See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Step 1 - Core concepts: - Dipole moment depends on vector addition of bond dipoles in cis vs trans isomers. - Melting point: trans isomers are more symmetric, pack better in crystal lattice, so trans generally has higher melting point than cis. - Boiling point: cis isomers generally have higher dipole moments (for most substituents), leading to stronger intermolecular forces and higher boiling points than trans. Step 2 - Analyze each compound: (a) H3C-CH=CH-CH3 (2-butene): Both substituents on each carbon are the same type (CH3 on one side, H on the other for each sp2 carbon). In the cis isomer, the two CH3 groups are on the same side, giving a net dipole (small but nonzero due to hyperconjugation/induction). In the trans isomer, the molecule has a center of symmetry, so dipole = 0. Therefore dipole: cis > trans (p). Trans-2-butene is more symmetric, so melting point: trans > cis (r). Since cis has higher dipole, boiling point: cis > trans (s). So (a) matches (p, r, s). (b) H3C-CH=CH-CN (but-2-enenitrile / crotononitrile): CH3 donates electrons; CN withdraws electrons. In the cis isomer, CH3 and CN are on the same side - their dipoles partially cancel (CH3 pushes toward double bond, CN pulls away from double bond, and in cis configuration the vectors partially oppose each other more). In the trans isomer, CH3 and CN are on opposite sides, and since they pull in opposite directions along the chain but the geometry means their contributions add rather than cancel, trans has a larger net dipole. Thus dipole: trans > cis (q). Trans is more symmetric in terms of shape, so melting point: trans > cis (r). So (b) matches (q, r). (c) H3C-CH=CH-Cl (1-chloropropene / but-1-en-... actually 1-chloro-2-butene): CH3 is electron-donating, Cl is electron-withdrawing. Similar to CN case, in the trans isomer the opposing groups (CH3 on one carbon, Cl on the other) create a larger net dipole than in the cis isomer where partial cancellation occurs. Thus dipole: trans > cis (q). Trans has higher melting point (r). So (c) matches (q, r). (d) Cl-CH=CH-Cl (1,2-dichloroethene): Both substituents are identical (Cl). In the cis isomer, both Cl atoms are on the same side, so their dipoles add up, giving a large net dipole. In the trans isomer, the molecule has a center of symmetry, so dipole = 0. Therefore dipole: cis > trans (p). Trans is more symmetric, so melting point: trans > cis (r). Since cis has higher dipole, boiling point: cis > trans (s). So (d) matches (p, r, s). Step 3 - Summary of matches: (a) -> p, r, s (b) -> q, r (c) -> q, r (d) -> p, r, s Therefore, the correct answer is {"a": ["p", "r", "s"], "b": ["q", "r"], "c": ["q", "r"], "d": ["p", "r", "s"]}.