See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: In electrophilic addition of HBr to alkenes, the rate-determining step is protonation of the alkene to form a carbocation intermediate. The activation energy (ΔG‡) for this step depends on the stability of the carbocation formed (Hammond's postulate: more stable carbocation → lower activation energy for its formation). Step 1 - Identify the carbocation formed in each reaction: (a) Ethylene (unsubstituted alkene) → protonation gives a primary carbocation (CH3-CH2+ or simply primary). This is the least stable carbocation. (b) A monosubstituted terminal alkene (like propene) → Markovnikov protonation gives a secondary carbocation. More stable than primary. (c) A disubstituted or branched terminal alkene → Markovnikov protonation gives a tertiary carbocation. Even more stable. (d) A trisubstituted/highly branched terminal alkene where the internal carbon upon protonation gives a tertiary carbocation with more alkyl stabilization, or the alkene is such that protonation gives the most stabilized (tertiary) carbocation among all options. Looking at structure (d): it shows CH2=C with two alkyl chains (ethyl and propyl or similar), meaning the substituted carbon already bears two carbon groups, so protonation of CH2= end gives a tertiary carbocation at the more substituted carbon. This tertiary carbocation is the most stable among all four options. Step 2 - Apply Hammond's postulate: The transition state for carbocation formation resembles the carbocation (endothermic, late transition state). The more stable the carbocation, the lower the energy of the transition state, and therefore the lower the ΔG‡. Step 3 - Rank activation energies: (a) primary carbocation → highest ΔG‡ (b) secondary carbocation → lower ΔG‡ (c) tertiary carbocation → even lower ΔG‡ (d) tertiary carbocation with maximum alkyl substitution / hyperconjugation → lowest ΔG‡ Option (d) involves addition to a terminal alkene where the resulting carbocation is tertiary and maximally stabilized by alkyl groups (two ethyl/propyl groups), making it the most stable carbocation and thus giving the lowest activation energy. Step 4 - Why other options fail: (a) Forms primary carbocation — highest barrier. (b) Forms secondary carbocation — lower than (a) but not the lowest. (c) Forms tertiary carbocation — lower than (b), but the alkene in (d) gives an equally or more stabilized tertiary carbocation due to greater branching/hyperconjugation. (d) The terminal alkene in (d) has two large alkyl substituents on the internal carbon, so protonation at the terminal CH2 gives the most substituted and most stabilized tertiary carbocation, resulting in the lowest ΔG‡. Therefore, the correct answer is D.