organic chemistry chapter 5 stereoisomers h. d....

Organic Chemistry Chapter 5 Stereoisomers H. D. Roth

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LECTURE POSTING V Stereoisomerism

A type of isomerism; two compounds are stereoisomers when they differ only in

the spatial relationship of their parts. In order to discuss isomerism we use the following

terms (with which you are familiar):

Composition: the type and number of atoms in a molecule;

Constitution: the way in which these atoms are connected

(we also call this connectivity);

Configuration: the arrangement of the atoms in three-dimensional

space;

The term Conformation: also describes the arrangement of atoms in 3D

space;

A. Conformational isomerism

The conformation can be changed by free rotation about a single bond (or two); Br

Br

axial Br equatorial Br B. Configurational isomerism I: cis-trans or geometric isomerism

The configuration cannot be changed; it is “fixed”, by restricting free rotation.

1) Rotation can be restricted if two carbons of an alkane chain are tied up forming

a ring. You have seen examples in the cycloalkanes: substituents can be on the same side

(cis) or on opposite sides (trans) of the ring plane; two examples are shown below.

CH3H

CH3

H

H

CH3

CH3

H


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2) Rotation can be restricted also when two adjacent carbon atoms are connected

by a π bond in addition a σ bond (see chapter 1); the π bond holds the substituents in one

plane; they can be on the same side (cis) or the opposite side (trans) of the double bond

(more in chapter 11).

H3C CH3

H3C

CH3

HH H

H

cis-2-butene trans-2-butene

Geometric isomers could be interconverted by breaking and reforming a bond,

either a σ bond of a cycloalkane or π bond of an alkene. These processes do occur, but

they require high energies – therefore the geometric isomers of cycloalkanes and alkenes

are stable at room temperature (unlike conformers, which are readily interconverted).

When we compare structural and geometric isomers we note:

Structural Isomers Geometric Isomers

identical composition identical

different connectivity identical

different arrangement in 3D space different

different heat of combustion different

C. Configurational isomers, II: compounds with carbon stereocenters

There is another way to arrange atoms in 3D generating isomers. You have long

been familiar with this type of relationship: look at your hands. Your two hands are


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identical in many ways, but they differ in their 3D arrangement: they are related as

mirror images, they are different because they are not superimposable.

Objects that are different in their 3D arrangement, but related as mirror images,

are called chiral (from Greek χειρ, hand). The difference lies in their “handedness”

(analogy to your left and right hands).

1) We recognize chirality in a molecule by an absence of symmetry. We

examine a molecule for elements of symmetry; objects that have a plane of symmetry

are not chiral (achiral); most compounds that lack a plane of symmetry are chiral. One

exception: compounds with a center of symmetry are achiral (not in book; example

below).

Cl

BrH3C

CH3

Br

BrH3C

CH3

•

Achiral due to plane of symmetry Achiral due to center of symmetry

CAUTION: It is not always easy to recognize symmetry, because molecules are

three-dimensional and their representations are two-dimensional.

2) Do we have a positive way to show that a molecule is chiral? An unmistakable,

structural feature identifying chirality: an asymmetrical carbon, or stereocenter. This is

an sp3 hybridized tetrahedral carbon (angle ~ 109.5°), with four different substituents.


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Compounds containing a asymmetrical carbon exist in two isomeric forms that

cannot be superimposed; they are "mirror images" of each other. We call such molecules

enantiomers (the left molecule is the enantiomer of the right molecule and vice versa).

C

R2R3

R4

R1C

R2R3

R4

R1

(where R1, R2, R3, R4 are all different)

CAUTION: if a compound has two stereocenters that are mirror images of each

other, they are symmetrical and, therefore, achiral.

C*

C*

H

CH3

CH3

H

The starred carbons of cis-1,2-dimethylcyclopentane have four different

substituents: these carbons are stereocenters. However, because the two carbons are

related as mirror images, the molecule is symmetrical, that is, achiral.

3) Drawing chiral molecules requires that you represent this 3-D feature clearly

and unmistakably in two dimensions. You are familiar with the wedge and dash (dotted-

line) method, which I have used above; this method is usually best for showing the 3-D

relationships in chiral compounds.

