lesson for CHM 213 interp

# lesson for CHM 213 interp -...

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Unformatted text preview: http://www.chem.hmc.edu/chem/nmr/page7.htm Problem 1 C 3 H 5 BrO 2 1 H (s) broad > 10 probably COOH MW = 152 1 double bond/ring H 3 C O OH Cl or C H 2 H 2 C O OH Cl H q d t t Problem 3 C 9 H 12 MW = 120 C 9 H 20 – C 9 H 12 = 4 rings/double bonds – think aromatic if protons show up at 7.2-7.8 Possible? With an aromatic system, you can determine the number of substitutions by counting the number of aromatic H’s = 5 so it is mono-substituted. What about the side chain? Subtract C 6 H 5 from the formula of the starting material: C 9 H 12 - C 6 H 5 = C 3 H 7 Two possible side chains: 1 carbon with two methyl groups branching off if it: 1 m @ 2.5 (1 H) , 1 large d @ 1.8 (6 H) In a linear system, the splitting pattern would be: 3 signals: 1 t @ 2.8 (2 H), 1 m @ 1.8 (2 H), 1 t @ 1.1 (3 H) What do you see? Problem 7 C 6 H 10 C 6 H 14- C 6 H 10 = H 4 /2 = 2 rings or double bonds With only three signals, this molecule is HIGHLY symmetrical. What does this mean? The peak at 5.7 is a single peak and the only electronegative What does this mean?...
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## This note was uploaded on 06/29/2011 for the course CHEM 2130 taught by Professor Silverman during the Spring '11 term at Missouri (Mizzou).

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lesson for CHM 213 interp -...

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