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Stereoisomerism: A Model Exercise

In this experiment you will construct models with your molecularmodel set that illustrate the concepts of chirality, chiral center(stereogenic center, asymmetric carbon atom), enantiomers,diastereomers, and meso forms. You will also learn about twoconventions, R-S and Fischer, for designating the configurations ofchiral molecules.

You will be asked several questions related to each constructionexercise. Record the answers and then when you have finished all ofthe exercises, take the assessment listed in this folder. Thequestions will be repeated so that you can submit your answerselectronically for grading.

Chiral centers

Procedure: Construct a model* in which a tetrahedral (sp 3)carbon atom (black) has four different model atoms attached to it.Use the light blue ball, red, blue and, green polyhedrons torepresent four different atoms or groups attached to the centralatom (black).

1. Does the model have a plane of symmetry?

a. Yes

b. No

A carbon atom that has four different groups attached to it is achiral center, or an asymmetric carbon atom. The carbon marked witha # in 3-methylhexane is a chiral center.

2. What groups are attached to the chiral center in3-methylhexane?

a. Hydrogen

b. Methyl

c. Ethyl

d. n-propyl

e. n-butyl

Replace the green atom (or group) in your model with a secondred atom. Now two of the groups attached to the carbon atom areidentical:

3. Does the model now have a plane of symmetry?

a. Yes

b. No

4. �� Describe it.

a. Plane through the light blue ball, black polyhedron and bluepolyhedron

b. Plane through a red polyhedron, black polyhedron and bluepolyhedron

c. Plane through the light blue ball, black polyhedron and redpolyhedron

d. Plane through a red polyhedron, black polyhedron and redpolyhedron

Draw structural formulas for the following compounds, and markany chiral centers (asymmetric carbons) with an asterisk:

l-bromobutane, 2- bromobutane, 1,2-dibromobutane,

1,3-dibromobutane, l, 4-dibromobutane, 2,3-dibromobutane.

5. Which of the compounds contain chiral centers?

a. 1-bromobutane

b. 2-bromobutane

c. 1,2-dibromobutane

d. 1,3-dibromobutane

e. 1,4-dibromobutane

f. 2,3-dibromobutane

Chirality and Enantiomers

A center of chirality (from the Greek cheir, hand) imparts theproperty of handedness to a molecule. In this part of theexperiment, the left- or right- handedness of molecules with achiral center will be illustrated with models.

A molecule is said to be chiral (that is, to have the propertyof handedness) if its mirror image is not superimposable. Themirror image of a left hand, for example, is a right hand. Amolecule that is achiral has a mirror image that is superimposible.We shall see that any molecule with a plane of symmetry isachiral.

Procedure: Reconstruct the original model (carbon (black) withlight blue ball, red, blue and, green polyhedrons attached). Setthe model on the desktop so that the substituent light blue ballatom points toward the ceiling.

Looking down on the model and proceeding clockwise from thegreen atom, record the colors of the three atoms that rest on thedesktop.

Now construct a second model that is the mirror image of thefirst, and place it on the desktop with the light blue ball atomup:

6. In which direction, clockwise or counterclockwise, must youproceed in order to list the same sequence of colors of the threeatoms resting on the desk's surface?

a. Clockwise

b. Counterclockwise

Try to superimpose the two models.

7. The models are

a. Superimposable

b. Not superimposable

The two models that you have just constructed represent chiralmolecules- they lack a plane of symmetry and have mirror imagesthat are notsuperimposable. Two substances, the molecularstructures of which are related as an object and itsnonsuperimposable mirror image are called enantiomers. They differfrom each other only in properties that have a direction or\"handedness,\" such as, for example, the direction(clockwise orcounterclockwise) in which they rotate a beam of plane-polarizedlight. Because of this latter property, such substances aresometimes called optical isomers. They are optically active.

Now we will examine the consequence of having at least twoidentical atoms or groups attached to a tetrahedral carbonatom.

Replace the green atom in each model with a red atom, so thateach model has two identical groups attached to the central carbonatom:

8. Are the models still mirror images?

a. Yes

b. No

9. Does either of the models have a plane of symmetry?

a. Yes

b. No

10. Are the models superimposable?

a. Yes

b. No

11. Do the models represent identical molecules or differentmolecules?

a. Identical

b. Different

Place each model on the desk so that the light blue ballsubstituent points up.

