Chapt3&4 problems - 6.95 OCH 3 CO 2 H Br H H H 7.27 +...

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NMR Problems 3.6, 4.17 and 4.20
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3.6 (a)
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3.6 (a) (IHD = 2) COOH 2 x alkene CHs -CH3 COOH Alkene CHs
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3.6 (a) H 3 C H H CO 2 H
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3.6(b)
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3.6(b) (IHD = 0) 2nd order pattern 1H 2H 2H 3H No symmetry
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3.6(b) H 3 C Cl Cl
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3.6 (d)
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3.6 (d) (IHD = 1) carbonyl 1H 2H 2H 3H
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3.6 (d) H O O CH 3
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3.6 (e)
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3.6 (e) IHD = 5 OH X Y X CH 2 CH 2 Y O CH 3 C=O X Y
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3.6 (e) HO CH 2 CH 2 O O CH 3
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3.6 (f)
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3.6 (f) IHD = 5 1H each -O-CH3 -COOH Six unique carbons -COOH -O-CH3
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3.6 (f) CO 2 H Br H 3 CO H H H 7.27 + 0.8-0.43-0.13 = 7.51 7.27 + 0.22+0.14-0.09 = 7.54 7.27 - 0.43+0.20-0.13 = 6.91 Br CO 2 H H 3 CO H H H 7.27 + 0.22-0.43+0.14 = 7.2 7.27 + 0.80+0.14-0.13 = 8.08 7.27 - 0.43+0.14-0.03 =
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Unformatted text preview: 6.95 OCH 3 CO 2 H Br H H H 7.27 + 0.22-0.43+0.14 = 7.2 7.27 + 0.80+0.14-0.13 = 8.08 7.27 + 0.22 -0.37 + 0.14 = 7.26 7.6ppm 7.38ppm 7.04ppm 3.6(g) 3.6(g) 3.6(g) N H 3 C CHO 4.17 4.17 (a) OH Cl Cl 4.17(b) 4.17(b) OH Cl Cl OH Cl Cl 4.17(c) 4.17(c) AX2 pattern OH Cl Cl 4.17(d) 4.17(d) OH Cl Cl 4.20 4.20 H5 Two large Js J 56 > J 45 J 56 = 10.0Hz J 45 = 7.0 Hz 4.20 H4 H6 4.20 H1 (more shielded) H2 4.20 H1 H2 H5 H4 H6 O O O O less shielded than H2...
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This note was uploaded on 01/18/2012 for the course CHEM 460AB taught by Professor Sasaki during the Fall '11 term at University of Washington.

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Chapt3&4 problems - 6.95 OCH 3 CO 2 H Br H H H 7.27 +...

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