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5B Parkinson’s Disease and L-Dopa

5B Parkinson’s Disease and L-Dopa

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462



THE THREE-DIMENSIONAL SHAPE OF MOLECULES



15.6 FISCHER PROJECTIONS

The chirality centers in some organic compounds, most notably carbohydrates (Chapter 20), are

often drawn using a different convention than is used for other chirality centers.

Instead of drawing a tetrahedron with two bonds in the plane, one in front of the plane, and one

behind it, the tetrahedron is tipped so that both horizontal bonds come forward (drawn on

wedges) and both vertical bonds go behind (on dashed lines). This structure is then abbreviated by a cross formula, also called a Fischer projection formula. In a Fischer projection

formula, therefore,

Draw a tetrahedron as:



Abbreviate it as a cross formula:



W



W



Horizontal bonds come

forward, on wedges.



Z



C



=



X



Z



X

Y



Y

Vertical bonds

go back, on dashes.



chirality center



Fischer projection formula



• A carbon atom is located at the intersection of the two lines of the cross.

• The horizontal bonds come forward, on wedges.

• The vertical bonds go back, on dashed lines.



For example, to draw the chirality center of 2-butanol [CH3CH(OH)CH2CH3, Section 15.3B] using

this convention, draw the tetrahedron with horizontal bonds on wedges and vertical bonds on

dashed lines. Then, replace the chirality center with a cross to draw the Fischer projection.



H



H

= CH3



C



CH3CH2



CH2CH3



C



CH3 =



OH



OH

one enantiomer of

2-butanol



mirror image



Replace the chirality

center with a cross.

H

CH3



H

CH2CH3



CH3CH2



CH3



OH

OH

Fischer projection formulas for both enantiomers of 2-butanol



Sample Problem 15.2 illustrates another example of drawing two enantiomers using this

convention.



SAMPLE PROBLEM 15.2



Draw both enantiomers of glyceraldehyde, a simple carbohydrate, using Fischer projection

formulas.

HOCH2CHCHO

OH

glyceraldehyde



ANALYSIS



smi26573_ch15.indd 462



• Draw the tetrahedron of one enantiomer with the horizontal bonds on wedges, and the

vertical bonds on dashed lines. Arrange the four groups on the chirality center—H, OH,

CHO, and CH2OH—arbitrarily in the first enantiomer.

• Draw the second enantiomer by arranging the substituents in the mirror image so they are a

reflection of the groups in the first molecule.

• Replace the chirality center with a cross to draw the Fischer projections.



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COMPOUNDS WITH TWO OR MORE CHIRALITY CENTERS



SOLUTION



463



Draw the compound.



chirality center



Then, draw the mirror image.



CHO



HOCH2CHCHO

H



OH



C



glyceraldehyde



CHO



OH



HO



CH2OH



C



H



CH2OH



• Horizontal bonds come

forward.

• Vertical bonds go back.



CHO

H



CHO



OH



HO



CH2OH



CH2OH



Fischer projection of

one enantiomer



PROBLEM 15.13



a.



H



C



COOH



b.



Br



H



Cl



C



OH



CH2Cl



Convert each Fischer projection formula into a representation that uses wedges and dashes.

CH3



a.



CH3CH2



H



b.



Br



CH3CH2



CH3

NH2



Cl



PROBLEM 15.15



Fischer projection of the

second enantiomer



Convert each molecule into a Fischer projection formula.

CH3



PROBLEM 15.14



H



Draw Fischer projections of both enantiomers for each compound.

a.



b.



CH3CHCH2CH2CH3



CH3CH2CHCH2Cl



OH



Cl



15.7 COMPOUNDS WITH TWO OR MORE CHIRALITY CENTERS

Although most chiral compounds we have seen thus far have one chirality center, molecules may

have two, three, or hundreds of chirality centers. Two examples—tartaric acid and ephedrine—

were shown in Section 15.3.

• A compound with one chirality center has two stereoisomers—a pair of enantiomers.

• A compound with two chirality centers has four possible stereoisomers.



Let’s examine the four stereoisomers of the amino acid threonine. The two chirality centers in

threonine are labeled in the given structure. Threonine is one of the 20 naturally occurring amino

acids needed for protein synthesis. Since humans cannot synthesize threonine, it is an essential

amino acid, meaning that it must be obtained in the diet.

CO2H

H2N



C



H



H



C



OH



[The chirality centers are labeled in red.]



CH3

carbon skeleton of the

amino acid threonine



smi26573_ch15.indd 463



12/9/08 5:26:39 PM



464



THE THREE-DIMENSIONAL SHAPE OF MOLECULES



There are four different ways that the groups around two chirality centers can be arranged, shown

in structures A–D.

The four stereoisomers of a compound with two chirality centers

CO2H



CO2H



H2N



C



H



H



C



OH



CO2H



CO2H



H



C



NH2



H



C



NH2



H2N



C



H



HO



C



H



H



C



OH



HO



C



H



CH3

A



CH3



CH3



B



C



CH3

D



enantiomers



enantiomers



A and B are mirror images, but they are not superimposable, so they are enantiomers. C and D

are also mirror images, but they are not superimposable, so they are enantiomers as well. In other

words, with two chirality centers, there can be two pairs of enantiomers.

Because these structures are drawn with the horizontal bonds of the chirality centers on wedges

and the vertical bonds on dashed lines, Fischer projection formulas can be drawn for each stereoisomer by replacing each chirality center by a cross.

