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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2$\begingroup$ Hint: consider H-bonding in o-nitrophenol $\endgroup$– Aniruddha DebJun 20, 2020 at 17:31
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$\begingroup$ 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 $\endgroup$– ParteekJun 20, 2020 at 19:56
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2 Answers
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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.
answered Jun 21, 2020 at 0:23
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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? $\endgroup$– ParteekJun 22, 2020 at 0:45
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$\begingroup$ 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. $\endgroup$ Jun 22, 2020 at 2:06
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$\begingroup$ @OscarLanzi Can you please explain why p-nitrophenol is less acidic than carbolic acid? Despite that the conjugate base of p-nitrophenol is more stable due to electron withdrawing nature? $\endgroup$ Aug 27, 2022 at 9:42
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1$\begingroup$ Carbonic acid, $\ce{H2CO3}$, is not to be confused with carbolic acid which is another name for phenol $\ce{C6H5OH}$. $\endgroup$ Aug 27, 2022 at 9:47
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1$\begingroup$ @OscarLanzi Oops, I am extremely sorry. $\endgroup$ Aug 27, 2022 at 9:56
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o-Nitrophenol won't react with $\ce{NaHCO3}$ because when $\ce{-NO2}$ group is attached at ortho-position, hydrogen bonding is observed between $\ce{-H}$ of $\ce{-OH}$ and $\ce{-O}$ of $\ce{-NO2},$ due to which acidic strength of o-nitrophenol decreases.
Since o-nitrophenol becomes less acidic than $\ce{H2CO3},$ the equilibrium shifts in backward direction. In other words, it doesn't react with $\ce{NaHCO3}.$
