Why Does Sorbitol Have Four Chiral Centers Instead of Two?

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I wonder why the number of chiral centres in sorrbital is 4? Why isn't it 2? Is it because the H-C-OH bond cannot be rotated? Otherwise isn't the two chiral centres above identical to the 2 at the bottom?
 

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If the atoms on the carbons at the end are switched around, the carbon will be able to rotate to bring back the orignial configuration. If the other carbons have the atoms they are attached to switched around, they will not be able to rotate any which way to find the original configuration. Does that sound about right?
 
Continuity said:
If the atoms on the carbons at the end are switched around, the carbon will be able to rotate to bring back the orignial configuration. If the other carbons have the atoms they are attached to switched around, they will not be able to rotate any which way to find the original configuration. Does that sound about right?
Uhhm, I don't understand what you are saying.Are you saying that the whole chain has to rotate and not just a part of it?
 
each of the carbon atoms can rotate individually. If you rotate the carbon at either end you will see that all the different configurations are equivalent. That is, if you switch the OH with one of the H's, it will look different but it will be the same molecule because the carbon atom can just rotate back. Don't forget that the bonds can always rotate and they are constantly rotating when there is any energy present at all.
 
Continuity said:
each of the carbon atoms can rotate individually. If you rotate the carbon at either end you will see that all the different configurations are equivalent. That is, if you switch the OH with one of the H's, it will look different but it will be the same molecule because the carbon atom can just rotate back. Don't forget that the bonds can always rotate and they are constantly rotating when there is any energy present at all.
If so, shouldn't there be only two chiral centres? Why 4? Since if they can rotate aren't the top 2 same as the bottom 2?
 
Ok I think I understand your question. I think you're forgetting about the tetrahedral configuration of each of the carbons. The third carbon from the top can rotate but that does not mean it becomes equivalent to the ones around it. Remember that in this representation of the molecule the atoms that come out to the sides are actually intended to be coming out of the paper towards you. If it rotates around it will be going into the paper away from you. Get out a plastic model kit or visualize the tetrahedral structures in your head.
 
Continuity said:
Ok I think I understand your question. I think you're forgetting about the tetrahedral configuration of each of the carbons. The third carbon from the top can rotate but that does not mean it becomes equivalent to the ones around it. Remember that in this representation of the molecule the atoms that come out to the sides are actually intended to be coming out of the paper towards you. If it rotates around it will be going into the paper away from you. Get out a plastic model kit or visualize the tetrahedral structures in your head.
Ok thanks. The third and fourth carbon from the top are also chiral centres right? Even if you rotate them, their numberings just change, that's all.
 
yeah I'm pretty sure there's no symmetry between the third and fourth carbon. Even though they have the same groups attached to them, they are attached in a different configuration.
 
Continuity said:
yeah I'm pretty sure there's no symmetry between the third and fourth carbon. Even though they have the same groups attached to them, they are attached in a different configuration.
Ok, thanks.