12. To define the same sequence of colors for the three atomsresting on the desk top, must you proceed

a. Clockwise

b. Counterclockwise

c. Either way

13. Are the models chiral (handed)?

a. Yes

b. No

The models you have just studied represent achiral molecules.Their mirror images are identical. They are optically inactive. Anymolecule that has a plane of symmetry is achiral.

Diastereomers and Meso Forms

For any molecule that has two or more chiral centers, it ispossible to have stereoisomers that are not mirror images.Stereoisomers that are not related as enantiomers arediastereomers. Diastereomers differ in all properties, chiral andachiral.

Procedure: Construct a model with four different groups (lightblue ball, blue, red, and green) attached to a central carbon atom(black). Construct another model identical to the first. (Be surethey are identical by making sure the models are superimposable.)Now remove the green substituent from each model and connect thetwo carbon atoms with a bond. Use this model with the twostereogenic centers for the next series of questions (14 -26).

14. How many chiral centers (asymmetric carbons) does this modelhave?

a. 1

b. 2

c. 3

d. 4

e. 5

Note that there are four different groups attached to eachchiral center and that each chiral center has the same four groupsattached.

15. Does the model have a plane of symmetry in any of itsconformations?

a. Yes

b. No

Construct the mirror image of the first model.

16. Is the mirror image identical to or different from the firstmodel?

a. Identical

b. Different

17. What term describes the two models?

a. Enantiomer

b. Diastereomer

18. Is each model chiral or achiral?

a. Chiral

b. Achiral

Now interchange a red and blue atom on the same carbon in one ofthe models.

19. Are the models identical or different now?

a. Identical

b. Different

20. Are they mirror images (enantiomers)?

a. Yes

b. No

21. Are they stereoisomers?

a. Yes

b. No

22. What term describes the two models?

a. Enantiomer

b. Diastereomer

Carefully examine the conformations of the model in which youinterchanged the red and blue atoms.

23. Does the model have a conformation with a plane ofsymmetry?

a. Yes

b. No

24. Would the mirror image of this model be identical(superimposable) or different from the model itself?

a. Identical

b. Different

Verify your prediction by constructing the mirror-imagemodel.

25. This model is

a. Chiral

b. Achiral

26. Would a molecule corresponding to this model be opticallyactive?

a. Yes

b. No

The last model studied here represents a meso form. The modelpossesses two chiral centers, but they are of equal and oppositechirality. This situation arises when a molecule has two identicalchiral centers. Because the molecule has a readily accessibleconformation with a plane of symmetry, it is achiral and opticallyinactive.

Tartaric acid is a molecule that corresponds to the modelsconstructed in this section of the experiment.

Tartaric Acid

It exists in three forms; two are optically active enantiomers,and the third is an optically inactive meso form that is adiastereomer of the optically active forms. ,

27. Draw Newman projection formulas (looking at the bond betweenC-2 and C-3) for the three tartaric acids. Label pairs ofenantiomers and diastereomers, as well as the meso form.

When a molecule has two different chiral centers, it may existin four optically active forms (two pairs of enantiomers). Toillustrate this with the models you just used, replace one of thecolored atoms on one of the carbon atoms with a black atom. Thereare now four distinct ways of constructing the models: two pairs ofenantiomers. Construct the two pairs of enantiomers.

28. What name is given to a pair of molecules consisting of onemolecule from each of the two pairs of enantiomers you justconstructed?

a. Enantiomers

b. Diastereomers

c. Meso forms

The R-S Convention

The letter R (from rectus, right) or S (from sinister, left) isused to designate the configuration at a chiral center. The fouratoms or groups attached to the chiral center are arranged in apriority order according to atomic number: the higher the atomicnumber, the higher the priority. If two atoms have the same atomicnumber, we move to the next atoms out from the chiral center, oreven further, until we observe a difference in atomic number. Wethen view the molecule from the side opposite the group with thelowest priority. If the remaining three groups in order fromhighest to lowest priority form a clockwise array, theconfiguration is R; if they form a counterclockwise array, theconfiguration is S.

Procedure: Construct a model of 2-chlorobutane.

