ORGANIC CHEMISTRY:
`
`STRUCTURE, MECHANISM, AND SYNTHESIS
`
`ROBERT J. OUELLETTE
`Professor Emeritus, Department
`of Chemistry, The Ohio State University
`
`AND
`
`J. DAVID RAWN
`
`Professor Emeritus,
`Towson University
`
`
`
`kg
`
`
`I AMSTER
`PARIS -
`
`- BOSTON - HEIDELBERG - LONDON - N
`DIEGO - SAN FRANCISCO - SINGAPORE -
`
`YORK - OXFORD
`DNEY - TOKYO
`
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`

`
`Elsevier
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`
`Copyright © 2014 Elsevier Inc. All rights reserved.
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`Notice
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`No responsibility is assumed by the publisher for any injury and/or damage to persons or
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`of any methods, products, instructions or ideas contained in the material herein. Because of rapid
`advances in the medical sciences, in particular, independent verification of diagnoses and drug
`dosages should be made.
`
`Library of Congress Cataloging-in-Publication Data
`Rawn, J. David, 1944-
`Organic chemistry / J. David Rawn, Robert J. Ouellette. — First edition.
`pages cm
`Includes index.
`ISBN 978—0—12—800780—8
`
`1. Chemistry, Organic—Textbool<s. I. Ouellette, Robert J., 1938- II. Title.
`QD251.3.R39 2014
`547—dc23
`
`British Library Cataloguing in Publication Data
`A catalogue record for this book is available from the British Library
`
`For information on all Elsevier publications
`visit our Web site at store.elsevier.com
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`Printed and bound in China
`1415161718 10987654321
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`ISBN: 978—0—12—800780—8
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`6% Working together
`W , ; § to grow libraries in
`“L‘”"““
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`E§Z31l§f}.§§i1 developing countries Www.elsevier.co1n o WWw.bookaid.org
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`

`
`STEREOCHEMISTRY
`
`
`
`M. C. ESCHER, DRAWING HANDS, 1948
`
`8_’|
`S-I-EREOISO M E R5
`
`Molecules that have different arrangements in three—dimensional space are stereoisomers. Stereoiso—
`mers have different configurations. In Chapter 4, we considered the structures of geometric isomers
`—E,Zisomers—an important class of stereoisomers that have different configurations. Another type
`of stereoisomerism, which is based on mirror image relationships between molecules, is the subject of
`this chapter. The mirror image relationships of stereoisomers are not as easily visualized as the relation
`between geometric isomers, and a bit of practice is likely to be needed before we can “see” their relation-
`ship in three—dimensional space. This is a case where molecular models are very handy.
`Changes in molecular configuration that occur in a reaction provide a valuable tool for
`probing many reaction mechanisms. Stereochemistry can also play an important role in organic syn-
`thesis since it is not an easy task to synthesize only a “right—handed” or “left—handed” molecule when
`both could potentially be formed in a chemical reaction. The chemical synthesis of molecules with
`precisely the right three—dimensional structure is often a huge experimental challenge; in practical
`terms, “chiral synthesis” is an essential component of virtually all drug synthesis.
`A molecule’s configuration also plays a major role in its biological function. We will see many
`examples of stereoisomerism in biological systems in this chapter and beyond.
`
`We are all familiar with mirror image objects. Every object has a mirror image, but this refiected image
`8_2
`need not be identical to the actual object. Thus, when we look into a mirror, we see someone who does
`E 0 5'
`R IM
`M I R
`M I RR0 R IMAG E M 0 LECU LES not actually exist, namely, our mirror image.
`AN D C H I RALITY
`'
`A simple wooden chair looks exactly like its mirror image (Figure 8.1a). Similarly, the mirror
`images ofa teacup or a hammer are identical to the objects themselves. When an object and its mirror image
`are identical, they are superimposaéle. Superimposable objects can be “placed” on each other so that
`each feature of one object precisely coincides in space with an equivalent feature in the mirror image
`object.
`
`Some objects cannot be superimposed upon their mirror images: They are nomuperimposable.
