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First carboxylic acid is more acidic than the second

  1. Oct 20, 2004 #1
    Hello

    Here are a few questions:

    1.

    [tex]O_{2}N-CH_{2}-COOH[/tex] pKa = 1.68
    [tex]Me_{3}N^{+}-CH_{2}-COOH[/tex] pKa = 1.83

    According to my book (Guidebook to Mechanism in Organic Chemistry), the first carboxylic acid is more acidic than the second. I fail to understand the reason for this, since in class we were told that groups which have actual positive charge (such as the N,N,N-trimethyl amino group) have greater -I effect than those with lone pairs or partial positive charges (or polarizable groups). With this line of reasoning, the second acid should be more acidic. Can someone explain why this is not so? Has it got something to do with +I effect/steric effects of three Me groups?

    2. Why is o-bromobenzoic acid more acidic than o-chlorobenzoic acid? (pKa of o-bromobenzoic acid = 2.85, pKa of o-chlorobenzoic acid = 2.94)

    I understand that both acides will be stronger than benzoic acid on account of Ortho Effect. As ortho effect is active in both acids, we must find some other parameter to compare relative acidity. If -I effect is chosen as a parameter, then Cl is a stronger inductively withdrawing group than Br so the chloro substituted benzoic acid should be stronger contrary to experimental data.

    I would be very grateful if someone could help me with these problems.

    Thanks and cheers
    Vivek
     
  2. jcsd
  3. Oct 20, 2004 #2

    chem_tr

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    Hello, your questions are very challenging, I like them very much indeed.

    Well, the reason is somewhat simple; trimethylammonium is still electron-donating, since the nitrogen is at 3- oxidation state. However, I cannot say the same for nitro group, as its oxidation state is 4+ for just NO2 group. I'm trying to say that the oxidation state is important here.

    We may explain it in an alternative way; nitrogen in nitro group is very positive, thus wants to find some other electrons since it has "lost" them to oxygen. It results in an increased pulling of the electrons in CH2. The carboxylic acid carbon is unhappy about it, thus weakens against carboxylic oxygens, leading to a harder dissociation, i.e., a greater pKa value. The trimethylammonium-type acid has a more balanced electron sharing within the molecule, so dissociation becomes easier.

    About your second question, you may also recall your knowledge about hydrogen halogenides; HBr is more acidic than HCl, since the valence shell of Br is weaker than Cl's, thus making it more susceptible to lose proton. We may use a similar approach here; ortho-bromobenzoic acid is more powerful than ortho-chlorobenzoic one, since chlorine is more electronegative.

    As a last word, please consider experimental details absolutely correct, and devise your reasonings according to them; it would be better.
     
  4. Oct 20, 2004 #3

    movies

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    The tetraalkylammonium ion is still inductively withdrawing! Conventionally, I think NO2 is considered a better withdrawing group than R3N+, at least in Hammet type correlations. I believe that this can be rationalized through "group electronegativities" as well, but I am not very familiar with those calculations.

    For your second question I think it is important to consider back-donation of the chlorine lone pairs into the aromatic system. The orbitals on chlorine are about the right size to donate back into the aromatic system, so imagine a chlonrine-carbon double bond and then push the electrons up into the carboxylate at the ortho position. The orbitals on bromine, however, are too large for such an interaction with the aromatic system.
     
  5. Oct 20, 2004 #4
    Thanks movies and chem_tr for the explanations. I have one more question:

    Why is p-acetyl benzoic acid a stronger acid than o-acetyl benzoic acid?

    My reasoning: at the para position, the acetyl group is strongly electron withdrawing but at the ortho position, this effect is decreased somewhat by intramolecular hydrogen bonding. Is this correct? Also, why is the ortho effect not valid here?

    Thanks and cheers
    Vivek
     
  6. Oct 21, 2004 #5

    movies

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    Yeah, I think that your hydrogen bonding explanation probably works best. There might be some sort of dipole effect as well, though.

    How big is the pKa difference in that case?
     
  7. Oct 22, 2004 #6
    Hi movies

    Well I don't know the pKa values of p and o acetyl benzoic acids but thats the explanation I found in a cookbook as well. But I was wondering why the ortho effect didn't take over. Isn't it an important factor in deciding acidities of benzoic acids?

    Thanks and cheers
    Vivek
     
  8. Oct 22, 2004 #7

    chem_tr

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    I think para-substitution causes the molecule to be more stable because of lower transition energies in a reaction (activated complex formation is easier than ortho-substituted one). The steric hindrance lowers the stability of the molecule.

    Yes, you are right about deciding the acidities according to ortho-effect and para-effect. Even in NMR predictions, I suppose that J values for para-substituted compounds are greater than ortho-substituted ones, this may be another explanation for the observed stability.

    Take care.
     
  9. Oct 22, 2004 #8

    movies

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    The ortho effect is observed when you are addind a substituent to the aromatic ring, right? You' aren't doing that in this case.

    As chem_tr mentioned there may be a steric effect as well. Once you deprotonate the acid the carboxylate is charged and needs to be stabilized by solvent or something. If there are bulky substituents adjacent to the charged atoms then that can hinder solvation. I wouldn't expect that to be a big effect in this case though since there is only moderate steric hindrance.
     
  10. Oct 22, 2004 #9
    Oh thats interesting...I haven't heard of anything called para effect. Thanks anyway.
     
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