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[answered] Models and Stereochemistry Introduction One of the most int


hi?andrew, thanks for the other answered question. Pls can you help answer this organic questions?


Models and Stereochemistry

 

Introduction

 

One of the most interesting aspects of carbon chemistry is a very subtle form of isomerism

 

that was predicted by the brilliant chemist Louis Pasteur in 1849 as he was studying the tetravalent

 

nature of carbon (that is, ability to form four covalent bonds). Pasteur correctly reasoned that if a

 

carbon atom bonds to four different types of atoms to create a tetrahedral geometry about itself,

 

mirror-image isomers that are not identical or superimposable, should be possible. Isomers of this

 

type, called stereoisomers, are based on sp3 hybridized carbon atom. Mirror-image isomers of this

 

type are called enantiomers. In order for this type of isomer to exist, it must be based on what is

 

known as an asymmetric or ?chiral carbon.? The word chiral comes from a Latin root meaning

 

?handedness.? This description is apt, since these molecules form mirror images of one another, just

 

as your left hand is a mirror-image of your right.

 

The factor which makes a chiral carbon special is a lack of symmetry. If a plane of

 

symmetry can be passed through a carbon, then it is not be chiral and enantiomers will not be

 

possible. A plane of symmetry splits the molecule (including the carbon of interest) into two mirrorimage halves. If a plane of symmetry cannot be found for the carbon and its attached groups, then it

 

will be chiral. An easier method for recognizing a chiral carbon is to inspect the groups or atoms that

 

are attached directly to the carbon. If the carbon has atoms of four different elements or groups

 

attached to it, it will be chiral. Note that it will only be possible to attach four different groups if

 

carbon is sp3 hybridized.

 

One of the interesting characteristics of enantiomers is that they have identical physical

 

properties with one exception?the rotation of plane-polarized light. Although a discussion of planepolarized light is beyond the scope of this exercise, you should be aware that this property can be

 

measured and used to identify and characterize these molecules. Your Textbook describes plane

 

polarized light. When placed in a polarimeter, one of the isomers will rotate the plane of polarized

 

light a characteristic number of degrees in a clockwise direction. The mirror-image isomer will

 

rotate the plane of polarized light an equal amount in the opposite or counterclockwise direction.

 

When such a compound exhibits the ability to rotate the plane of polarized light, it is said to be

 

optically active. Isomers that rotate plane polarized light in the clockwise (dextrorotatory)

 

direction are given the designation (+). Conversely, isomers that rotate plane polarized light in the

 

counterclockwise (levorotatory) direction are given the designation (-). The terms dextrorotatory

 

and levorotatory were the basis of an older designation, where the letters D and L were used to

 

represent carbohydrate isomers. When both isomers are present in equal concentrations (a racemic

 

mixture), they will optically counteract one another and the mixture will not be optically active and

 

will have no effect on plane-polarized light.

 

Research has shown that compounds produced and used in biological systems are usually

 

optically active. Enzymes, which catalyze all biological reactions, function by recognizing the three

 

dimensional characteristics of molecules, and are therefore capable of recognizing and/or producing

 

only one of the two possible enantiomers. Compounds produced synthetically (in the laboratory),

 

however, are formed in random fashion, and even if they contain chiral carbons, they will be formed

 

as racemic mixtures and will be optically inactive. The importance of optical isomers to living

 

systems cannot be over-emphasized. Biological systems are extremely sensitive to the differences in

 

these isomers. For example, biological systems are capable of utilizing only the L-amino acids and

 

D-glucose. Many pharmaceuticals and nutritional supplements exist as enantiomers, where one

 

isomer may be much more potent than the other. The sensory organs are also capable of

 

distinguishing the differences between enantiomers. For example, the enantiomers (+)-carvone and

 

(-)-carvone produce the aromas associated with spearmint and caraway seeds respectively. Models and Stereochemistry, Page 1 Although it is possible to measure the optical rotation that an enantiomer exhibits, until 1949

 

nobody had determined the arrangement of atoms or groups about a central atom. When this

 

information about a compound has been established, its absolute configuration is said to be known.

