See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Acid-catalyzed isomerization of alkenes proceeds via protonation to form the most stable carbocation intermediate, followed by deprotonation to give the thermodynamically most stable alkene. The driving force is conjugation and substitution of the double bond. Step 1 - Identify compound A: Compound A is a tricyclic compound containing a decalin-type bicyclic system with a cyclohexene ring. It bears an allyl group (–CH2–CH=CH2) attached to a carbon that is allylic to an endocyclic double bond. The terminal double bond (vinyl) is unconjugated with the ring double bond. Step 2 - Mechanism of acid-catalyzed isomerization: Traces of H2SO4 protonate the terminal alkene (or the ring alkene) to generate a carbocation. Through a series of 1,2-hydride shifts or proton transfers, the double bonds migrate to their most thermodynamically stable positions. The key driving force here is the formation of a conjugated diene system (or extended conjugation), which is significantly more stable than isolated double bonds. Step 3 - Determining product B: Starting from A, the allyl side chain (–CH2CH=CH2) can isomerize under acid catalysis. The terminal vinyl group migrates inward. Simultaneously, the endocyclic double bond can shift. The net result is that the side chain becomes a propenyl group (–CH=CH–CH3, i.e., prop-1-en-1-yl) that is conjugated with the endocyclic double bond of the ring. This gives a conjugated diene system where the exocyclic double bond (=CHCH2CH3 in trans configuration) is in conjugation with the ring double bond, maximizing thermodynamic stability through conjugation. Step 4 - Why option C is correct: Option (c) shows a tricyclic structure where the side chain has become a trans-1-propenyl group (–CH=CHCH3) conjugated with an endocyclic double bond, creating a conjugated system. This is the most thermodynamically stable isomer because: (1) the double bonds are conjugated (lower energy), (2) the internal double bond is more substituted (Zaitsev-type stability), (3) trans geometry of the side chain double bond is preferred. Step 5 - Why other options fail: - Option (a): Shows an exocyclic ethylidene with only a 2-carbon unit on the side, inconsistent with the 3-carbon allyl group in A. The carbon count does not match. - Option (b): Shows a vinyl group still present (unconjugated terminal alkene), which is not the thermodynamically favored product; also the ring system appears to have gained extra unsaturation inconsistently. - Option (d): Shows a fully saturated propyl side chain with no conjugation benefit from the side chain; acid-catalyzed isomerization would not saturate the side chain, and this represents unnecessary loss of conjugation stability. Therefore, the correct answer is C.