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Organic Chemistry: acid pKa of a substituted phthalic acid

  1. May 1, 2008 #1
    1. The problem statement, all variables and given/known data

    Two protons have different pKas: 3.50 and 4.25. The molecule is 2-bromo-3-nitrophthalic acid (if I'm naming it correctly). Assign the correct pKa to each proton.

    [​IMG]


    2. Relevant equations

    A lower pKa means a stronger acid, which means a more stable conjugate acid.

    Carboxylates are stabilized by resonance and by induction.


    3. The attempt at a solution

    I say the carboxylate ortho to bromine and meta to nitro is the strongest acid, pKa = 3.50. The inductive effect of an adjacent bromo is a strong stabilizing effect on the conjugate base.

    A resonance form may be drawn such that a positive charge is situated adjacent to the other carboxylate: the carboxylate meta to bromine and para to nitro. The resonance stabilization is compromised by the adjacent substituted positions to the carboxylate, which force the carboxylate out of planarity with the ring.

    My professor disagrees. His solution:
    [​IMG]




    Please weigh in with your answer. Also can anyone help me find a literature value for these? I think the professor made them up. I haven't been able to find pKa's for this compound myself. Where is a good place to look, that would have data for an obscure compound like this?
     
    Last edited: May 1, 2008
  2. jcsd
  3. May 1, 2008 #2

    GCT

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    Bromine has a comparable electronegativity with oxygen - maybe even lower - not quite certain at the moment in addition it is separated by three degrees from the anion oxygen.

    Not quite certain with your second claim - planarity with the ring is required even for your unstable positively charged resonance state.

    Your answer is imaginative yet it isn't the classic textbook answer your professor is requiring - take notes in class and practice upon his selected problems-in class it's not about discovery it's about how much you understand the history of chemistry. Your imagaination belongs in the lab.
     
  4. May 3, 2008 #3

    epenguin

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    1. I could not find the ones you asked for but here found some that are relevant to the problem. When googling look for acid dissociation constants not pK’s.

    2. AFAIR your Prof’s explanation is the one given for dissociation constants of phenols. At least it explains (well fits :biggrin:) the orto meta para effects there. Nice pics with the explanations in http://www.chem.wwu.edu/lampman/chap7

    3 But his account for your case is IMHO:shy: unconvincing theoretically; a double bond from O to the ring C would not be able to be formed in this case. And it does not fit the facts either and your inductive effect one does!:approve: For a number of analogous compounds in tables below you see the electronegative atom/group having most effect, and typically a pK shift of -1 next door, the ortho position and the meta and para not so affected and not so different from each other!

    4 Did you notice the little word ‘measured’ pK values? Could they have quoted values that had never been measured? I think what they have in mind is that the pK of a di-acid reflects two processes. With obvious notation

    [tex]H_2mp \rightleftharpoons H^+ + Hm^-p[/tex] and
    [tex]H_2mp \rightleftharpoons H^+ + Hmp^-[/tex] .

    If the dissociation constants of these is [tex]k_1[/tex] and [tex]k^\prime_1[/tex] then the measured (by titration) K is

    [tex]K_1 = k_1 + k^\prime_1[/tex]

    For disociation to of the overall monodisociated to [tex]m^-^-[/tex] we have a measured diss const

    [tex]K_2 = \frac{k_2k^\prime_2}{(k_2 + k^\prime_2)}[/tex]

    The k's are called microscopic and the Ks measured or macroscopic diss consts.

    You can only obtain the Ks and not the ks, of which there are 4 but 3 independent, from titration alone. You might however form an idea of them from structure/pK comparisons. In any case note that the 2nd dissociation is from and already dissociated and charged species, unlike any in my tables. Again it might be possible to get an idea of any effects of this charge on pK by comparing data. Think about. At least the direction of how the microscopic ones relate to the measured you can predict.

    5 I would not dig long for those pK’s – it is an excuse to go see your prof. Usually does student less than no harm to be seen as thinking and reactive. However work out and clarify your ideas based on facts above etc. first.

    6 some data but there must be other sources:

    http://ifs.massey.ac.nz/resources/chemistry/dissociation/orgacids.htm#B


    Benzoic, 2-bromo 25 1.45 x 10–3 2.84
    3-Bromobenzoic 25 1.37 x 10–4 3.86


    http://www.zirchrom.com/organic.htm

    C6H5NO3 2-Nitrophenol 25 7.17
    C6H5NO3 3-Nitrophenol 25 8.28
    C6H5NO3 4-Nitrophenol 25 7.15


    C7H5BrO2 2-Bromobenzoic acid 25 2.84
    C7H5BrO2 3-Bromobenzoic acid 25 3.86
    C7H5CIO2 2-Chlorobenzoic acid 25 2.92
    C7H5CIO2 3-Chlorobenzoic acid 25 3.82


    C7H5CIO2 4-Chlorobenzoic acid 25 3.98
    C7H5IO2 2-Iodobenzoic acid 25 2.85
    C7H5IO2 3-Iodobenzoic acid 25 3.80


