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    Hi there, I'm new here. Just would like some help with my chem practical write up. In the practical I had to make, purify and determine the mpt. of aspirin. One of my questions on the sheet is "Attempt to draw a mechanism for the reaction that occurs". I am not too sure about this, I made the aspirin using 2-hydroxybenzoic acid and ethanoic anhydride using H2PO4. Please could someone help me out. Thanks in advance.
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    (Original post by ukmd)
    Hi there, I'm new here. Just would like some help with my chem practical write up. In the practical I had to make, purify and determine the mpt. of aspirin. One of my questions on the sheet is "Attempt to draw a mechanism for the reaction that occurs". I am not too sure about this, I made the aspirin using 2-hydroxybenzoic acid and ethanoic anhydride using H2PO4. Please could someone help me out. Thanks in advance.
    Look up aspirin structure on wikipedia. http://en.wikipedia.org/wiki/Aspirin

    What is different from your starting 2-hydroxy benzoic acid is that the hydroxy group is now an ester group.

    Anhydride is an activated form of carboxylic acid, but not as reactive as acyl chloride, but that will do.

    So all this spells out esterification - if you know how to do mechanism for normal esterification, you can do this!
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    Hi. thank you for the help. I know about acid anhydrides and acyl chlorides being acid derivatives of carboxylic acids. But, i;d like to know where does the H off the 2-hydroxybenzoic acid disappear to and why (from the hydroxy group)? How many steps is the reaction mechanism? Please could you explain what's going on in ech of the steps, as I don't really understand why part of the ethanoic anhydride bonds to the O on the benzene structure, and the other part does not...
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    (Original post by ukmd)
    Hi. thank you for the help. I know about acid anhydrides and acyl chlorides being acid derivatives of carboxylic acids. But, i;d like to know where does the H off the 2-hydroxybenzoic acid disappear to and why (from the hydroxy group)? How many steps is the reaction mechanism? Please could you explain what's going on in ech of the steps, as I don't really understand why part of the ethanoic anhydride bonds to the O on the benzene structure, and the other part does not...
    The aromatic benzene is not involved in the esterification process, think instead of it as ROH where R is an aryl group(something with benzene in it).

    I would assume that it is the lone pair on the middle oxygen of anhydride that protonates first, then O on alcohol group attacks the carbonyl, and then leaves the other groups(ie oxygen with the positive charge). The anhydride is merely an activated carboxylic acid, and a by product of this esterification with this anhydride happens to be a carboxylic acid.

    Think of anydride as 2 x RCOOH where R in this case is just let say alkyl group.
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    thank you again. is there a diagram or something out there to show me what's going on, or any text books you know of? A little confused still...
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    (Original post by ukmd)
    thank you again. is there a diagram or something out there to show me what's going on, or any text books you know of? A little confused still...
    chemguide is an excellent site to learn school chemistry

    http://www.chemguide.co.uk/physical/.../esterify.html
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    (Original post by shengoc)
    chemguide is an excellent site to learn school chemistry

    http://www.chemguide.co.uk/physical/.../esterify.html
    thanks... so where do each of the arrows go to/from in each step. I would really appreciate it if somebody could describe "step 1, step 2, and so on..." and what is happening in each, and where exactly each of the arrows is going. Will rep to the most helpful person.
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    (Original post by ukmd)
    thanks... so where do each of the arrows go to/from in each step. I would really appreciate it if somebody could describe "step 1, step 2, and so on..." and what is happening in each, and where exactly each of the arrows is going. Will rep to the most helpful person.
    All the steps are self explanatory, just read the whole page, try drawing the same for your molecules. You aren't going to learn with people telling you everything.
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    (Original post by shengoc)
    All the steps are self explanatory, just read the whole page, try drawing the same for your molecules. You aren't going to learn with people telling you everything.
    I know what you mean, but I have my 2-hydroxybenzoic acid drawn out, and the ethanoic anhydride, and I know the first is acylated by the second compound, and I know how to draw acylation reactions with acyl chlorides, but when I look at this I have no idea how to go about it.
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    I have something like this written down in front of me. Now what happens?
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    (Original post by ukmd)
    I know what you mean, but I have my 2-hydroxybenzoic acid drawn out, and the ethanoic anhydride, and I know the first is acylated by the second compound, and I know how to draw acylation reactions with acyl chlorides, but when I look at this I have no idea how to go about it.
    Esterification is usually done with little bit of acid. oxygen in the middle of anhydride protonates, leaving a positive charge on that oxygen.

