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?
Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers, and students in the field of chemistry. It only takes a minute to sign up.
Sign up to join this communityBoth 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?
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.
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 $\ce{NaHCO3}.$