 

Although the absolute configuration of a compound may not be known, it is still important to

 

be able to identify a particular configuration. In order for organic chemists to communicate with one

 

another about the precise configuration around a chiral carbon, a method has been devised to specify

 

a configuration. The R/S designation refers to the actual configuration about a chiral carbon, but it

 

tells the chemist nothing about the molecule's effect on polarized light. The R/S designation is just a

 

convenience used to communicate a specific configuration. In order to designate a molecule's

 

configuration, perform the following steps:

 

1. Draw a representation of a chiral carbon that shows the three dimensional arrangement of the

 

molecule. Several methods are shown below. Their meaning is described in more detail in

 

your textbook.

 

A E E A A c B B

 

D

 

D Ball and Stick

 

Model Line and Wedge E B D Dash and Wedge

 

Projection Fischer Projection Each of these representations has advantages and disadvantages. However, you should

 

become familiar with the three dimensional meaning of each of them. The wedge-bonds are

 

used to create a perspective of depth where necessary, and the line-bonds are considered to

 

be in the plane of the paper. The third representation utilizes wedges and dashes. The dash

 

bonds represent bonds that angle away from the reader (below the plane of the paper). The

 

wedges represent bonds that angle forward, toward the reader (up out of the plane of the

 

paper). The Fischer projection on the right has the same orientation of bonds as the dash and

 

wedge representation. Although it appears to show less detail, if you understand its meaning,

 

it is probably the simplest and most convenient representation to use.

 

2. Once you have correctly represented the molecule using one of the models or representations

 

above, you will assign a priority to each group or atom attached to carbon. Priority is based

 

on the atomic number of each atom directly attached to carbon. If two identical atoms are

 

directly attached to carbon, the priority of those atoms will be based on what is attached to

 

them. Whichever of the atoms is bonded to atoms of greater atomic number will be given a

 

higher priority. More complete details for prioritization, including rules for multiple bonds,

 

can be found in your textbook.

 

3. Once priorities are established, you must re-orient the molecule in space to point the lowest

 

priority group away from you. When this done properly, the low priority group should be

 

behind the central carbon atom from your perspective (see Figure 1 below). Once this

 

perspective has been created, you will examine groups with priorities 1, 2 and 3. These

 

groups should be directed slightly toward you and will appear to be 120? apart. If, in

 

proceeding from group 1 to group 2 to group 3, you trace out a clockwise arc, the enantiomer

 

is designated R. If, on the other hand, in going from group 1 to group 2 to group3, you trace

 

out a counter- clockwise arc, the enantiomer is designated S. An example of the procedure

 

Models and Stereochemistry, Page 2 described above is shown in Figures 1 and 2 below. You will probably find that this

 

procedure is most easily performed with models at first. However, you should be able to

 

designate R or S configurations with any of the representations shown previously. Be sure

 

that in thinking of these molecules that you do not confuse R and S designations with the

 

actual optical rotations that these molecules can exhibit. The atom with f ourth

 

priority has been moved

 

to the back.

 

4

 

3 C 3

 

2 Reorientation 4C 2 1 1 Molecule with priorities assigned

 

Figure 1: Re-orientation of molecule for configuration assignment 3 4C 2 1 Moving from the atom with the first priority to the second and on to the third is

 

counterclockwise movement. Therefore, the molecule has the S configuration

 

Figure 2: Assignment of "S" configuration to a molecule

 

Biological systems, as discussed previously, tend to produce optically active compounds.

 

These compounds often contain more than one chiral carbon. In general, n chiral carbons will exhibit

 

2n stereoisomers. Ordinarily, however, only one of those stereoisomers will be recognized or utilized

 

by an organism. For example, open-chain glucose contains four chiral carbons. There are therefore,

 

sixteen different stereoisomers in this carbohydrate family. However, the enzymes which process

 

glucose will only process glucose. Among that family of sixteen isomers, one of them will be a

 

mirror-image isomer (enantiomer) of glucose. The others, although similar, will not be enantiomers

 

of glucose. They will be stereoisomers called diastereomers. Glucose will then be a diastereomer of

 

these other fourteen isomers in the family.

 

Occasionally, when a compound contains multiple chiral carbons, it will be possible to draw

 

a plane of symmetry through the molecule. If that is the case, the compound will be known as a

 

meso compound. The meso compound will be identical to its own enantiomer, and it will be

 

optically inactive. You should suspect the possibility of a meso compound if a molecule exhibits a

 

form of symmetry (i.e. it looks the same on both ends). Models and Stereochemistry, Page 3 Name ____________________________________ Problems; Chiral Carbon

 

In this laboratory exercise you will use models to help you understand this type of subtle

 

isomerism. You will also learn what type of structural features make this type of isomerism possible.