`One example is the sidearm chair shown in Figure 8.1b. When a chair with a “right—handed arm” is re-
`flected in a mirror, it becomes a chair with a “left—handed arm” (Figure 8.1b). We can convince ourselves
`of this by imagining sitting in the chair or its mirror image. Or, we could stop by a classroom, which
`usually has chairs for both right— and left—handed persons, and do the experiment.
`
`Organic Chemistry. http://dx.doi.org/1 0. 1 01 6/B978—0— 1 2—800780—8.00008—5
`Copyright © 2014 Elsevier Inc. All rights reserved.
`
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`
`Figure 8.1 Objects and
`Their Mirror Images
`In (a), the chair and its mirror
`
`image are identical. They can be
`superimposed. In (b), the mirror
`image, side—arm chairs cannot be
`superimposed. One chair has a
`“right—handed” arm, and the other
`has a “left—handed” arm. (These
`
`particular chairs were designed by
`the renowned woodworker George
`Nakashima.)
`
`L
`
`iilililiiir
`
`‘ipil
`
`ll“! _‘
`
`j.
`
`iiiii
`
`I
`
`Our hands are related as nonsuperimposable mirror images. We know that we cannot superimpose
`our hands, as is deftly shown by the M. C. Escher lithograph found at the beginning of this chapter and
`in Figure 8.2. An object that is not superimposable on its mirror image is chiral (Greek ebimn, hand).
`Objects such as gloves and shoes also have a “handedness,” and they are also chiral. An object that can
`be superimposed on its mirror image is achiral.
`We can determine whether or not an object is chiral without trying to superimpose it on its
`mirror image. If an object has a plane ofgmmetry, it is not chiral. A plane of symmetry bisects an object
`so that one half is the mirror image of the other half. For example, a cup has a plane of symmetry that
`divides it so that one half is the mirror image of the other half The chair in part (a) of Figure 8.3 is
`achiral because it has a plane of symmetry. 77Jepre.ve7zee 07 absence ofLl plane ofsymmetry tells us whether
`an object is third! or at/Jiml.
`
`F i g u re 8. 2
`Nonsuperimosable Mirror
`I m a g es
`A left and a right hand are
`nonsuperimposable mirror images.
`(M.C. Escher’s “Drawing Hands”
`© 2014 The M.C. Escher
`
`Company—The Netherlands. All
`rights reserved. wvvw.mcescher.com)
`
`
`
`Chiral Molecules
`
`We can extend the concept of chirality from macroscopic objects to molecules. A molecule is third!
`zfit contain: at least one carbon atom attae/Jed tofmr difierent atams or gmups. Such a carbon atom is
`a stereogenic center. A stereogenic center is sometimes called a chiral center, and the carbon atom
`is sometimes called a chiral carbon atom, although it is the molecule that is chiral, not a single
`carbon atom within it. Most molecules produced in living organisms are chiral, nearly all drugs
`are chiral, and the synthesis of chiral molecules in the laboratory is a significant part of organic
`synthesis.
`
`The four atoms or groups at a stereogenic center can be arranged in two ways to
`give two stereoisomers. The stereoisomers of bromochlorofluoromethane provide an example.
`Bromochlorofluoromethane does not have a plane of symmetry. Figure 8.3 shows that it can exist as a
`pair of nonsuperimposable mirror image isomers. Therefore, bromochlorofluoromethane is chiral.
`
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`
`F i g u re 8.3
`Nonsuperimosable Mirror
`Image Molecules
`Bromochlorofluoromethane does not
`
`have a plane of symmetry. Therefore,
`it is chiral, and it exists as a pair of
`nonsuperimposable mirror image
`isomers. (a) Schematic diagram;
`(b) Ball—and—stick molecular models.
`
`Figure 8.4 Planes of Symmetry
`in Dichloromethane
`
`Dichloromethane, which has not one,
`
`but two planes of symmetry, can be
`superimposed on its mirror image. It is
`achiral.
`
`Figure 8.5 Plane of Symmetry
`in Bromochloromethane
`
`Bromochloromethane has a plane
`of symmetry, and therefore, it can be
`superimposed on its mirror image. It is
`achiral.
