In This Chapter
^ Assigning Cis – And Trans – Configurations
^ Reflecting on enantiomers and diastereomers
W^tow About this: Two organic molecules have identical chemical formulas. Each atom in Ґ m One molecule is bonded to the same groups as in the other. They’re identical molecules, right? Wrong! (Mischievous chemistry gods point and snicker.) Many organic molecules are Isomers, Compounds with the same formula and types of bonds, but with different structural or spatial arrangements. Who cares about such subtle differences? Well, you might. Consider thalidomide, a small organic molecule widely prescribed to pregnant women in the late 1950s and early 1960s as a treatment for morning sickness. Thalidomide exists in two isomeric forms that rapidly switch from one to the other in the body. One isomer is very effective at combating morning sickness. The other isomer causes serious birth defects. Isomers matter.
Isomers can be confusing. They fall into different categories and subcategories. So, before committing your brain to a game of isomeric Twister, peruse the following breakdown:
Structural isomers Have identical molecular formulas but differ in the arrangement of bonds.
Stereoisomers Have identical connectivities — all atoms are bonded to the same types of other atoms — but differ in the arrangement of atoms in space.
• Diastereomers Are stereoisomers that are Not Non-superimposable mirror images of each other. Two types of diastereomers exist: Geometric isomers (or Cis-trans isomers) Are diastereomers that differ in the arrangement of groups around a double bond, or the plane of a ring. Conformers And Rotamers Are diastereomers that differ because of rotations about individual bonds (we don’t cover them in this book because they’re beyond the scope of general chemistry).
• Enantiomers Are stereoisomers that Are Non-superimposable mirror images of each other.
Chapter 21 neatly handles structural isomers, describing how to recognize and name them appropriately. This chapter focuses on the trickier category: stereoisomers.
Picking Sides with Geometric Isomers
Geometric isomers Or Cis-trans isomers Are a good place to start in the world of stereoisomers because they’re the easiest of the stereoisomers to understand. In the following sections, we explain how isomers relate to alkenes, alkanes that aren’t straight-chain, and alkynes (see Chapter 21 for an introduction to these structures).
Alkenes: Keen on cis-trans configurations
Straight-chain alkanes are immune from geometric isomerism because their carbon-carbon single bonds can rotate freely. Unsaturate (or add another bond to) one of those bonds, however, and you’ve got a different story. Alkenes have double bonds that resist rotation. Furthermore, the Sp2 Hybridization of double-bonded carbons gives them trigonal planar bonding geometry (see Chapter 5 for an introduction to hybridization). The result is that groups attached to these carbons are locked on one side or the other of the double bond. Convince yourself of this by examining Figure 22-1.
Figure 22-1:
Cis And Trans Isomers of an alkene.
H3C
3 ^\
CH
YC=C\
H3C
H
/C = C\
H CH

In Figure 22-1a, the carbon chain continues along the same side of the carbon-carbon double bond. Both methyl (-CH3) groups lie on the same side of the unsaturation. This is called a Cis Configuration. In Figure 22-1b, the carbon chain swaps sides as it proceeds across the double bond. The methyl groups lie on opposite sides of the unsaturation. This is called a Trans Configuration.
Naming Cis-trans Isomers is simple. Attach the appropriate Cis – Or Trans – Prefix before the number referring to the carbon of the double bond (refer to Chapter 21 if you don’t know how to name alkenes). For example, Figure 22-1a is Cis-2-butene, while Figure 22-1b is Trans-2-butene.
A)
B)
H
Alkanes that aren’t straight-chain: Making a ringside bond
Although straight-chain alkanes happily avoid isomerism by rotating merrily about their single bonds, the four bonds of sp3-hybridized carbons assume tetrahedral geometry. Detailed representations like the one shown for methane in Figure 22-2 reveal this geometry. In the structure of methane, the bonds depicted as straight lines run in the plane of the page. The bond drawn as a filled wedge projects outward from the page. The bond drawn as a dashed wedge projects behind the page. These filled and dashed wedge symbols are known as Stereo bonds Because they’re helpful in identifying stereoisomers.
Figure 22-2:
Stereo bonds in methane.
C
HH
»H
H
When alkanes close into rings, they can no longer freely rotate about their single bonds, and the tetrahedral geometry of sp3-hybridized carbons creates Cis-trans Isomers. Groups bonded
To ring carbons are locked above or below the plane of the ring, as shown in Figure 22-3. The figure depicts two different versions of trans-1,2-dimethylcyclohexane. In both versions, the adjacent methyl substituents are locked in Trans Positions, on opposite sides of the ring.
The upper set of structures shows the plane of the ring as seen from above and highlights the Trans – Configuration of the methyl groups with stereo bonds.
The lower set of structures shows the same rings rotated 90 degrees downward and toward you.
H, C,
CH, ____• ,
CH CH,
,
CH CH,
Figure 22-3:
Two isomers of Trans-1,,-Dimethylcy-clohexane.
H3C
3
CH CH,
H3C
3
CH CH,
CH

