# How can bromine water be used to distinguish between benzoic acid and salicylic acid?

I have come to learn that bromine water can distinguish benzoic acid from salicylic acid. I know that salicylic acid on reaction with bromine water forms 2,4,6-tribromophenol. I think that benzoic acid will form m-bromobenzoic acid. So with this respect how can bromine water distinguish the two compounds?

Salicylic acid is partially activated towards electrophilic aromatic substitution by the hydroxyl group, and slightly deactivated by the carboxylic acid. Overall, it is activated towards the substitution. Benzoic acid is deactivated by the carboxylic acid, so overall, it is deactivated towards substitution. Hence, salicylic acid will react without the need of a catalyst (and upon decarboxylation, the reaction will actually become faster), whereas benzoic acid needs a catalyst, such as $\ce{FeBr3}$, to react.
Hence, salicylic acid will react with bromine water, but benzoic acid won't. Physically, bromine water is orange due to $\ce{Br2}$, but if the bromine reacts, then the orange color will no longer be present. Thus, the solution will turn colorless upon contact with salicylic acid, and will stay orange in contact with benzoic acid. Furthermore, depending on the scale and concentrations, you may also observe the evolution of $\ce{CO2}$.
You can clearly distinguish Salicylic acid from Benzoic acid after treatment with $\ce{Br_2/H_2O}$.
In Salicylic acid, when bromine water is added due to polarity of the modium, Oxygen atom of the $\ce{OH}$ group attached to the benzene ring becomes highly activated in Electrophilic aromatic substitution. It forms ortho-para substittued product and also ipso -substitution occurs where the $\ce{-COOH}$ group leaves the ring as $\ce{CO_2}$ and also $\ce{Br }$ group gets attached to the ortho position.So, in the polar medium, Salicylic acid reacts with bromine water to give 2,4,6-tribromophenol, and in you will get a white precipitate of the product formed in case of Salicylic acid.