`
`
`
`
`
`mirror
`
`(b)
`
`afi:J
`
`L:
`
`%
`
`plane of symmetry
`
`Mirror Image Isomers
`
`Two stereoisomers related as nonsuperimposable mirror images are called enantiomers (Greek mantios,
`opposite + meros, part). We can tell that a substance is chiral and predict that two enantiomers exist by
`identifying the substituents on each carbon atom. A carbon atom with four different substituents is a
`stereogenic center, and a molecule with a stereogenic center is chiral. It can exist as either of a pair of
`enantiomers. For example, 2—bromobutane is chiral because C—2 is attached to four different groups
`(CH3—, CH3CH2—, Br—, and H—). In contrast, no carbon in 2—bromopropane is bonded to four dif-
`ferent groups; C—2 is bonded to two methyl groups. Thus, 2—bromopropane is not chiral (Figures 8.4
`and 8.5).
`
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`
`stereogenic center \ l|3r
`
`I’10t 3. St€I‘€Og€l’1lC
`
`C€I’1t€I‘
`
`Br/
`
`H
`
`H
`
`2—bromobutane
`(a chiral molecule)
`
`2—bromopropane
`(an achiral molecule)
`
`The existence of a stereogenic center in a complex molecule may not be immediately apparent. This
`situation occurs when the groups bonded to a chiral carbon atom differ at sites not immediately ad-
`jacent to the stereogenic center. The difference between a methyl group and an ethyl group is readily
`apparent in 2—bromobutane. However, in some molecules, the difference is less obvious. For example,
`5—bromodecane and 5—bromo—1—nonene both have a stereogenic center.
`
`pentyl group
`
`butyl group
`A
`
`l|3r
`
`H
`
`5—bromodecane
`
`butenyl group
`butyl group
`A
`l|3r
`A
`CH3: CH2:CH2:CH2:CjCH2:CH2:CH:CH2
`
`H
`
`5—bromo—1—nonene
`
`Problem 8.1
`
`Which of the following molecules are chiral? Explain your answer.
`
`OH
`
`(3)
`
`(b)
`
`(0)
`
`CH3—CH2-CH2—CH2—d—CH2—CH2—CH2—CH2—CH3
`$14.
`
`Br
`
`CH2=cH—CH2—CH2—d—CH2—CH2—CEcH
`IL
`
`OH
`
`CH3
`
`CH3—CH2—CH2—CH2—d—CH2—CH2—CH2—dH—CH3
`54.
`
`Problem 8.2
`The structure of nicotine is shown below. Is nicotine chiral?
`
`244
`
`Ii
`CH3
`
`\
`/
`
`N
`
`nicotine
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`

`
`Pro blem 8.3
`
`Phenytoin has anticonvulsant activity. Is phenytoin chiral or achiral? Determine your answer by identi-
`fying the number of different groups bonded to its tetrahedral carbon atoms; then determine whether
`or not it has a plane of symmetry.
`
`
`
`phenytoin
`
`Sample Solution
`Phenytoin does not contain a carbon bonded to four different groups. It has a plane of symmetry that
`lies in the plane of the page. One of the benzene rings of phenytoin is above and the other below the
`symmetry plane. It is achiral.
`
`Properties of Enantiomers
`
`We can regard hands as analogous to the enantiomers of a chiral molecule. Let’s consider the interaction
`of hands with a symmetrical object such as a pair of tweezers. The tweezers are symmetrical. They can
`be used equally well with either hand because there is no preferred way to pick up or manipulate a pair
`of tweezers. However, even if blindfolded, we could easily use our hands to distinguish right— and left-
`handed gloves. Our hands are “a chiral environment,” and in this environment, mirror image gloves do
`not interact with hands in the same way. The right glove will fit only the right hand. V72 can distinguish
`chiral objects only because we are c/Jimi.
`Pairs of enantiomers have the same physical and chemical properties: they have that same
`heats of formation, density, melting point, and boiling point. They also have the same chemical
`properties, and undergo the same reactions in an achiral environment. However, enantiomers can be
`distinguished in a chiral environment. This difference is important in many processes in living cells.
`Only one of a pair of enantiomers fits into a specific site in a biological molecule such as an enzyme
`catalyst because the site on the enzyme that binds the enantiomer is chiral. The binding of this enan-
`tiomer is stereospecific.
