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    Hi! I'm new here

    I wonder if anyone could help me, I have a few questions that i'm unsure about regarding an experiment I did.

    We dissolved the product in methanol and recrystallized it by suction filtration, but why is the minimum amount of methanol used to dissolve it in the first place?

    And also, if the product contains impurities that also dissolve in the methanol how is it still possible to purify the product? - Is this because the impurities are smaller molecules and will pass through the filter paper with the water? :confused:

    Many thanks to anyone who takes the time to help!
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    Hi, welcome to TSR

    The minimum amount of solvent is used so that when you try and grow the crystals the solution is saturated and therefore the product drops out of solution easier. If you have too much solvent then it can be difficult to get crystallisation to occur and you may need to remove some solvent.

    If its a single solvent recrystallisation then your product is only weakly soluble in the solvent (only dissolves at higher temperatures)- and so when cooled down only your product crystallises out whilst the impurities remain dissolved in the solvent. It has nothing to do with the filter paper.
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    (Original post by EierVonSatan)
    Hi, welcome to TSR

    The minimum amount of solvent is used so that when you try and grow the crystals the solution is saturated and therefore the product drops out of solution easier. If you have too much solvent then it can be difficult to get crystallisation to occur and you may need to remove some solvent.

    If its a single solvent recrystallisation then your product is only weakly soluble in the solvent (only dissolves at higher temperatures)- and so when cooled down only your product crystallises out whilst the impurities remain dissolved in the solvent. It has nothing to do with the filter paper.
    Thank you very much! So once it's cooled, the crystals will form, but the rest will remain dissolved in the solvent and just pass into the flask below? Do you mind if I ask another question?
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    (Original post by hidden088)
    Thank you very much! So once it's cooled, the crystals will form, but the rest will remain dissolved in the solvent and just pass into the flask below?
    Yeah essentially, you do suction filtration to collect the pure crystals on the filter paper whilst the filtrate contains the unwanted material.

    Do you mind if I ask another question?
    Not at all
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    We used a quickfit condenser as an air condenser in this experiment, and my teacher told us not to connect the apparatus to the cold water tap. The question asks why it would not be sensible to use cold running water through a standard Liebig condenser to cool the vapour in the experiment?
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    (Original post by hidden088)
    We used a quickfit condenser as an air condenser in this experiment, and my teacher told us not to connect the apparatus to the cold water tap. The question asks why it would not be sensible to use cold running water through a standard Liebig condenser to cool the vapour in the experiment?
    That's...interesting. As methanol boils well below 100 degrees I would say that it was essential to connect a water condenser to cool the vapour so that you didn't lose any solvent whilst heating. If you were trying to boil off the solvent then you wouldn't need a condenser at all. I'm not sure what the question is getting at. :confused:
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    Sorry, probably my fault for lack of details, the methanol is not added until later once the product is formed to help purify it. The reagent (boiling point of 204-207 degrees) is added with iodine and then refluxed. So i'm guessing the high boiling point of the liquid is the reason why? As the water would just boil aswell...or? :confused:
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    Oh right, I thought it was for the recrystallisation :p: So, you start with an organic material (in water?) and add iodine and heat to boiling point?
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    (Original post by hidden088)
    I will pm you the experiment method, easier for you to see
    Okay
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    Sorry for yet another question! But how could I tell the ratio between the two isomers on the experiment method using an NMR?

    Would the two CH3 groups on the cis isomer interact with each other as there closer then I could look at the size of the peaks? And on the trans they would not interact as they are further away?
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    What are your expected products?
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    (Original post by EierVonSatan)
    What are your expected products?
    Dimethyl maleate and dimethyl fumarate (even though I thought the experiment converts it all into fumarate), and the question reads:

    'What is the ratio of cis to trans isomers, as calculated on your NMR spectra?'

    Edit: maybe its getting at that only 1 is made and only one would be on the spectra. Not sure...
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    (Original post by hidden088)
    Dimethyl maleate and dimethyl fumarate (even though I thought the experiment converts it all into fumarate), and the question reads:

    'What is the ratio of cis to trans isomers, as calculated on your NMR spectra?'

    Edit: maybe its getting at that only 1 is made and only one would be on the spectra. Not sure...
    Ah right okay - look at the integration of the alkene protons - the resonance with the higher ppm will be the trans isomer (more conjugation)
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    (Original post by EierVonSatan)
    Ah right okay - look at the integration of the alkene protons - the resonance with the higher ppm will be the trans isomer (more conjugation)
    Mind explaining this to me a little bit more? Just confused the hell outta me
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    (Original post by hidden088)
    Mind explaining this to me a little bit more? Just confused the hell outta me
    Can I assume that this is for university and not A-level?
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    (Original post by EierVonSatan)
    Can I assume that this is for university and not A-level?
    Yeah first year uni, I can't recall learning anything about that sort of stuff with NMR, just the basics at a-level and we haven't covered any NMR stuff here yet. I think they just assume we know
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    (Original post by hidden088)
    Yeah first year uni, I can't recall learning anything about that sort of stuff with NMR, just the basics at a-level and we haven't covered any NMR stuff here yet.
    Ahh okay - well you will at some point learn that the aromatic protons have higher ppm values than double bonds (and double bonds higher than single bonds). The reason is due to delocalisation - electrons spread around further makes the protons less shielded (more exposed to the magnetic field).

    As conjugation (delocalisation) is greater in the trans than the cis (better orbital overlap - see cross-conjugation also) it has a higher ppm value
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    (Original post by EierVonSatan)
    Ahh okay - well you will at some point learn that the aromatic protons have higher ppm values than double bonds (and double bonds higher than single bonds). The reason is due to delocalisation - electrons spread around further makes the protons less shielded (more exposed to the magnetic field).

    As conjugation (delocalisation) is greater in the trans than the cis (better orbital overlap - see cross-conjugation also) it has a higher ppm value
    Ahh ok! Thanks! It doesn't help that our lectures/workshops about spectroscopy are after we have finished our spectroscopy/organic labs :rolleyes:
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    (Original post by hidden088)
    Ahh ok! Thanks! It doesn't help that our lectures/workshops about spectroscopy are after we have finished our spectroscopy/organic labs :rolleyes:
    yeah, I can pretty much guarantee that this won't be the only time a module will expect you to know things that are in courses later in the year.
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    (Original post by EierVonSatan)
    yeah, I can pretty much guarantee that this won't be the only time a module will expect you to know things that are in courses later in the year.
    Yeah i'm beginning to notice
 
 
 
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