1-(2,4-Dinitrophenyl)-3-(2-hydroxy-5-methylphenyl)pyrazole-4-carboxaldehyde

Abstract

Key indicatorsSingle-crystal X-ray study T = 293 K Mean '(C±C) = 0.004 A Ê R factor = 0.050 wR factor = 0.176 Data-to-parameter ratio = 12.0 For details of how these key indicators were automatically derived from the article, see

3-(3-Chloro-2-hydroxyphenyl)-1-phenyl-1H-pyrazole-4-carbaldehyde

“…The phenol hydroxyl is hydrogen bonded to the pyrazole nitrogen, N2, and the formyl oxygen atom is directed towards the phenol ring to make an intramolecular C-H•••O hydrogen bond with C5-H5. In contrary, in the crystal structures of the related compounds (Jeyakanthan et al, 2001;Shanmuga Sundara Raj et al, 1999) (Table 1).…”

“…For structures of similar compounds, see: Jeyakanthan et al (2001);Shanmuga Sundara Raj et al (1999). Hydrogen-bond geometry (Å , ).…”

Copper(II) ions mediated crystal formation of 3-(3-hydroxy phenyl)-1-phenyl-1h-pyrazole- 4-carbaldehyde

“…However, the presence of hydroxy group enables the formation of different hydrogen bonded networks compared to the other reported derivatives. This is also the case when compared with 2-hydroxy substituent, since intramolecular hydrogen bond between 2-hydroxy and the pyrazole N atom is observed [21,23,24].…”

“…Formyl group is almost in plane with the pyrazole ring with the dihedral angle C3-C2-C16-O2 of 5.5(4)°. Crystal structures of phenyl derivatives of pyrazole-4-carbaldehyde with several different substituents on phenyl ring, like 2-hydroxy, 4-bromo, 4-chloro, 4-methoxy, 4-methyl and 3,4dimethoxy have been reported [14,[21][22][23][24][25][26][27][28][29][30]. However, the presence of hydroxy group enables the formation of different hydrogen bonded networks compared to the other reported derivatives.…”

The Structure ofN‐phenyl‐pyrazoles and Indazoles: Mononitro, Dinitro, and Trinitro Derivatives

“…51 : torsion = 125.8°; N1 = 111.5°, N2 = 105.6°, Δθ = 5.9°. 52 : torsion = 24.0°; N1 = 111.5°, N2 = 105.8°, Δθ = 5.7°. 53 : torsion = 125.6°; N1 = 112.5°, N2 = 103.9°, Δθ = 8.6°.…”