See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The starting material is a cyclohexadienyl methanol (specifically, the structure shown is a cyclohexa-2,4-dien-1-yl carbinol, i.e., 2,4-cyclohexadien-1-yl methanol), which has a CH2-OH group attached to a non-aromatic cyclohexadiene ring. Step 2 - Formation of (A): Under acid (H+) and heat (delta), the alcohol undergoes dehydration. The OH is protonated and water leaves to form a carbocation or the reaction proceeds to give a conjugated diene with an exocyclic methylene group. The product (A) is a methylenecyclohexadiene - specifically 1-methylene-2,4-cyclohexadiene (option a structure), formed by elimination of water giving an exocyclic alkene. Step 3 - Isomerisation of (A) to (B): (A), which is 1-methylene-2,4-cyclohexadiene (also known as fulvene-type or more precisely it is the non-aromatic isomer), undergoes isomerization on heating. The isomerization involves a [1,5]-H shift or acid-catalyzed proton migration that shifts the exocyclic double bond into the ring system. 1-methylene-2,4-cyclohexadiene can isomerize to toluene (methylbenzene) because toluene is the aromatic, thermodynamically most stable isomer with the same molecular formula C7H8. The driving force is aromatization - gaining the large resonance stabilization energy of the benzene ring. Step 4 - Why other options fail: (a) is compound (A) itself, not (B). (c) is another non-aromatic diene isomer, less stable than toluene. (d) phenol (C6H5OH) has a different molecular formula (C6H6O vs C7H8) and cannot be formed from this isomerization. Step 5 - Conclusion: The thermodynamically driven isomerization of the non-aromatic C7H8 isomer (A) gives toluene (B), the fully aromatic and most stable C7H8 isomer. Therefore, the correct answer is B.