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Organic Chemistry c3444y Problem Set 2 - Answer Key 1. All bonding orbitals filled and no electrons in non-bonding or antibonding orbitals. Therefore: Aromatic! More accurate Non-bonding level representation. 2. Indicate which is(are) the most acidic proton(s) of indene: H H B Aromatic If you were to deprotonate indene at this site would there be any special stability associated with the anion you have left behind? Explain. 3. Recall our in class analysis of the structure of naphthalene and anthracene. Use resonance theory to explain the greater resonance energy of phenanthrene relative to anthracene. In other words why is phenanthrene "more aromatic" than anthracene? Resonance Energy Benzene = 36 kcal/mol Naphthalene = 61 kcal/mol = 84 kcal/mol Anthracene = 92 kcal/mol Phenanthrene Anthracene Phenanthrene The rings on the ends are aromatic in 4/5 or 80% of the resonance structures. The ring in the middle is aromatic in 2/5 or 40% of the resonance structures. Overall, this is more "aromaticity per ring" than anthracene where we concluded that each ring is aromatic in 50% of its resonance structures. 4. a. H2, Catalyst In doing this reduction, we've gone from a phenanthrene to two separate benzene rings, and we've lost 92 - (36 x 2) = 20 kcal/mol of resonance energy. Reduction of any other of the double bonds of phenanthrene would lead to a much greater loss of resonance energy. Try it! b. In class we saw that anthracene adds Br2 in a 1,4 rather than a 1,2 sense: Br Br2 Br Br Br Br Because of aromaticity, this resonance structure is clearly the most important contributor. Br Br 5. a. CH Most acidic site. Even without the ring these protons are allylic. All other 3 protons are alkenyl - very difficult to deprotonate. B H b. CH2 CH2 Like any allyl anion, we can draw these resonance structures. In this case the resonance structure on the right is aromatic: 6. For the following pair of reactions, which one should be faster? Why? ii is faster. In i, an antiaromatic cation is produced. In ii, a stable doubly allylic cation is produced. CH3 CH3 7. CH3 O O CH3 CH3 CH3 CH3 CH3 O O Very strong contributor, because both rings are aromatic.Therefore, much single bond character in the middle, and rotation is facile. 8. H 10 π electrons - Aromatic!.

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