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the question was whether the following compounds have the center of symmetry or not .the question was : enter image description here

And the answer is that there is no center of symmetry .how? according to me, there should be the centre of symmetry. I have made it like this:enter image description here

on asking my teacher how there is no center of symmetry he told me that since it is a Fischer projection, therefore, the horizontal lines are above the planes. therefore there will be no center of symmetry. I understand that if the horizontal lines are out of the plane then there will be no COS. but what I don't understand how come the horizontal lines are above the plane/out of the plane. please explain.

Garima Singh
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    Paper is 2D. But real world is 3D. Models are like that. – Ivan Neretin Sep 06 '19 at 13:35
  • obviously! but how come he know or we will know that all the groups present on the horizontal line are out of the plane and not few of them are inside and few are outside the plane? – Garima Singh Sep 06 '19 at 13:41
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    The angle is supposed to be tetrahedral, it is not possible on a single 2Dplane. – Tojra Sep 06 '19 at 14:44
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    Problem 1: understand what a Fischer projection represents. Problem 2: if you still don't get it, build a 3D model. – matt_black Sep 06 '19 at 15:12

1 Answers1

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One way to quickly get a visualization without building a 3D model (having a model is more instructive, though):

  1. Go to https://chemapps.stolaf.edu/jmol/jsmol/simple.htm
  2. Click on Load by name and type "(2S,3S)-Butane-2,3-diol"
  3. Click on console and type "rotate branch @2 @4 180" to get a conformation that is easy to compare with the Fisher projection.

enter image description here

Karsten
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