The vanillin residue makes dihedral angles of 4.84 (18) and 11.96 (7) with the planes of the butyl group and the terminal benzene ring, respectively. Intermolecular N-HÁ Á ÁO and C-HÁ Á ÁO hydrogen bonds help to consolidate the crystal packing.
N′-(4-Hydroxy-3-methoxybenzylidene)-2-methoxybenzohydrazide ethanol hemisolvate
✍ Scribed by Jing, Zuo-Liang ;Zhao, Yan-Li ;Chen, Xin ;Yu, Ming
- Publisher
- International Union of Crystallography
- Year
- 2006
- Tongue
- English
- Weight
- 209 KB
- Volume
- 62
- Category
- Article
- ISSN
- 1600-5368
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✦ Synopsis
In the title compound, C 16 H 16 N 2 O 4 Á0.5CH 3 CH 2 OH, the Schiff base is approximately planar. An intramolecular N-HÁ Á ÁO hydrogen bond stabilizes the molecular structure. The molecules are linked by O-HÁ Á ÁO hydrogen bonds to form a chain along the b axis.
📜 SIMILAR VOLUMES
In the title compound, C 17 H 18 N 2 O 3 , the isovanillin group makes a dihedral angle of 9.63 (11) with the plane of the terminal phenyl ring. Intermolecular N-HÁ Á ÁO hydrogen bonds help to consolidate the crystal packing.
The asymmetric unit of the title compound, C~14~H~13~N~3~O~3~·H~2~O, contains one molecule of __N__′-(4-hydroxy-3-methoxybenzylidene)isonicotinohydrazide and one solvent water molecule. The crystal packing is stabilized by O—H...N and O—H...O hydrogen bonds.
The title compound, C 16 H 16 N 2 O 4 , was synthesized by the reaction of 3-hydroxy-4-methoxybenzaldehyde with hydrazine hydrate. The molecule possesses a crystallographically imposed centre of symmetry. An intramolecular O-HÁ Á ÁO hydrogen bond promotes planarity of the molecular backbone. In the