6
$\begingroup$

Which one of the following substituents at para-position is least effective in stabilizing the phenoxide ion $\ce{PhO-}$

(a)$\ce{-CH3}$

(b)$\ce{-OCH3}$

(c)$\ce{-COCH3}$

(d)$\ce{-CH2OH}$

According to me it should be $\ce{-OCH3}$ because of its mesomeric effect which will counter the mesomeric effect of $\ce{O-}$.

$\ce{-CH3}$ has a positive inductive effect but, the mesomeric effect of $\ce{-OCH3}$ should be stronger and destabilize the phenoxide ion to a greater extent

Improve this question
$\endgroup$
2
  • $\begingroup$ Seems to me the answer has already been given by Ron here. $\endgroup$
    – Nicolau Saker Neto
    Commented Mar 28, 2015 at 15:57
  • $\begingroup$ @NicolauSakerNeto Thanks for pointing that out. I had completely forgotten about that question I had asked. $\endgroup$
    – Binary Geek
    Commented Mar 28, 2015 at 16:25

1 Answer 1

Reset to default
4
$\begingroup$

Which one of the following substituents at para-position is least effective in stabilizing the phenoxide ion $\ce{PhO^{−}}$

Yes, the p-methoxy group is the least stabilizing of the choices. I'm not quite sure what you mean by

its mesomeric effect which will counter the mesomeric effect of $\ce{O^{−}}$ .

so let me add a few words of explanation for the choice.

In the figure below, I've drawn a few of the resonance structures possible for p-methoxyphenoxide (A-C). The methoxy group donates electrons through resonance and resonance structure B illustrates this effect. However, note that in resonance structure B we have placed two negative charges close to each other - this is very destabilizing. Therefore, as the methoxy group donates electrons through resonance it destabilizes the system. In resonance structure C, the electron withdrawing inductive effect of the methoxy group is stabilizing. However the resonance effect outweighs the inductive effect and overall the p-methoxy group is strongly destabilizing in the case of phenoxide.

Resonance structure D illustrates the p-methyl case. Methyl is very mildly electron releasing both through resonance and inductively, so it also destabilizes the phenoxide anion, but since its effect is weak, it is not as destabilizing as the p-methoxy substituent.

Finally, notice in the p-acetyl case we can draw resonance structure E which actually stabilizes the phenoxide anion via resonance. The p-acetyl group is also inductively stabilizing in the phenoxide example.

enter image description here

Improve this answer
$\endgroup$
13
  • $\begingroup$ I meant that $\ce{O-}$ will try to donate electrons to the ring and to the methoxy group (and will be very successful in doing so).The methoxy group will try counter the effects of donation of electrons by $\ce{O-}$ by its mesomeric effect. $\endgroup$
    – Binary Geek
    Commented Mar 28, 2015 at 16:12
  • $\begingroup$ In this answer, you explain why methoxy at para position will stabilize phenoxide more than methyl at para position. Doesn't this contradict the above answer? $\endgroup$
    – Binary Geek
    Commented Mar 28, 2015 at 16:14
  • $\begingroup$ I would agree Ron, however the pKa's then seem counter intuitive, since pKa of p-cresol = 10.26 and pKa of p-Methoxyphenol = 10.16. (pubchem.ncbi.nlm.nih.gov/compound/… and research.chem.psu.edu/brpgroup/pKa_compilation.pdf). Or am I not seeing something? There differences are very small anyhow. $\endgroup$
    – Jori
    Commented Mar 28, 2015 at 16:15
  • $\begingroup$ Thanks for clarifying. That is correct, the phenoxide oxygen does donate to the ring, but it can't donate to the methoxy oxygen. Try to draw a resonance structure to show this and you wind up with 10 electrons around the methoxy oxygen. $\endgroup$
    – ron
    Commented Mar 28, 2015 at 16:16
  • $\begingroup$ No, I explain why methoxy destabilizes. Resonance structures like B, where the methoxy donates electron density into the ring, place 2 negative charges close to one another - destabilizing. $\endgroup$
    – ron
    Commented Mar 28, 2015 at 16:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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