`An exainple of a stereospecific process is the conversion of the drug levodopa to dopamine,
`a neurotransmitter in the brain. Levodopa (or L—dopa), the precursor of dopamine, is administered to
`treat Parkinson’s disease. Levodopa has one chiral carbon atom. Therefore, it exists as either of two en-
`antiomers. Only the enantiomer with the configuration shown below is transformed into dopamine.
`
`OH
`
`HO
`
`OH
`
`HO
`
`T»
`
`+ CO;
`
`(EH2
`H““"Ck\CO2H
`
`NH2
`
`CH2
`(:3H2
`
`NH;
`
`levodopa
`
`dopamine
`
`The reaction occurs because a stereospecific decarboxylase catalyzes the loss of a carboxyl group by for-
`mation of carbon dioxide (decarboxylation). This enzyme has a chiral binding site for levodopa, but it
`does not bind the enantiomer of levodopa.
`
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`

`
`83
`0 P.“ CAL ACTIVITY
`
`Although enantiomers have identical chemical properties in achiral environments, they differ in one
`important physical property: Enantiomers behave differently toward plane—polarized light. This differ-
`ence allows us to distinguish a chiral molecule from its enantiomer in the laboratory.
`
`Plane-Polarized Light
`
`A beam of light consists of electromagnetic waves oscillating in an infinite number of planes at right
`angles to the direction of propagation of the light. When a light beam passes through a polarizing filter,
`it is converted to plane—polarized [cg/Jt whose electromagnetic waves oscillate in a single plane. We are
`familiar with this phenomenon in everyday life: Plane—polarized light can be produced by certain sun-
`glasses, which reduce glare by acting as a polarizing filter. They partly block horizontally oscillating light
`refiecting off the surfaces of various objects. Some camera lenses also have polarizing filters to reduce
`glare in brightly lit photographs.
`Plane—polarized light interacts with chiral molecules. This interaction can be measured by an
`instrument called a polarimeter (Figure 8.6). In a polarimeter, light with a single wavelength—that
`is, monoc/Jromatic Zig/1t—passes through a polarizing filter. The polarized light then traverses a tube
`containing a solution of the compound to be examined. Plane—polarized light is not affected by achiral
`molecules. However, the plane of polarized light rotates when it is absorbed by chiral molecules. When
`the plane—polarized light leaves the sample tube, it passes through a second polarizing filter called an
`analyzer. The analyzer is rotated in either clockwise or counterclockwise direction to match the rotated
`polarization plane so that it passes through the filter with maximum intensity. An angle, 05, is read off
`the analyzer. This angle is called the ooscrz/ca’ optical rotation, ocnbs. It equals the angle by which the light
`has been rotated by the chiral compound. Because chiral molecules rotate plane—polarized light, they are
`optically active. Achiral molecules do not rotate plane—polarized light, so they are optically inactive.
`
`Analyzer rotated to pass
`light with scale to read angle
`
`Polarizer
`
`
`
`Sample tube
`
`
`
`
`
`Rotad plane
`of polarized light
`
`
`
`T
`
`‘.~
`
`-'
`Plane of
`
`Sodium lamp
`
`polarized light
`
`Figure 8.6 Schematic Diagram of a Polarimeter
`Plane—polarized light is obtained by passing light through a polarizing filter. Any chiral compound in
`the sample tube rotates the plane—polarized light. The direction and magnitude of the rotation are deter-
`mined by rotating the analyzer to allow the light to pass through with maximum brightness. In a modern
`instrument, this is all done electronically, but the basic principle is the same.
`
`Specific Rotation
`
`The amount of rotation observed in a polarimeter depends on the structure of the substance and on its
`concentration. The optical activity of a pure chiral substance is reported as its specific rotation, symbolized
`by [0L]D. It is the number of degrees of rotation of a solution at a concentration measured in g mL"1 in a
`tube 1 dm (10 cm) long. The standard conditions selected for polarimetry measurements are 25 °C, and
`a wavelength of 589 nm. This yellow light is the D line of a sodium vapor lamp.