The Trans Configuration of the methyl groups is most clear in the lower structures. The front-most methyl group is highlighted with explicit hydrogen atoms to emphasize its position above or below the plane of the ring. (Astute readers may ponder, if both are Trans, Why are they different? The answer to this mystery lies in the upcoming "Staring into the Mirror with Enantiomers and Diastereomers" section.)
Alkynes: No place to create stereoisomers
Alkynes also contain carbon-carbon bonds that can’t rotate freely. However, the Sp Hybridization of the carbons in these bonds leads to linear bonding geometry. The two-carbon alkyne, ethyne, is shown in Figure 22-4. Each carbon locks three of its valence electrons into the axis of the triple bond. Each has only one valence electron remaining with which to bond to hydrogen. No Cis-trans Isomerism is possible in this scenario.
Figure 22-4:
No iso-merism is possible at
The triple H-CC-H
Bonds of alkynes, such as ethyne.

Q. Draw the structure of Cis-3-hexene.
A. The name helpfully informs you that
You’re dealing with a six-carbon alkene, and that the double bond occurs between the third and fourth carbon atoms. The Cis – Prefix tells you that the carbon chain continues along the same
Side of the double bond, as drawn in the following figure:
HH
/C=C\
H3C
3
CH
CH
CH
1. Name the structure shown in the following figure:
/
CH, CH3
,3
H3C C H, C H,
CH2 CH2
22
/ \
H3C
3
2. Draw the structure of Trans-2-pentene. Solve It
H
Solve It
3. Draw the Cis- And Trans – Isomers of 3-heptene.
Solve It
4. Can geometric isomerism occur across the first and second carbons in a chain? Draw the structure of 1-butene. Are different geometric isomers possible across the double bond in this molecule? Explain why or why not.
Solve It
Staring into the Mirror with Enantiomers and Diastereomers
Geometric isomers are really just prominent types of Stereoisomers, Compounds that differ only by the arrangement of groups in space. Geometric isomers belong in the category of Diastereomers, Stereoisomers that aren’t non-superimposable mirror images. This section introduces a devilish category of stereoisomers called Enantiomers, Isomers that Are Non-superimposable mirror images of each other. Telling the difference between enantiomers and diastereomers can take some practice. You can practice here.
Getting a grip on chirality

