An organic compound $$(C_8H_{10}O_2)$$ 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?

Degree of unsaturation:
$$\text{DU}= \dfrac{2\times 8+2-10}{2}=4$$
Four units of unsaturation correspond exactly to one benzene ring, so every isomer must contain a benzene nucleus.

Ferric-chloride test:
A pink/violet colour with neutral $$FeCl_3$$ is given by phenolic $$-OH$$. Hence one oxygen atom is present as a phenolic group directly attached to the ring.

Locating the second oxygen atom and the two extra carbon atoms:
With one phenolic $$-OH$$ fixed on the ring, the remaining stoichiometry is $$C_2H_4O$$.
The only way to fit $$C_2H_4O$$ so that (i) the molecular formula $$C_8H_{10}O_2$$ is satisfied, and (ii) a chiral centre is generated, is to attach a 1-hydroxyethyl group $$-CH(OH)CH_3$$ to the ring. That group provides

$$C_2H_4O = CH(OH)CH_3$$  and  contains the chiral carbon $$^\ast CH(OH)CH_3$$ (bonded to $$\{H,\;OH,\;CH_3,\;\text{aryl}\}$$).

Thus every optically active isomer has the general structure
$$\mathrm{HO\!-\!C_6H_4\!-\!CH^\ast(OH)CH_3}$$

Positional (constitutional) isomers of the two substituents on the ring:

Case 1: ortho (1,2-disubstitution)
Case 2: meta (1,3-disubstitution)
Case 3: para (1,4-disubstitution)

Each constitutional isomer possesses one chiral centre, so it exists as an enantiomeric pair (R and S).

Total optical isomers:
number of positions (3) $$\times$$ enantiomers per position (2) $$= 6$$.

Hence the compound can have 6 optically active isomers in all.