See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1: Analyze each compound for unique hydrogen environments (excluding stereoisomerism for Column II, including for Column III). (a) The structure shown as two fused rectangles is bicyclo[1.1.0]butane or more likely two fused cyclobutane rings (bicyclo[2.2.0]hexane / bicyclohexane). However, given the answer a -> q-x (2 structural, 3 including stereo), this is consistent with a structure having 2 sets of unique H environments structurally but one set giving a chiral center producing stereoisomers, yielding 3 total including stereoisomers. (b) 2-methylbutane: CH3-CH(CH3)-CH2-CH3 H environments (excluding stereo): C1 (CH3 on branch, 6H equivalent by symmetry? No — C1 is (CH3)2CH-, so the two CH3 on C2 are equivalent = 1 type; C2 H = 1 type; C3 H2 = 1 type; C4 CH3 = 1 type). That gives 4 structural monochlorides. Including stereoisomerism: Chlorination at C3 gives a chiral center (CHClCH3 next to CH(CH3)2), so C3 gives R and S = 2; C2 already has 4 different groups when Cl replaces H giving chiral center = 2 stereoisomers; C1 (either of equivalent CH3 groups) = 1; C4 = 1. Total = 1+2+2+1 = wait, let me recount. Positions: C1 (the two equivalent methyls on C2, i.e., (CH3)2 group) -> 1 product; C2-H -> 1 product (chiral, gives 2 stereoisomers: R+S); C3-H2 -> 1 structural product (chiral center created, gives R+S = 2); C4-CH3 -> 1 product. Structural = 4 (s), stereoisomers: C1=1, C2=2, C3=2, C4=1 = 6 total (z). So b -> s-z. ✓ (c) 2,2,3,3-tetramethylbutane: (CH3)3C-C(CH3)3 All 18 H atoms are equivalent (all in equivalent CH3 groups). Only 1 structural monochloride (p), and no chiral center is created (the product is (CH3)2ClC-C(CH3)3 which has no chiral center due to two identical CH3 groups on same carbon... actually CCl(CH3)2-C(CH3)3: the carbon bearing Cl has CH3, CH3, C(CH3)3 and Cl — two identical CH3 groups so not chiral). So 1 including stereo (w). So c -> p-w. ✓ (d) n-butane: CH3-CH2-CH2-CH3 H environments (excluding stereo): C1/C4 equivalent CH3 = 1 type; C2/C3 equivalent CH2 = 1 type. So 2 structural monochlorides (q). Including stereoisomerism: C1 chlorination -> CH2Cl-CH2-CH2-CH3, no chiral center = 1; C2 chlorination -> CH3-CHCl-CH2-CH3, chiral center created = 2 (R and S). Total = 1+2 = 3 (x). So d -> q-x. ✓ (a) revisited: The fused rectangle structure — given a -> q-x (same as d), this must also have 2 structural and 3 including stereo. This is consistent with bicyclo[1.1.0]butane or a similar small bicyclic compound with 2 types of H (bridgehead vs bridge) giving 2 structural products, and one of them generating a stereocenter giving 3 total. Therefore, the correct answer is {"a": "q-x", "b": "s-z", "c": "p-w", "d": "q-x"}.