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    Hellooo
    Ive come across a question in one of the past papers and im really stuck!
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    a is referring to the second carbon on the right in molecule J. It is attached to two carbons-one with 2 H's attached to it and one carbon with 0 H's.

    So using the n+1 rule do I do :
    2 H's on adjacent carbon atom + 1=3 i.e triplet

    or 0+1=singlet

    ??????????????

    im sooo confusedddd

    Can anyone help please?
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    (Original post by Mel6542)
    Hellooo
    Ive come across a question in one of the past papers and im really stuck!
    Name:  kk.JPG
Views: 407
Size:  19.3 KB

    a is referring to the second carbon on the right in molecule J. It is attached to two carbons-one with 2 H's attached to it and one carbon with 0 H's.

    So using the n+1 rule do I do :
    2 H's on adjacent carbon atom + 1=3 i.e triplet

    or 0+1=singlet

    ??????????????

    im sooo confusedddd

    Can anyone help please?
    There are only two protons on an adjacent carbon so, yes, it's a triplet.
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    Further to what charco said, it seems like in cases like this you're confused about which adjacent carbon to work from?

    If so, it's both - the total number of hydrogen on all adjacent carbons determines the splitting pattern - so if one of those CH3's had been a hydrogen, then it would be 1 hydrogen on one adjacent carbon, 2 on the other - hence a quartet.
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    (Original post by charco)
    There are only two protons on an adjacent carbon so, yes, it's a triplet.
    (Original post by Stiff Little Fingers)
    Further to what charco said, it seems like in cases like this you're confused about which adjacent carbon to work from?

    If so, it's both - the total number of hydrogen on all adjacent carbons determines the splitting pattern - so if one of those CH3's had been a hydrogen, then it would be 1 hydrogen on one adjacent carbon, 2 on the other - hence a quartet.
    Oh I didnt know that !! Thank youuu

 
 
 
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