For drawing 3-D projections of substituted alkanes, it is convenient to place the

main carbon chain as a horizontal zig zag in the plane of the paper (the book does NOT

always do that). We will use this convention whenever possible. Also, if possible we


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place a H on a stereocenter on a dotted line and the other substituent on the wedge. The

projection of the two enantiomers (mirror images) of 2-bromopentane are shown below.

H Br HBr

Please, note that the dash and the wedge should always point away from the chain

(this is a consequence of the carbon being tetrahedral).

4) Naming enantiomers - absolute configuration

In order to properly name a stereocenter we have to take two steps:

“rank” the four substituents; a (high) – b – c – d (low)

determine their arrangement in 3-D space as either R (rectus) or S (sinister)

4a) We rank substituents according to a convention by Cahn, Ingold, Prelog,

following these rules:

Rule 1 atomic number of substituent

High atomic number has preference over low

F > O > N > C > > > > H

I > Br > Cl > S > O

Rule 2 For groups with the same atomic number (if there is a tie in the atomic

numbers) break the tie by ranking its substituents to the first point of difference: the

substituent with the first point of difference in priority wins. Since we are dealing with

compounds containing many carbon atoms, there is almost always a tie. For example,

compare methyl (C with 3 H’s) to ethyl (the attached C has 2 H’s and one C), propyl (the

attached C has 2 H’s and one C) and isopropyl (the attached C has one H and two C’s;

the 2 C’s are the point of difference).


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CH2–CH3 > CH2–H

CH(CH3)2 > CH2– CH3

C(CH3)2 > CH–CH2–CH3

Note: we don’t “weigh” the entire group, we look for the first difference; higher ranking

substituent further away do not matter.

CH2-CH(CH3)2 > CH2-CH2-CH2-CH3

O–CH3 > O-H

2-methylpropyl > butyl

CH2-Cl > CH2-CH2-Br

CH2NH2 > C(t-Bu)3

Rule 3 multiple (double or triple) bonds

H(C = C) treated as H(CC2)

C≡C treated as CC3

HC=O treated as HCO2

HC=S treated as HCS2

Rule 4 isotopes (they really thought of everything )

heavier isotope has priority

D > H

13C > 12C (not often encountered)

4b)Once you have ranked the substituents by priority, determine the arrangement

of the four substituents in 3-D space (R or S)

i) Direct (point) the lowest priority group, R4 away from you:


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ii) Draw an arrow connecting the substituents in order of decreasing priority:

C

R2

R3

R1

C

R2

R3

R1

R Rectus S Sinister

The arrow in the example on the left is clockwise – we call this 3D arrangement R

(for Latin rectus); the mirror image (shown on the right) requires a counterclockwise

arrow – we call this S (for Latin sinister).

Let’s look at 2-bromopentane.

H BrH lowest: d

CH3 second lowest: c

Br highest: a

C3H7 second ranking: b

a

c

H Br

b

clockwise – R

The bromopentane shown (with a clockwise arrow) is the R-enantiomer, or R-2-

bromopentane.

Its enantiomer, which requires a counterclockwise arrow to connect the

substituents in order of decreasing priority, is the S-enantiomer, S-bromopentane.


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HBra

cbcounterclockwise – S

5. Detecting stereoisomers

Is there a property that allow us to distinguish between enantiomers? Enantiomers

are identical in every respect except for their “handedness” (their asymmetry or chirality).

If we compare some properties of geometric isomers with stereo isomers, we note

Geometric Isomers Stereo Isomers


identical connectivity identical


different heat of combustion identical (enantiomers)

Differences between enantiomers are revealed only by asymmetric probes. The

best-known probe is plane polarized light – therefore stereo isomerism also is called

optical isomerism. How does this work?

6. Optical Rotation

Light is electromagnetic radiation with an electric and a magnetic component

perpendicular to one another. Light has a dual nature, two different ways how light

manifests itself: a) a quantized nature (light = photons, hν); you learned about this nature

of light in the initiation step of free radical halogenation); b) light has wave character (we

use this concept here).

i. Ordinary light vectors oscillate randomly (in all directions); it is symmetrical.

When ordinary light is passed through a Nicol prism (polarizer), only light whose


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electric and magnetic vectors oscillate in one specific direction (plane) can pass: the light

becomes plane polarized light (only one vector shown; electric and magnetic vectors are

still ⊥).

ii. When plane polarized light passes through a solution of chiral molecules the

plane of polarization is rotated by a certain angle in a certain direction, either clockwise

or counterclockwise (that’s why chiral compounds are said to be optically active).

