Chapter 4 Introduction to Alkenes. Structure and Reactivity

1. Chapter 4Introduction to Alkenes.Structure and Reactivity

2. Alkenes • Hydrocarbons containing one or more carbon-carbon double bonds • Sometimes called olefins • Classified as unsaturated hydrocarbons • Ethylene is the simplest alkene

5. Physical Properties • Very similar to corresponding alkanes: • Flammable • Non-polar • Insoluble in Water • Low melting and boiling points • Low density compared to water 5

7. Bond Length • Carbon-carbon multiple bonds are shorter than their carbon-carbon single bond counterparts 133.9 pm 153.6 pm

8. Dipole moments and polarization

9. Structure and Bonding in Alkenes • Carbons involved in the carbon-carbon double bond exhibit trigonal planar geometry

10. Carbon Hybridization in Alkenes • sp2-hybridization  Trigonal planar geometry

11. sp2 = 33% s character, 66% p character • More s character → electrons are held closer to the nucleus

12. Carbon Hybridization in Alkanes

13. Carbon-carbon single bond in propene = 150 pm • Carbon-carbon single bond in propane = 154 pm • Due to sp2 vs. sp3 hybridization • Bonds with more s character are shorter

14. Hybrid Orbital Picture of Ethylene 4.1 Structure and Bonding in Alkenes

15. s bond: head-to-head overlap of orbitals • Cylindrically symmetric

16. bond: side-to-side overlap of p orbitals

17. p Orbital System in Ethylene 4.1 Structure and Bonding in Alkenes

18. EPM of Ethylene • The filled  molecular orbital is the  bond • Most of the important reactions of alkenes involve the electrons of the  bond 4.1 Structure and Bonding in Alkenes

19. Find the longest chain that includes the double bond and name this as the parent chain by changing the ending from “-ane” to “-ene” Hexane Hexene If there are two possibilities, choose the principle chain to be the one with the greatest number of double bonds Number the chain so that the double bond has the lowest number possible and place that number before the parent chain The double bond takes precidence over the substituents when numbering the chain Identify the substituents with number and name Rules for Naming Alkenes

20. Problems • Name the following:

21. If there are more than one double bonds in the parent chain, identify each double bond’s position by number and indicate how many are present using Greek prefixes preceded by the letter “a.” • Example: -adiene, -atriene, etc.

22. Problems • Name the following:

23. Substituents Containing Alkenes • Some common groups with alkenes • Nonsystematic traditional names 4.2 Nomenclature of Alkenes

24. Problems • Name the following compounds:

25. Problems • Name the following compounds: CH3CH=CHCH3

26. Stereoisomers: Molecules that have the same molecular formula, same atom connectivity, but a different spatial arrangement • Cis-trans isomers: • Recall: Constitutional/Structural Isomers: same molecular formula, different atom connectivity

27. Restricted Rotation of Alkenes 4.1 Structure and Bonding in Alkenes

28. Properties of 2-butene • Cis-2-Butene • M.P. = -139°C • B.P. = 3.7°C • Density = 0.6213 g/mL • Trans-2-Butene • M.P. = - 105°C • B.P. = 1°C • Density = 0.6041 g/mL

29. Requirements for Cis-Trans Isomerism • Only when both carbons are bonded to two different groups are cis-trans isomers possible • Compounds that have one of their carbons bonded to two identical groups can’t exist as cis-trans isomers

30. Problems • Which of the following compounds can exist as pairs of cis-trans isomers? Draw each cis-trans pair. • CH3CH=CH2 • CH3CH2CH=CHCH3 • (CH3)2C=CHCH3 • ClCH=CHCl • There are two isomers for 3-methyl-2-pentene. Draw both of them. Which one is cis and which one is trans?

31. E,Z Nomenclature System • cis vs trans is not always clear • Cahn-Ingold-Prelog system: Assign priorities • Z (zusammen): “together” • On Zee Zame Zide • E (entgegen): “across” • E = Enemies = On Opposite Sides 4.2 Nomenclature of Alkenes

33. Priority Assignment Rules • Examine each of the double bond carbons separately. Identify the two atoms directly attached to the C and rank them according to atomic #. • Higher atomic # = higher priority • Higher isotopic mass = higher priority

34. If the first atoms connected to the double bond carbon are the same, continue moving outward until the first point of difference

35. Multiple-bonded atoms are equivalent to the same number of single bonded atoms

36. Problems • Using the Cahn-Ingold-Prelog rules, identify whether the following molecules are E or Z:

37. Problems • Give the IUPAC names for each of the following molecules, including the E, Z designation.

39. What’s the Structure of C6H10? • Saturated is C6H14 • Therefore 4 H's are not present • So, what does it look like? • Double bond(s)? • Triple bond(s)? • Ring(s)? • Ring and double bond?

40. Degree of Unsaturation • Gives info on number of rings and/or p bonds • Maximum # of H’s in a hydrocarbon: CnH2n+2 • Each ring and/or p bond reduces number of hydrogens by 2 • aka: unsaturationnumber

41. For C6H10 the unsaturation # is 2 • Two double bond • Two rings • One double bond and a ring • One triple bond

42. Calculating Degree of Unsaturation for Compounds Containing Elements Other than Carbon and Hydrogen • Halogens are counted as 1 H: 4.3 Unsaturation Number

43. Nitrogen increases H count by 1:

44. Ignore Oxygens • O’s form two bonds, don’t affect the formula of an equivalent hydrocarbon • No change in number of H atoms if insert an O into a hydrocarbon compound

45. Problems • Calculate the degree of unsaturation for the following molecules • C12H20 • C4H6 • C6H5N • C6H5NO2 • C8H9Cl3 • Draw as many structures as you can for #2

46. Alkene Stability • Who’s more stable?

48. Product Ratio • Interconversion between cis and trans isomers can be made to happen using a strong acid catalyst • More stable product is favored

49. The cis configuration is more strained = higher energy Heats of combustion ΔH°combustion = -2685.5 kJ/mol ΔH°combustion = -2682.2 kJ/mol

50. Heats of Hydrogenation • More stable alkene gives off less heat