arenes and aromaticity benzene n.
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Arenes and Aromaticity ( Benzene ). All C—C bond distances = 140 pm. 140 pm. 140 pm. 140 pm. 140 pm. 140 pm. 140 pm. Benzene empirical formula = CH. All C—C bond distances = 140 pm. 146 pm. 140 pm. 140 pm.

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Arenes and Aromaticity ( Benzene )


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    1. Arenes and Aromaticity(Benzene)

    2. All C—C bond distances = 140 pm 140 pm 140 pm 140 pm 140 pm 140 pm 140 pm Benzene empirical formula = CH

    3. All C—C bond distances = 140 pm 146 pm 140 pm 140 pm 140 pm is the average between the C—C single bond distance and the double bond distance in 1,3-butadiene. 140 pm 140 pm 134 pm 140 pm 140 pm

    4. Resonance & Benzene

    5. Thermochemical Measures of Stability heat of hydrogenation: compare experimentalvalue with "expected" value for hypothetical"cyclohexatriene" Pt + 3H2 H°= – 208 kJ

    6. Cyclic conjugation versus noncyclic conjugation 3H2 Pt heat of hydrogenation = -208 kJ/mol (-49 kcal/mol) 3H2 Pt heat of hydrogenation = -337 kJ/mol (-85.8 kcal/mol)

    7. H H H H H H H H H H H H Resonance & Benzene express the structure of benzene as a resonancehybrid of the two Lewis structures. Electrons arenot localized in alternating single and double bonds,but are delocalized over all six ring carbons.

    8. Resonance & Benzene Circle-in-a-ring notation stands for resonance description of benzene (hybrid of two resonance structures)

    9. Question • Which of the following compounds has a double bond that is conjugated with the p system of the benzene ring? • A) p-Benzyltoluene • B) 2-Phenyl-1-decene • C) 3-Phenylcyclohexene • D) 3-Phenyl-1,4-pentadiene • E) 2,4,6-trichloroanisole

    10. Question • Which of the following has the smallest heat of combustion? • A) B) • C) D)

    11. The  Molecular Orbitalsof Benzene

    12. Orbital Hybridization Model of Bonding in Benzene High electron density above and below plane of ring

    13. Hückel's Rule Frost's circle is a mnemonic that allows us to draw a diagram showing the relative energies ofthe p orbitals of a cyclic conjugated system. 1) Draw a circle. 2) Inscribe a regular polygon inside the circle so that one of its corners is at the bottom. 3) Every point where a corner of the polygon touches the circle corresponds to a p electron energy level. 4) The middle of the circle separates bonding and antibonding orbitals.

    14. Antibonding Bonding p MOs of Benzene Frost's Circle

    15. Energy Benzene MOs Antibondingorbitals 6 p AOs combine to give 6  MOs 3 MOs are bonding; 3 are antibonding Bondingorbitals

    16. Energy Benzene MOs Antibondingorbitals All bonding MOs are filled No electrons in antibonding orbitals Bondingorbitals

    17. Benzene MOs

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

    19. Aromatic vs. “anti-aromatic”A. Deniz, et. al., Science, 5 November 1999

    20. Question • How many p electrons does the compound shown have? • A) 5 • B) 8 • C) 10 • D) 12

    21. Hückel's Rule Among planar, monocyclic, completely conjugated polyenes, only those with 4N + 2 p electrons have resonance stability (i.e. They are aromatic; and they are also planar.) N 4N+2 0 2 1 6 2 10 3 14 4 18

    22. Hückel's Rule Among planar, monocyclic, completely conjugated polyenes, only those with 4N + 2 p electrons have resonance stability (i.e. They are aromatic; and they are also planar.) N 4N+2 0 2 1 6 benzene! 2 10 3 14 4 18

    23. Question • What is the value of N of Huckel's rule for the cyclopentadienyl cation (shown)? • A) N=1/2 • B) N=1/4 • C) N=1 • D) N=2

    24. Question I II • Which is a true statement based on Huckel’s Rule. • (Assume that both are planar and that vacant p-orbitals do not interupt conjugation.) • A) I is aromatic and II is not. • B) II is aromatic and I is not. • C) I and II are aromatic. • D) I and II are not aromatic.

    25. Hückel's Rule&molecular orbitals Hückel”s rule applies to: cyclic, planar,conjugated, polyenes the p molecular orbitals ofthese compounds have a distinctive pattern one p orbital is lowest in energy, another is highest in energy, and the others are arranged in pairs between the highestand the lowest

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

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

    28. Question • How many isomers of dibromophenol are aromatic? • A) 3 • B) 4 • C) 6 • D) 8 • E) none

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

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

    31. 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 Non-bonding Cyclo-octatetraene Bonding

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

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

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

    35. Question • The compound shown is classified as • A) non-aromatic • B) aromatic • C) anti-aromatic • D) none of the above

    36. [10]Annulene predicted to be aromatic by Hückel's rule,but too much angle strain when planar and all double bonds are cis (therefore non-planar) 10-sided regular polygon has angles of 144°

    37. [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

    38. [10]Annulene van der Waalsstrain betweenthese two hydrogens incorporating two trans double bonds intothe ring relieves angle strain but also introducesvan der Waals strain into the structure andcauses the ring to be non-planar

    39. [14]Annulene 14 p electrons satisfies Hückel's rule van der Waals strain between hydrogens insidethe ring & thererfore non-planar H H H H

    40. [16]Annulene 16 p electrons does not satisfy Hückel's rule alternating short (134 pm) and long (146 pm) bonds not aromatic

    41. H H H H H H [18]Annulene 18 p electrons satisfies Hückel's rule resonance energy = - 418 kJ/mol