Compounds that rotates light clockwise are called dextrorotatory (dexter is

Greek for right); they are designated by (+) or d in front of their names. Compounds that

rotate light to the left are called levorotatory (Greek for left); they are designated by a (–

) or l in front of their names. Please, note that there is no direct connection between the

direction of rotation (d or l) and the absolute configuration (R and S).

iii. Measuring Optical Rotation


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We can measure optical rotation, the direction and the degree (angle) of rotation,

using an instrument called a polarimeter. The measured angle of rotation, α, depends on

several factors, including the type of molecule and the number of molecules in the light

path; this quantity is given by concentration, c, of the chiral substance and the distance

light travels through the solution, the cell length, d. Other factors include the temperature

and wavelength of the polarized light.

We combine these factors in defining the specific rotation, [α], which is

measured in a solution of concentration, c = 1.0g/mL, and a path length, l = 1.0 dm. The

specific rotation is calculated from the measured rotation, α, by

observed rotation

Tλ

[α] = l × cα

specific rotation length concentration

The temperature and wavelength are indicated by superscripts and subscripts,

respectively.

iv. Absence of Optical Activity.

When we detect optical activity, we can be sure that a chiral compound is present.

But, failure to observe optical activity does not mean that no chiral compound is present.

Each compound with an asymmetric C atom has two enantiomers, each rotating light

with the same magnitude of specific rotation, but rotating light in opposite directions,

one to the right (+) and one to the left (–). If we have a 50-50 mixture of the pair of

enantiomers, we observe NO rotation at all, because the two rotations, equal in magnitude

but opposite in direction, cancel each other.

We call a 50-50 mixture of two enantiomers a racemic mixture or a racemate,

represented by (±). Racemic mixtures are found very often: reactions generating a chiral


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carbon in a symmetrical environment form a racemic mixture. Why? The energies are

identical, so entropy prevails. The conversion of one enantiomer into a racemic mixture is

called racemization.

7. Optical purity

In addition to 50-50 (racemic) mixtures of two enantiomers we may have

mixtures that are not racemic. This may happen if an environment is not completely

asymmetric or because racemization was stopped before completion. In such cases the

observed rotation is due to the excess of one enantiomer over the other. We can

determine the composition of the mixture (if we know the specific rotation of the

enantiomers) by comparing the rotation observed for the mixture to the rotation of the

pure enantiomers. We define the optical purity of the mixture as:

[α] observed × 100% % optical purity = –––––––––––––––––––

[α] for pure enantiomer

Assume that the optical purity is 60%; this means that 60% is one pure

enantiomer and that the remaining 40% is a racemic (50:50) mixture. 40% racemate

means that one half of this (= 20%) is the same as the dominant enantiomer; we have a

total of 60% + 20% = 80% of the dominant isomer in the nonracemic mixture.

8. Compounds with more than one asymmetric carbon

If there are no other factors, the maximum number of stereoisomers for a

compound with n asymmetric carbons is 2n. This means that for 2, 3, 4 asymmetric

carbons there are 4, 9, 16 stereoisomers (this is getting out of hand fast!).


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Let’s begin with the case where n =2 and two adjacent asymmetric atoms in a

chain, bearing two different substituents. We consider 2-bromo-3-chlorobutane. First we

draw one isomer and its mirror image:

H Br

Cl H

HBr

ClH

2-R,3-R 2-S,3-S

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By applying our “rank/assign3D” procedure we determine that these isomers are

the 2-R,3-R (left) and 2-S,3-S (right) stereoisomers; they are enantiomers.

If we exchange H and Br in these stereoisomers we change (invert) the absolute

configuration at that carbon, but we do not change the other carbon. The structures we

have drawn are the 2-S,3-R (left) and 2-S,3-R (right) stereoisomers.

Br H

Cl H

BrH

ClH

2-S,3-R 2-R,3-S

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You can see that these structures are mirror images of each other. But what is

their relationship to the two stereoisomers above? One stereocenter has the same absolute

configuration, but the second one is different. We call such compounds diastereomers.

