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4. Draw the structures of the two enantiomers of 2,3-dibromo-3-phenylpropanoic acid that were produced by the reaction. Assign the absolute configuration (R or S) at the chiral centers in each of the product structures.

4. Draw the structures of the two enantiomers of 2,3-dibromo-3-phenylpropanoic acid that were produced by the reaction. Assign the absolute configuration (R or S) at the chiral centers in each of the product structures.

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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4. Draw the structures of the two enantiomers of 2,3-dibromo-3-phenylpropanoic acid that were produced by the reaction. Assign the absolute configuration (R or S) at the chiral centers in each of the product structures.
Transcribed Image Text:4. Draw the structures of the two enantiomers of 2,3-dibromo-3-phenylpropanoic acid that were produced by the reaction. Assign the absolute configuration (R or S) at the chiral centers in each of the product structures.
Molecular Weight Milligrams Millimoles Concentration Volume Trans-cinnamic acid 148.1586 g/mol 159.808 (149x1000) = 149mg •128 x1000 = 149. 128 1.006 = 148. 1586 mmol 159.808 n/a Bromine in acetic acid g/mal n/a Equivalents 1.006.801.125 •801 Imey 2 1 = .801 128mg bodonge our /02.55 mg an DC. 1 M mmol om 10 dan M 2,3-dibromo-hydrocinnamic acid = .801 mL 307.97 g/mol 333 x 307.97= 801.4156)=333 •333 .801 n/a pla =.42
Transcribed Image Text:Molecular Weight Milligrams Millimoles Concentration Volume Trans-cinnamic acid 148.1586 g/mol 159.808 (149x1000) = 149mg •128 x1000 = 149. 128 1.006 = 148. 1586 mmol 159.808 n/a Bromine in acetic acid g/mal n/a Equivalents 1.006.801.125 •801 Imey 2 1 = .801 128mg bodonge our /02.55 mg an DC. 1 M mmol om 10 dan M 2,3-dibromo-hydrocinnamic acid = .801 mL 307.97 g/mol 333 x 307.97= 801.4156)=333 •333 .801 n/a pla =.42
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