Carey - Organic Chemistry - sgchapt07

Carey - Organic Chemistry - sgchapt07

(Parte 1 de 3)

CHAPTER 7 STEREOCHEMISTRY

1,1,3-Trimethylcyclobutane however, has no stereogenic centers.

H3CHNot a stereogenic center; two of its substituents are the same.

CH3

A stereogenic center; the four substituents to which

BrCH2 CH2CH3C CH3

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7.3(b)There are twoplanes of symmetry in (Z)-1,2-dichloroethene, of which one is the plane of the molecule and the second bisects the carbon–carbon bond. There is no center of symmetry. The molecule is achiral.

(c)There is a plane of symmetry in cis-1,2-dichlorocyclopropane that bisects the C-1@C-2 bond and passes through C-3. The molecule is achiral.

(d)trans-1,2-Dichlorocyclopropane has neither a plane of symmetry nor a center of symmetry. Its two mirror images cannot be superposed on each other. The molecule is chiral.

The concentration cis expressed in grams per 100 mLand the length lof the polarimeter tube in decimeters. Since the problem specifies the concentration as 0.3 g/15 mLand the path length as 10cm, the specific rotation [ ] is:

7.5From the previous problem, the specific rotation of natural cholesterol is [ ] 39°. The mixture of natural ( )-cholesterol and synthetic ( )-cholesterol specified in this problem has a specific rotation [ ] of 13°.

The mixture is two thirds natural ( )-cholesterol and one third synthetic ( )-cholesterol.

Cl H H Cl H Cl and are nonsuperposable mirror imagesClH

ClH Cl H

Plane of symmetry

C ClCl

Planes of symmetry

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7.6Draw the molecular model so that it is in the same format as the drawings of ( ) and ( )-2-butanol in the text.

Reorient the molecule so that it can be compared with the drawings of ( ) and ( )-2-butanol.

The molecular model when redrawn matches the text’s drawing of ( )-2-butanol.

The lowest ranked substituent (H) at the stereogenic center points away from us in the drawing. The three higher ranked substituents trace a clockwise path from CH2Ft oC H2CH3 to CH3.

The absolute configuration is R; the compound is (R)-( )-1-fluoro-2-methylbutane.

The lowest ranking substituent (H) is directed toward you in the drawing, and therefore the molecule needs to be reoriented so that H points in the opposite direction.

H CH3 BrCH2

CH2CH3

( )-1-Bromo-2-methylbutane turn 180

C CH3H

CH2Br CH3CH2

C H3C CH2F

CH2CH3

C HH3C which becomesReorient

Redraw as

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The three highest ranking substituents trace a counterclockwise path when the lowest ranked substituent is held away from you.

The absolute configuration is S, and thus the compound is (S)-( )-1-bromo-2-methylbutane. (d)The highest ranked substituent at the stereogenic center of 3-buten-2-ol is the hydroxyl group, and the lowest ranked substituent is H. Of the remaining two, vinyl outranks methyl.

The lowest ranking substituent (H) is directed away from you in the drawing. We see that the order of decreasing precedence appears in a counterclockwise manner.

When the lowest ranked substituent points away from you, the remaining three must appear in descending order of precedence in a counterclockwise fashion in the S enantiomer. (S)-1, 1- difluoro-2-methylcyclopropane is therefore

7.9(b)The Fischer projection of (R)-( )-1-fluoro-2-methylbutane is analogous to that of the alcohol in part (a). The only difference in the two is that fluorine has replaced hydroxyl as a substituent at C-1.