Not all mirror images are superimposable on one another. This fact is so fundamental that it may have escaped your attention. If you doubt the truth of it, just try this: Extend your fingers and thumbs so that each hand makes an L-shape. (You’ve just synthesized two L-shaped molecules — well done.) Now, try to orient your L-shaped hand-molecules so both palms face upwards and so your fingers and thumbs all point in the same directions. . . At the same time. It can’t be done without serious injury. That’s because your L-shaped hand-molecules are Chiral, Meaning they have the property of non-superimposability with their mirror images. Molecules must be chiral in order to be enantiomers.
Carbon atoms can be chiral, too. When sp3-hybridized carbons bond to four different groups, those carbons have chiral geometry and can form Chiral centers Within molecules. Compare the two molecules shown in Figure 22-5, remembering to visualize the projections of the stereo bonds drawn as wedges (see the earlier section "Alkanes that aren’t straight-chain: Making a ringside bond" for more details). Rotate the molecules in your mind, trying to superimpose them. Although this is potentially a safer experiment than trying to superimpose your hands, it stands no greater chance of success. Chirality is what it is.
Figure 22-5:
Each carbon center bonds to the same set of four different partners, but these two chiral carbon atoms are not superim-posable.
Br
Cl
Cl
C
Br F
Molecules with chiral centers are often — not always — chiral. Chiral molecules often — always — possess chiral centers. Chiral centers are important enough that you should understand what they are and seek them out.
Not
C
F
Depicting enantiomers and diastereomers in Fischerprojections
Chains of tetrahedral carbon atoms show up so frequently in organic molecules that chemists have devised shorthand methods for drawing their structures. One such method is the Fischer projection. Fischer projections are a simple way to condense the three-dimensional reality of tetrahedral carbon onto a two-dimensional page.
.jjSiJj^ff In a Fischer projection, a bonded chain of tetrahedral carbon atoms is depicted as a vertical «Ґ\ line. Horizontal lines represent other bonds projecting from these central carbon atoms.
Each intersection between lines represents one carbon atom. Vertical lines symbolize bonds that project away from you. Horizontal lines represent bonds that project toward you. Examples of two molecules drawn as Fischer projections are shown in Figure 22-6.
Figure 22-6′
Enantiomers of 2-bromo-3-chlorobu-tane drawn as Fischer projections.
CH,
-Cl
- Br
CH
Cl Br
CH
CH
Fischer projections are convenient, but you have to be careful when using them to visualize bonds to make decisions about whether something is chiral.
First, you can only rotate these structures In the plane of the page. Don’t try to rotate them out of the page, or they’ll lose all their meaning.
Second, Only consider one carbon center at a time. If you try to simultaneously visualize the three-dimensional bonding of two adjacent carbon atoms on the vertical chain, you’ll only get yourself into trouble, and you may well burst a blood vessel.
With those caveats in mind, take a closer look at the two structures in Figure 22-6. Do you see how no amount of sliding or rotating them in the plane of the page can superimpose them? That’s because these two molecules are enantiomers.
Compare that situation with the one presented by the two molecules shown in Figure 22-7. Do you see how rotating one of the molecules 180 degrees in the plane of the page allows you to superimpose the two? That’s because these two molecules are diastereomers.
H
H
Figure 22-7′
Diastereomers of 2,3-dichlorobu-tane drawn as Fischer projections.
CH
Cl
Cl
CH
Cl Cl
CH
CH
H
H

You protest — how can it be? Both sets of molecules contain chiral centers! That’s true, but remember: Not all molecules with chiral centers are themselves chiral. In fact, a special term exists for these chiral-but-not-chiral molecules. They are called Meso compounds. Many meso compounds pull off this trick by having an internal plane of symmetry. In other words, they are their own mirrors. Narcissistic little buggers.

