Evan B. answered • 06/07/23
Professional Chemist with Extensive Teaching and Research Experience
To approach this question, consider a fisher projection as a 2D representation of a molecule with bonds rotated to resemble a ring. Sugars are especially known to open and close in the body constantly, so I imagine biochemists figured it was easier to draw them in their "ring like" orientation for the sake of ease when drawing mechanisms.
To do this problem specifically, we can use a method known as the "Superman" visualization. Start by drawing the carbonyl and imagine a man flying above the molecule in the direction toward the carbonyl. It is important to make sure the carbon you are evaluating is "pointed at" the observer you draw above the molecule. Since the Hydroxyl (OH) group on carbon 2 is represented by a wedge, it would be located to the left of the man when looking downward from above. Because of this, we will say it's on the left and draw it on that side of our fisher projection.
Moving to carbon 3, we now fly underneath the molecule toward the direction of the carbonyl while ensuring that the carbon being evaluated is facing the observer. In this orientation, the Hydroxyl group is on a dash and therefore to the left of said observer. Because of this, we will draw the Hydroxyl on the left of the fisher projection.
Lets do carbon 4. We are flying above again. Since the OH is on a dash, it would be to the right of the observer. We therefore draw this on the right side when placing it on the fisher projection.
If you follow this procedure, you will always get these questions correct. It's also important to note that since stereocenters are either R or S, we can verify our answers by comparing the configuration of each carbon on a fisher projection to the original zip zag model. Additionally, the direction in which you choose to fly (i.e. left to right vs right to left) does not matter as long as each carbon is evaluated the same way.
Dont hesitate to reach out if you need more assistance.
