1. Organic Chemistry M. R. Naimi-Jamal Faculty of Chemistry Iran University of Science & Technology

2. Chapter 3. Continue Alkenes: Structure and Reactivity Based on McMurry’s Organic Chemistry, 6 th edition

3. 3 Alkene - Hydrocarbon With Carbon- Carbon Double Bond Includes many naturally occurring materials Flavors, fragrances, vitamins Important industrial products These are feedstocks for industrial processes

4. 4 Alkenes

5. 5 6.2 Degree of Unsaturation Relates molecular formula to possible structures Degree of unsaturation: number of multiple bonds or rings Formula for saturated a acyclic compound is C n H 2n+2 Each ring or multiple bond replaces 2 H's

6. 6 Degree of Unsaturation

7. 7 Example: C 6 H 10 Saturated is C 6 H 14 Therefore 4 H's are missing This has two degrees of unsaturation Two double bonds? or triple bond? or two rings or ring and double bond

8. 8 Compounds with the same degree of unsaturation can have many things in common and still be very different Degree of Unsaturation and Variation

9. 9 Add number of halogens to number of H's (X equivalent to H) Don't count oxygens (oxygen links H) Summary - Degree of Unsaturation

10. 10 6.3 Naming of Alkenes Find longest continuous carbon chain for root Number carbons in chain so that double bond carbons have lowest possible numbers Rings have “cyclo” prefix

11. 11 Alkene Nomenclature: longest chain must contain the C=C

12. 12 Alkene Nomenclature

13. 13 Cycloalkene nomenclature

14. 14 Problem: nomenclature these alkenes

15. 15 Problem: nomenclature these alkenes

16. 16 Many alkenes are known by common names Ethylene = ethene Propylene = propene Isobutylene = 2-methylpropene Isoprene = 2-methyl- 1,3-butadiene

17. 17

18. 18 Alkene Group Names

19. 19 6.4 Electronic Structure of Alkenes Carbon atoms in a double bond are sp 2 - hybridized Three equivalent orbitals at 120º separation in plane Fourth orbital is atomic p orbital Combination of electrons in two sp 2 orbitals of two atoms forms  bond between them

20. 20 6.4 Electronic Structure of Alkenes Additive interaction (overlap) of p orbitals creates a  bonding orbital Subtractive interaction creates a  anti-bonding orbital Occupied  orbital prevents rotation about  -bond Rotation prevented by  bond - high barrier, about 268 kJ/mole in ethylene

21. 21 Rotation of  Bond Is Energetically Costly This prevents rotation about a carbon-carbon double bond (unlike a carbon-carbon single bond). Creates possible geometric isomers (cis/trans)

22. 22 6.5 Cis-Trans Isomerism in Alkenes The presence of a carbon-carbon double can create two possible structures cis isomer - two similar groups on same side of the double bond trans isomer similar groups on opposite sides Each carbon must have two different groups for these isomers to occur

23. 23 Cis, Trans Isomers Require That End Groups Must Differ in Pairs 180° rotation results in identical (superimposable) structures Bottom pair cannot be superimposed without breaking C=C (ie. rotating around the double bond) X

24. 24 Sequence Rules: The E, Z Designation Neither compound is clearly “cis” or “trans” Substituents on C1 are different than those on C2 We need to define “similarity” in a precise way to distinguish the two stereoisomers Cis, trans nomenclature only works for similarly disubstituted double bonds

25. 25 Develop a System for Comparison of Priority of Substituents Assume a valuation system If Br has a higher “value” than Cl If CH 3 is higher than H Then, in A, the higher value groups are on opposite sides In B, they are on the same side Requires a universally accepted “valuation”

26. 26 E, Z Stereochemical Nomenclature Priority rules of Cahn, Ingold, and Prelog Compare where higher priority group is with respect to bond and designate as prefix E -entgegen, opposite sides Z - zusammen, together on the same side

27. 27

28. 28 Ranking Priorities: Cahn-Ingold-Prelog Rules Must rank atoms that are connected at comparison point Higher atomic number gets higher priority Br > Cl > O > N > C > H In this case,The higher priority groups are opposite: (E )-1-bromo-1-chloro-propene

29. 29 2-chloro-2-butenes

30. 30 If atomic numbers are the same, compare at next connection point at same distance Compare until something has higher atomic number Do not combine – always compare Extended Comparison

31. 31

32. 32 Dealing With Multiple Bonds Substituent is drawn with connections shown and no double or triple bonds Added atoms are valued with no ligands themselves

33. 33

34. 34 Some examples

35. 35 Practice problem:

36. 36 Solution

37. 37 Problem:

38. 38 Reaction of alkenes: Electrophilic Addition of HX to Alkenes General reaction mechanism: electrophilic addition Attack of electrophile (such as HBr) on  bond of alkene produces carbocation and bromide ion Carbocation is itself an electrophile, reacting with nucleophilic bromide ion

39. 39 Writing Organic Reactions

40. 40 Two step process First transition state is high energy point Electrophilic Addition Energy Diagram:

41. 41 Chapter 3-2, Questions 27, 31, 32, 37, 42, 45, 46