See image — Reaction Mechanism Chemistry Question

💡 Solution & Explanation

Concept: The reaction of an alkane with diazomethane (CH2N2) under heat (Delta) is a carbene insertion reaction. Diazomethane decomposes thermally to generate carbene (:CH2), which inserts into C–H bonds of the substrate alkane (n-pentane, CH3–CH2–CH2–CH2–CH3). Step 1: Identify the substrate. The starting material is n-pentane (a 5-carbon straight-chain alkane) with the structure CH3–CH2–CH2–CH2–CH3. Step 2: Understand the reaction. CH2N2 under heat generates :CH2 (methylene carbene), which inserts into any C–H bond of n-pentane, effectively inserting a –CH2– group between a carbon and one of its hydrogens, increasing the chain by one carbon (forming a hexane isomer). Step 3: Identify the distinct C–H environments in n-pentane: - C1 (methyl, CH3–): insertion here gives CH3–CH2–CH2–CH2–CH2–CH3 = n-hexane - C2 (–CH2– adjacent to end): insertion here gives CH3–CH(CH3)–CH2–CH2–CH3 ... wait, let us re-examine. Inserting –CH2– into a C2–H bond of n-pentane: C1–C2 becomes C1–CH2–C2, giving CH3–CH2–CH2–CH2–CH2–CH3 (n-hexane) if inserted between C1 and C2... Actually, carbene insertion into a C–H bond at C2 places an extra CH2 at that position: the hydrogen on C2 is replaced by CH2H, effectively making C2 into a branch point or extending the chain. Let me redo carefully. n-Pentane: C1H3–C2H2–C3H2–C4H2–C5H3. Carbene :CH2 inserts into a C–H bond, meaning it goes between the C and the H, so that C–H becomes C–CH2–H (i.e., a new CH2 is inserted). - Insertion at C1 (terminal CH3): one H of C1 is replaced, giving C1 now bonded to CH2H (a new CH2), producing CH3CH2–C2H2–C3H2–C4H2–C5H3 extended by one carbon at the end = n-hexane (CH3–CH2–CH2–CH2–CH2–CH3). - Insertion at C2: the CH2 inserts into a C2–H bond. This gives a structure where C2 has an extra CH2 attached: CH3–CH(CH2H)–CH2–CH2–CH3... i.e., CH3–CH(CH3)... no. Inserting :CH2 into C2–H: the carbon C2 keeps its other bonds (to C1 and C3 and one remaining H) and gains a new CH2H group. Result: CH3–C(H)(CH2H)–CH2–CH2–CH3 = CH3–CH(CH3)–CH2–CH2–CH3... wait that is only possible if CH2H = CH3. Yes, the inserted :CH2 takes the H from C2 giving –CH3 branch. So product: CH3–CH(CH3)–CH2–CH2–CH3 = 2-methylpentane (isopentane skeleton but 6 carbons) = 2-methylpentane. - Insertion at C3 (middle CH2): similarly, :CH2 inserts into C3–H, giving CH3–CH2–CH(CH3)–CH2–CH3 = 3-methylpentane. Step 4: Check products against options: - n-hexane: not listed as an option. - 2-methylpentane (isopentane is 2-methylbutane, a 5-carbon compound): option (a) is isopentane = 2-methylbutane (5 carbons), which is NOT produced here (we get 6-carbon products). - 3-methylhexane: this would require a 7-carbon product, not possible from one carbene insertion into pentane. - n-pentane (option c): no insertion gives back starting material; this would only occur without reaction. - 3-methylpentane (option d): CH3–CH2–CH(CH3)–CH2–CH3, a 6-carbon compound. This IS produced by carbene insertion at C3 of n-pentane. This matches. Step 5: Why other options fail: (a) Isopentane = 2-methylbutane has only 5 carbons; carbene insertion adds one carbon giving 6-carbon products, so isopentane cannot be obtained. (b) 3-Methylhexane has 7 carbons; single carbene insertion on pentane gives only 6-carbon products. (c) n-Pentane is the starting material itself, not a product of carbene insertion. (d) 3-Methylpentane (6 carbons) is correctly formed by :CH2 insertion into the C3–H bond of n-pentane. Therefore, the correct answer is D.