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In the title compound, C14H16N2O3, an analog of the chromophore in green fluorescent protein, the meth­oxy­phenyl substituent and the imidazole N adopt a Z conformation with respect to the C=C bond. Aside from the hy­droxy­ethyl group, the mol­ecule is approximately planar, with the five- and six-membered ring planes forming a dihedral angle of 9.3 (1)°. An intra­molecular C—H...N contact occurs. In the crystal, O—H...N hydrogen bonds link the mol­ecules, forming chains along the b-axis direction. C—H...O hydrogen bonds are also observed.

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In the title mol­ecule, C14H12Cl2O2, the dihedral angle between the least-square planes of the benzene rings is 62.17 (6)°. Both meth­oxy groups are slightly out of the plane of the benzene rings to which they are attached, making dihedral angles of 4.22 (18) and 18.82 (16)°.

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In the title mol­ecule, C16H14Cl4O, the dihedral angle between the least-square planes of the benzene rings is 84.40 (7)°. No unusual intermolecular interactions are present.

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In the title compound, C19H22N2, both the benzimidazole unit and the 2,6-diiso­propyl­phenyl group are essentially planar [maximum deviations from the least-squares planes of 0.005 (1) and 0.009 (1) Å, respectively]. The dihedral angle between the two planes is 79.6 (7)°. In the crystal, mol­ecules are linked into chains along the a-axis direction by weak C—H...N inter­actions. The crystal structure also features C—H...π inter­actions, which link the chains into a three-dimensional network.

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In the title compound, C14H12N2O3, the pyridine ring is twisted with respect to the phenyl ring and the carb­oxy­lic acid group at angles of 37.1 (5) and 8.1 (3)°, respectively; the phenyl ring forms a dihedral angle of 41.4 (1)° with the mean plane of the C—NH—C=O fragment. An intra­molecular O—H...O hydrogen bond occurs between the carb­oxy­lic acid and carbonyl groups. In the crystal, N—H...O hydrogen bonds link mol­ecules into a supra­molecular chain running along the a-axis direction.
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