An organic compound ( C 8 H 10 O 2 ) rotates plane-polarized light. It produces pink color with neutral FeCl 3 solution. What is the total number of all the possible isomers for this compound?

Chemistry · Isomerism

JEE Advanced 2020 — Paper 2 — Question 22

An organic compound $\left(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{O}_{2}\right)$ rotates plane-polarized light. It produces pink color with neutral $\mathrm{FeCl}_{3}$ solution. What is the total number of all the possible isomers for this compound?

Answer: 6

Numerical answer — enter this value.

Step-by-step solution

Since given compound produces pink colour with neutral $\mathrm{FeCl}_{3}$ .

So, it must have at least one - OH group attached to benzene ring.

Also, given compound is optically active, it means it should have a chiral centre, which is only possible when it has

$-\mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{3}$

So, based on the above observations, compound $\left(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{O}_{2}\right)$ is expected to possess the following isomeric

structures.

So, total isomeric structures $=6$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2020
Paper: Paper 2
Subject: Chemistry
Chapter: Isomerism
Topic: Optical Isomerism