See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

The starting material is 2-methylbutane: CH3-CH(CH3)-CH2-CH3. It has four distinct types of hydrogen atoms: 1. C1 hydrogens: The methyl group attached to C2, i.e., the (CH3) branch — 3H (equivalent). Chlorination here gives 1-chloro-2-methylbutane. C2 is a chiral center in the product (4 different groups), so this gives a pair of enantiomers (R and S) = 2 stereoisomers. 2. C2 hydrogen: The CH at C2 — 1H. Chlorination here gives 2-chloro-2-methylbutane. C2 in the product has two methyl groups, so no chiral center = 1 compound (no stereoisomers). 3. C3 hydrogens: The CH2 at C3 — 2H. Chlorination here gives 3-chloro-2-methylbutane. C3 becomes a chiral center (4 different groups), giving R and S enantiomers = 2 stereoisomers. 4. C4 hydrogens: The terminal CH3 at C4 — 3H (equivalent). Chlorination here gives 1-chloro-2-methylbutane... wait, this is actually the other end: it gives 1-chloro-2-methylbutane as well? No — C4 is -CH3 at the end of the chain, giving 4-chloro-2-methylbutane (i.e., CH3-CH(CH3)-CH2-CH2Cl). C2 is still a chiral center here (attached to CH3, H, CH(CH3), and CH2CH2Cl — wait, let me re-examine). The structure is CH3-CH(CH3)-CH2-CH3 numbered as: C1=CH3 (the branch), C2=CH, C3=CH2, C4=CH3 (end of main chain). Chlorination at C4 gives CH3-CH(CH3)-CH2-CH2Cl. C2 is a chiral center (groups: CH3, H, CH3 branch, CH2CH2Cl) — actually C2 has two CH3 groups? No: C2 is bonded to the branch CH3, the main chain C1 (CH3), H, and C3. So C2 has: H, CH3 (branch), CH3 (C1), and CH2CH2Cl — two identical CH3 groups, so NOT a chiral center = 1 compound. Let me recount carefully. 2-methylbutane: main chain C1-C2-C3-C4 with a methyl branch on C2. Actually: CH3(a)-CH(CH3(b))-CH2-CH3(c) - Ha type: 6H on the two equivalent methyls (C1 and the branch CH3) if they are equivalent... Are CH3(a) and CH3(b) equivalent? C1 (CH3) is at position 1, branch CH3 is also on C2 — in 2-methylbutane these two methyls ARE equivalent by symmetry (both are directly bonded to C2 which bears H, and the chain continues to CH2CH3). So Ha: 6H, gives 1-chloro-2-methylbutane. Product: ClCH2-CH(CH3)-CH2-CH3. C2 is chiral (H, CH3, CH2Cl, CH2CH3) — 2 enantiomers. - Hb type: 1H on C2, gives 2-chloro-2-methylbutane. C2: Cl, CH3, CH3, CH2CH3 — not chiral = 1 product. - Hc type: 2H on C3, gives 3-chloro-2-methylbutane. CH3-CH(CH3)-CHCl-CH3. C2 and C3 both potentially chiral. C3: Cl, H, CH3, CH(CH3)2 — chiral. C2: H, CH3, CH3, CHClCH3 — two CH3 groups, NOT chiral. So only C3 is chiral = 2 enantiomers. - Hd type: 3H on C4, gives 4-chloro... wait that's the same as above since C3=CH2 and C4=CH3. Chlorination at the terminal CH3 (C4 in the longest chain numbering if we consider 2-methylbutane as having 4 carbons in main chain): product is CH3-CH(CH3)-CH2-CH2Cl. C2 has H, CH3, CH3, CH2CH2Cl — two methyls, not chiral = 1 product. Total: 2 + 1 + 2 + 1 = 6 stereoisomers (counting each stereoisomer as a separate product). Therefore, the correct answer is C.