Although other Fischer projections may be drawn by rotating the perspective view in other directions, the one shown is preferred because it has the longest chain of carbon atoms oriented on the vertical axis with the lowest numbered carbon at the top. (c) As in the previous parts of this problem, orient the structural formula of (S)-( )-1-bromo-

CH3CH2

CH2BrH CH3C

CH2CH3

CH2Br

CH2Br His the same aswhich becomes the Fischer projection

C H3C

CH3CH2

CH2FH C CH3

CH2CH3

CH2F His the same aswhich becomes the Fischer projection

CF2CH2 CH2CF2  CH3 H

Highestrank

Lowest rank

C HH3C CH2C

H3C CH

CH2CH3

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7.10In order of decreasing rank, the substituents attached to the stereogenic center in lactic acid are

@OH, @CO2H, @CH3, and @H. The Fischer projection given for ( )-lactic acid (a) corresponds to the three-dimensional representation (b), which can be reoriented as in (c). When (c) is viewed from the side opposite the lowest ranked substituent (H), the order ofdecreasing precedence is anticlockwise, as shown in (d). ( )-Lactic acid has the Sconfiguration.

in this case OH and NH2, are on the same side when the carbon chain is vertical. There are two erythro stereoisomers that are enantiomers of each other:

Analogous substituents are on opposite sides in the threo isomer:

7.12There are four stereoisomeric forms of 3-amino-3-butanol:

In the text we are told that the (2R,3R) stereoisomer is a liquid. Its enantiomer (2S,3S) has the same physical properties and so must also be a liquid. The text notes that the (2R,3S) stereoisomer is a solid (mp 49°C). Its enantiomer (2S,3R) must therefore be the other stereoisomer that is a crystalline solid.

CH3

Threo CH3

CH3 HO

Threo

CH3

OH NH2

Erythro CH3

CH3 HO

Erythro

HO CO2H HC

HO CO2H H

HO CO2H H

OHHO2C

HO CH CH2

CH CH2 CH CH2

C OH CH3

H OH CH3 His the same aswhich becomes the Fischer projection

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7.13Examine the structural formula of each compound for equivalently substituted stereogenic centers. The only one capable of existing in a meso form is 2,4-dibromopentane.

None of the other compounds has equivalently substituted stereogenic centers. No meso forms are possible for:

7.14There is a plane of symmetry in the cis stereoisomer of 1,3-dimethylcyclohexane, and so it is an achiral substance—it is a meso form.

The trans stereoisomer is chiral. It is not a meso form.

7.15Amolecule with three stereogenic centers has 23, or 8, stereoisomers. The eight combinations of R and S stereogenic centers are:

Stereogenic centerStereogenic center

Isomer 1R R RIsomer 5SSS Isomer 2R R SIsomer 6SSR Isomer 3R S RIsomer 7SRS Isomer 4S R RIsomer 8RSS

7.162-Hexuloses have three stereogenic centers. They are marked with asterisks in the structural formula.

No meso forms are possible, and so there are a total of 23, or 8, stereoisomeric 2-hexuloses.

HOCH2CCHCHCHCH2OH OHO OH

CH3

Plane of symmetry passes through C-2 and C-5 and bisects the ring.

2,3-Dibromopentane

CH3CHCHCH2CH3

Br Br 4-Bromo-2-pentanol

CH3CHCH2CHCH3

OH Br 3-Bromo-2-pentanol

CH3CHCHCH2CH3 OH

2,4-Dibromopentane

Equivalently substituted stereogenic centers

CH3CHCH2CHCH3

Br* Br* CH3

Br H

H BrH

Fischer projection of meso-2,4-dibromopentane

Plane of symmetry

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7.17Epoxidation of (Z)-2-butene gives the meso (achiral) epoxide. Oxygen transfer from the peroxy acid can occur at either face of the double bond, but the product formed is the same because the two mirror-image forms of the epoxide are superposable.

Epoxidation of (E)-2-butene gives a racemic mixture of two enantiomeric epoxides.

7.18The observed product mixture (68% cis-1,2-dimethylcyclohexane: 32% trans-1,2-dimethylcyclohexane) contains more of the less stable cis stereoisomer than the trans. The relative stabilities of the products therefore play no role in determining the stereoselectivity of this reaction.

7.19The tartaric acids incorporate two equivalently substituted stereogenic centers. ( )-Tartaric acid, as noted in the text, is the 2R,3Rstereoisomer. There will be two additional stereoisomers, the enantiomeric ( )-tartaric acid (2S,3S) and an optically inactive meso form.