CO2H



CO2H

H2N



H



H

OH



H



HO



CH3

A



CO2H



CO2H



NH2



H



NH2



H2N



H



H



H



OH



HO



H



CH3



CH3



B



C



CH3

D

enantiomers



enantiomers



Thus, there are four stereoisomers for threonine: enantiomers A and B, and enantiomers C and

D. What is the relationship between two stereoisomers like A and C? A and C are stereoisomers,

but if you arrange a mirror plane between A and C you can see that they are not mirror images of

each other. A and C represent a second broad class of stereoisomers, called diastereomers.

• Diastereomers are stereoisomers but they are not mirror images of each other.



A and B are diastereomers of C and D, and vice versa. Naturally occurring threonine is a single

stereoisomer corresponding to structure A. The other stereoisomers do not occur in nature and

cannot be used to synthesize biologically active proteins.



SAMPLE PROBLEM 15.3



Considering the four stereoisomers of 2,3-pentanediol (E–H): (a) Which compound is an

enantiomer of E? (b) Which compound is an enantiomer of F? (c) What two compounds are

diastereomers of G?

CH3

Four stereoisomers

of 2,3-pentanediol:



smi26573_ch15.indd 464



CH3



CH3



CH3



H



C



OH



H



C



OH



HO



C



H



HO



C



H



H



C



OH



HO



C



H



HO



C



H



H



C



OH



CH2



CH2



CH2



CH2



CH3



CH3



CH3



CH3



E



F



G



H



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FOCUS ON THE HUMAN BODY: THE SENSE OF SMELL



ANALYSIS



465



Keep in mind the definitions:

• Enantiomers are stereoisomers that are nonsuperimposable mirror images of each

other. Look for two compounds that when placed side-by-side have the groups on both

chirality centers drawn as a reflection of each other. A given compound has only one

possible enantiomer.

• Diastereomers are stereoisomers that are not mirror images of each other.



SOLUTION



a. E and G are enantiomers because they are mirror images that do not superimpose on each

other.

b. F and H are enantiomers for the same reason.

CH3



CH3



CH3



CH3



H



C



OH



HO



C



H



H



C



OH



H



C



OH



HO



C



H



HO



C



H



HO



C



H



H



C



OH



CH2



CH2



CH2



CH2



CH3



CH3



CH3



CH3



G



F



E

enantiomers



H

enantiomers



c. Any compound that is a stereoisomer of G but is not its mirror image is a diastereomer.

Thus, F and H are diastereomers of G.



PROBLEM 15.16



Answer the following questions about the four stereoisomers of 3-bromo-2-chloropentane

(W–Z). (a) Which compound is the enantiomer of W? (b) Which compound is the enantiomer

of X? (c) What two compounds are diastereomers of Y? (d) What two compounds are

diastereomers of Z?

CH3

Four stereoisomers of

3-bromo-2-chloropentane:



PROBLEM 15.17



CH3



CH3



CH3



H



C



Cl



H



C



Cl



Cl



C



H



Cl



C



H



H



C



Br



Br



C



H



H



C



Br



Br



C



H



CH2



CH2



CH2



CH2



CH3



CH3



CH3



CH3



W



X



Y



Z



Convert each stereoisomer (W–Z) in Problem 15.16 into a Fischer projection.



15.8 FOCUS ON THE HUMAN BODY

THE SENSE OF SMELL

Research suggests that the odor of a particular molecule is determined more by its shape than

by the presence of a particular functional group (Figure 15.4). For example, hexachloroethane

(Cl3CCCl3) and cyclooctane have no obvious structural similarities. However, both molecules

have a camphor-like odor, a fact attributed to their similar spherical shape, which is readily shown

using space-filling models.

H



C

C



=



Cl



Cl Cl

=



Cl



C



C



Cl



Cl Cl

cyclooctane



smi26573_ch15.indd 465



space-filling model



space-filling model



hexachloroethane



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466



THE THREE-DIMENSIONAL SHAPE OF MOLECULES







FIGURE 15.4 The Shape of Molecules and the Sense of Smell



brain

olfactory

nerve cell

airflow



mucus

receptor on an

olfactory hair



lining of the olfactory

bulb in the nasal passage



olfactory

hairs



nasal passage



cyclooctane bound

to a receptor site



Cyclooctane and other molecules similar in shape bind to a particular olfactory receptor on

the nerve cells that lie at the top of the nasal passage. Binding results in a nerve impulse that

travels to the brain, which interprets impulses from particular receptors as specific odors.



Since enantiomers interact with chiral smell receptors, some enantiomers have different odors.

There are a few well-characterized examples of this phenomenon in nature. For example, one

enantiomer of carvone is responsible for the odor of caraway, whereas the other carvone enantiomer is responsible for the odor of spearmint.



CH3



CH3



O



O



C



H



CH3



CH3



H 2C

caraway seeds



C



H

CH2



carvone

enantiomer A



carvone

enantiomer B



A has the odor of caraway.



spearmint leaves



B has the odor of spearmint.



Thus, the three-dimensional structure of a molecule is important in determining its odor.



PROBLEM 15.18



Limonene is similar to carvone in that each enantiomer has a different odor; one enantiomer

occurs in lemons and the other occurs in oranges. Identify the chirality center in limonene and

draw both enantiomers of limonene.

CH2

CH3



C

CH3

limonene



smi26573_ch15.indd 466



12/9/08 5:26:40 PM



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