# Kaplan Book - Fischer Projections

Discussion in 'DAT Discussions' started by skyisblue, May 13, 2007.

1. ### skyisblue

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I've learned that if you rotate a fischer projection by 180 degrees, then you get the exact same compound. If you rotate it by 90 degrees, you get it's enantiomer.

Question 1: If you rotate it by 270 degrees, do you still get it's enantiomer?

Question 2: Kaplan states that interchanging any pairs of substituents is the same as rotating it 180 degrees. For an example, Kaplan states that a fischer projection of Cl on top, CH3 on the bottom, H to the left and Br to the right is equal to another fischer projection of Br on top, H on the bottom, Cl to the right and CH3 to the left.

I don't see how these two fischer projections are equal after interchanging the two pairs of substituents from the first fischer projection.

If I were to rotate the first projection by 180 degrees, I would get Cl on the bottom, CH3 on top, H to the right and Br to the left. Then they would be equal.

Is there another way to figure this?

3. ### andyjl

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For Question 2: I had this problem too.. This is how I figured this out. First found out the absolute configuration by assigning priority numbers and finding out if it's R or S enantiomers. Then try switching any two pairs of substituents and find the absolute configurations by assigning priority number sand finding out if it's R or S enantiomers. It should come out the same as the first time which means they are the same compounds. It may be a bit hard to visualize how they are the same, but doing the absolute configurations and doing the way I said should give you a bit of confidence that switching any two pairs should give you the same compound.

4. ### jeezy

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1. rotating by 270 is the same as rotating by 90 in the opposite direction, so yes.. enantiomer

5. ### skyisblue

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I guess just to make sure, you just figure out the R or S configuration from the fischer projection. Yes?

6. ### andyjl

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yes. from the original fischer projection and from the rotated projection (or the projection found by switching of the substituents) would always work. I think just finding out the absolute configurations is probably the best because it is quick and you should always get the answer you are looking for

7. ### lor

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stop rotating things, you'll need that for your PAT section so save it for then plus what if you get a bigger molecule, ie, two chiral centers...
any how, the two molecules you mentioned are mirror immage of each other but not supper imposable so they are enantiomers, I use the right and left hand rule, my right hand is the mirror image of my left hand but not super imposable just like kaplan says
use R and S, don't forget to rank them by atomic number and you are always looking at the first atom attached to the chiral carbon, use your bow technique and you are on your way
speaking of enantiomers, know the following:
-racemic mixture and meso show no optical activity.
-the specific rotation of racemic mixture is equal to aero
-2^n posible sterioisomers for n chiral centers
-enantiomers have the same properties
-chair cyclohexane is the most stable

thought I add those things in
best of luck to you