 

You will also discover that it is possible to have more than one chiral carbon in a single molecule.

 

You will also learn how to use some of the more common designations and representations for these

 

compounds. Some of the exercises involve model construction, and some of the exercises will

 

require you to represent these isomers on paper.

 

1. Construct a model with a black carbon atom at the center, and to it attach red, green, grey and

 

white atoms. The different colored atoms will be used to represent four different groups

 

attached to the carbon. After you have constructed the model, place it on the bench in front of

 

you and construct another model that is the mirror image of the first. These two models

 

represent enantiomers. Have your instructor inspect and approve your models.

 

Instructor initials: ______________

 

2. Examine your enantiomer models closely. Is it possible for you, by rotating and turning them

 

in space, to superimpose them?

 

Ans: ________________________ 3. Is it possible for a plane of symmetry to bisect either of your models?

 

Ans: ________________________ 4. Now replace the red atom on each model with another white atom. possible now to imagine a

 

plane of symmetry that would bisect the molecules into mirror-image halves?

 

Ans: ________________________ Ans: ________________________ 5. Is it possible now to superimpose the two models? 6. Is it possible for an sp3 carbon to be bonded to two identical atoms or groups and be chiral?

 

Ans: _________________________ Models and Stereochemistry, Page 4 1. Mark with an asterisk all chiral carbons in the structures below: H 3C OH OH CH3

 

H3C

 

OH Br Cl H 3C CH CH3 H 3C CH CH2 CH3 Br Problems; R and S Designations

 

1. Restore the models you constructed in the previous exercise to their original forms. Use the

 

models to help assign R or S configurations to the following structures. Note: c and d are

 

Fischer projections.

 

a) Cl

 

Br C I H

 

Ans: b) H H3C C OH CH2 CH3 Ans: Models and Stereochemistry, Page 5 COOH

 

H CH3 c) OH Ans:

 

d) H C

 

3

 

H CH CH 3 CH3

 

CH2 CH3 Ans:

 

2. Use the Fischer projection diagrams to correctly draw the following compounds. If you don't

 

know the structure by name, look it up in one of the references.

 

a) (S)-2-bromobutane b) (R)-3-methylhexane c) (R)-alanine (an amino acid) Models and Stereochemistry, Page 6 d) (S)-lactic acid Problems; Multiple Chiral Carbons

 

1. Using the enantiomer models you created originally, remove the white atoms, and connect the

 

two carbon atoms together. You will note that you now have two chiral carbons in your

 

molecule. Now construct a mirror image molecule of the two-carbon molecule you just created.

 

Have your instructor inspect and initial your two enantiomers.

 

Instructor initials: ______________ 2. Are the two molecules superimposable?

 

Ans: ________________________ 3. . Is it possible to pass a plane of symmetry through either of these isomers?

 

Ans: ________________________ 4. Are these two molecules enantiomers or are they really identical?

 

Ans: ________________________ Ans: ________________________ 5. What is the name associated with this type of molecule? 6. Now construct two identical (not enantiomers) models like the ones you originally made (i.e.

 

attach red, green, grey and white atoms to a central black carbon atom). Now remove the white

 

atoms and connect the carbon atoms together as you did before. Do you still have two chiral

 

carbons in the molecule?

 

Ans: ________________________ 7. What is the relationship of this two-carbon molecule to the two carbon molecule that you

 

constructed previously?

 

Ans: ________________________ 8. Is it possible to pass a plane of symmetry through this molecule?

 

Ans: ________________________ Models and Stereochemistry, Page 7 9. Now construct a mirror image molecule of the two-carbon molecule you just created. Does this

 

molecule exhibit a plane of symmetry?

 

Ans: ________________________ Ans: ________________________ 10. Is this isomer superimposable with its mirror image? 11. Are these molecules identical or are they enantiomers?

 

Ans: ________________________

 

12. Assign R or S configurations to each of the chiral carbons in open chain glucose below.

 

CHO

 

H

 

HO OH

 

H H OH H OH

 

CH 2OH 13. Draw Fischer projections of all possible stereoisomers of tartaric acid and label the chiral

 

carbons R or S. Be sure not to draw the same structure twice. Models and Stereochemistry, Page 8

 


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