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Both ortho- and para-nitrophenol are less acidic than carbonic acid. But still paranitrophenol reacts with sodium bicarbonate to give carbonic acid and sodium paranitrophenoxide.

However, this is not true for ortho-nitrophenol. Any explanations?

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    $\begingroup$ Hint: consider H-bonding in o-nitrophenol $\endgroup$ – Aniruddha Deb Jun 20 '20 at 17:31
  • $\begingroup$ Thanks for your reply. All it shows is that orthonitrophenol is less acidic but my point is that why does paranitrophenol even react with sodium bicarbonate despite being less acidic than carbonic acid. Try checking the reaction and pka values to satisfy yourself $\endgroup$ – Parteek Jun 20 '20 at 19:56
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p-Nitrophenol does have a higher $\mathrm{p}K_\mathrm{a}$ than carbonic acid (7.15 vs 6.3), the two logarithms differ by less than one unit, if we render the acidic species in carbonic acid as $\ce{CO2}$ rather than $\ce{H2CO3}$. So the equilibrium constant in the reaction

$$\ce{p-C6H4(NO2)OH + HCO3- <=> p-C6H4(NO2)O- + CO2 + H2O}$$

is close to $1$ and a large percentage of the p-nitrophenol could react with bicarbonate.

o-Nitrophenol actually has $\mathrm{p}K_\mathrm{a} = 7.23,$ about the same as the para isomer. It should, in fact, react similarly with sodium bicarbonate. It is the meta isomer that has higher $\mathrm{p}K_\mathrm{a}$ and may not visibly react.

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  • $\begingroup$ Thanks Oscar, I too had the same views but the textbook i referred says that ortho nitrophenol does not give sodium bicarbonate test. Should I consider that wrong? $\endgroup$ – Parteek Jun 22 '20 at 0:45
  • $\begingroup$ Likely the book missed. Based on duissociation constants the meta isomer is the one that drags its feet. The proton is indeed hydrogeb-bonded in the ortho isomer, but the closer proximity of the nitro group to the hydroxyl group in that isomer. $\endgroup$ – Oscar Lanzi Jun 22 '20 at 2:06

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