`
`l(XlD =
`
`aobs
`
`Z X c
`
`246
`
`If a chiral substance rotates plane—polarized light to the right—that is, in a positive (+) or clockwise direc-
`tion—the substance is dcxtrorotatory (Latin dcxtra, right). If a chiral substance rotates plane—polarized
`light to the left—in a negative (—) or counterclockwise direction—the substance is Zcz/orotatory (Latin
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`

`
`laez/us, left). The enantiomers of a chiral substance—called dextrorotatory and levorotatory isomers—
`rotate polarized light the same number of degrees, but in opposite directions. Therefore, they are some-
`times called optical isomers.
`
`We often refer to an enantiomer by prefixing the sign of the optical rotation at 589 nm to the name of
`the compound. For example, one of the enantiomers of 2—iodobutane has [0c]D = -15.15. It is called
`(—)—2—iodobutane. The other enantiomer is (+)—2—iodobutane,
`[0C]D = +15.15.
`
`mirror
`\
`
`H
`_.~= CHZCH3
`1—C
`\
`CH3
`
`H
`CH3CH2 K2,
`C—1
`/
`CH3
`
`(+)—2—iodobutane
`
`(—) —2—iodobut;1r1e
`
`The (+) isomer is sometimes called the if form because it is dextrorotatory; the (—) isomer is sometimes
`called the l form because it is levorotatory. Earlier, we encountered levodopa, so named because it is
`levorotatory. It is also called L—dopa and (—)—dopa. The specific rotation of L—dopa is -13.1“. Table 8.1
`lists the specific rotations of some common substances.
`
`Table 8.1
`
`Specific Rotations of
`Common Compounds
`
`Compound
`
`Azidothymidine (AZT)
`
`Cefotaxin (a cephalosporin)
`Cholesterol
`
`Cocaine
`
`Codeine
`
`Epinephrine (adrenaline)
`
`Levodopa
`
`Monosodium glutamate (MS G)
`
`Morphine
`
`Oxacillin (a penicillin)
`
`Progesterone
`Sucrose
`
`Testosterone
`
`[06]D
`
`+99°
`
`+ 5 5°
`—3 1 5°
`
`—1 6°
`
`—1 36°
`
`-5 .0“
`
`—13.1°
`
`+2 5 . 5°
`
`-132“
`
`+201“
`
`+ 172°
`+66“
`
`+109"
`
`Circularly Polarized Light and Optical Rotation
`
`We have said that chiral molecules cannot be distinguished in a symmetric environment. In the next
`sentence, we said that chiral molecules interact differently with plane—polarized light. However, by defi-
`nition, plane—polarized light has a plane of symmetry! Is there not a massive contradiction here?
`The answer is no because we can interpret plane—polarized light in terms of circularly polarized light
`(Figure 8.7). One form of circularly polarized light has a right—handed helical sense, and the other form
`has a left—handed helical sense. A helix is a chiral object, and right— and left—handed helices are related as
`mirror images. If we superimpose the two, we obtain plane—polarized light. So this helicity is “hiding” in
`plane—polarized light, which has inherent chiral components.
`
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`
`Figure 8.7
`Schematic Diagrams of Plane-
`and Circularly Polarized Light
`(a) In plane—polarized light, the electric
`field Vectors of the light all oscillate in
`a single plane.
`(b) In circularly polar—
`ized light,
`the electric field Vector can
`rotate in a right—handed (clockwise) or
`left—handed
`(counterclockwise)
`direc-
`tion. (c) If right—handed and left—handed
`phases of circularly polarized light are
`superimposed,
`the electric field Vectors
`in the +96 to —x directions cancel, and the
`
`y—components are additive, and directed
`along the y—axis. The net result is plane—
`polarized light.
`
`(a)
`
`B
`
`x
`
`E
`
`(b)
`
`Pl3“€‘P0l3fiZ€d
`
`light Side View)
`
`
`
`Mirror image helices
`
`(C)
`
`J’
`
`left—handed
`
`r
`
`ight—handed
`hase
`
`
`
`A chiral center is bonded to four different groups, and each of these bonds has an electric field. There—
`fore, the net electric field around a chiral center is chiral, and it absorbs one phase of circularly polarized
`light more than the other. As a result, the Vectors no longer cancel, and the light is rotated in either a
`clockwise or counterclockwise direction.