Note that enantiomers have RR vs. SS or RS vs. SR coformation whereas

diastereomers have RR vs. RS or SR or SS vs RS or SR. We can summarize the

relationship of the four stereoisomers as follows:


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We have learned above that pairs of enantiomers have identical energies. In

contrast, diastereomers have different energies. So if we compare the different features

of stereoisomers, as we did above, we have to make one significant amendment:

Geometric Isomers Stereo Isomers


identical connectivity identical


identical (enantiomers) different heat of combustion different (diastereomers)

Diastereomers also have different physical and chemical properties so they can be

separated by crystallization or chromatography.

9. Compounds with two identically substituted stereocenters

A compound with two equally substituted carbons has three stereoisomers. We

consider 2,3-dibromobutane, related to the 2-bromo-3-chlorobutane discussed above.


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H Br

Br H

HBr

BrH

2-R,3-R 2-S,3-S

Br H

Br H

BrH

BrH2-S,3-R 2-S,3-R

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enantiomers

identical rotateC-2 by 180°2

BrHBrH

2

internal mirror plane

The R,R- and S,S- stereoisomers clearly are enantiomers (mirror images); in

contrast, the presumed diastereomers, S,R- and R,S- are actually identical. You can verify

this by rotating one of the stereocenters by 180°, as shown. The resulting structure has an

internal mirror plane: it is symmetrical. We call this stereoisomer a meso compound.

Thus, a compound with two equally substituted carbons has two enantiomers (forming a

racemate) and their one diastereomer, a meso compound.

10) Fischer projection

The German chemist Emil Fischer was a pioneer of the chemistry of

carbohydrates (sugars) that contain multiple –OH functions and stereocenters. In order to

visualize these compounds Fischer used a dash and wedge projection without actually

showing the dashes and wedges. The projection consists of intersecting perpendicular

lines representing the four bonds of an asymmetric carbon located at their intersection.

By convention, the two vertical lines are dotted, the two horizontal lines are wedges; the

main carbon chain is oriented vertically with the lowest numbered carbon at the top. This

means that all substituents are eclipsed. We can derive the Fischer projection of R,R-2,3-

dibromobutane in two steps as shown below.


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H

CH3

Brrotate

HC

CH3

C

Br

H Br

Br H

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CH3 CH3

BrH BrH

We will not use the Fischer projection; but note that an in plane rotation of a

Fischer projection by 180° will leave the absolute configuration unchanged, whereas a

rotation by 90° will convert S to R or R to S.

11. Ring compounds

Monosubstituted cycloalkanes cannot have an asymmetric carbon in the ring,

because there is a plane of symmetry. Disubstituted cycloalkanes are different; we

consider them on a case-by-case basis, depending on a) the pattern of substitution, b) the

ring size, and c) the particular geometric isomer we are considering with. For the purpose

of evaluating symmetry you may assume that the rings are flat.

i. 1,2-Disubstituted cyclohexanes have two stereocenters; if the substituents are

different, neither of the two geometric (cis- and trans-) isomers has a plane of symmetry.

The trans-isomers are enantiomers of each other; the cis-isomers are enantiomers of each

other; the two cis-stereoisomers are diastereomers of the two trans-isomers. These

features are general for all 1,2-disubstituted cycloalkanes.

H3C CH3Cl Cl

H3C CH3Cl Cl

HH H

H


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For 1,2-disubstituted cyclohexanes with two identical substituents, the cis-isomer

is a meso form, whereas the trans-isomer exists as a pair of enantiomers; there are three

stereoisomers (see section 9, above). This feature is general for all 1,2-disubstituted

cycloalkanes.

ii. 1,3-Disubstituted cyclohexanes, with different substituents also have two cis-

enantiomers and two trans-enantiomers. If the two substituents are identical the cis-

isomer has a plane of symmetry; it is a meso form and is optically inactive. trans-1,3-

disubstituted cyclohexanes have two stereo-isomers, they are enantiomers of each other;

they lack a plane of symmetry; each of them shares one stereocenter with the cis-isomer:

both are diastereomers of the cis-isomer. Can you assign the absolute configuration of the

two trans-1,3-dimethylcyclohexanes?

H3CCH3

H3C CH3

CH3H3C

iii. 1,4-disubstituted cyclohexanes have two geometric isomers; both are optically

inactive (achiral) because of a plane of symmetry that bisects C-1 and C-4 and their

substituents.

BrOH Br

OH

This feature is generally found for even-numbered cycloalkanes substituted in

opposite positions, for example, 1,3-disubstituted cyclobutanes or 1,5-disubstituted

cyclooctanes.


