Tutorial 7 - Solutions

Tutorial 7 - Solutions - CHEMISTRY 1AA3 March 1-5, 2010...

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CHEMISTRY 1AA3 March 1-5, 2010 TUTORIAL PROBLEM SET 7 SOLUTIONS ________________________________________________________________________ Chemistry 1AA3 1. Indicate whether or not you would expect the following molecules to be aromatic and indicate the total number of π -electrons present in each compound. SOLUTION: (i) Aromatic, 6 π -electrons (ii) Not aromatic (not a cyclic compound), 6 π -electrons (iii) Not aromatic, 2 π -electrons (iv) Not aromatic (not alternating single and double bonds), 6 π -electrons (v) Not aromatic (non-cyclic), 4 π -electrons (vi) Not aromatic, 8 π -electrons (vii) Aromatic, 14 π -electrons (viii) Not aromatic (not alternating single and double bonds), 8 π -electrons 2. It is possible to create cyclic carbocations and carbanions that are aromatic. If an ionic species has alternating single and double bonds and fits Hückel's rule, it will likely be aromatic and therefore much more stable than non-aromatic ionic species. Cyclic systems containing alternating single and double bonds that do not follow the Hückel rule are typically very reactive (often called antiaromatic because they are less stable than their open chain counterparts). Indicate which of the following ions would be expected to be favourably formed (particularly if their precursors are not aromatic). SOLUTION:
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CHEMISTRY 1AA3 March 1-5, 2010 TUTORIAL PROBLEM SET 7 SOLUTIONS ________________________________________________________________________ Chemistry 1AA3 (i) Aromatic, 4n+2, n = 0 π -electrons (favourable). (ii) Aromatic, 6 π -electrons (the carbon bearing the anion is sp 2 hybridized so that the negative charge can be delocalized around the ring) (favourable) (iii) Not aromatic, 4 π -electrons (not favourable) (iv) Aromatic, 6 π -electrons (the carbon bearing the cation is sp 2 hybridized so that the charge can be delocalized around the ring) (favourable) 3. The Stinker: A researcher was interested in preparing a series of cyclic ketones. Two of the targets are cycloheptatrienone (I) and cyclopentadienone (II). Compound (I) was found to be much less reactive and therefore an easier synthetic target. Explain why? SOLUTION:
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Tutorial 7 - Solutions - CHEMISTRY 1AA3 March 1-5, 2010...

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