`
`Optical Purity
`
`Most naturally occurring molecules that contain one stereogenic center exist as one enantiomer. Samples
`that contain only one enantiomeric form are optically pure. Naturally occurring cholesterol, for ex-
`ample, exists only as the (—) form. It rotates light in a counterclockwise direction. However, compounds
`synthesized in the laboratory may not all have the same handedness, and the reaction yields a mixture
`of two enantiomers.
`
`What is the Optical rotation of a mixture of enantiomers, and how is it related to the percent-
`age of each enantiomer in the mixture? When plane—polarized light interacts with a single enantiomer
`of a chiral molecule, the plane is rotated in one direction. If the plane—polarized light interacts with the
`other enantiomer, the plane is rotated in an equal and opposite direction. If a solution contains equal
`amounts of two enantiomers, the clockwise and counterclockwise rotations resulting from all molecules
`
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`248
`
`

`
`cancel, and there is no net rotation. Mixtures containing equal amounts of enantiomers are called ra-
`cemic mixtures. A racemic mixture is represented in the name of a compound with a (1) prefix, as in
`(:)—2—iodobutane. The word “racemic” is derived from the Latin word mcemm, a cluster of grapes. It is
`so named because racemic mixtures were first found in tartaric acid, which precipitates from many wines
`as they age. See Figure 8.13.
`Now consider a circumstance in which the percent ratio of a mixture of enantiomers is not
`50:50. The percent enantiomeric excess of the enantiomer present in the larger amount is calculated as
`follows.
`
`% enantiomeric excess = % of one enantiomer — % of other enantiomer = optical purity
`
`The percent enantiomeric excess is the optical purity of the sample. For example, a 60:40 ratio of
`(+)—2—iodobutane and (—)—2—iodobutane is 20% optically pure. This value indicates that the rotation of
`the (—) isomer (40% of the total) cancels the rotation of some of the (+) isomer (40% of the total). The
`
`remaining 20% of the sample, which is (+)—2—iodobutane, is responsible for the observed rotation, so
`the sample is 20% optically pure.
`
`observed rotation
`_
`_
`optical purity =T X 100%
`rotation of pure enantiomer
`
`Problem 8.4
`
`What is [0c]D of the enantiomer of naturally occurring testosterone? (See Table 8.1.) What is the name
`of this enantiomer?
`
`Problem 8.5
`
`A sample of a solution of 1.5 g of cholic acid, a bile steroid, in 10 mL of alcohol is placed in a 10.0—cm
`sample tube. The observed rotation is +5.5. Calculate [0C]D for cholic acid.
`Problem 8.6
`
`A sample of epinephrine prepared in the laboratory has ocnbs = —0.5°. What is the optical purity of the
`sample? What is the percentage of each enantiomer in the sample? [0c]D for epinephrine is -5.0“.
`
`Sample Solution
`The specific rotation of epinephrine is —5.0°. We calculate the optical purity using the following
`equation.
`
`observed rotation
`
`optical purity:
`
`rotation o pure enantiomer
`
`X 100%
`
`optical purity:
`
`-5.0
`
`X 100% = 10%
`
`The enantiomeric excess is equal to the optical purity. The sum of the two enantiomers is 100%. Let the
`percent of the enantiomer with the negative optical rotation be x. The percent of the other enantiomer
`is 100 — x. Use the following equation and substitute the algebraic quantities.
`
`% enantiomeric excess = % of one enantiomer — % of other enantiomer = optical purity
`
`10% = x— (100% — x)
`
`x = 55%
`
`Thus, the percentages of the two enantiomers are 55% and 45%.
`
`249
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`

`
`8_4
`F '50 H E R p ROJ ECTIO N
`
`F0 R M U LAS
`
`Drawing molecules in three dimensions is time consuming. Furthermore, it is not easy to “read” the
`resulting perspective structural formulas, especially for compounds that contain several chiral centers
`(Section 8.6). However, the structural formula of a chiral substance can be conveniently drawn as a
`.
`.
`.
`.
`.
`.
`.
`.
`Fischer projection, which was introduced by the German chemist Emil Fischer more than a century
`ago. The configuration of a chiral substance in a Fischer projection formula is obtained by comparing
`it to the configuration of a reference compound whose common name is glyceraldehyde.
