The total number of structural isomers possible for the substituted benzene derivatives with the molecular formula $$C_9H_{12}$$ is ________.

First compute the double-bond (or index of) unsaturation for the molecular formula $$C_9H_{12}$$.

The formula for a hydrocarbon is
$$\text{DBE}= \frac{2n+2-m}{2}$$
where $$n$$ and $$m$$ are the number of carbon and hydrogen atoms respectively.

For $$C_9H_{12}$$,
$$\text{DBE}= \frac{2(9)+2-12}{2}= \frac{20-12}{2}=4$$

Exactly four elements of unsaturation correspond to one benzene ring (1 ring + 3 π bonds). Hence every isomer must contain a single benzene nucleus and no other rings or multiple bonds. All variations will therefore come only from the nature, number and relative positions of the alkyl substituents attached to that ring.

Let the ring be $$C_6H_6$$. Each hydrogen on the ring that is replaced by an alkyl group reduces the ring hydrogens by 1 and adds the atoms of the substituent in their place.

We now distribute the three surplus carbons ($$9-6=3$$) and the six surplus hydrogens ($$12-6=6$$) among substituents so that the overall formula stays $$C_9H_{12}$$. The possibilities are:

One three-carbon substituent (propyl).
The side chain can be straight (n-propyl, $$CH_3CH_2CH_2-$$) or branched (isopropyl, $$CH_3CH(CH_3)-$$). In a monosubstituted benzene all ring positions are equivalent, so each chain gives exactly one isomer.

Total isomers from this case = $$2$$.

One ethyl ($$C_2H_5-$$) and one methyl ($$CH_3-$$) substituent.
With two different groups there are three distinct relative positions on the ring: ortho (1,2), meta (1,3), and para (1,4).

Total isomers from this case = $$3$$.

Three methyl substituents (trimethylbenzenes).
The possible substitution patterns are
$$1,2,3$$-trimethylbenzene,
$$1,2,4$$-trimethylbenzene, and
$$1,3,5$$-trimethylbenzene.

Total isomers from this case = $$3$$.

Adding all distinct structural isomers:
$$2 + 3 + 3 = 8$$

Hence, the total number of structural isomers of aromatic hydrocarbons having the molecular formula $$C_9H_{12}$$ is 8.