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12. Reactions of Compounds with Asymmetric Carbons

We need to consider two cases:

A. No bonds of the C* are broken in the reaction; the relative configuration of the

four substituents at the stereocenter is maintained, but the absolute configuration (the R,S

designation) may change because the priority of the substituents may change.

B. One or more bonds at the asymmetric carbon are broken and new bonds are

formed. The stereochemical outcome depends on the mechanism of the reaction; it may

either be specific or unspecific (random).

a) Some reactions, such as a free radical halogenation, proceed through a planar

intermediate. Even if the product has a stereocenter, it will be a racemic mixture. Such

reactions are said to be stereorandom.

b) Other reactions have stereospecific mechanisms, meaning a given stereoisomer

forms another given stereoisomer; the spatial relations of all the participants in the

reaction are specified (and predictable) without options. We will see examples of such a

reaction in Chapter 6. Stereospecific reactions can occur in two ways: i) in the same

position as the bond being broken; ii) the new bond is formed from the opposite side.

H XH Y Y Hsame side

replacement

opposite side

replacement

C) Examples

We illustrate both cases with a free-radical chlorination, a reaction you have

already studied; first we consider chlorination of an achiral substrate, butane; then, to

make things more interesting , we look at a chiral substrate, S-2-bromobutane. Since

chlorine atoms are not very selective, we expect chlorination to occur at all four centers.


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Chlorination of butane

The reaction of butane with Cl• generates a primary and a secondary free radical;

abstraction of Cl by the primary free radical forms 1-chlorobutane, an achiral product.

Simple. The case of the secondary free radical is more interesting because its reaction

generates a new stereocenter. The intermediate is planar; both sides are equally

accessible; both sides abstract Cl with the same probability: we obtain 2-chlorobutane as

a racemic mixture.

Cl•

•

H Cl

Cl• •

Cl

Cl H

enantiomers

R- S-

Chlorination of S-2-bromobutane

• Chlorination at C-4 leaves the chiral center unaffected (case A, above); the

priority of the substituents remains unchanged; we have formed S-3-bromo-1-

chlorobutane. • Reaction at C-1 also leaves the chiral center unchanged (case A), but the priority

of the groups is changed: in the starting material CH3 is the third-ranking substituent, in

the product CH3Cl ranks second. We have generated R-2-bromo-1-chloro-butane.

Br HBr H Br H

ClCl


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• Reaction at C-3 also leaves the chiral center unchanged (case A), but we are

facing two problems. First, substituting an H at C-3 will create a new stereocenter,

resulting in a pair of diastereomers. Because we start with the S-enantiomer the product is

a mixture of (2-S,3-S)- and (2-S,3-R)-2bromo-3-chlorobutane. We call a pair of

hydrogens that give rise to diastereomers upon replacement of one or the other

diastereotopic. Please, verify the assignment (see also Figure 5.14).

Second, the free radical generated by hydrogen abstraction has two different faces

with different steric hindrance. For this reason, the two products will be formed in

unequal yields. and a reaction giving diastereomers in unequal amounts

diastereoselective.

Br HCl•

H-Cl

Br H

•CH3

Br H

Br H Br H

Cl HH Cl

H3C

H

diastereomers

• Chlorination at C-2 is an example of case B: hydrogen abstraction from C-2

produces an achiral free radical (C–2 is planar); reaction of the two faces forms two

different enantiomers: they are said to be enantiotopic. The planar intermediate (only

one conformation is shown below) allows each face to react equally likely with Cl2.

Therefore chlorine abstraction by the intermediate produces a 50:50 mixture of the two

enantiomers, a racemate; the product is optically inactive.


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Br HCl•

H-Cl

Br

•

Br Cl Cl Br

H

H3C H

H3C

Br

•

enantiomers

The above reaction is an example of a mechanism that is not stereospecific. Even

though the product, 2-bromo-2-chlorobutane, has an asymmetric carbon, it is formed as a

racemic mixture (see also Figure 5.13 and pg 196).

Effect of enantiomers on living creatures can be very different; thalidomide was

used as a sedative; the (+)-isomer has no side effects whereas the (–)-isomer is a

teratogen, causing serious birth defects in babies.

NH

O O

N

O

O

Do you recognize the stereocenter in this molecule?

NH

O O

N

O

ONH

O O

N

O

O

HH

organic chemistry chapter 5 stereoisomers h. d....

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