JEE Mains Chemistry Past PapershardNUMERICAL

Isomeric hydrocarbons negative Baeyer's test (Molecular formula C9H12) The total number of isomers fJEE Mains Chemistry Past Papers Chemistry Question

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: .

💡 Solution & Explanation

# Solution: C₉H₁₂ Isomers Failing Baeyer's Test **Step 1: Interpret the conditions** - Molecular formula: C₉H₁₂ - Negative Baeyer's test means NO C=C double bonds or rings - Must be saturated hydrocarbons (alkanes only) - Need exactly 4 different non-aliphatic (aromatic) substitution sites **Step 2: Identify the core structure** - C₉H₁₂ with only aromatic rings and single bonds - This indicates a benzene ring (C₆H₄) as the base with C₃H₈ as substituents - The compound must be **trimethylbenzene (xylene derivatives with one additional methyl)** **Step 3: Determine substitution pattern** - For "four different non-aliphatic substitution sites," we need 4 positions on the benzene ring to be distinct - This requires the substituent pattern to create 4 magnetically/chemically different aromatic positions **Step 4: Count isomers with 4 different aromatic positions** - 1,2,3-trimethylbenzene has 4 different aromatic hydrogen positions ✓ - 1,2,4-trimethylbenzene has 4 different aromatic hydrogen positions ✓ - 1,3,5-trimethylbenzene has only 2 different positions (symmetric) ✗ - 1,2,3,4-tetramethylbenzene (C₁₀H₁₄) - wrong formula **Step 5: Identify valid isomers** Only **1,2,3-trimethylbenzene** and **1,2,4-trimethylbenzene** have exactly four different non-aliphatic substitution sites. **Therefore, the answer is 2**

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