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organic compounds
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The asymmetric unit of the title compound, C25H22N2O, contains two independent molecules (A and B), with significantly different conformations. In molecule A, the central imidazole ring makes dihedral angles of 88.26 (10) and 12.74 (11)° with the two phenyl rings, and 22.06 (9)° with the benzene ring. In molecule B, one of the phenyl rings is disordered over two sites, each having an occupancy of 0.5. Here the central imidazole ring forms dihedral angles of 79.24 (10)° with the ordered phenyl ring, and 3.5 (5) and 22.6 (5)° with the two parts of the disordered phenyl ring. The dihedral angle involving the benzene ring is 67.49 (10)°. The —N—C(H2)—C(H)—C(H2) torsion angles of the prop-1-ene group in the two molecules are very similar, 0.5 (3) and 1.3 (4)° for molecules A and B, respectively. The crystal structure is stabilized by C—H
π interactions.
π interactions.organic compoundsOpen access
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Except two F atoms of the –CF3 group, the title compound, C14H8BrF3N2O3, has an almost planar conformation, the dihedral angle between the aromatic rings being 3.60 (16)°. The molecule adopts the enol–imine tautomeric form, with an intramolecular O—HN hydrogen bond, which generates an S(6) ring motif. In the crystal, face-to-face π–π stacking [centroid–centroid distances = 3.669 (2) and 3.732 (2) Å] between the aromatic rings of the molecules, which lie in sheets parallel to (202), help to establish the packing.
N hydrogen bond, which generates an S(6) ring motif. In the crystal, face-to-face π–π stacking [centroid–centroid distances = 3.669 (2) and 3.732 (2) Å] between the aromatic rings of the molecules, which lie in sheets parallel to (202), help to establish the packing.
