Isomeric hydrocarbons $$\rightarrow$$ negative Baeyer's test (Molecular formula $$C_{9}H_{12}$$) The total number of isomers from above with four different non-aliphatic substitution sites is -

Molecular formula $$C_9H_{12}$$: degree of unsaturation = $$\frac{2(9)+2-12}{2} = \frac{8}{2} = 4$$.

Since it gives a negative Baeyer's test (no alkene/alkyne), the 4 degrees of unsaturation come from a benzene ring. So these are alkylbenzenes with formula $$C_6H_5-C_3H_7$$.

The isomeric alkylbenzenes with $$C_9H_{12}$$ are:

1. n-propylbenzene

2. isopropylbenzene (cumene)

3. 1,2,3-trimethylbenzene (hemimellitene)

4. 1,2,4-trimethylbenzene (pseudocumene)

5. 1,3,5-trimethylbenzene (mesitylene)

6. 1-ethyl-2-methylbenzene

7. 1-ethyl-3-methylbenzene

8. 1-ethyl-4-methylbenzene

We need isomers with exactly 4 different non-aliphatic (aromatic) substitution sites, meaning 4 different types of aromatic H atoms.

Looking at 1,2,4-trimethylbenzene: positions on ring are C1(CH₃), C2(CH₃), C3(H), C4(CH₃), C5(H), C6(H). The three H atoms at C3, C5, C6 are all in different environments. So 3 types of aromatic H, not 4.

For 1-ethyl-2-methylbenzene: C1(Et), C2(Me), C3(H), C4(H), C5(H), C6(H). The H's at C3, C4, C5, C6 are in 4 different environments. This has 4 different aromatic substitution sites. ✓

For 1-ethyl-3-methylbenzene: C1(Et), C2(H), C3(Me), C4(H), C5(H), C6(H). The H's at C2, C4, C5, C6 are in 4 different environments. ✓

So 2 isomers have 4 different non-aliphatic substitution sites.

The answer is 2.