`
`CH0
`H_(I:_OH
`|
`CH 201-1
`glyceraldehyde
`
`Figure 8.8 Fischer Projection
`Structures of Glyceraldehyde
`(a) Perspective structures of glycer—
`aldehyde.
`(b) Projection structures.
`(c) Fisher projection structures of
`the enantiomers glyceraldehyde. The
`chiral center is located at the point
`where the bond lines intersect. The
`carbon atom is not usually shown.
`The Vertical lines extend away from
`the viewer, behind the plane of the
`page; horizontal lines extend toward
`the viewer, out. of the plane of the
`page, as shown in part (b).
`
`Glyceraldehyde contains a carbon atom bonded to four different groups, so it can exist as either of two
`enantiomers (Figure 8.8). The enantiomers ofglyceraldehyde in a Fischer projection are drawn accord-
`ing to the following conventions:
`1. Arrange the carbon chain vertically with the most oxidized group (—CHO in glyceraldehyde) at
`the “top”,
`Place the carbon atom at the chiral center in the plane of the paper. It is C-2 in glyceraldehyde.
`Because C-2 is bonded to four groups, the CH0 group and the CHZOH group extend behind the
`plane of the page, and the hydrogen atom and the hydroxyl group extend up and out of the plane.
`Project these four groups onto a plane. The carbon atom at the chiral center is usually not shown
`in this convention. It is located at the point where the bond lines cross. The vertical lines project
`away from the viewer. The horizontal lines project toward the viewer.
`
`2.
`3.
`
`4.
`
`(3)
`
`perspeeeive semeeures
`
`CH0
`l
`C..,,
`HOCH/ \ "OH
`H
`A
`
`T
`
`CH0
`|
`,,..C
`
`\CH2OH
`
`HO“
`
`H
`
`B
`
`Mirror
`
`EH0
`H>OH
`
`EH0
`HO—<H
`
`(b)
`
`(':H2oH
`
`__
`
`A
`
`(':H2oH
`
`B
`
`(c)
`
`Fischer projection structures
`
`CHO
`
`H+OH
`
`CHZOH
`
`CHO
`
`HO+H
`
`CHZOH
`
`A
`
`B
`
`A Fischer projection formula is a two—dimensional representation. It might appear that if we lifted
`one formula out of the plane and rotated it 180° around the carbon backbone, we would obtain the
`structure of the enantiomer. However, if this were done for molecule A in Figure 8.8, the carbonyl
`group and the hydroxymethyl group, originally behind the plane, would be in front of the plane.
`These groups would not occupy identical positions with respect to the carbonyl group and hydroxy-
`methyl group of molecule B, which are behind the plane. Therefore, to avoid the error of apparently
`achieving a two—dimensional equivalence of nonequivalent three—dimensional molecules, we cannot
`lift the two—dimensional representations out of the plane of the paper.
`Fischer projection formulas can be drawn for any pair of enantiomers. These formulas imply
`that we know the configuration at the chiral carbon atom. However, the true configuration could not be
`determined by early chemists because there was no way to determine the arrangement of the atoms in
`space. Therefore, Fischer arbitrarily assigned a configuration to one member of the enantiomeric pair of
`
`250
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`
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`
`

`
`glyceraldehydes. The dextrorotatory enantiomer of glyceraldehyde, which rotates plane—polarized light
`in a clockwise direction (+13.5°), was assigned to the Fischer projection with the hydroxyl group on the
`right side. Fischer called the compound D—glyceraldehyde. The mirror image compound, (—)—glyceralde—
`hyde, corresponds to the structure in which the hydroxyl group is on the left. It rotates plane—polarized
`light in a counterclockwise direction (—13.5°). Fischer called the compound L—glyceraldehyde.
`
`mirror plane
`
`CH0
`
`HO+H
`
`CHZOH
`
`
`
`CH0
`
`H+OH
`
`CHZOH
`
`L-(_)-g1ycera1dehyde
`[0c]D=—13.5°
`
`T D—(+)—glyceraldehyde
`[0tlD=+13.5°
`
`Fischer projection structures
`
`Problem 8.7
`
`Write the Fischer projection formula of each of the following compounds.
