Drawing Fischer Projections

Fischer projections are yet another way to draw molecules. They are commonly used in Biochemistry as a simple way to depict sugars. However, drawing Fischer projections can be tricky at first. Here is what a Fischer projection looks like:


This figure is used for drawing Fischer projections. Specifically, glucose is shown in this example going from Fischer projection to bond line structure.

The horizontal bonds are coming out of the page as wedges, while the vertical bonds are going back into the page as dashes. Because all the horizontal bonds are wedges, the Fischer projection is drawn in the eclipsed configuration.


Many questions about drawing Fischer projections in Organic Chemistry require you to go from a Fischer projection to a bond line drawing. They can be tricky, but here is a method that vastly simplifies the process. This procedure works very well and requires almost no mental rotation, one of the hardest things about Orgo for many.

a. Start with a small turn of the molecule. It can be in either direction, but here I have shown it in only one orientation. This is the only step that requires some mental rotation:



There is a carbon atom at each point where there is a dash and a wedge, so this molecule has 6 carbons (including the 2 explicitly shown).

Some people have trouble seeing the way the molecules flip. Likewise, some have trouble seeing how the atoms that were originally on the left side of the Fischer projection become wedges, while those on the right side become dashes. The following figure shows a right hand making precisely the same flip as our molecule. Note that the thumb is pointing to the left at the beginning, and after the flip, it points up.




b. Rotate the molecule 90 degrees. Then, number each carbon.



c. In step b, we saw that we had a 6-carbon chain (this can change of course depending on the question asked), so we draw a normal, 6-carbon chain in bond line form. Note the orientation after drawing the normal chain. Some are in their “up” conformation, while others are in their “down” conformation.


If we apply this up/down system to the final structure in part “b,” notice that 1 and 6 are down, while 2, 3, 4, and 5 are up.

d.  Now we fill in the groups on each carbon. Notice that groups 1, 2, and 4 are in the same orientation for the drawings in b and c (1 is down in each, 2 is up, and 4 is up). This means that the way the atoms are arranged on this carbon stays the same (dashes stay dashes; wedges stay wedges). In 3, 5, and 6, the orientations are flipped (3 and 5 go from up to down, while 6 goes from down to up), so we must flip the dash and the wedge (wedges turn into dashes; dashes turn into wedges). So on carbon 3, the –OH is a wedge, but after flipping it downward, the –OH now becomes a dash and the –H becomes a wedge. No mental rotation is required here. Just remember that if the configuration changes, then the dashes and wedges change.


From here, you can find whether each stereogenic center is R or S.