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    draw the structure of an optically active carboxylic acid having the molecular formula (C6H12O2). with five peaks in its 13C n.m.r. spectrum.

    mark scheme shows Name:  Capture.PNG
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    but why it cant be drawn like that
    Attachment 730114730116
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    (Original post by Qer)
    draw the structure of an optically active carboxylic acid having the molecular formula (C6H12O2). with five peaks in its 13C n.m.r. spectrum.

    mark scheme shows Name:  Capture.PNG
Views: 23
Size:  1.2 KB

    but why it cant be drawn like that
    Attachment 730114730116
    Your diagram only has four environments ...
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    Name:  Capture.PNG
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    how?
    (Original post by charco)
    Your diagram only has four environments ...
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    (Original post by Qer)
    Name:  Capture.PNG
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    how?
    sorry I was looking at 1H NMR

    BUT your molecule is not optically active
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    (Original post by charco)
    sorry I was looking at 1H NMR

    BUT your molecule is not optically active

    thanks
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    (Original post by charco)
    sorry I was looking at 1H NMR

    BUT your molecule is not optically active

    would you please also help in this part
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    (Original post by Qer)
    would you please also help in this part
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    There are two equivalent protons on the carbon to the RHS which will split the "a" protons into a triplet.
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    (Original post by charco)
    There are two equivalent protons on the carbon to the RHS which will split the "a" protons into a triplet.

    thanks
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    hydroxyl protons are usually hydrogen bonded to other hydroxy groups and do not tend to split or be split.
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    Attachment 730144730146
    Attachment 730144730146730100

    This is the route that mark scheme suggest

    But why we cant do in that way

    react with HCN to get phenylnitrile and then use LiAlH4 in ether to reduce it into amine?





    Spoiler:
    Show

    sorry for too many questions but I have mocks next week so I am doing preparation for it.
    Attached Images
       
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    (Original post by Qer)
    Attachment 730144730146
    Attachment 730144730146730100

    This is the route that mark scheme suggest

    But why we cant do in that way

    react with HCN to get phenylnitrile and then use LiAlH4 in ether to reduce it into amine?





    Spoiler:
    Show


    sorry for too many questions but I have mocks next week so I am doing preparation for it.


    question 8 c
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    will post in a min
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    Name:  Capture.PNG
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    the possible way to do it is to first react with HCL by electrophilic addition and then react by NH3 by nucleophilic addition to form the amine.

    Is this possible that we first react it with HCN and then reduce HCN by LIALh4 in ether to amine ????
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    (Original post by Qer)
    Attachment 730144730146
    Attachment 730144730146730100

    This is the route that mark scheme suggest

    But why we cant do in that way

    react with HCN to get phenylnitrile and then use LiAlH4 in ether to reduce it into amine?





    Spoiler:
    Show



    sorry for too many questions but I have mocks next week so I am doing preparation for it.


    alkenes do not react with HCN ...

    and anyway it's cyclohexene not benzene
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    (Original post by Qer)
    Name:  Capture.PNG
Views: 26
Size:  13.5 KB


    the possible way to do it is to first react with HCL by electrophilic addition and then react by NH3 by nucleophilic addition to form the amine.

    Is this possible that we first react it with HCN and then reduce HCN by LIALh4 in ether to amine ????
    No,

    see your other thread ...
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    (Original post by charco)
    No,

    see your other thread ...

    I dont know why I cant be able to open that now
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    (Original post by Qer)
    I dont know why I cant be able to open that now
    I've merged the threads as they were asking the same question ...
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    (Original post by charco)
    I've merged the threads as they were asking the same question ...
    ohh that's fine


    Can you explain to me why HCN cant undergo reaction?
 
 
 
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