1. Write down the rules for the Fisher Convention.

ANS: Rule 2.1, 2.2 ,2.3 and 2.6 of CBM-2 (1.5  points for each =6 points)

2. Consider the following compounds:





a. Assign the configuration using the Fisher Convention.

b. Explain how you made the assignment.

ANS: They are both L. The H and OH are oriented correctly in the glycerol phosphate structure. The hydroxymethyl group can be oxidized to an aldehyde without involving  the chiral carbon. Glyceraldehyde 3 phosphate is the product. The aldehyde is on top as required. The OH is on the left so the compound is L. (5 points)

The H and OH in the lactic acid structure point away from us.  This violates Rule 1 of the Fisher convention.  It can be reoriented:




All the groups are now oriented correctly. The carboxyl can be reduced to an aldehyde  without involving  the chiral carbon. This gives L-glyceraldehyde. (15 points)

3. Assign the configuration of the structures in question 2 using the R/S Convention.


a. We are dealing with three dimensional objects so we have to be careful when we convert them to two dimensions.  Glycerol 3-phosphate is oriented correctly, i.e. the H and OH point towards us, so we can assign priorities and then  flatten the structure into two dimensions:






Carrying out two switches to put the d on the bottom and maintain  the original configuration gives:





The structure we were given is R.  (5 points)

b.  The structure of glyceric acid is oriented with the H and OH pointing  away.  We must reorient the structure before flattening  it to two dimensions. Then, carrying out the double switch gives:





( 6 points)

Let us compare the original  and the reoriented structures. Note what happens when we flatten them into 2D:








Analysis of the two structures is going to give opposite results even though we are dealing with a single compound. The orientation of the two dimensional structures is critical. For both the Fisher and the R/S Conventions, the H and OH or their equivalents, must point toward us. Of course, if we think in 3D, there is no problem. Can you see the “steering wheel” in 3D? That is the safest way to proceed.

4. Isocitric acid is (2R,3S)-1-hydroxy-1,2,3-tricarboxypropane. Draw the structure of isocitric acid.










(10 points)



5. Define

a. Enantiomers

ANS: Enantiomers are mirror images. (5 points)

b. Diasteriomers

ANS: Diasteriomers are stereoisomers that are not mirror images. (5 points)

6. Write the open chain structures for the aldopentoses.

ANS: Note: The names are included, but they are not a necessary part of your answer.











( 2 points for each structure = 16 points)

a. Identify the enantiomers.

ANS: The enantiomers are D and L with same name. There are 4 enantiomeric pairs (D,L). . (2 points)

b. How many diasteriomers does each aldopentoses have?

ANS: Each aldopentose has  6 diesteriomers. (2 points)

7. Identify the meso carbon in the structure below:





(3 points)

8. Number the glycerol carbons using the stereospecific numbering system.








(10 points)


9. Name the following structure:












(10 points)


Grade Yourself: 90-100 = A, 75-89 = B, 50-74 = C, 25-49 = D, <25 = F













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