7.20No. Pasteur separated an optically inactive racemic mixture into two optically active enantiomers. A meso form is achiral, is identical to its mirror image, and is incapable of being separated into optically active forms.

(2S,3S)-Tartaric acid (optically active) meso-Tartaric acid (optically inactive) (mp 140°C)

CO2H OH

OH Plane of symmetry

CH3

O H3C

H3CH H

CH3COOHO CH3COOH

E H3C

CH3

CH3 H meso-2,3-Epoxybutane

H3C H H meso-2,3-Epoxybutane H3C

CCH3COOHO CH3COOH

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7.21The more soluble salt must have the opposite configuration at the stereogenic center of 1-phenylethylamine, that is, the Sconfiguration. The malic acid used in the resolution is a single enantiomer, S.In this particular case the more soluble salt is therefore (S)-1-phenylethylammonium (S)-malate.

7.22In an earlier exercise (Problem 4.23) the structures of all the isomeric C5H12O alcohols were presented. Those that lack a stereogenic center and thus are achiralare

The chiral isomers are characterized by carbons that bear four different groups. These are: 7.23The isomers of trichlorocyclopropane are

CH3CH CHCH2BrAchiral

CH3CHCH CH2 *

Br Chiral

ClCH2CHCH2OH*

3-Chloro-1,2-propanediol Chiral

HOCH2CHCH2OH

2-Chloro-1,3-propanediol Achiral cis-1,2,3-Trichlorocyclopropane (achiral—contains a plane of symmetry)

ClHH Cl trans-1,2,3-Trichlorocyclopropane (achiral—contains a plane of symmetry)

ClHH

ClCl Cl

HCl Cl ClH

Enantiomeric forms of 1,1,2-trichlorocyclopropane (both chiral)

3-Methyl-2-butanol2-Pentanol

CH3CHCH2CH2CH3 OH

2-Methyl-1-butanol

CH3CH2CHCH2OH CH3

3-Pentanol 2-Methyl-2-butanol

CH3CH2COH CH3

CH3

CH3CH2CHCH2CH3 OH

1-Pentanol 3-Methyl-1-butanol 2,2-Dimethyl-1-propanol

CH3CH2CH2CH2CH2OH (CH3)3CCH2OHCH3CHCH2CH2OH CH3

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(c)Both stereoisomers have two equivalently substituted stereogenic centers, and so we must be alert for the possibility of a meso stereoisomer. The structure at the left is chiral. The one at the right has a plane of symmetry and is the achiral meso stereoisomer.

(d)The first structure is achiral; it has a plane of symmetry. The second structure cannot be superposed on its mirror image; it is chiral.

7.25There are four stereoisomers of 2,3-pentanediol, represented by the Fischer projections shown. All are chiral.

There are three stereoisomers of 2,4-pentanediol. The meso form is achiral; both threo forms are chiral.

meso-2,4-Pentanediol CH3

CH3

CH3

CH3

CH3

CH3

Enantiomeric threo isomers

Enantiomeric erythro isomers CH2CH3

CH3

OH CH2CH3

CH3

Enantiomeric threo isomers CH2CH3

CH3

CH3

Cl Cl Cl

Reference structureMirror imageReoriented mirror image

Cl 1

Plane of symmetry passes through C-1, C-4, and C-7.

NH2 CH3

CH3

Chiral

Plane of symmetry

NH2 NH2

CH3

CH3

Meso: achiral

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7.26 Among the atoms attached to the stereogenic center, the order of decreasing precedence is Br

Cl F H. When the molecule is viewed with the hydrogen pointing away from us, the order Br→ Cl →F appears clockwise in the Renantiomer, anticlockwise in the Senantiomer.

The molecule is oriented so that the lowest ranking substituent is directed away from you and the order of decreasing precedence is clockwise.

