Rotation about C-C bond Cl H Ph Ph H D Cl H Ph D H Ph B Cl Ph D H H Ph B B Conformational Equilibria Diastereoisomers Conformers Low Energy Transition State High Energy Transition State Cl Ph H D H Ph B Ph Ph Minor Product D H Ph Ph Major Product H H
- 14. Cl Ph D H H Ph B Steric clash of the two phenyl groups raises the energy of the transition state, as the two carbon centres become sp2 hybridised Cl Ph H D H Ph B High Energy Transition State Low Energy Transition State
- 15. 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
- 16. Cyclohexane Rings – E1 No C-H bonds are antiperiplanar to the C-Cl bond Cl H H EtO H H H H H EtO H Cl CH3 H3C CH3 CH3 H3C CH3 CH3 H3C CH3 H2O EtOH Rate = k[R-Cl] 32% 68% Polar Solvent Supports Carbocation Formation Cl H H
- 17. – 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–
- 19. Exercise 1: Substitution/Elimination Reactions Rationalise 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 H Cl 1 NaOEt EtOH H H H H H H Cl 2 H H OEt NaOEt EtOH
- 20. Answer 1: Substitution/Elimination Reactions Rationalise 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 H Cl 1 NaOEt EtOH H H H H H H Cl 2 H H OEt NaOEt EtOH 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
- 21. Cl Cl NaOEt EtOH MODERATE Temperature rate = k[R-Cl][NaOEt] NaOEt EtOH HIGH Temperature rate = k[R-Cl][NaOEt] Exercise 2: Elimination Reactions Rationalise the following Cl Cl NaOEt EtOH MODERATE Temperature rate = k[R-Cl][NaOEt] NaOEt EtOH HIGH Temperature rate = k[R-Cl][NaOEt]
- 22. Cl Cl NaOEt EtOH MODERATE Temperature rate = k[R-Cl][NaOEt] NaOEt EtOH HIGH Temperature rate = k[R-Cl][NaOEt] Answer2: Elimination Reactions Rationalise the following Cl Cl NaOEt EtOH MODERATE Temperature rate = k[R-Cl][NaOEt] NaOEt EtOH HIGH Temperature rate = k[R-Cl][NaOEt] H Cl Cl H Cl H H Cl H Cl H Cl H H EtO 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.