Q.
In the structures shown in the following figure, fill in the missing substituents so the two molecules become enantiomers.
H, C -
3
CH2CH2CH3
, , 3
-CH, CH,
CH2CH3
,3
A. To make the two molecules enantiomers, simply fill in the missing pieces so the
Molecules are mirror images of each other, as shown in the following figure. The pre-existing substituents guarantee that no pesky Meso Symmetry lurks in the shadows.
H3CH2C -
3 ,
H3C
3
CH2CH2CH3
, , 3
CH2CH2CH3
, , 3
-CH2CH3
, 3
CH
CH2CH3
, 3
CH2CH3
, 3
Redraw the structure shown in the following figure as a Fischer projection, identify the chiral centers, and draw the corresponding enantiomer.
OH O
! II
HO CH C
\ / \ / \ CCH OH
II I
OOH
SotVe It
6. Which of the structures in the following figure are enantiomers?
CH2CH3
, 3
H3CH2C -
3 ,
CH
CH
■OH H3CH2C -
3 ,
H
OH
CH2CH3
, 3
A)
H
H
H
CH2CH3
, 3
OH OH
CH2CH3
, 3
HO HO
CH2CH3
, 3
CH2CH3
, 3
B)
Solve It
7. Which of the structures in the following figure are Meso Compounds?
A)
C)
H HO H
H
Br -
CH2OH
OH H
OH
CH2OH
COOH
OH OH
COOH CH
Br H
CH3
OH H OH
OH OH
Br H
CH2OH
H
OH H
CH2OH
COOH
COOH CH
H Br
CH3
B)
Solve It
8. Identify the structures in the following figure as geometric isomers, enantiomers, or Meso Compounds. Name any geometric isomers (including Cis – Or Trans – Designations), identify all chiral carbon atoms, and draw any internal planes of symmetry.
A)
H
\
C
CC
,CH2CH2CH3
H3C
3
CH
OH OH
CH3
OH OH
CH
CH3
CH
H’Si
F

CH
II ^xV——- Si.

COOH H — C— NH
COOH
H N C H
H
H
B)
C)
D)
H
Solve It
Answers to Questions on Stereoisomers
D 4-methyl-frans-4-nonene. The structure contains a nine-carbon parent chain, with a double
Bond occurring between carbon atoms 4 and 5. The main chain swaps sides of the unsaturation as it proceeds. The molecule is also substituted at the fourth carbon with a methyl group.
Mb 7Vans-2-pentene is a five-carbon alkene with the double bond occurring between the second and third carbon atoms. The parent chain proceeds along opposite sides of the carbon-carbon double bond, as shown in the following figure:
H
H3C
3
,CH..CH3
/ \
H
MM The Cis – And Trans – Isomers of 3-heptene differ only in the orientation of the carbon chain about the double bond, as shown in the following figure:
HC
3
Cis
CH CH
/ \ HH
CH
/ CH
Trans
H
CH
H3C
3
\ / C^=C
/ \ ■ CH2 H
CH3
3
CH
MM Consider the structure of 1-butene, shown in the following figure:
H
\ / C^=C
/ \ HH
CH2CH3
23
The chain terminates at the double bond. The first carbon atom is bonded to two identical hydrogen atoms. No unique substituent group lies Cis Or Trans To the continuing parent chain.
EM This is partially a trick question. (Hey! At least we admitted it.) Although the structure has
Chiral centers, proper drawing of the Fischer projection (pay attention to stereo bonds!), seen in the following figure, makes it clear that the compound is Meso. The molecule has an internal plane of symmetry.
CO2H
-OH – OH
CO2H
MM The two structures shown in Figure A are enantiomers, Although they don’t appear to be at first glance. To make the enantiomeric relationship more obvious, rotate one of them 180 degrees in the plane of the page. The two structures in Figure B are not enantiomers. In fact, this is a Meso Compound.
MM The structures shown in Figures A and B are Meso Compounds. The structures shown in
Figure C are not Meso. Although the structures in Figure A don’t have an internal plane of symmetry, the two are superimposable mirror images. The structures in Figure B have an internal plane of symmetry and, in fact, are the same compound. The two structures in Figure C are non-superimposable mirror images, and are therefore enantiomers.
MM The structure in Figure A is a geometric isomer, Frans-2-hexene. The pair of molecules
Shown in Figure B are different orientations of a Meso Compound with two chiral centers and an internal plane of symmetry, as shown in the following figure. The plane cuts the molecule in half, across the bond connecting the second and third carbon atoms. The pair of structures in Figure C are enantiomers. The chiral center in this case isn’t carbon! Rather, it’s silicon, which sits right below carbon on the periodic table, and also bonds with tetrahedral geometry. The pair of molecules shown in Figure D aren’t isomers at all, But are simply different orientations of the amino acid glycine. The central carbon of the projection bonds to two identical hydrogen atoms, so no isomerism is possible.
CH,
CH,
■ OH
■ OH
OH OH
CH
CH
H
H
H
H