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Drawing trans -

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    The compound [Co(ho202(Nh2ch2ch2Nh2)2]2+
    can form 3 different isomers.
    label the cis/trans isomers



    I've drawn these two but wasn't sure if it is right
    (I know how to do cis but not sure for trans) please somone clarify whether my drawing is right or not


    also wondered why only cis version can form optical isomerism (the trans versions cannot)


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    You've drawn the trans isomer twice.
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    Yeah, it's a mistake you make by trying to be too cute. When you've drawn one, just draw the exact same thing but with two of the groups in swapped positions, rather than draw everything in a different position and start from scratch, and potentially get confused.
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    (Original post by EierVonSatan)
    You've drawn the trans isomer twice.
    If you read my post- I mentioned I could draw cis but wasn't sure about trans

    the fact that you say I've drawn trans twice that means both are acceptable for trans isomerism..

    Also could you indicate why only cis isomerism can form optical isomerism
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    (Original post by arvin_infinity)
    If you read my post- I mentioned I could draw cis but wasn't sure about trans
    I did and it's not all that clear what you were trying to draw?

    Also could you indicate why only cis isomerism can form optical isomerism
    because here the cis isomer has a non-superimposable mirror image, the trans isomer does
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    (Original post by EierVonSatan)



    because here the cis isomer has a non-superimposable mirror image, the trans isomer does
    I know what non-superimposable mirror image is but still don't understand what you said


    I did and it's not all that clear what you were trying to draw?
    My bad..basically wanted to see if both drawing are acceptable if I was asked to draw the trans isomer
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    (Original post by arvin_infinity)
    I know what non-superimposable mirror image is but still don't understand what you said
    Well, if you draw and reflect the trans-isomer you'll see that the reflection is the same as the original - so it's not optically active.

    If you do that with the cis-isomer, you'll see that the two are different - so it is optically active.

    My bad..basically wanted to see if both drawing are acceptable if I was asked to draw the trans isomer
    then yeah, either of those are fine
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    (Original post by EierVonSatan)
    Well, if you draw and reflect the trans-isomer you'll see that the reflection is the same as the original - so it's not optically active.

    If you do that with the cis-isomer, you'll see that the two are different - so it is optically active.

    Well if we draw the mirror image of my drawing (the one on the left) , the mirror images are non-superimposable

    For the other trans I drew and it's fine
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    (Original post by arvin_infinity)
    Well if we draw the mirror image of my drawing (the one on the left) , the mirror images are non-superimposable

    For the other trans I drew and it's fine
    I think you're just confusing yourself with the way you're drawing them, as it's the same compound if you can reflect one and not get a non-superimposable image you can't do it with the other :nah:
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    (Original post by EierVonSatan)
    I think you're just confusing yourself with the way you're drawing them, as it's the same compound if you can reflect one and not get a non-superimposable image you can't do it with the other :nah:

    Well am not so hung up about just answering this question..
    It's just I'am struggling to spot the complex compounds which they can form optical isomerism


    do I have to look for line of symmetry and if there was one then it cannot form optical isomerism


    also is it always the cis isomers that can form optical isomerism
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    (Original post by EierVonSatan)
    I think you're just confusing yourself with the way you're drawing them, as it's the same compound if you can reflect one and not get a non-superimposable image you can't do it with the other :nah:
    So I went to look for more questions and so far no luck..It's weird how every year they ask the same question about the optical isomerism..hopefully they are gonna caryy on



    Also I did another question on a different paper and this is the MS and my answer is also attached ..pls clarify if they are the same thing
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    (Original post by arvin_infinity)
    So I went to look for more questions and so far no luck..It's weird how every year they ask the same question about the optical isomerism..hopefully they are gonna caryy on



    Also I did another question on a different paper and this is the MS and my answer is also attached ..pls clarify if they are the same thing
    Yep looks good :yy:
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    (Original post by EierVonSatan)
    Yep looks good :yy:
    Thanks for clearing that up

    it's like x axis which is perpendicular to y axis
    also x axis perpendicular to z axis

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Updated: June 10, 2012
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