Abstract
The results of an X-ray crystal structure analysis of the crystalline complex, $(CH_3)_3Pt(CH_3.CO.CH.CO.CH_3)C_10H_8N_2,$ formed by trimethylplatinum with acetylacetone and 2:2'-bipyridyl are presented. There are four monomeric molecules in a monoclinic unit cell having a = 14.863, b = 8.480, c = 13.749$\overset{\circ}{\mathrm A}$, $\beta$ = 99$^\circ24'$ at 110 $\pm$ 10$^\circ$K and space group P2$_1$/c. The atomic positions have been determined by three-dimensional analysis from observations collected at 100 to 120$^\circ$K. Refinement, including allowance for anisotropic thermal motion, was carried out by the method of least squares until the value of the residual R was 0.084. The platinum atom is octahedrally co-ordinated to three methyl groups in the cis configuration (Pt-C = 2.05$\overset{\circ}{\mathrm A}$), to the two nitrogen atoms of the bipyridyl (Pt-N = 2.15$\overset{\circ}{\mathrm A}$) and to the central or `active methylene' carbon atom of acetylacetone (Pt-C = 2.36$\overset{\circ}{\mathrm A}$). The oxygen atoms of the acetylacetone take no part in the co-ordination; this is the first example of a $\beta$-diketone acting as an unidentate ligand solely through a carbon atom. Comparison of the bond lengths in this and in the dimeric trimethyl 4:6-dioxononyl platinum suggests that in both compounds the $\beta$-diketone is in the keto and not the enol form; in the present compound the oxygen atoms are not chelated and the C=O bonds are not parallel.