Given information

The compound has molecular formula: $C_8H_{10}O_2$

It:

• rotates plane-polarised light $\Rightarrow$ it is optically active,
• gives pink colour with neutral $FeCl_3$ $\Rightarrow$ it contains a phenolic $-OH$ group.

Degree of unsaturation

$\text{DBE} = \frac{2C+2-H}{2} = \frac{2(8)+2-10}{2} = \frac{18-10}{2} = 4$

DBE $=4$ suggests a benzene ring. Since one oxygen is in phenolic $-OH$, the other oxygen is likely in an alcohol/ether type substituent.

Need for optical activity

Since the molecule is optically active, it must contain a chiral center (no symmetry-based optical activity is expected here).

Also, because it gives $FeCl_3$ test, one substituent must be directly attached to benzene as $-OH$.

Possible carbon skeleton

Start from phenol ring: $C_6H_5OH$ This is $C_6H_6O$.

Remaining part to reach $C_8H_{10}O_2$ is: $C_2H_4O$

So the second substituent on the ring must contribute effectively as a two-carbon oxygen-containing group.

To get chirality, the suitable substituent is: $-CH(OH)CH_3$

Then the full structure is: $HO-C_6H_4-CH(OH)CH_3$

Check formula:

• ring with two substituents: $C_6H_4$
• one $-OH$
• one $-CH(OH)CH_3 = C_2H_5O$

Total: $C_6H_4 + C_2H_5 + O_2 + H_{?} \Rightarrow C_8H_{10}O_2$ Correct.

The carbon bearing $OH$ in side chain is attached to:

• $H$
• $OH$
• $CH_3$
• aryl group

Hence it is a chiral center.

Count positional isomers on benzene ring

Relative positions of $-OH$ and $-CH(OH)CH_3$ on benzene can be:

• ortho
• meta
• para

So there are 3 positional isomers.

Count optical isomers

Each positional isomer has one chiral center, so each exists as two enantiomers: $R \text{ and } S$

Therefore total isomers: $3 \times 2 = 6$

Final answer

Total number of possible isomers: $\boxed{6}$