1. Interpret the question

We need isomeric hydrocarbons with molecular formula $C_9H_{12}$ that give negative Baeyer’s test.

• Negative Baeyer’s test means no alkene/alkyne type unsaturation.
• Since the compound is a hydrocarbon and still unsaturated by formula, the unsaturation must be due to an aromatic ring.

Now, for $C_9H_{12}$:

$\text{Degree of unsaturation} = \frac{2C+2-H}{2} = \frac{2(9)+2-12}{2} = \frac{20-12}{2} = 4$

A benzene ring itself accounts for 4 degrees of unsaturation, so the compounds are alkyl benzenes.

2. List all aromatic hydrocarbon isomers of formula $C_9H_{12}$

Start from benzene, $C_6H_6$. We need to add $C_3H_6$ overall as alkyl substitution.

Possible alkyl-benzene isomers are:

(a) Mono-substituted benzene with a $C_3H_7$ group

• n-Propylbenzene
• isopropylbenzene (cumene)

(b) Di-substituted benzene with $C_2H_5 + CH_3$

Ethyl methyl benzenes:

• o-ethyltoluene
• m-ethyltoluene
• p-ethyltoluene

(c) Tri-substituted benzene with $CH_3 + CH_3 + CH_3$

Trimethylbenzenes:

• 1,2,3-trimethylbenzene
• 1,2,4-trimethylbenzene
• 1,3,5-trimethylbenzene

So total isomers are 8.

3. Meaning of “four different non-aliphatic substitution sites”

“Non-aliphatic substitution sites” means substitution positions on the aromatic ring where a ring hydrogen is present.

We count the number of chemically distinct ring positions still bearing H, i.e. distinct positions where electrophilic substitution on the ring can occur.

We now inspect each isomer.

4. Check each isomer

(a) Propylbenzene and isopropylbenzene

These are monosubstituted benzenes. A monosubstituted benzene has only 3 distinct ring positions:

• ortho
• meta
• para

So these do not have 4 different non-aliphatic substitution sites.

(b) Ethyl methyl benzenes

1. o-Ethyltoluene (1-ethyl-2-methylbenzene)

Available ring H positions are 3, 4, 5, 6. Because substituents are different ($Et \neq Me$), symmetry is lost and these positions can be distinct. But check carefully:

• There is no symmetry making all 4 equivalent in pairs.
• Hence it has 4 different ring substitution sites.

2. m-Ethyltoluene (1-ethyl-3-methylbenzene)

Available ring H positions are 2, 4, 5, 6. Again, since substituents are different, no symmetry equates these positions. So it also has 4 different ring substitution sites.

3. p-Ethyltoluene (1-ethyl-4-methylbenzene)

Here there is a plane of symmetry through positions 1 and 4:

• positions 2 and 6 equivalent
• positions 3 and 5 equivalent Thus only 2 different ring substitution sites.

So among ethyl toluenes, 2 isomers qualify.

(c) Trimethylbenzenes

1. 1,2,3-Trimethylbenzene

Remaining H positions: 4, 5, 6. By symmetry, 4 and 6 are equivalent, 5 is different. So only 2 distinct ring substitution sites.

2. 1,2,4-Trimethylbenzene

Remaining H positions: 3, 5, 6. All three are not equivalent, so 3 distinct ring substitution sites.

3. 1,3,5-Trimethylbenzene

Remaining H positions: 2, 4, 6. All are equivalent, so only 1 distinct ring substitution site.

No trimethylbenzene has 4 different non-aliphatic substitution sites.

5. Final count

The isomers satisfying the condition are:

• $o$-ethyltoluene
• $m$-ethyltoluene

Hence, the total number of such isomers is

$\boxed{2}$

6. Comparison with stored answer

Stored correct answer = 2.

Our derived answer also = 2, so they agree.