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Anti-coplanar once AND FOR ALL! lol

Discussion in 'DAT Discussions' started by wired202808, Jul 26, 2011.

  1. wired202808

    wired202808 Membership Revoked
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    I keep seeing this and every time I see it I get it and then i forget it the next day.

    so there is one particular question that asks which one of the following reaction undergoes E2 fastest?

    I was left with 2 choices:

    Cyclohexane with a T-butyl group at Carbon 1 and a halide at Group 3 (cis to each other)

    OR

    Cyclohexane with a T-butyl group at Carbon 1 and a halide at Group 4 (cis to each other)

    Which one of them work and please explain why? Im confused as to what I should be looking at and how to properly do this! Thanks guys.
     
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  3. flin5845

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    Ok first with rings, the leaving group must be in the axial position. Since The t butyl is so big it will take the equatorial position to give the more stable structure. So now you need to figure out where the halide lies in relationship to the t butyl group (axial or eq) if it's eq it will not do an elimination. If it's axial you need to look at the beta hydrogens and see which one will get eliminated.


    Hopefully this helps


    Also I'll take the 1 and 3 example. Instead of drawing chair conformations if you just put the t butyl as a wedge on carbon 1 and call that eq, then on C2 the eq position will be dashed and on the C3 position(the one with halide) the eq will be wedged. And that means that both t butyl and halide are cis (both wedged) and eq. Therefore this wil not do elimination.

    For the other one, same concept but this tome you have a C1 and C4 . If we put the t butyl as a wedge and call that eq then dashed on C2 is eq, wedge on C3 is eq, dash on C4 is eq. But since they are cis to eachother that means that the halide is axial on C4 (wedged) and will do elimination..

    I know it is confusing but hopefully makes sense
     
    #2 flin5845, Jul 26, 2011
    Last edited: Jul 26, 2011
  4. wired202808

    wired202808 Membership Revoked
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    ok so would this make sense? when doing this I can assume that both groups are always anti to each other so one is always axial and the other is always equatorial, this way when you put the t-butyl on the equatorial the halide HAS to be axial and is therefore unable to be on C-3 because it can only equatorial there (since its a cis formula)

    Please confirm and thanks for the write up!
     
  5. Jab1113

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    dont always assume that both groups are going to be anti. the reason why the t-butyl group is in the equatorial position is because it is a big, bulky base and cannot go axial.

    if you were given methyl for example, you cannot just assume that it will be anti and put it in the equatorial position.

    the halide must be in the axial position and there must be an adjacent alpha hydrogen that is also axial for elimination to occur.

    i hope this makes sense, good luck bro
     
  6. flin5845

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    you cant always assume that if something is di-subsituted that they will be anti. Lets use the same example you provided but this time make it tans 1-3 and trans 1-4.

    Here is the order in which you need to think about it.

    1. Which substituent will take up more room. ( I assume you know this but for example if you had a -CH3 and -Br, the -CH3 takes up more room, therefore will be equatorial over the bromine. It has nothing to do with molecular weight but with SIZE)

    2. Place the largest substituent equatorial. Then based on your nomenclature provided, place the other leaving group as needed.

    3. See if the LG is axial, if it is, elimination can occur...If LG is equatorial then elimination will NOT occur.


    So for example. Take 1-3 TRANS T-buyl on C1 and bromine on C3. In this case we do the same thing as before.

    I am going to call the wedge on C1 equatorial. C2 equatorial will be dashed, and C3 equatorial will be wedged. now in this case, since they are TRANS the bromine will be dashed on C3 (since the T butyl is wedged on C1, if the Bromine is dashed on C3 then we know they are TRANS). Since we established that the equatorial position on C3 is wedged, that means that the dashed is axial. Therefore in this case Bromine can leave and elimination can occur.

    I leave it up to you to try TRAS 1-4. But it should work out that the Bromine is going to be Equatorial and therefore CAN NOT do elimination.

    I sorry if this is confusing it is hard to explain over text.
     
  7. Bis-GMA111

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    is the LG being axial a rule for cycloalkanes?



    *EDIT*. nevermind, found it on chad's videos
     

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