    C7H5NO4 2-Nitrobenzoic acid 18 2.16
    C7H5NO4 3-Nitrobenzoic acid 25 3.47
    C7H5NO4 4-Nitrobenzoic acid 25 3.41

    C8H6O4 o-Phthalic acid 1 25 2.89
    2 25 5.51
    C8H6O4 m-Phthalic acid 1 25 3.54
    2 18 4.60
    C8H6O4 p-Phthalic acid 1 25 3.51


    C6H5CIO 2-Chlorophenol 25 8.49
    C6H5CIO 3-Chlorophenol 25 8.85
    C6H5CIO 4-Chlorophenol 25 9.18

    C6H6O Phenol 20 9.89

    And
    http://www.digischool.nl/sk/Binas/KzKb_org.pdf
     
    Last edited: May 3, 2008
  5. May 4, 2008 #4

    Borek

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    And not

    [tex]K_1 = \frac{k_1 + k^\prime_1}{2}[/tex]?

    Can you elaborate? I don't understand the notation here - extending the idea from the first step I am arriving at two identical k2 and k2' dissociation constants, both already equal to the second overall dissociation constant. That fits your other remark

    3 independent being k1, k1', and k2 = k2'. But it doesn't fit your formula for K2.



     
  6. May 5, 2008 #5

    epenguin

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    The notation was m, p stands for meta, para, so [tex]m^-[/tex] for meta group dissociated, negatively charged. For the doubly ionised molecule I should have written [tex]m^-p^-[/tex] .


    For the dissociation formulae you are quite right about the formula for [tex]K_1[/tex]. It can be confusing and is something a student needs to grasp. I would have had to write out a page to be completely clear, which I will later if anyone is confused. But your link should do that job. The constants need to be defined in such a way that the first and second ionisations can be directly compared. (It sometimes has been done otherwise as I was doing, and I did find the formulae I gave in a book too, but it was a confusing convention best forgotten.)

    I do not follow the rest of your deductions. When I say 3 independent, I mean that [tex]k_1k_2 = k\prime_1k\prime_2[/tex] is a thermodynamic requirement, reducing the 4 constants to 3 indepdent ones.

    My point remains that if the pKs are not too separate, then they neither represent a single process.

    On the explanation of the pKs in the question originally asked, I have thought further and looked at the data, and now think we have all been on the wrong track so far and missed an important point, of which more later! :smile:
     
  7. May 5, 2008 #6

    epenguin

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    The notation was m, p stands for meta, para, so e.g. [tex]Hm^-p[/tex] for meta group dissociated, negatively charged. For the doubly ionised molecule I should have written [tex]m^-p^-[/tex] .


    For the dissociation formulae you are quite right about the formula for [tex]K_1[/tex]. It can be confusing and is something a student needs to grasp. I would have had to write out a page to be completely clear, which I will later if anyone is confused. But your link should do that job. The constants need to be defined in such a way that the first and second ionisations can be directly compared. (It sometimes has been done otherwise as I was doing, and I did find the formulae I gave in a book too, but it was a confusing convention best forgotten.)

    I do not follow the rest of your deductions. When I say 3 independent, I mean that [tex]k_1k_2 = k\prime_1k\prime_2[/tex] is a thermodynamic requirement, reducing the 4 constants to 3 indepdent ones.

    My point remains that if the pKs are not too separate, then neither of the 'measured' ones represents a single process.

    On the explanation of the pKs in the question originally asked, I have thought further and looked at the data, and now think we have all been on the wrong track so far and missed an important point, of which more later! :smile:
     
  8. May 5, 2008 #7

    Borek

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    OK, it clicked now. For some reason I thought k2 is already the overall dissociation constant and I couldn't see why there are two, but one is for m-p -> m-p- and the other for mp- -> m-p-.

    Never thougt about it this way. Could make interesting addition to the pH lectures. In general it is "only" another formalism, but such "parallel" approaches often give interesting insight into what is going on in the solution.

    Do you know of any references? Preferably web based, as my access to literature is limited?

    That's obvious regardles of the approach - as long as you calculate correctly what is going on :smile:
     
  9. May 5, 2008 #8

    epenguin

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    I work it out myself evey time needed! It was late at night and I was slightly confused! The book I took my confirmation of formula from was 'Enzymes: Physical Princples' H. Gutfreund p 40, but I will post my take on it, right now I have not time, there is all the tex formulation involved. It was a point fairly secondary to the question at issue under discussion, though more important didactically.

    In fact it is even more secondary in the light of the following considerations, where the dissociated structure is a single one, though I suppose the proton resides more in one place than another.
     