(b)In order of decreasing sequence rule precedence, the four substituents at the stereogenic center of monosodium L-glutamate are

When the molecule is oriented so that the lowest ranking substituent (hydrogen) is directed away from you, the other three substituents are arranged as shown.

The order of decreasing rank is counterclockwise; the absolute configuration is S.

Orient with lowest ranked substituent away from you.

CH3

TD H3C T OH

H CH3T C DHO

CH3 biological oxidation

CH2CH2CO2

CH2CH2CO2

CO2

CH2CH2CO2

CO2

Cis the same as

CH2CH2CH3 HO H

ClCl H F

Br H

ClF C Br

C BrH

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The order of decreasing rank in the reactant is anticlockwise; the configuration is S. The order of decreasing rank in the product is clockwise; the configuration is R. (b)Retention of configuration means that the three-dimensional arrangement of bonds at the stereogenic center is the same in the reactant and the product. The Rand Sdescriptors change because the order of precedence changes in going from reactant to product; for example, CH3 is the highest ranked substituent in the reactant, but becomes the second-highest ranked in the product.

7.29Two compounds can be stereoisomers only if they have the sameconstitution. Thus, you should compare first the constitution of the two structures and then their stereochemistry. The best way to compare constitutions is to assign a systematic (IUPAC) name to each molecule. Also remember that enantiomers are nonsuperposable mirror images, and diastereomers are stereoisomers that are not enantiomers.

(a)The two compounds are constitutional isomers. Their IUPAC names clearly reflect this difference.

(b)The two structures have the same constitution. Test them for superposability. To do this we need to place them in comparable orientations.

and

The two are nonsuperposable mirror images of each other. They are enantiomers.

To check this conclusion, work out the absolute configuration of each using the Cahn– Ingold–Prelog system.

and is equivalent to CH2CH3

Br CH3 HC

Br CH2CH3 CH3 is equivalent to

CH2CH3

Br CH3 HC

CH3

CH3CH2

Br H

(S)-2-Bromobutane

CH3CH2

Br H

(R)-2-Bromobutane

C Br

CH3CH2

BrH is equivalent to

CH2CH3

H CH3 BrC

CH3CH2

Br H

CH2CH3

Br CH3 His equivalent to

CH3CHCH2Br

OH 1-Bromo-2-propanol

CH3CHCH2OH

Br 2-Bromo-1-propanol and

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The two structures represent the same compound, since they are superposable. (As a check, notice that both have the Sconfiguration.) (d)If we reorient the first structure, which is the enantiomer of

As a check, the first structure is seen to have the Sconfiguration, and the second has the Rconfiguration. (e)As drawn, the two structures are mirror images of each other; however, they represent an achiral molecule. The two structures are superposable mirror images and are not stereoisomers but identical.

(g)The two structures are enantiomers, since they are nonsuperposable mirror images. Checking their absolute configurations reveals one to be R, the other S.Both have the Econfiguration at the double bond.

(h)These two structures are identical; both have the Econfiguration at the double bond and the R configuration at the stereogenic center.

Alternatively, we can show their superposability by rotating the second structure 180° about an axis passing through the doubly bonded carbons.

(i)One structure has a cis double bond, the other a trans double bond; therefore, the two are diastereomers. Even though one stereogenic center is Rand the other is S, the two structures are

Reference structureIdentical to reference structureRotate 180 around this axis

HO H OHH (2R,3E)-3-Penten-2-ol (2S,3E)-3-Penten-2-ol

H3C H H cis-1-Chloro-2-methylcyclopropane

H3C HC l trans-1-Chloro-2-methylcyclopropane andare both identical

CH2OH

OH CH2OH H

CH2OH

H CH2OH HO

CH2CH3

H CH3 Br becomes CH2CH3

Br CH3 HC

CH3

CH3CH2

Br H

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(j)Here it will be helpful to reorient the second structure so that it may be more readily compared with the first.

The two compounds are enantiomers.

Examining their absolute configurations confirms the enantiomeric nature of the two compounds.

(Parte 1 de 3)

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