ORGANIC CHEMISTRY (LOOSELEAF)
ORGANIC CHEMISTRY (LOOSELEAF)
6th Edition
ISBN: 9781260475630
Author: SMITH
Publisher: MCG
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Chapter 5, Problem 68P
Interpretation Introduction

(a)

Interpretation: Enantiomeric excess of the given solution is to be calculated.

Concept introduction: Enantiomeric excess is the excess of major enantiomer over the minor enantiomer. Enantiomeric excess is equivalent to the optical purity. Enantiomeric excess is the fraction of observed rotation of the mixture to specific rotation of the pure enantiomer.

Interpretation Introduction

(b)

Interpretation: The percent of each enantiomer in the given solution is to be calculated.

Concept introduction: Enantiomeric excess is the excess of major enantiomer over the minor enantiomer. Enantiomeric excess is equivalent to the optical purity. Enantiomeric excess is the fraction of observed rotation of the mixture to specific rotation of the pure enantiomer.

Interpretation Introduction

(c)

Interpretation: The specific rotation [α] for the enantiomer of quinine is to be predicted.

Concept introduction: Enantiomers are stereoisomers, which are non-superimposable images of each other. They have identical physical and chemical properties in symmetric environment. They rotate the plane-polarized light in equal amounts and in opposite directions.

Interpretation Introduction

(d)

Interpretation: Enantiomeric excess of the given solution is to be calculated.

Concept introduction: Enantiomeric excess is the excess of major enantiomer over the minor enantiomer. Enantiomeric excess is equivalent to the optical purity. Enantiomeric excess is the fraction of observed rotation of the mixture to specific rotation of the pure enantiomer.

Interpretation Introduction

(e)

Interpretation: The [α] value for the solution is to be calculated.

Concept introduction: Enantiomeric excess is the excess of major enantiomer over the minor enantiomer. Enantiomeric excess is equivalent to the optical purity. Enantiomeric excess is the fraction of observed rotation of the mixture to specific rotation of the pure enantiomer.

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A J то گای ه +0 Also calculate the amount of starting materials chlorobenzaldehyde and p-chloroacetophenone required to prepare 400 mg of the given chalcone product 1, 3-bis(4-chlorophenyl)prop-2-en-1-one molar mass ok 1,3-bis(4-Chlorophenyl) prop-2-en-1-one = 277.1591m01 number of moles= 0.400/277.15 = 0.00144 moles 2 x 0.00 144=0.00288 moves arams of acetophenone = 0.00144 X 120.16 = 0.1739 0.1739x2=0.3469 grams of benzaldehyde = 0.00144X106.12=0.1539 0.1539x2 = 0.3069 Starting materials: 0.3469 Ox acetophenone, 0.3069 of benzaldehyde 3
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