Question

An organic compound $({C_8}{H_{10}}{O_2})$ rotates plane polarised light. It produces a pink colour with a neutral $FeC{l_3}$ solution. What is the total number of all the possible isomers for this compound?

Answer 1

Hint: The molecular formula of the compound is given in the question, we can determine the structural formula of the compound by calculating the value of its double bond equivalence to ensure the type of chain/ring present within a molecule and further we can check the presence of functional group with the help of test given with $FeC{l_3}$ solution. After knowing the structure we can easily calculate the number of possible isomers for the given compound.

Complete Step by Step Solution:
DBE (Double bond equivalent): It is also known as the degree of unsaturation that predicts the nature of the ring or chain within a compound i.e., whether the ring or chain is saturated or unsaturated. The value of DBE can be calculated as per the following formula:
$DBE = \frac{{2C + N - H - X + 2}}{2}$

Substituting values as per the molecular formula given in the question, the value of DBE can be calculated as follows:
$DBE = \frac{{2 \times 8 - 10 + 2}}{2}$
$ \Rightarrow DBE = 4$

The value of DBE represents the presence of benzene. Therefore, we can say that the base ring of the molecule is benzene. On moving further, reaction of a compound with neutral $FeC{l_3}$ solution gives a pink colour i.e., gives a positive result for $FeC{l_3}$ test. Thus, we can say that a phenolic group is present in the compound.

As per the question, the organic compound rotates plane polarised light which means it is optically active i.e., it must consist of at least one chiral centre in the compound. Based on the observations made, the structures possible for the given compound are as follows:

Therefore, the total number of all the possible isomers for the given compound is 6.

Note: It is important to note that in general cases, the students may get confused with carboxylic group due to presence of double oxygen atoms within the molecule but as per the value of DBE which is 4, no double bond is present outside the benzene ring and also, the carboxylic group gives deep red colour with neutral $FeC{l_3}$ solution.