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    Hey, im stuck on this question.


    Im pretty sure A is Propan-2-ol and B is Ethanoic Acid, but can someone confirm this, and what about C


    Thanks
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    A has to be a primary alcohol (or it cant be oxidised fully to acids) so has to be propan-1-ol I assume and hence B is propanoic acid. C is is 1(?)-chloropropanoic acid I think.

    Step 2 is free radical substitution.
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    B must be propanoic acid not ethanoic acid
    C would be CH3CH(Cl)COOH or CH2(Cl)CH2COOH
    It's the free radical substitution
    I don't understand the question from C -> D
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    ok thanks all. ive not come across a Carboxlyic Acid with Chlorine Reaction before - soo do you just apply the alkane + chlorine reaction to that then
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    Not sure what exam board you're with but if you get a complex organic compound with lots of functional groups, you usually assume they react independently of eachother unless there's a reagent used that you know will react with more than one functional group (e.g. if benzene ring -hot alkalione potassium manganate will oxidise all side chains) which isn't many tbh.
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    ah ok thanks - edxcel im with.
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    Is there are question following on from that one asking about D?

    As said before, A must be a primary alcohol, so propan-1-ol and is oxidised all the way to propanoic acid in B.

    Whenever you have u/v you can pretty much say it's a free-radical reaction (don't know if there are any exceptions). Can't see why it stops when there's only one Cl on there, and I don't know how you know where it will add, so I suppose you can get the mark for either 2-chloropropanoic acid or 3-chloropropanoic acid.

    I think what's happening when you make D is first the OH- adds by nucleophilic addition and replaces the chlorine. I think the only reason it's acidified after is to return the proton that will have been removed by the alkali. So D is either 2-hydroxypropanoic acid or 3-hydroxypropanoic acid, depending where you put the chlorine. (If it hadn't been acidified, it would have been 2/3-hydroxypropanoate). Have you got the actual answers?
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    A is a primary alcohol (propanol...or propan-1-ol to be pedantic),
    B is formed from A using acidified Pot..dichromate..which oxidises the primary alcohol into a carboxylic acid.. (propanoic acid)

    B to C is free radical substitution, producing, i think, 2-chloropropanoic acid,
    then the reaction with NaOH..in my opinion would form 2-chloropropanoate and water..ooo..i suppose that would be the reason for the HCl..aha..

    so a nucleophilic substitution could occur, substituting the chlorine on the 2nd carbon for the OH- group, forming the product in the question, and the byproduct, salt, (NaCl)..
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    There isnt any questions that ask about the step C-->D


    Although this Question mentions D




    Ive worked out the Ka for D. Ka = 8.32 x 10^-3

    So would the answer for the last part be something like - Ka for D is higher than Ka for prop.acid. So less enery is needed to dissociate D than prop acid. Therefore more engergy released so D is more exothermic than Propanoic acid.


    [nope dont have answers for these am afraid]
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    How could you get Ka for D like that?
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    I got a Ka of 8.31 * 10-4

    ((10-2.04)2) / 0.1

    Hm.. next part.. without talking about an inductive effect, the reaction of D with NaOH(aq) will be more exothermic, as it has a larger Ka value than propanoic acid, so will dissociate more easily than propanoic acid, less energy from the neutralisation going into further dissociation.
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    [H+] = 10^-pH

    [H+]^2 / 0.1 I think it is, which comes out to that
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    i did:

    antilog of 2.04 = 9.120x10^-3 = [H+]

    Ka=[H+][A-]/[HA]

    organic acid with weak(ish) dissociation, so [H+]=[A-]

    therefore, Ka=[H+]^2/[HA]

    Substitute the numbers in

    Ka= (8.317x10^-5)/0.100
    =8.317x10^-4
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    ok - yeah well still the Ka for D is higher than Ka for Prop Acid,
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    Yeah, that would be right, as the OH group on compound D with draw more negative charge towards it, from the COOH group, weakening the -O-H (acidic) bond, so it is more likely to dissociate.

    This is the inductive effect, jus' so's ya know!
    hehe
    J.
 
 
 
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