    O on ROH attacks the carbonyl(C=O), either one is fine,as your anhydride is symmetric, forming a tetrahedral intermediate. This then kicks out CH3COOH giving you what looks like ester.

    Final step is deprotonation of the H on the oxygen of ester. Hence the catalyst, typically H2SO4 is regenerated, you get acetic acid as by product and your ester.

    please don't tell me you don't get this without even lifting your pencil/pen trying to draw out these.
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    Alternatively, the lone pair on carbonyl can protonate, then ROH attacks that carbonyl, giving a tetrahedral intermediate. Transfer of proton to the central oxygen on the anhydride, making that oxygen positively charged, hence a good leaving group.

    Then lone pair on OH(near ester look part) push electrons down, giving you C=O and makes the CH3COOH leave. you get the same products.
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    When you say the acid ahnydride protonates (the O in the middle), youa re left with H3COOH and H3COO- ? then the H3C00- takes part in the first step?
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    the anhydride doesn't directly give off CH3COOH, but instead, it does after the ROH attacks the carbonyl and some proton shift here and there.
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    (Original post by shengoc)
    the anhydride doesn't directly give off CH3COOH, but instead, it does after the ROH attacks the carbonyl and some proton shift here and there.
    where does the H disappear on the 2-hydroxybenzoic acid (on the hydroxy?)
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    (Original post by ukmd)
    where does the H disappear on the 2-hydroxybenzoic acid (on the hydroxy?)
    The oxygen on the hydroxy is now positively charged, the H is therefore deprotonated to return the oxygen to its neutral charge, basically simple proton transfer, that you should know even at a level
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    (Original post by shengoc)
    The oxygen on the hydroxy is now positively charged, the H is therefore deprotonated to return the oxygen to its neutral charge, basically simple proton transfer, that you should know even at a level
    never mind I think I have finally got it... it was staring me in the face all along. It is pretty much EXACTLY THE SAME as with acyl chlorides.... duhhh.... sorry for wasting your time. :rolleyes:
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    (Original post by ukmd)
    never mind I think I have finally got it... it was staring me in the face all along. It is pretty much EXACTLY THE SAME as with acyl chlorides.... duhhh.... sorry for wasting your time. :rolleyes:
    anhydride pretty much reacts in the same way, except that i think it is better, at least it is not generating HCl fumes which it would if acyl chloride is used. Either of them is more reactive than carboxylic acid to do this anyway.

    remember, there is often more than one ways to draw mechanisms, as long as they are sensible and logically possible.
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    (Original post by shengoc)
    anhydride pretty much reacts in the same way, except that i think it is better, at least it is not generating HCl fumes which it would if acyl chloride is used. Either of them is more reactive than carboxylic acid to do this anyway.

    remember, there is often more than one ways to draw mechanisms, as long as they are sensible and logically possible.
    thanks. have "repped" you.
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    (Original post by ukmd)
    Hi there, I'm new here. Just would like some help with my chem practical write up. In the practical I had to make, purify and determine the mpt. of aspirin. One of my questions on the sheet [b]is [\b]"Attempt to draw a mechanism for the reaction that occurs". I am not too sure about this, I made the aspirin using 2-hydroxybenzoic acid and ethanoic anhydride using H2PO4. Please could someone help me out. Thanks in advance.
    You can take the paper home (get a lot of help) and complete it???
 
 
 
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