PS 6 Answers, Spring 2003, Prof

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Chem 332, PS 6 answers, Spring 2003, Prof. Fox 1. Which of the following are aromatic. Draw "Frost Circles" that support your answers 2– 2+ 2+ + all bonding MO's although all electrons all bonding MO's this is an interesting case: this is are in bonding orbitals, are filled are filled a 4n+2 molecule, and it is flat-- two bonding MO's are aromatic aromatic characteristic of aromatic only half-filled molecules. However, it has four anti-aromatic electrons in non-bonding orbitals, so we might have considered this to be non-aromatic. So it comes down to how you define aromaticity. For now we will call it aromatic with a question mark. 2. Which of the following are aromatic. Use the Huckel rule to support your answer 2– P H B 14 electrons, aromatic H 6 electrons 10 electrons 2 6 electrons sp3 center (filled sp on aromatic (empty sp2 on not conjugated phosphorus) boron) and not aromatic aromatic aromatic 3. Compound 1 has an unusually large dipole moment for an organic hydrocarbon. Explain why (consider resonance structures for the central double bond) We need to remember that for any double bond compound, we can draw two polar resonance structure. Normally, these resonance structures are unimportant unimportant unimportant However, for 1 the story is different because there is a polar resonance form that has two aromatic components. In other words, 'breaking' the double bond gives rise to aromaticity, and therefore the polar resonance structure is important and leads to the observed dipole. dipole 1 8 e– 4 e– 6 e– 6 e– anti- anti- aromatic aromatic aromatic aromatic important resonance unimportant resonance structure structure 5. Provide a mechanism. This problem is easier than it looks! HCl = 13C H+ alkyl 1,2 shift back the shift other alkyl H H Cl– 4. Provide mechanisms for the formation of 2 and 3. O OH O AlCl 3 CO2H H3PO4 + O 2 O 3 OH O O Cl3Al O O OAlCl2 OAlCl2 O H Cl– + HCl Cl3Al O O O H2O O O O H+ CO H O 2 H3PO4 O OH2 + O O OH acid catalysed enolization H O O OH 3 – H2PO4 6. Devise syntheses of the following, using benzene and any other materials containing 3 carbons or less. F O NO2 Br I OH O NO2 NH2 I HNO 3 Zn 1) HONO, HCl H SO HCl 2 4 2) KI SO3 H2SO4 I I NO2 NO2 H2O HNO3 H2SO4 cat H2SO4 I SO3H SO3H NH2 AcCl, Et3N see above NHAc O 1) NaCN O Br Zn/Hg 2) H O+ 3 Cl HCl 3) SOCl2 AlCl3 NHAc NHAc NaOH Br 1) HONO, HCl Br2 2) HBF4 FeBr3 F F NH2 O O O AlCl3 Cl HNO3 H2SO4 AlCl NO2 Br 3 H2/Pd/C O O 1) HONO, HCl Zn/Hg HCl OH OH 2) H2O NH2 1) O Cl Cl O O 1) Br2, FeBr3 O AlCl3 O 2) NaOEt, CuBr (cat) 2) AlCl3.

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