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Chapter 6

Chapter 6. Alkenes: Structure and Reactivity. Introduction. An alkene is a hydrocarbon that contains a carbon-carbon double bond It is also called an olefin but alkene is better It is abundant in nature It is an important industrial product.

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Chapter 6

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  1. Chapter 6 Alkenes: Structure and Reactivity

  2. Introduction • An alkene is a hydrocarbon that contains a carbon-carbon double bond • It is also called an olefin but alkene is better • It is abundant in nature • It is an important industrial product

  3. It includes many naturally occurring materials • flavors, fragrances, vitamins

  4. I. Alkenes: An Overview • Industrial Preparation and Use of Alkenes • Degree of Unsaturation • Naming Alkenes

  5. Ethylene and propylene are the simplest alkenes the most important organic chemicals produced industrially A.Industrial Preparation and Use of Alkenes

  6. Compounds derived from ethylene and propylene

  7. Industrial Synthesis of Ethylene, Propylene and Butene • Ethylene, propylene, and butene are synthesized industrially by thermal cracking of light (C2-C8) alkanes:

  8. Industrial Synthesis of ethylene, propylene and butene • The high-temperature conditions cause spontaneous homolytic breaking of C-C and C-H bonds, forming smaller fragments: • Thermal cracking is an example of a reaction whose energetics is governed by TDSo rather than DHo DGo = DHo – TDSo TDSo > DHo

  9. B.Degree of Unsaturation • Alkenes - are unsaturated hydrocarbons. • They have fewer hydrogens than alkanes with same number of carbons

  10. Acyclic Alkanes - have the general formula CnH2n+2 where n is an integer • Acyclic Alkenes - have the general formula CnH2n where n is an integer

  11. Degree of Unsaturationis the number of p bonds and/or rings in the molecule. • Each ring or multiple bond replaces 2 H's in the alkane formula CnH2n+2 • It relates molecular formula to possible structures

  12. Degree of Unsaturationis the number of p bonds and/or rings in the molecule. (2n+2)-x Degree of Unsaturation = 2 where n is the number of carbons x is the number of hydrogens

  13. Example:A hydrocarbon with a molecular weight (82 g/mol) C6H10 • Saturated is C6H14 • Therefore 4 H's are not present • This has two degrees of unsaturation • Two double bonds? • or triple bond? • or two rings? • or ring and double bond?

  14. Degree of Unsaturation With Other Elements • Organohalogens (C, H, X where X: F, Cl, Br, I) • Halogen replaces hydrogen • Add the number of halogens to the number of hydrogens • C4H6Br2 and C4H8 have one degree of unsaturation

  15. Organooxygens (C, H, O) • Oxygen forms two bonds. It does not affect the total count of H's • Ignorethe number of oxygens • C5H8O and C5H8 have two degrees of unsaturation

  16. Organonitrogens (C, H, N) • Nitrogen has three bonds. So if it connects where H was, it adds a connection point • Subtract the number of nitrogens from the number of hydrogens

  17. Summary: Degree of Unsaturation • Degree of Unsaturation can be calculated: • Count pairs of H's below CnH 2n+2 • Add number of halogens to number of H's (X equivalent to H) • Ignore or don't count oxygens (O links H) • Subtract number of nitrogens from H's (N has two connections)

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

  19. Practice Problem: Calculate the degree of unsaturation in the following hydrocarbons: • C8H14 • C5H6 • C12H20 • C20H32 • C40H56 (b-carotene)

  20. Practice Problem: Calculate the degree of unsaturation in the following formulas, and then draw as many structures as you can for each: • C4H8 • C4H6 • C3H4

  21. Practice Problem: Calculate the degree of unsaturation in the following formulas: • C6H5N • C6H5NO2 • C8H9Cl3 • C9H16Br2 • C10H12N2O3 • C20H32ClN

  22. C.Naming Alkenes • Like alkanes, alkenes are named according to the system devised by the International Union of Pure and Applied Chemistry (IUPAC).

