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    Here's a dye, Lawsone:


    It's first reacted with NaOH solution to give the R-O- ion(bonded with the carbon with the double bond) where R represents the rest of the molecule. It's then reacted with ethanoyl chloride to give , which is named B. A question then asks the structure of an isomer of B, which is also formed during the reaction of Lawsone with ethanoyl chloride, has the same functional groups as B, but with different arrangements.

    My problem is that will the isomer be a cis-trans isomer, because there's no rotation allowed by the carbon-carbon double bond? And if it's cis-trans isomer, can it be shown by drawing the new "ester linkage" within the second ring?
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    It's not a geometric (cis-trans) isomer given the symmetry. Where did you get this question from, out of interest? :p:

    I think it's probably wanting you to notice the keto-enol tautomerism :yes:
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    (Original post by EierVonSatan)
    It's not a geometric (cis-trans) isomer given the symmetry. Where did you get this question from, out of interest? :p:

    I think it's probably wanting you to notice the keto-enol tautomerism :yes:
    It's from a past paper of A Level chemistry.

    Never heard about "keto-enol tautomerism"!!:eek3: How could an A Level paper ask such a question then? :woo:
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    (Original post by Zishi)
    It's from a past paper of A Level chemistry.

    Never heard about "keto-enol tautomerism"!!:eek3: How could an A Level paper ask such a question then? :woo:
    I don't know, that's why I'm asking where you got it from :mmm:

    I'm not seeing what else it could be referring to, maybe someone else can.
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    (Original post by EierVonSatan)
    I don't know, that's why I'm asking where you got it from :mmm:

    I'm not seeing what else it could be referring to, maybe someone else can.
    Ah, thanks for your help.

    I hope someone else spots something related to A Level.
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    (Original post by Zishi)
    Ah, thanks for your help.

    I hope someone else spots something related to A Level.
    I guess you're not being asked about how the isomer is formed, so could just not worry about the details and just drawn an isomer with the same groups but in different positions.
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    (Original post by EierVonSatan)
    I guess you're not being asked about how the isomer is formed, so could just not worry about the details and just drawn an isomer with the same groups but in different positions.
    Alright, so it means that I can attach CH3CO- group at one of the Oxygen atoms in the ketone in that dye?
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    (Original post by Zishi)
    Alright, so it means that I can attach CH3CO- group at one of the Oxygen atoms in the ketone in that dye?
    Yeah, try that.
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    Impossible! the question has it wrong. Surely the R-[O-] will form ion-dipole bonds with the Na+ from NaOH. So you get:

    R - Na+O-.

    And of course, water is eliminated.
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    (Original post by Ilyas)
    Impossible! the question has it wrong. Surely the R-[O-] will form ion-dipole bonds with the Na+ from NaOH. So you get:

    R - Na+O-.

    And of course, water is eliminated.
    But wouldn't the O-Na+ bond dissociate in solution as it's ionic?
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    (Original post by EierVonSatan)
    I don't know, that's why I'm asking where you got it from :mmm:

    I'm not seeing what else it could be referring to, maybe someone else can.
    My thoughts were keto-enol tautomerisation as well.

    (Original post by thegodofgod)
    But wouldn't the O-Na+ bond dissociate in solution as it's ionic?
    Indeed. He is wrong.

    If you're interested, you can form the second product by pushing a few arrows around. If you take the C=C-O-, and push the negative charge down to form a C=O, push an arrow from the C=C down towards the cabonyl - double bond onto the next carbon, and then push the C=O electrons up onto the O you get a resonance form of the deprotonated structure (same idea as resonance structures in benzene). Then form the ester on this different O-.
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    (Original post by illusionz)
    My thoughts were keto-enol tautomerisation as well.


    Indeed. He is wrong.

    If you're interested, you can form the second product by pushing a few arrows around. If you take the C=C-O-, and push the negative charge down to form a C=O, push an arrow from the C=C down towards the C=O, and then push the C=O electrons up onto the O you get a resonance form of the deprotonated structure (same idea as resonance structures in benzene). Then form the ester on this different O-.
    So are you suggesting: (attachment)
    Attached Images
     
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    (Original post by thegodofgod)
    So are you suggesting: (attachment)
    Not quite. You will still have a enolate (C=C with an O- bound to one of the carbons).
    You simply need to move the double bond 1 carbon round the ring and then push the arrows on the carbonyl up to the O so you don't get a pentavalent carbon.



    Just re-read what I wrote and I can see how you got confused if you're not used to this sort of thing. When I said down towards the C=O i just meant down on the diagram, not actually onto the C=O, sorry!
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    (Original post by illusionz)
    Not quite. You will still have a enolate (C=C with an O- bound to one of the carbons).
    You simply need to move the double bond 1 carbon round the ring and then push the arrows on the carbonyl up to the O so you don't get a pentavalent carbon.



    Just re-read what I wrote and I can see how you got confused if you're not used to this sort of thing. When I said down towards the C=O i just meant down on the diagram, not actually onto the C=O, sorry!
    See what you mean now
    Attached Images
     
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    (Original post by thegodofgod)
    See what you mean now
    :yes:
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    (Original post by illusionz)
    :yes:
    Fun stuff!

    Spoiler:
    Show
    Not being sarcastic.


    Spoiler:
    Show
    I need to get a life :sadnod:
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    (Original post by thegodofgod)
    Fun stuff!

    Spoiler:
    Show
    Not being sarcastic.


    Spoiler:
    Show
    I need to get a life :sadnod:
    I found stuff like that much more interesting than most of the stuff you learn at A level. I honestly have no idea why some of the stuff at A level is even on the syllabus. Making people learn colour changes and chemical tests seems completely pointless and I've never come across a book/lecturer even mention chemical/flame tests apart from saying how crap they are.

    Anyway, I should get to bed, should really do a few hours revision before brunch tomorrow. Finals and all
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    (Original post by illusionz)
    I found stuff like that much more interesting than most of the stuff you learn at A level. I honestly have no idea why some of the stuff at A level is even on the syllabus. Making people learn colour changes and chemical tests seems completely pointless and I've never come across a book/lecturer even mention chemical/flame tests apart from saying how crap they are.

    Anyway, I should get to bed, should really do a few hours revision before brunch tomorrow. Finals and all
    Me too - it's almost 2! :eek:

    Anyway, best of luck for your exams! :yy:
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    Oh, my goodness. I wish I knew about that. :cry2:
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    (Original post by EierVonSatan)
    It's not a geometric (cis-trans) isomer given the symmetry. Where did you get this question from, out of interest? :p:

    I think it's probably wanting you to notice the keto-enol tautomerism :yes:
    So hold up, are you telling me that this would be one of the products :

    Name:  IMG_0258.jpg
Views: 207
Size:  146.5 KB

    Because that's what I drew.
 
 
 
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