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Identify the most acidic hydrogen in the compound below:

2‐nitro‐[1,1'‐biphenyl]‐4,4'‐diol

Can someone please explain why is $\ce{H^\mathrm{a}}$ in the compound below more acidic than $\ce{H^\mathrm{b}}$?

This is what is written in the answer key although the conjugate base obtained from losing $\ce{H^\mathrm{b}}$ is stabilized by both resonance, and the inductive electron withdrawing nitro group.

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    $\begingroup$ Anion without Hb is not stabilized by resonance. NO2 is in the wrong position with respect to that OH. $\endgroup$ – Ivan Neretin Apr 1 at 10:39
  • $\begingroup$ It is stabilized by resonance due to the phenol ring. The stabilizing role of the nitro group is inductive. $\endgroup$ – user208973 Apr 1 at 10:48
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    $\begingroup$ Well, actually the stabilizing role of NO2 is not just inductive. $\endgroup$ – Ivan Neretin Apr 1 at 10:58
  • $\begingroup$ Well, another way of looking at it is that the Nitro group is ortho to the ring at the right, and thus decreases electron density in that ring. $\endgroup$ – William R. Ebenezer Apr 1 at 11:54

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