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Aromatic Compounds and Acidity

  1. Jul 5, 2007 #1
    In the oxidation of the side chain in toluene with potassium permanganate and water as the reagents, does the acidity increase as the redox reaction is carried out?

    Thank you.
     
  2. jcsd
  3. Jul 6, 2007 #2
    Actually, you would never carry this reaction out without protecting the benzene ring, because of the methyl group on the ring, it gets activated, and acts as a reducing agent. Thus, the benzene ring undergoes the possibility of breaking apart.

    If just the methyl group is oxidized, then it gets converted to -COOH, so yes the acidity will increase. Any alkyl group attached to the benzene ring when oxidized gets converted to acid and co2.
     
  4. Jul 6, 2007 #3
    For the final product, does the number of mL of HCl added to the reaction mixture affect the purity of the benzoic acid obtained? (Does it matter if I use 10 mL of HCl or 15 mL of HCl?)

    Thanks.
     
  5. Jul 6, 2007 #4
    Yeah, but thats stoichiometry. Sometimes, depending on the compound you're converting, the product and the initial reactant might react to give you a third, unwanted product, so the HCl must be in excess..
     
  6. Jul 6, 2007 #5
    Do you know what kind of undesired product might form in this reaction? Would it be distinguishable from the crystallized benzoic acid?

    I was trying to work out the reaction mechanism for the oxidation of toluene just to understand the reaction more, but I am unsure of the steps since I was following an example, the oxidation of benzaldehyde. Instead of arriving at PhCOOH from PhCH3, I got PhCHO. How do 2 oxygen atoms replace the side chain from PhCH3? (Does the mechanism show PhCH3 oxidized to PhCHO, which is then oxidized again by MnO4- to finally get PhCOOH? I’m sorry if this is a foolish question. ) How exactly does this mechanism look?

    Thanks.
     
    Last edited: Jul 6, 2007
  7. Jul 7, 2007 #6
    Yeah. Thats it. I don't remember the exact mechanism, but I think MnO4- forms a closed chain which is replaced by -OH groups. Since a carbon atom cannot have more than 2 -OH groups at the same time, one pair of -OH molecules give water and a =O at that position. So the three hydrogens are replaced by -OH, two of which leave as water and one of the oxygens of the two remains as a =O, hence your -COOH.

    I may be wrong where the mechanism is concerned, but I am sure about the products and their order of appearance. Just google the mechanism once.
     
  8. Jul 7, 2007 #7
    I tried using a search engine, but I mostly get reaction equations and not a mechanism or published papers that you must have an account for.

    I also have another question pertaining to this redox reaction.

    After KMnO4 is added to toluene and is refluxed, benzoic acid forms. However, after MnO2 is removed but before acidification occurs (before HCl is added to filtrate), is the benzoic acid, the product in the clear filtrate, in an aqueous form? (Did I use "aqueous" correctly? I know the product does not exist in a solid form as yet.)

    This the the equation before acidification:

    PhCH3 + 2MnO4- --> PhCOO- + 2MnO2 (precip.) + H2O + OH-

    Did I mix up the aqueous state with the liquid state? I thought that if ions exist in water it would be labeled “aqueous.”

    Thanks again.
     
  9. Jul 8, 2007 #8
    I actually have no idea about that... Ill have to look it up.
     
  10. Jul 9, 2007 #9

    chemisttree

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    These oxidations are usually conducted under basic conditions. The product (benzoic acid) is converted to the water-soluble salt, potassium (or sodium) benzoate. This is in the clear filtrate.
     
  11. Jul 9, 2007 #10
    Even though the reaction mixture occurs under basic conditions, the reaction mixture's pH still decreases as the redox reaction transpires, right?
     
  12. Jul 9, 2007 #11

    chemisttree

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    I believe the balanced theoretical equation is:

    [tex]2KMnO_4 + PhCH_3 \rightarrow\ PhCOOK + H_2O + KOH + 2MnO_4[/tex]

    You will note that as the reaction progresses, for each mole of toluene reacted, one mole of potassium benzoate and one mole of potassium hydroxide is produced.

    Will this raise or lower the pH?
     
  13. Jul 9, 2007 #12
    With the formation of the benzoate and hydroxide, won't the pKa values be lower than that of toluene (pKa = 41), so the pH is lowered?
     
    Last edited: Jul 9, 2007
  14. Jul 9, 2007 #13

    chemisttree

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    The pKa of toluene has no bearing here. The likelihood of toluene deprotonating to form benzylic anion is vanishingly small under these conditions. Anyway, there are two phases in this reaction... the aqueous (K2MnO4/OH-) and the organic (toluene). pH is not really relevant to the organic phase and the aqueous phase has a very strong base present.
     
  15. Jul 10, 2007 #14
    So the pH actually does increase? (It's only after acidification, not oxidation, that the pH decreases?)

    In the theoretical equation, doesn't the MnO4 get reduced to MnO2?
     
    Last edited: Jul 10, 2007
  16. Jul 10, 2007 #15

    chemisttree

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    The oxidation produces a molecule of benzoate and a molecule of hydroxide. What will the pH do?

    Write out the half reaction for that transformation. Pay attention to the oxidation state of manganese.
     
  17. Jul 10, 2007 #16
    I still don't understand. If the aqueous phase is basic and 1 mol of benzoate is produced along with one mol of hydroxide, will the reaction mixture become neutralized, so the pH lowers?
     
    Last edited: Jul 10, 2007
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