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11.18 Cyclobutadiene and Cyclooctatetraene

11.18 Cyclobutadiene and Cyclooctatetraene. Requirements for Aromaticity. cyclic conjugation is necessary, but not sufficient. not aromatic. not aromatic. aromatic. Heats of Hydrogenation. to give cyclohexane (kJ/mol).

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11.18 Cyclobutadiene and Cyclooctatetraene

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  1. 11.18Cyclobutadiene and Cyclooctatetraene

  2. Requirements for Aromaticity cyclic conjugation is necessary, but not sufficient notaromatic notaromatic aromatic

  3. Heats of Hydrogenation to give cyclohexane (kJ/mol) heat of hydrogenation of benzene is 152 kJ/mol less than 3 times heat of hydrogenation of cyclohexene 120 231 208

  4. Heats of Hydrogenation to give cyclooctane (kJ/mol) heat of hydrogenation of cyclooctatetraene is more than 4 times heat of hydrogenation of cyclooctene 97 205 303 410

  5. C(CH3)3 (CH3)3C 138 pm 151 pm (CH3)3C CO2CH3 Structure of Cyclobutadiene structure of a stabilized derivative is characterizedby alternating short bonds and long bonds

  6. Structure of Cyclooctatetraene cyclooctatetraene is not planar has alternating long (146 pm)and short (133 pm) bonds

  7. Conclusion there must be some factor in additionto cyclic conjugation that determines whether a molecule is aromatic or not

  8. 11.19Hückel's Rule:Annulenes the additional factor that influences aromaticity is the number of p electrons

  9. Hückel's Rule among planar, monocyclic, completely conjugated polyenes, only those with 4n + 2 p electrons possess special stability (are aromatic) n 4n+2 0 2 1 6 2 10 3 14 4 18

  10. Hückel's Rule among planar, monocyclic, completely conjugated polyenes, only those with 4n + 2 p electrons possess special stability (are aromatic) n 4n+2 0 2 1 6 benzene! 2 10 3 14 4 18

  11. Hückel's Rule Hückel restricted his analysis to planar,completely conjugated, monocyclic polyenes he found that the p molecular orbitals ofthese compounds had a distinctive pattern one p orbital was lowest in energy, another was highest in energy, and the others were arranged in pairs between the highestand the lowest

  12. p-MOs of Benzene Antibonding 6 p orbitals give 6 p orbitals 3 orbitals are bonding; 3 are antibonding Benzene Bonding

  13. p-MOs of Benzene Antibonding 6 p electrons fill all of the bonding orbitals all p antibonding orbitals are empty Benzene Bonding

  14. p-MOs of Cyclobutadiene(square planar) Antibonding 4 p orbitals give 4p orbitals 1 orbital is bonding, one is antibonding, and 2 are nonbonding Cyclo-butadiene Bonding

  15. p-MOs of Cyclobutadiene(square planar) Antibonding 4 p electrons; bonding orbital is filled; other 2p electrons singly occupy two nonbonding orbitals Cyclo-butadiene Bonding

  16. p-MOs of Cyclooctatetraene(square planar) Antibonding 8 p orbitals give 8 p orbitals 3 orbitals are bonding, 3 are antibonding, and 2 are nonbonding Cyclo-octatetraene Bonding

  17. p-MOs of Cyclooctatetraene(square planar) Antibonding 8 p electrons; 3 bonding orbitals are filled; 2nonbonding orbitals are each half-filled Cyclo-octatetraene Bonding

  18. p-Electron Requirement for Aromaticity 6 p electrons 4 p electrons 8 p electrons notaromatic notaromatic aromatic

  19. H H Completely Conjugated Polyenes 6 p electrons;not completelyconjugated 6 p electrons;completely conjugated notaromatic aromatic

  20. Annulenes Annulenes are planar, monocyclic, completely conjugated polyenes. That is, they are the kind of hydrocarbons treated by Hückel's rule.

