0
$\begingroup$

I have to complete the equation

4-nitrophenol + phenol --->

and then label the acid, the base, the conjugate acid and the conjugate base.

I think the answer might be

4-nitrophenol + phenol ---> 4-nitrophenoxide + [C6H5OH2]+

I don't know what the name of the last compound would be and it looks unlikely, but it's the only option I can think of.

If this is right, what would the name of the resulting conjugate acid be? If it's wrong, why? and what would the right answer be?

Thank you in advance

$\endgroup$
3
  • 5
    $\begingroup$ I would not expect neutral phenol molecules to be protonated to a large extent by an acid as weak as 4-nitrophenol. Second reactant should be phenolate ion maybe? $\endgroup$ Nov 22 '21 at 20:42
  • $\begingroup$ @OscarLanzi would you know of any alternate solutions? $\endgroup$
    – Gallus
    Nov 22 '21 at 20:46
  • 1
    $\begingroup$ @Gallus Construction in ChemDraw JS of the protonated phenol (for reference, SMILES string this version assigns is [H][O+]([H])C1=CC=CC=C1) is auto-named phenyloxonium. For future reference, among ACS' organicchemistrydata.org are Reich's compilations of pKa, e.g., here including the two phenols in question. $\endgroup$
    – Buttonwood
    Nov 22 '21 at 21:00
-2
$\begingroup$

You can find the stronger acid by comparing the stability of their conjugate bases.

4-nitrophenol has the NO2 group (showing both electron withdrawing inductive and resonance effect) at the para position which will pull the electron pair away from the O- group. As you can see there is an additional resonance structure in its conjugate base giving it more stability.

enter image description here

Phenoxide ion shows one resonance form less making it less stable. Also it has no group which will inductively pull away the negative charge (which NO2 does very well), therefore we can conclude that 4-nitrophenol will in fact be the the stronger acid and protonate phenol.

pKa(4-nitrophenol) = 7,15

pKa(phenol) = 9,95

$\endgroup$
2
  • 1
    $\begingroup$ Unfortunately, this counting method, while it has been incredibly popular, is not based on any scientific rigor. You also only show the most prominent structures, but not all of them; the ionic ones are all missing. Nevertheless, while you can compare acidity based on $\mathrm{p}K_\mathrm{a}$ values, the remaining question is still: to what extent would/will a proton transfer happen. You also completely neglect whether or not the potonated forms will be stable enough to account for ionisation. $\endgroup$ Nov 24 '21 at 17:37
  • $\begingroup$ Some of the resonance structures in the first diagram need work. $\endgroup$
    – user55119
    Nov 24 '21 at 18:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.