  10. May 5, 2008 #9

    epenguin

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    Dicarboxylic acids, especially when carboxyl groups are adjacent, are moderately strong acids. Starting with the comparison of benzoic, pK 4.19 with o-phthallic, 2.16. So this led me to suppose the monoionised species is stabilized by hydrogen bonding in a structure such as [tex]O=C_\mid-O-H...^-O-C_\mid=O[/tex] possibly involving other water molecules. From glimpses of literature it appears that this is accepted. An even more striking effect in a somewhat analogous structure is maleic acid pK 1.83 compared with say acetic, 4.86. The –COOH s are a little differently positioned in phthallic than in maleic, evidently such as to allow a more favorable interaction in the latter case.

    The same stabilisation, as well perhaps as an electrostatic contribution would explain why the second dissociation is disfavoured in the dicarboxylic acids, 5.51 in phthallic cf 4.19 benzoic, 6.07 in maleic cf 4.86 acetic.

    As regards the effect of the substituents, electron withdrawing could explain both the effects by destabilizing the H-bonded structure, but note that we are explaining the opposite of what we thought were explaining previously – not that the Br decreases the first pK but that it increases it!

    To repeat, I exclude the resonance explanation here because the phenomenon (similarity of effects of ortho- and para- substituents in phenols) is just not the one we have in benzoic acid derivatives. I would not exclude that what effect m- and p- nitro substitution does have on them (and the Br substitution to an extent) is not a hydrogen-bonding or solvent-mediated effect. To tell the truth I have always felt uncomfortable with these purely internal-to-to-the molecule explanations and that they underestimated solvent effects, but I hope that somebody who knows what they are talking about, unlike us and kallery's prof. :smile: happens along.
     
  11. May 12, 2008 #10
    Thank you for responding. I read over these thoughtful replies several times. Interesting comments! I still think the professor's reasoning for his answer is specious and incorrect. Without verifiable published pKa's, it's my word against his *sigh*
     
  12. May 18, 2008 #11

    epenguin

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    I don't think there is any reason to think his pK values have been made up! They are not out of line with anything or unexplained it seems now.

    I had not remembered when answering before that p- and m- substuted benzoic acids are used as a kind of standard of the Hammett sigma function. I.e. where there are just these straight inductive effects only. What I mean better explained in http://www.auburn.edu/~deruija/pda1_acids1.pdf which I just found.

    which also gives the same H-bonding explanation as I did. Everything in the various tables seems to fit nicely into these explanations. The absence of resonance stabilisation effects supposed by your prof. is explictly mentioned in the quotation below with the same reasoning I gave. It is sometimes too easy to semi-convince oneself by things one can draw, but I suspect he has some supressed uneasiness and would be glad of a clarification and to stop teaching a mistake!?

    "It is important to realize that aromatic substituents DO NOT stabilize carboxylate base
    charge by resonance effects! This is because that the charge formed upon ionization of an
    acid is not "conjugated" with the aromatic system. Because of the basic atomic structure
    of carboxylic acids, the charge generated upon ionization is "insulated" from the aromatic
    ring by the presence of two single bonds as shown below. Thus aromatic rings, and
    substituents on aromatic rings, have only a modest effect on carboxylic acid acidity:
    This is very different from phenols (see "Phenol Tutorial") where the negative charge
    formed in the conjugate base could be delocalized into the aromatic ring, and properly
    positioned substituents could dramatically enhance acidity. In the case of phenols the
    charge formed on an atom (oxygen) directly conjugated to the aromatic ring. This is
    illustrated by the figure below:
    However, acids are generally more acidic than their corresponding phenols due to the
    delocalization of negative charge by resonance over two adjacent electronegative oxygen
    atoms in the case of acids (see discussion above).
    Another interesting example of the influence of structure on relative acidity is illustrated
    by the dicarboxylic acids. Consider the structures of the four diacids listed in the table
    below and their pKa values:" .....

    "The first pKa for oxalic and malonic acid is significantly lower than a typical
    monocarboxylic acid, indicating that these compounds form a mono-anion more readily.
    The enhanced ease of ionization of one carboxyl group in these two diacids results from
    the ability of the “other” adjacent acid group to stabilize the negative charge of the
    conjugate base by hydrogen bonding as shown in the figure below. Note also that the
    second pKa of these acids is significantly higher than a typical mono carboxylic acid.
    This is because that “second” acid proton is held more tightly by the mono-ionized
    complex via intramolecular hydrogen bonding."
     
  13. May 28, 2008 #12
    Here's why I think he may have made up the pKa values:

    The pKa's for phthalic acid (the 1,2-dicarboxylic acid of benzene) are pKa1 = 2.98 and pKa2 = 5.28. What this would mean is that somehow phthalic acid is more acidic than 3-bromo-4-nitrophthalic acid (pKa1 = 3.50), which we know cannot be true.

    Is it really possible that my professor has the para nitro resonance effect for phenols confused with the effect for benzoic acids? Hopefully I will have another discussion with my professor before the term's up, and hopefully he will listen to what I have to say about the problem.
     
  14. May 28, 2008 #13

    epenguin

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    Why do you say that cannot be true?
    However I do think on the second point he has confused, according to explicit statement about this in one of my quotations.
     
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