See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: A molecule has a non-zero dipole moment when its individual bond dipoles do not cancel due to asymmetry in geometry or constitution. Step 1 - Analyze option (b) 1,3,5-tricyanobenzene: The three CN groups are symmetrically placed at 1, 3, and 5 positions of benzene. By C3v symmetry, all three C-CN bond dipoles point outward and cancel vectorially. Net dipole = 0. Step 2 - Analyze option (c) C(CH2Cl)4 (neopentane with four CH2Cl groups): The molecule has tetrahedral symmetry (Td). All four CH2Cl arms are identical and arranged symmetrically. The dipole moments of all four C-CH2Cl units cancel. Net dipole = 0. Step 3 - Analyze option (d) piperazine: Piperazine has two NH groups at 1,4-positions. In the chair conformation, the two N-H bonds can be both equatorial or both axial, giving a centrosymmetric arrangement. The molecule has an inversion center, so the two N-H dipoles cancel. Net dipole = 0. Step 4 - Analyze option (a) trans-1,3-cyclobutanediol: In the trans isomer, one OH is on the wedge (above the ring plane) at C1 and one OH is on the dash (below the ring plane) at C3. Although in some cyclic systems trans-disubstitution creates a center of symmetry, cyclobutane is a four-membered ring. In trans-1,3-cyclobutanediol, C1 and C3 are not related by an inversion center in the same way as in cyclohexane. The ring is planar or nearly so; with OH up at C1 and OH down at C3, the vertical components of the two O-H dipoles point in opposite directions (canceling vertically) but the in-plane components also need consideration. However, because the ring is not flat and the two OH groups are on opposite faces at non-centrosymmetric positions in a four-membered ring, the dipole moments do not completely cancel, giving a non-zero net dipole moment. In contrast, the cis isomer would have both OH groups on the same side and their dipoles would add, but the trans isomer in a cyclobutane (unlike cyclohexane trans-1,4) still has a net dipole because the geometry does not permit complete cancellation. Why other options fail: Options (b), (c), and (d) all possess sufficient molecular symmetry (C3v, Td, and Ci respectively) to cause complete cancellation of bond dipoles, resulting in zero net dipole moment. Therefore, the correct answer is A.