  23. Steps to naming alkenes1 • Name the parent hydrocarbon • Find the longest continuous carbon chain containing the double bond • Name using the suffix -ene

  24. Steps to naming alkenes2 • Number the carbon atoms in the main chain • The double-bond carbons receive the lowest possible numbers • The correct sequence is when the substituents have the lowest possible number

  25. Steps to naming alkenes3 • Write the full name • Name, number and list the substituents alphabetically • Indicate the position of the double-bond

  26. Steps to naming alkenes3 • Write the full name • If more than one double bond is present: • indicate the position of each and • Use the suffixes -diene, -triene, …

  27. Naming Cycloalkenes • Like alkenes, cycloalkenes are also named by the rules devised by the International Union of Pure and Applied Chemistry (IUPAC).

  28. Steps to naming cycloalkenes1 • Name the parent hydrocarbon • Number the cycloalkene so that the double bond is between C1 and C2 • Name using the prefix cyclo- and the suffix -ene

  29. Steps to naming cycloalkenes2 • Number the substituents and write the name • The correct sequence is when the substituents have the lowest possible number

  30. Common Names of Alkenes • Many alkenes are known by their common names. • These common names are recognized by the International Union of Pure and Applied Chemistry (IUPAC).

  31. Practice Problem: Give IUPAC names for the following compounds:

  32. Practice Problem: Draw structures corresponding to the following IUPAC names: • 2-Methyl-1,5-hexadiene • 3-Ethyl-2,2-dimethyl-3-heptene • 2,3,3-Trimethyl-1,4,6-octatriene • 3,4-Diisopropyl-2,5-dimethyl-3-hexene • 4-tert-Butyl-2-methylheptane

  33. Practice Problem: Name the following cycloalkenes:

  34. II. Alkenes: Structure • Electronic Structure • Cis-Trans Isomerism • Sequence Rules: The E, Z Designation • Stability of Alkenes

  35. A.Electronic Structure of Alkenes • Carbon atoms in a double bond are sp2-hybridized • Three equivalent orbitals are in a plane at 120º angle • Fourth orbital is atomic p orbital (perpendicular to the plane)

  36. Two sp2-hybridized Carbon atoms form: •  bond • Head-on overlap of two sp2 orbitals forms a  bond •  bond • Side-to-side overlap of two p orbitals forms a  bond

  37. Molecular Orbitals (MO) • Additive interaction of p orbitals (combining p orbital lobes with the same algebraic sign) creates a  bonding orbital • Subtractive interaction (combining lobes with opposite signs) creates a  anti-bonding orbital

  38. Only bonding MO is occupied. • Occupied  orbital prevents rotation. • Rotation prevented by  bond - high barrier, about 268 kJ/mole in ethylene

  39. Rotation of  Bond is Prohibitive • The  bond does not have circular cross-section. • The  bond must break for rotation to take place around a carbon-carbon double bond (unlike a carbon-carbon single bond). • It creates possible alternative structures

  40. 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 B.Cis-Trans Isomerism

  41. Cis-Trans Isomerism requires that end groups must differ in pairs Each carbon must havetwo different groups for these isomers to occur

  42. Cis-Trans Isomerism requires that end groups must differ in pairs • 180°rotation superposes the upper pair • Bottom pair cannot be superposed without breaking C=C X

  43. Practice Problem: Which of the following compounds can exist as pairs of cis-trans isomers? Draw each cis- trans pair, and indicate the geometry of each isomer. • CH3CH=CH2 • (CH3)2C=CHCH3 • CH3CH2CH=CHCH3 • (CH3)2C=C(CH3)CH2CH3 • ClCH=CHCl • BrCH=CHCl

  44. Practice Problem: Name the following alkenes, including the cis or trans designation:

  45. 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 disubstituted double bonds C.Sequence Rules: The E, Z Designation

  46. Develop a System for Comparison of Priority of Substituents • Assume a valuation system • If Br has a higher “value” than Cl • If CH3 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”

  47. E,Z Stereochemical Nomenclature • Priority rules of Cahn, Ingold, and Prelog are used. • Compare where higher priority group is with respect to C=C bond and designate a prefix • E - entgegen:opposite sides • Z - zusammen: together on the same side

  48. E - entgegen Z - zusammen

  49. Cahn-Ingold-Prelog Rules • Ranking Priorities • Look at the atoms directly attached to each double-bond carbon • Rank them according to atomic number. Higher atomic number gets higher priority.

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