`(a) D—lactic acid, CH3CH(OH)CO2H
`(b) L—serine, HOCH2CH(NH2)CO2H
`(c) D—valine, (CH3)2CHCH(NH2)CO2
`
`We began this chapter by saying that the arrangement of atoms in space determines the configura-
`8_5
`ABSOLUTE CON FIGURATION tion of a molecule. When we know the exact positions of these atoms in space, we know the mol-
`ecules absolute configuration. The absolute configuration of an enantiomer cannot be established
`by measuring the direction or magnitude of its optical rotation. Optical rotation depends on both the
`configuration and the identity of the four groups around the central carbon atom. One “left—handed”
`molecule could be levorotatory, whereas another “left—handed” molecule with different groups could
`be dextrorotatory. For example, in spite of the similarity of three of the groups (CH3CH2, CH3, and
`H), the following structures of 2—methyl—1—butanol and l—chloro—2—methylbutane, which have the same
`configuration, have different directions of optical rotation.
`
`H
`CH3"-----\C—CH2OH
`
`H
`CH3"-----C—CH2Cl
`
`CH3CH2
`
`CH3CH2
`
`(—)—2—methyl—l—butanol
`
`(+)—1—chloro—2—methylbutane
`
`[0c]D=—5.8
`
`[oc]D=+1.7
`
`To determine the absolute configuration, we require a method that ca_n specify the positions of all
`atoms in the molecule. One way to do this is by X—ray crystallography. The absolute configuration
`of an optically active substance was first determined in 1950. The arrangement of its atoms in space
`corresponds to the arrangement of atoms in (+)—glyceraldehyde arbitrarily assigned by Fischer. His
`original choice was correct! As a result, all configurations that had been deduced by using (+)—glyc-
`eraldehyde as the reference compound are also correct, and this includes all the amino acids isolated
`from proteins, all carbohydrates, and many other compounds.
`The absolute configuration of a compound can be determined by comparing it to a reference
`compound of known absolute configuration. This structure proof sometimes requires an elaborate
`series of reactions. However, the principle is easily illustrated with the conversion of 2—methyl—l—butanol
`to l—chloro—2—methylbutane. Alcohols can be converted into chloroalkanes by thionyl chloride (SOCl2).
`The reaction does not affect any of the bonds at the stereogenic center of 2—methyl—l—butanol. Hence,
`the configuration is unchanged. If the absolute configuration of the alcohol is known, the groups
`bonded to the stereogenic center in the chloroalkane must be arranged in the same configuration. If the
`absolute configuration of the alcohol were not known, we would still know that the haloalkane would
`have the same relative configuration.
`
`251
`
`Janssen Ex. 2046
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`Lupin Ltd. v. Janssen Sciences Ireland UC
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`
`

`
`H
`
`H
`
`CH3"-----C—CH2OH
`
`
`SOCIZ
`
`CH3"-----C—CH2Cl + so, + HCl
`
`CH3CH2
`
`CH3CH2
`
`R,S Configurations: The Kahn—|ngold—Prelog System of Configurational Nomenclature
`
`The configurations of some molecules, such as aniino acids and carbohydrates, ca_n easily be com-
`pared to reference compounds such as 1)—glyceraldehyde. But this procedure is not easily applied to
`molecules whose structures differ considerably from the reference compound. To circumvent this
`difficulty, R. S. Kahn, K. C. Ingold, and V. Prelog established a set of rules in 1964 that describe the
`absolute configuration of any chiral molecule.
`The R,S system of configurational nomenclature for describing absolute configurations is
`related to the method we described in Chapter 6 to assign the E,Z configuration of alkenes. In the
`R,S system, the four groups bonded to each chiral carbon atom are ranked from highest to lowest
`priority. The highest priority group is assigned the number 1, the lowest priority group is assigned
`the number 4. Then, the molecule is oriented so that the bond from the carbon atom to the group of
`lowest priority is arranged directly along our line of sight pointing downward (Figure 8.9). When this
`has been done, the three higher priority groups point up and lie on the circumference of a circle. (It
`may help to imagine holding the lowest priority group in your hand like the stem of a flower as you
`examine the petals.)