Use the data table to answer the question. 4. Draw the structures of the two enantiomers of 2,3-dibromo-3-phenylpropanoic acid thatwere produced by the reaction. Assign the absolute configuration (R or S) at the chiralcenters in each of the product structures. MolecularWeightMilligramsMillimolesConcentrationVolumeTrans-cinnamic acid148.1586 g/mol159.808(149x1000) = 149mg •128 x1000 =149.1281.006=148. 1586 mmol 159.808n/aBromine in acetic acidg/maln/aEquivalents 1.006.801.125•801Imey 21= .801128mg bodonge our /02.55 mganDC.1 Mmmolom 10 dan M2,3-dibromo-hydrocinnamic acid= .801 mL307.97 g/mol333 x 307.97=801.4156)=333•333.801n/apla=.42

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.

Organic Chemistry

8th Edition

ISBN:9781305580350

Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote

Publisher:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote

Chapter23: Amines

Section: Chapter Questions

Problem 23.22P

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Concept explainers

Stereochemistry and Isomerism

Stereochemistry deals with the three-dimensional structure and properties of molecules. It explains the arrangement of atoms and their manipulation within a molecule in a 3D structure. Stereochemistry can be dealt with all types of molecules. Based on the stereochemistry of a molecule, the physical and chemical properties of the molecules vary.

Question

Use the data table to answer the question.

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

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