Join TSR now and get all your revision questions answeredSign up now
    • Thread Starter
    Offline

    3
    ReputationRep:
    http://www.thestudentroom.co.uk/atta...1&d=1362677872

    please help me!
    Offline

    9
    ReputationRep:
    Your very first step; look up in books about mechanism(s) on how amines usually act as nucleophile. The hydrogen on amine is not lost immediately when an amine protonates or attack a carbonyl.
    Offline

    3
    ReputationRep:

    (Original post by shengoc)
    Your very first step; look up in books about mechanism(s) on how amines usually act as nucleophile. The hydrogen on amine is not lost immediately when an amine protonates or attack a carbonyl.
    Yes, lone pair attacks giving a positively charged nitrogen species after having lost the acetate anion. You can then use this to deprotonate the positively charged amino nitrogen
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by JMaydom)
    Yes, lone pair attacks giving a positively charged nitrogen species after having lost the acetate anion. You can then use this to deprotonate the positively charged amino nitrogen
    ive read your feedbacks several times and i dont understand would you be able to draw the part you think ive done wrong if its not too much of a pain?

    :blush:
    Offline

    16
    ReputationRep:
    (Original post by asaaal)
    ive read your feedbacks several times and i dont understand would you be able to draw the part you think ive done wrong if its not too much of a pain?

    :blush:
    Attack the anhydride with the nitrogen atom to generate a 4 coordinate, positively charged nitrogen intermediate, which then loses a proton.
    Offline

    3
    ReputationRep:
    (Original post by asaaal)
    ive read your feedbacks several times and i dont understand would you be able to draw the part you think ive done wrong if its not too much of a pain?

    :blush:

    (Original post by illusionz)
    Attack the anhydride with the nitrogen atom to generate a 4 coordinate, positively charged nitrogen intermediate, which then loses a proton.
    Yeah, better worded by Illusionz. See if you understand now... If not i'll draw it out.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by JMaydom)
    Yeah, better worded by Illusionz. See if you understand now... If not i'll draw it out.
    i really dont understand im so sorry could you draw it out?
    Offline

    3
    ReputationRep:
    (Original post by asaaal)
    i really dont understand im so sorry could you draw it out?
    Hopefully the picture has successfully attached. Here is the mechanism!
    Sorry the picture is a bit blurred but I had to take the photo on my phone.
    Attached Images
     
    Offline

    16
    ReputationRep:
    (Original post by JMaydom)
    Hopefully the picture has successfully attached. Here is the mechanism!
    Sorry the picture is a bit blurred but I had to take the photo on my phone.
    I wouldn't use the double headed arrow in this case because the positively charged N is probably a better leaving group than the carboxylate. At least, that was what I was told by one of the academics in a 2nd year practical write up. I've always lost the proton before carbonyl reformation since then. That said, I don't know how important it actually is...
    Offline

    3
    ReputationRep:
    (Original post by illusionz)
    I wouldn't use the double headed arrow in this case because the positively charged N is probably a better leaving group than the carboxylate. At least, that was what I was told by one of the academics in a 2nd year practical write up. I've always lost the proton before carbonyl reformation since then. That said, I don't know how important it actually is...
    It's not an issue with our examiners. So long as it is feasible (seeing as the whole thing is one great equilibrium of many intermediates) they are happy. Obviously as the positive N is a better leaving group, mostly it will be kicked back out but that is not the productive pathway so you ca ignore it.

    Really I should have put equilibrium arrows!!!
    Offline

    16
    ReputationRep:
    (Original post by JMaydom)
    It's not an issue with our examiners. So long as it is feasible (seeing as the whole thing is one great equilibrium of many intermediates) they are happy. Obviously as the positive N is a better leaving group, mostly it will be kicked back out but that is not the productive pathway so you ca ignore it.

    Really I should have put equilibrium arrows!!!
    True. I remember thinking he was being a bit pedantic, but it's just something I've always done ever since
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by illusionz)
    I wouldn't use the double headed arrow in this case because the positively charged N is probably a better leaving group than the carboxylate. At least, that was what I was told by one of the academics in a 2nd year practical write up. I've always lost the proton before carbonyl reformation since then. That said, I don't know how important it actually is...

    what arrow should be used instead of the double headed one?
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by JMaydom)
    Hopefully the picture has successfully attached. Here is the mechanism!
    Sorry the picture is a bit blurred but I had to take the photo on my phone.
    you life saver !! thank you so much !!
 
 
 
Poll
If you won £30,000, which of these would you spend it on?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.