2. Part 4 Elimination Reactions

3. – 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–

4. Elimination Reactions DescriptorRate EquationStereochemical Outcome E2rate = k[R-Hal][Nu]Retension E1rate = 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 BHBH

6. The E2 Reaction Mechanism Compare to S N 2

8. The E1 Reaction Mechanism Compare to S N 1

9. Reactive Intermediate

10. Stereochemistry Compared H, C, C and Cl are antiperiplanar E2 E1

12. Alkene Stability Stability Increases

13. Constitutionally Different Eliminations Statistically favoured! Base Constitutional Isomers

14. Conformational Equilibria Diastereoisomers Conformers Low Energy Transition State High Energy Transition State

15. High Energy Transition State Low Energy Transition State

16. Cyclohexane Rings – E2 Two C-H bonds are antiperiplanar to the C-Cl bond

17. Cyclohexane Rings – E1 No C-H bonds are antiperiplanar to the C-Cl bond

18. – 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–

20. 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.

21. 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. As 2 undergoes an inversion of stereochemistry one must assume S N 2 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. E2 SN2SN2

22. Exercise 2: Elimination Reactions Rationalise the following

23. Answer2: Elimination Reactions Rationalise the following 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.