  21. [10]Annulene predicted to be aromatic by Hückel's rule,but too much angle strain when planar and all double bonds are cis 10-sided regular polygon has angles of 144°

  22. [10]Annulene incorporating two trans double bonds intothe ring relieves angle strain but introducesvan der Waals strain into the structure andcauses the ring to be distorted from planarity

  23. [10]Annulene van der Waalsstrain betweenthese two hydrogens incorporating two trans double bonds intothe ring relieves angle strain but introducesvan der Waals strain into the structure andcauses the ring to be distorted from planarity

  24. [14]Annulene 14 p electrons satisfies Hückel's rule van der Waals strain between hydrogens insidethe ring H H H H

  25. [16]Annulene 16 p electrons does not satisfy Hückel's rule alternating short (134 pm) and long (146 pm) bonds not aromatic

  26. H H H H H H [18]Annulene 18 p electrons satisfies Hückel's rule resonance energy = 418 kJ/mol bond distances range between 137-143 pm

  27. 11.20Aromatic Ions

  28. H H H H + H H H Cycloheptatrienyl Cation 6 p electrons delocalizedover 7 carbons positive charge dispersedover 7 carbons very stable carbocation also called tropylium cation

  29. H H H H H H H H + + H H H H H H Cycloheptatrienyl Cation

  30. + H Br Cycloheptatrienyl Cation Tropylium cation is so stable that tropyliumbromide is ionic rather than covalent. mp 203 °C; soluble in water; insoluble indiethyl ether Br– Ionic Covalent

  31. H H •• H H – H Cyclopentadienide Anion 6 p electrons delocalizedover 5 carbons negative charge dispersedover 5 carbons stabilized anion

  32. H H H H – •• H H H H – H H Cyclopentadienide Anion

  33. H H H H + H+ •• H H H H – H H H Acidity of Cyclopentadiene Cyclopentadiene is unusually acidic for a hydrocarbon. Increased acidity is due to stability of cyclopentadienide anion. pKa = 16 Ka = 10-16

  34. H H H H H Electron Delocalization in Cyclopentadienide Anion •• –

  35. H H H H – •• H H H H H H Electron Delocalization in Cyclopentadienide Anion •• –

  36. H H H H – •• H H H H H H H H • • H H H Electron Delocalization in Cyclopentadienide Anion •• – –

  37. H H H H – •• H H H H H H H H H H • – • • • H H H H H H Electron Delocalization in Cyclopentadienide Anion •• – –

  38. H H H H – •• H H H H H H H H H H • – • • • H H H H H H Electron Delocalization in Cyclopentadienide Anion •• – H H •• – H H H

  39. H H H H H H H H H H H H H H H Compare Acidities ofCyclopentadiene and Cycloheptatriene pKa = 16 Ka = 10-16 pKa = 36 Ka = 10-36

  40. H H H H H H •• H H – •• H H H H Compare Acidities ofCyclopentadiene and Cycloheptatriene – Aromatic anion 6 p electrons Anion not aromatic8 p electrons

  41. H H H H + + H H Cyclopropenyl Cation n = 0 4n +2 = 2 p electrons also written as

  42. Cyclooctatetraene Dianion H H H H n = 2 4n +2 = 10p electrons – H H H H •• alsowritten as 2– •• H H H H – H H H H

  43. 11.21Heterocyclic Aromatic Compounds

  44. •• •• •• O S N N •• •• •• H Examples Pyridine Pyrrole Furan Thiophene

  45. • N • N •• Examples Quinoline Isoquinoline

  46. 11.22Heterocyclic Aromatic CompoundsandHückel's Rule

  47. N •• Pyridine 6 p electrons in ring lone pair on nitrogen is in ansp2 hybridized orbital;not part of p system of ring

  48. •• N H Pyrrole lone pair on nitrogen must be part of ring p system if ring is to have6 p electrons lone pair must be in a p orbitalin order to overlap with ring psystem

  49. •• O •• Furan two lone pairs on oxygen one pair is in a p orbital and is partof ring p system; other is in an sp2 hybridized orbital and is notpart of ring p system

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