29. Which carbon in the chain is a chiral center?

a. 1

b. 2

c. 3

d. 4

There are four groups attached to this chiral center

.

30. Which group has the highest priority?

a. Cl

b. Methyl

c. Ethyl

d. H

31. Which group has the lowest priority?

a. Cl

b. Methyl

c. Ethyl

d. H

32. What is the priority order of the other two groups?

a. Ethyl > Methyl

b. Methyl> Ethyl

Set the model on the desktop so that it can be viewed from theside opposite the hydrogen. Put the chlorine atom at the top.

33. When viewing the model, the three remaining groups inpriority- order sequence form a

a. Clockwise array and the model has a R configuration.

b. Counterclockwise array and the model has a Rconfiguration.

c. Clockwise array and the model has a S configuration.

d. Counterclockwise array and the model has a Sconfiguration.

Interchange any two groups attached to the chiral center.

34. What configuration does the model have now?

a. The same as before.

b. The opposite configuration.

Note that to change configuration we must disconnect and remakebonds, whereas we can change conformation by rotating groups aroundsingle bonds.

Fischer Projection Formulas

It is sometimes convenient to have a two-dimensionalrepresentation for three-dimensional molecules, particularly whenwe are studying stereoisomerism. One common convention, devised bythe German organic chemist Emil Fischer and named after him, isdescribed in this section

Procedure: Construct a model of an asymmetric carbon atom [witha black, a blue, a green, and a red ball attached] that correspondsto the following three-dimensional drawing:

Set the model on the desk as shown. With your left hand, tip themodel to the left so that only the red and green balls rest on thetable. Viewed from the top, the black and blue balls project towardyou (left and right, respectively).

The convention for Fischer projection formulas is as follows:The asymmetric carbon lies in the plane of the page or paper,horizontal groups come out of the plane of the paper toward theviewer, and vertical groups recede behind the paper away from theviewer.

Save the model you have just constructed for comparison withother models that you will construct.

Let us first consider the effect of interchanging any two groupsin a Fischer projection formula. Your model corresponds to formulaA below:

Consider formula A', with groups blue and green interchanged.Construct a model corresponding to A'.

35. Is A' identical to A, or is it the enantiomer of A?

a. Identical

b. Enantiomer

Now construct a model corresponding to A\" (like A, but withgreen and red interchanged).

36. Is A\" identical to A, or is it the enantiomer of A?

a. Identical

b. Enantiomer

37. What generalization can you make about the interchange ofany two groups in a Fischer projection formula?

a. The new molecule is an enantiomer of the first.

b. The molecule is identical to the first.

38. If you were to make an even number of group interchanges ina Fischer projection formula, would you obtain the originalmolecule or its enantiomer?

a. Original molecule

b. Enantiomer of the original model

39. Do the following Fischer projections represent the samemolecule or enantiomers?

a. Same molecule

b. Enantiomers

Rotation of a Fischer projection formula in the plane of thepaper also affects the structure it represents.

.

Consider formula B, which corresponds to the Fischer projectionformula of A rotated 90° clockwise in the plane of the paper. Builda model of formula B. (Remember Fischer Projection Formulas are twodimensional diagrams written on paper of three dimensionalmolecules.)

40. What minimum number of group interchanges in the model arenecessary to convert formula B back to formula A?

a. Zero

b. One

c. Two

d. Three

41. Is model B identical to model A, or is it the enantiomer ofmodel A?

a. Identical

b. Enantiomer

Formula C results from a 90° counterclockwise rotation in theplane of the paper of the Fischer projection formula of A. Build amodel of formula C.

42. Is model C identical to model A, or is it the enantiomer ofA?

a. Identical

b. Enantiomer

Formula D results from a 180° rotation in the plane of the paperof the Fischer projection formula A . Build a model of formulaD.

43. What minimum number of group interchanges are necessary toconvert model D back to model A?

a. Zero

b. One

c. Two

d. Three

44. Is model D identical to model A, or is it the enantiomer ofA?

a. Identical

b. Enantiomer

45. What generalization can you make about the rotation of aFischer projection formula in the plane of the paper?

a. 90° rotation results in an enantiomer, 180° rotation resultsin the identical molecule.

b. 180° rotation results in an enantiomer, 90° rotation resultsin the identical molecule.