CHEM

What is the major product from an elimination reaction starting with 2-bromo-1-methylcyclohexane?

would it be:

a) 1-methyl-1-cyclohexene

b) a cyclohexane/ene with a double bond off of one C - basically the cyclohexane shape with =CH2 off of it (I don't remember the actual name for it)?

c) 3-methyl-1-cyclohexene

d) 4-methyl-1-cyclohexene

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  1. ( "B" is meant to read as a cyclohexane with a double bond off of one carbon & then a CH2 at the end of that)

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  2. <cyclohexane>=CH2

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  3. What do you think it is?

    The product that would be formed is the one that would make the bromine leave and would also form the double bond that would make the bromine fall off in the first place.

    Think about this:

    IF the CH3's hydrogen were attacked, would that form a double bond and cause the bromine to leave?

    For some of the answers I and you should also notice that the methyl group changes position..This is would definately not happen and this would lead to elimination of the choices by themselves.

    So what do you think is the product now?

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  4. so then "B" as the "Major" product? -the cyclohexane with the double bond off from it & then the "CH2" attached at the end? What is the correct name for this compound anyways?!

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  5. No, and why is that?
    let's analyze.

    like I said if the methyl is one the 1st carbon correct, then it has to stay on the first carbon. AND if the base (which is the negatively charged molecule [not shown by you but in general])attacks the H of the CH3 then it would form a double bond and then become CH2. The problem is that this would NOT cause the bromine to leave since what you DO need is a double bond between bromine and another carbon. Not only this, but in other cases you would have to see which is the most formed product which would be the most substituted product but it doesn't matter here if I'm not wrong since there is only 1 right answer to your problem.

    Now if the CH3 does NOT react, and stays as it is but a double bond does form...and note that the carbon where the CH3 is attached also has a H that is facing down. The carbon next to the bromine on the right, has 2 hydrogens as well.

    So it CAN form a double bond between the carbon connected to the CH3 by removal of the H connected to that carbon. OR it can remove one of the 2 hydrogens attatched to the other carbon attatched to the right of the carbon holding the BR. The thing is, your not forced to choose between these so you can draw both and then see which option is there on the choices

    The molecules stay as they are so the choice you have made can't be true since where was the methyl originally? (positions don't change)

    Which is it then?
    Analyze what I said especially what's in bold and I'll come back and take a look in a bit.

    ~c~

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  6. like I said if the methyl is one the 1st carbon correct, then it has to stay on the first carbon. AND if the base (which is the negatively charged molecule [not shown by you but in general])attacks the H of the CH3 then it would form a double bond and then become CH2. The problem is that this would NOT cause the bromine to leave since what you DO need is a double bond between carbon on which bromine is attatched to and another carbon. Not only this, but in other cases you would have to see which is the most formed product which would be the most substituted product but it doesn't matter here if I'm not wrong since there is only 1 right answer to your problem.

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  7. Thank you for all of your information!

    So then I believe that the answer "should" be: "C" (3-methyl-1-cyclohexene) b/c then the double bond would link to where the Br was originally & flip backwards..."a" (1-methyl-1-cyclohexene) could be a "minor" product then (correct?) & I don't think that it could be "d" (4-methyl-cyclohexene).

    what is the compound called with the cyclohexane with a double bond CH2 attached at the end called exactly?...I have had other questions showing it & I don't know the exact name of it.

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  8. I said that you were incorrect before so in essence, methyl should stay on the 1st carbon. I'll analyze each choice.

    a)1-methyl-1-cyclohexene.
    -well the methyl is on carbon 1 the original carbon that it was on.
    -the double bond is next to it and there is no bromine since it has left.

    b)a cyclohexane/ene with a double bond off of one C - basically the cyclohexane shape with =CH2 off of it (I don't remember the actual name for it)?

    -I said this was incorrect before as the methyl, having the H removed would not do anything to the molecule to make the Br leave

    c) 3-methyl-1-cyclohexene
    - you choose this but where is the methyl? carbon 3? where was it before?
    (carbon 1!) so it couldn't flip as you say. Bonds don't "flip", technically here on this problem anyhow. BUT the position of the double bond is CORRECT. So you understand that part.

    d) 4-methyl-1-cyclohexene
    -once again, methyl cannot change from carbon 1 to carbon 4 all of a sudden. NO change of configuration can occur.
    -bond position is correct though.

    IF I could draw it I would =/ but I can't here. I think you'd understand it better that way but I can only describe it unfortunatelyu and that leaves me limited.

    ****************************
    ***********CH3Br************
    *********H*|**|*************
    **********\C--C***H********
    **********/****\*/**********
    *********C******C-H**********
    **********\*****/***********
    ***********C--C**************

    I tried to draw it above and this is the molecule you have H's that can be removed on the side (right) and the 1 on the left. The left one is taken off so it becomes? 1-methyl-1-cyclohexane.

    If you really analyze this problem you should have less trouble analyzing these types of problems but of course there is much more you need to know.

    Hope this helps.
    Good luck

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  9. The Br should be on the other bond not connected to the CH3. It just looks that way.

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