Question

A solution of m-chloroaniline, m-chlorophenol and m- chlorobenzoic acid in ethyl acetate was extracted initially with a saturated solution of $NaHCO_3$ to give fraction $A$. The leftover organic phase was extracted with a dilute $NaOH$ solution to give fraction $B$. The final organic layer was labelled as fraction $C$. Fractions $A,B,C$ contain respectively:
A.m-chloroaniline,m-chlorobenzoic acid and m-chlorophenol
B.m-chlorophenol, m-chlorobenzoic acid and m-chloroaniline
C.m-chlorobenzoic acid,m-chlorophenol and m-chloroaniline
D.m-chlorobenzoic acid, m-chloroaniline and m- chlorophenol

Answer 1

To solve this question accurately we need to understand the structures of all the given compounds that are m-chloroaniline, m-chlorophenol and m-chlorobenzoic acid. Let’s have a look at the structures.

m-CHLOROANILINE m-CHLOROPHENOL m-CHLOROBENZOIC ACID
Also,$NaHCO_3$ is a weak base. M-chloroaniline we know is base and m-chlorophenol and m-chlorobenzoic acid these two both are acids. But m-chlorobenzoic acid is a strong acid as compared to m-chlorophenol and NaOH is a strong base.Now what the question says is that there is a solution of these three compounds and each compound needs to be extracted one after the other. Fraction $A$ to be extracted using concentrated $NaHCO_3$ then fraction $B$ to be extracted using dilute $NaOH$ from the remaining organic phase. The final organic compound that will be left is fraction $C$.

Complete step by step answer:
By now you must have guessed the correct answer. In case you haven’t let me explain it to you. We know that a weak base always reacts with a strong acid to form an acidic solution. A strong base always reacts with a weak acid to give a basic solution. So, here as I already said $NaHCO_3$ is a weak base which will react ( that is extract the H positive ion ) with a strong acid (here it is m-chlorobenzoic acid) , so fraction A is going to be m-Chlorobenzoic acid. Remaining solution has m-chloroaniline and m-chlorophenol. Next we have a strong dilute base that is $NaOH$ and a strong base reacts with a weak acid ( which is m-chlorophenol) in this case. So, even fraction $B$ is extracted which is m- chlorophenol. Now m-chloroaniline is left so the fraction $C$ is m-chloroaniline.
So, we have deduced that Fraction $A$ is m-chlorobenzoic acid, Fraction $B$ is m-chlorophenol and fraction $C$ is m-chloroaniline.
So, the correct answer is option (C)

Note:
To solve questions like these you should have a proper understanding of the structures of organic compounds, their acidic or basic character . Like carboxylic groups are highly acidic, then comes phenols and then comes alcohols. Another point in case you don’t know, the ‘m’ written in front of the names of compounds represents the position of chlorine that is meta.