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o-Nitrophenol is less acidic in comparison to p-nitrophenol due to intramolecular hydrogen bonding. This effect also is the reason for o-nitrophenol being less soluble in water.

At the same time, hydroquinone (para) is less acidic than catechol (ortho). Of course there is the inductive effect, but a reason mentioned quite often as well is hydrogen bonding. Isn't it contradictional that the reason given for o-nitrophenol being less acidic than p-nitrophenol is the same reason given for catechol being more acidic than hydroquinone?

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In the case of o-nitrophenol, the acidic hydrogen is hydrogen-bonded to the nitro group's oxygen atom, making it less acidic. In the case of catechol, one acidic hydrogen is hydrogen-bonded to the adjacent OH group's oxygen atom, but the other, more acidic hydrogen is not hydrogen bonded. This hydrogen is more acidic than that of hydroquinone. The resulting anion is even more strongly hydrogen-bonded than neutral catechol, stabilizing it even further. It is the second acidic hydrogen of catechol that is analogous to the acidic hydrogen of o-nitrophenol.

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  • $\begingroup$ So the oxygen in the adjacent OH group in catechol is also involved in hydrogen bonding with the first OH group, thus weakening its own O–H bond? $\endgroup$ – jona173 Jan 18 at 22:07
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    $\begingroup$ That's that basic idea. But that OH bond is not necessarily weaker (relative to homolysis), but rather more polarized and hence tending more to ionize. $\endgroup$ – iad22agp Jan 18 at 23:12

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