Organic Chemistry II - Midterm 2007

A 4 points planar molecule 1 pt all atoms are sp 2

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a. (4 points) - planar molecule 1 pt - all atoms are sp 2 hybridized 1 pt - 10 π electrons = 4n+2 1 pt (with n = 2) n = 2 YES (1 pt) , the molecule is aromatic because the 3 above criteria are satisfied . b. (5 points) O N sp 2 hybridized; lone pair is in an sp 2 orbital and not in a p orbital (the p orbital is already being used to form the π bond) lone pair is in a p orbital (resonates with the π bonds) 2 nd lone pair is in an sp 2 orbital; at 90° to the p orbitals Oxygen is sp 2 hybridized - The molecule is planar - All atoms are sp 2 hybridized - There are 6 π electrons (4 from the double bonds, 2 from the red lone pair in the p orbital) 6 = 4n + 2 n = 1 All 3 criteria are satisfied; the molecule is aromatic. 1 1 explanation of lone pairs on O 1 pt each + conclusion 8. Clearly show why the following proposal would not work. (3 points) 1 OH C C Li CH 3 C C CH 3 1. H + 2. O H H C C CH 3 × 1 Idea that deprotonation occurs rather than S N 2
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CHM 2120 midterm - ANSWERS 6 of 9 9. Circle the most acidic molecule below. Clearly justify your answer and be sure to show relevant structures. (8 points) OH O 2 N OH O 2 N vs O N O N O O O O O N O O O N O O H The conjugate base is stabilized by resonance with the aromatic ring AND with the nitro group The conjugate base is stabilized by resonance with the aromatic ring ONLY. Weaker conjugate base Stronger conjugate base Stronger acid Weaker acid 1 Draw each conjugate base 2 1 1 1 1 Clear resonance structure that shows that nitro can't stabilize negative charge OR words 1 comparison between both conjugate bases
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