Question

Isomeric hydrocarbons → negative Baeyer's test (Molecular formula C9H12 ) The total number of isomers from above with four different non-aliphatic substitution sites is

Answer 1

2

Answer 2

For C₉H₁₂ the index of hydrogen deficiency is   IHD = (2×9 + 2 – 12)/2 = 4, indicating an aromatic (benzene) ring. Thus the isomers are benzene derivatives.

One way to “make” C₉H₁₂ is to attach an alkyl group (with 3 C’s) to benzene. In fact: • Monosubstituted benzene (n‐propylbenzene or isopropylbenzene) gives only three types of (ring‐H) sites (ortho, meta, para).
• Disubstituted benzene of the type methyl + ethyl (since C₆H₄ + CH₃ + C₂H₅ yields C₉H₁₂) exists in three positional isomers (ortho, meta, para).
Among these, only the ortho (1‑ethyl‑2‑methylbenzene) and the meta (1‑ethyl‑3‑methylbenzene) arrangements break the ring symmetry enough so that all four remaining (non‐aliphatic) ring positions (the hydrogens) are in different environments. (In the para isomer the two pairs of ring positions are equivalent.)

Thus the total number of isomers having four different non‐aliphatic substitution sites is 2.