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Organic Chemistry stable conformation question

  1. Aug 29, 2009 #1
    1. In the chair conformation, why would 1-bromo-tetrahydropyran be more stable in the axial position, as opposed to the equitorial position? (Hint: consider how the bromo group affects the attached carbon and what the oxygen could do to alleviate that.)


    2. No relevant equations


    3. I am thinking this whole thing has to due with steric hinderance and charge sharing. I think that even though the bromine would be hindered with the pi bond electrons of the carbon in the axial position, it would be even more sterically hindered in the equitorial position with the oxygen. Also since bromine is a good leaving group, its relevant to think that the bromine could leave and there would be a resulting negative charge left with the molecule. Since the oxygen is highly electronegative, it could better handle the resulting negative charge. Also, with the bromine in the axial position, it would be easier for another molecule to attach to the carbon allowing for the bromine to leave.

    Any help or critique of my thinking would be a great help!
     
  2. jcsd
  3. Aug 31, 2009 #2
    Hi
    We can think or visualize it on the basis of electron pairs on O and Br .Both of them have lone pairs on them which can result in LP-LP repulsion .This can be the reason that Br is axial to O atom which is in the ring.You need to post the structure also so that I can think more clearly.Are you talking about 1 bromo or 2-bromo
    In 1-bromo ,bromine is bonded to oxygen and in 2- bromo it is bonded to carbon.
     
    Last edited by a moderator: Aug 31, 2009
  4. Aug 31, 2009 #3
    Sorry, I guess that would make it 2-bromo then. Some of us in the class talked about it today. We figured that the lone pairs from Br and O would have the repulsions as you mentioned. Also, we considered the fact that the EN from the O and Br would better be dispersed in the axial position as opposed to the equitorial. Any other thoughts???
     
  5. Aug 31, 2009 #4
    Okay than we are good with this explanation only.We can also think in terms of big size of Br .
    Since O is in the plane of the ring and all equatorial bonds are in the plane of the ring.So we can also say that when Br on second carbon is axial ,it minimizes steric repulsion .More suitable situation and favorable also.If you see this explanation again takes you on the first which we discussed on the basis of LP-Lp.What is steric repulsion?----repulsion among the electrons of the atoms or groups .
     
  6. Aug 31, 2009 #5
  7. Sep 2, 2009 #6
    The anomeric effect was a big help. Thanx a bunch!!!
     
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