See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Analyze molecule (a): 1,1,4,4-tetrafluorobutatriene with substituents F,H on C1 and H,F on C4. In a 4-carbon cumulene (even number of double bonds), the two end groups are coplanar. C1 has F and H; C4 has H and F. Looking at the drawing, C1 bears F (top) and H (bottom), C4 bears H (top) and F (bottom). Because the end groups are coplanar in an even cumulene and the arrangement shown is F,H on one end and H,F on the other end in the same plane, this molecule has a plane of symmetry (the plane of the molecule itself) and also a center of inversion. The dipole moments of the C-F bonds cancel due to the symmetry (the molecule is centrosymmetric), making it nonpolar. It is optically inactive. It possesses a plane of symmetry. Hence (a) matches r (optically inactive) and s (plane of symmetry). Step 2 - Analyze molecule (b): 1,3-difluoroallene (3-carbon cumulene, odd number of double bonds). In an odd cumulene, the two end groups are perpendicular to each other. C1 has F and H; C3 has H and F. This is the chiral allene: the two ends each bear two different substituents (F and H) and the planes are perpendicular, so there is no plane of symmetry, no center of symmetry, and no improper rotation axis. This molecule is chiral and optically active. Because the C-F bond dipoles are not cancelled (they are in perpendicular planes and the geometry does not allow cancellation), the molecule is polar. Hence (b) matches p (polar molecule) and q (optically active). Step 3 - Analyze molecule (c): trans-1,4-difluorocyclohexane (F on wedge at C1, F on dash at C4). In the trans isomer, one F is axial and one is equatorial (or considering the flat representation, they are on opposite faces). The trans-1,4-difluorocyclohexane has a center of inversion (or a plane of symmetry through C1 and C4 axis perpendicular to the ring), making it achiral. The two C-F dipoles are oriented in opposite directions along the ring axis and cancel, so the molecule is nonpolar. It is optically inactive. It has a plane of symmetry. Hence (c) matches r (optically inactive) and s (plane of symmetry). Step 4 - Analyze molecule (d): Fluorobenzene (monosubstituted benzene with one F). Fluorobenzene has a plane of symmetry (the plane of the ring, and the plane containing F-C1 axis bisecting the ring). The C-F bond is polar and since there is only one F with no opposing dipole of equal magnitude, the molecule is polar overall. It is optically inactive because it has planes of symmetry. Hence (d) matches p (polar molecule), r (optically inactive), and s (plane of symmetry). Summary of matches: - (a): r, s - (b): p, q - (c): r, s - (d): p, r, s Therefore, the correct answer is a-r,s; b-p,q; c-r,s; d-p,r,s.