1. Alkenes
CnH2n

2. Alkenes contain carbon - carbon double bonds
called unsaturated hydrocarbons
also known as olefins (oleum, latin, oil; facere, latin, make)
CnH2n
CnH2n + H2  CnH2n+2 - one degree of unsaturation

3. Degree of unsaturation
Degree of unsaturation = (2NC - NX + NN – NH + 2)/2
NC = number of carbons
NX = number of halogens
NN = number of nitrogens
NH = number of hydrogens

4. Nomenclature – the E/Z system
1. To name alkenes, select the longest carbon chain which includes the carbons of the double bond. Remove the -ane suffix from the name of the alkane which corresponds to this chain. Add the suffix -ene.
a derivative of octene not nonane

5. Nomenclature – the E/Z system
2. Number this chain so that the first carbon of the double bond has the lowest number possible.

6. Nomenclature – the E/Z system
trans
cis

7. cis/trans problems
This molecule is a 1-bromo-1-chloropropene but is it cis or trans!

8. Nomenclature – the E/Z system
(Z)-1-bromo-1-chloropropene
use the Cahn-Ingold-Prelog system to assign priorities to the two groups on each carbon of the double bond.
then compare the relative positions of the groups of higher priority on these two carbons.
if the two groups are on the same side, the compound has the Z configuration (zusammen, German, together).
if the two groups are on opposite sides, the compound has the E configuration (entgegen, German, across).

9. E-Z designations
Try problem 7.1, page 289 of Solomons and Fryhle.

10. Relative stabilities of alkenes
Cis isomers are generally less stable than trans isomers due to strain caused by crowding of the two alkyl groups on the same side of the double bond
Stabilities can be compared by measuring heats of hydrogenation of alkenes.

11. Overall relative stabilities of alkenes

12. Synthesis of alkenes by elimination reactions
dehydrohalogenation:
dehydration:

13. Dehydrohalogenation of alkyl halides
Reactivity: RX 3o > 2o > 1o
a 1,2 elimination reaction

14. Dehydrohalogenation

15. Alkoxide ions – bases used in dehydrohalogenation

16. Dehydrohalogenation of alkyl halides
- no rearrangement

17. The mechanism
In the presence of a strong base, the reaction follows second order kinetics:
rate = k[RX][B-]
However, with weak bases at low concentrations and as we move from a primary halide to a secondary and a tertiary, the reaction becomes first order.
There are two mechanisms for this elimination: E1 and E2.

18. E2 mechanism

19. E1 mechanism
slow
fast

20. Evidence for the E1 mechanism
Follows first order kinetics
Same structural effects on reactivity as for SN1 reactions - 3 > 2 > 1
Rearrangements can occur indicative of the formation of carbocations

21. Evidence for the E2 mechanism
The reaction follows second order kinetics
There are no rearrangements
There is a large deuterium isotope effect
There is an anti periplanar geometry requirement

22. Isotope effects
A difference in rate due to a difference in the isotope present in the reaction system is called an isotope effect.

23. Isotope effects
If an atom is less strongly bonded in the transition state than in the starting material, the reaction involving the heavier isotope will proceed more slowly.
The isotopes of hydrogen have the greatest mass differences. Deuterium has twice and tritium three times the mass of protium. Therefore deuterium and tritium isotope effects are the largest and easiest to determine.

24. Primary isotope effects
These effects are due to breaking the bond to the isotope.
Thus the reaction with protium is 5 to 8 times faster than the reaction with deuterium.

25. Evidence for the E2 mechanism - a large isotope effect

26. Further evidence for the E2 mechanism
RI > RBr > RCl > RF

27. Orientation and reactivity
KOH CH CH
CHCH CH
CH=CHCH 3 2 3 3 3 C H OH Cl 2 5
80%
The ease of alkene formation follows the sequence:-
R2C=CR2 > R2C=CHR > R2C=CH2, RHC=CHR > RHC=CH2
This is also the order of alkene stability. Therefore the more stable the alkene formed, the faster it is formed.
Why?

28. Orientation and reactivity
Let’s look at the transition state for the reaction:
The double bond is partially formed in the transition state and therefore the transition state resembles an alkene. Thus the factors which stabilize alkenes will stabilize this nascent alkene.
A Zaitsev elimination.

29. anti elimination

30. anti elimination
KOH ?

31. anti elimination

32. Formation of the less substituted alkene
Dehydrohalogenation using a bulky base favours the formation of the less substituted alkene:

33. Substitution vs elimination
SN2 v E2
substitution
elimination

34. Substitution vs elimination
SN1 v E1

35. Substitution vs elimination

36. Dehydration of alcohols

37. Dehydration of alcohols - the mechanism

38. Dehydration of alcohols - orientation

39. The Zaitsev product predominates

40. The Zaitsev product predominates
The transition state explains the orientation:

41. Dehalogenation of vicinal dihalides

42. Hydrogenation of alkynes

43. Synthesis of alkynes by elimination reactions

45. Problems
Try problems 7.19, 7.21, 7.22, 7.24, 7.26, 7.29, 7.36, 7.38, and 7.40 on pages 323 – 326 of Solomons and Fryhle.