See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Step 1: Identify the reaction. The reaction is: 3-methylbut-1-ene (branched alkene, C5H10) → 2-pentene (straight-chain alkene, C5H10). This is an isomerization reaction between two C5 alkenes. Step 2: Use Hess's Law to find delta H for the isomerization. We can construct a thermodynamic cycle using hydrogenation and combustion data. For 3-methylbut-1-ene: - Hydrogenation: 3-methylbut-1-ene + H2 → 2-methylbutane, delta H1 = -30 kcal/mol - Combustion of 2-methylbutane: delta H3 = -784 kcal/mol For 2-pentene: - Hydrogenation: 2-pentene + H2 → pentane, delta H2 = -28 kcal/mol - Combustion of pentane: delta H4 = -782 kcal/mol Step 3: Set up the Hess's Law cycle. We want: 3-methylbut-1-ene → 2-pentene, delta H = ? Route 1 (via 3-methylbut-1-ene → 2-methylbutane → combustion products): delta H_path1 = delta H1 + delta H3 = -30 + (-784) = -814 kcal/mol (hydrogenation then combustion, but we need to think carefully) Actually, use the following cycle: 3-methylbut-1-ene + H2 → 2-methylbutane (delta H = -30) 2-methylbutane → combustion products (delta H = -784) 2-pentene + H2 → pentane (delta H = -28) pentane → combustion products (delta H = -782) Since both alkanes combust to give the same products (CO2 and H2O), we can write: delta H (isomerization) = [delta H of hydrogenation of 3-methylbut-1-ene + delta H of combustion of 2-methylbutane] - [delta H of hydrogenation of 2-pentene + delta H of combustion of pentane] This is because: 3-methylbut-1-ene + H2 → 2-methylbutane → products (path A, total = -30 + (-784) = -814) 2-pentene + H2 → pentane → products (path B, total = -28 + (-782) = -810) For the isomerization 3-methylbut-1-ene → 2-pentene: Path A: 3-methylbut-1-ene → products (via hydrogenation + combustion) = -814 Path B: 2-pentene → products (via hydrogenation + combustion) = -810 delta H(isomerization) + (-810) = -814 delta H(isomerization) = -814 - (-810) = -814 + 810 = -4 kcal/mol Step 4: Verify. The isomerization is exothermic by 4 kcal/mol, meaning 2-pentene is more stable than 3-methylbut-1-ene by 4 kcal/mol (consistent with internal alkene being more stable than terminal branched alkene in this context). Options (a) 0, (c) -2, (d) 2 do not match the calculation. Therefore, the correct answer is B.