Structures of Iodophenyltellurium(II) and Diiododi-(β-naphtyl)tellurium(IV)

Abstract

The reactions between diphenyl ditelluride, (PhTe)2, or di(β‐naphtyl)ditelluride, (β‐naphtylTe)2, with equivalent amounts of iodine have been reinvestigated and the crystal and molecular structures of iodophenyltellurium(II), (PhTeI)4, and diiododi‐(β‐naphtyl)tellurium(IV), (β‐naphtyl)2TeI2, have been determined. The structure of iodophenyltellurium(II) (space group Cc, a = 13.850(5) Å, b = 13.852(3) Å, c = 16.494(6) Å and β = 101.69(2)°, Z = 4) is built up by four PhTeI units which are linked by weak Te–Te interactions with Te–Te distances between 3.152(5) Å and 3.182(4) Å. The angles between the tellurium atoms are approximately 90° giving an almost perfect square. Long range secondary bonds (Te–I: about 4.2 Å) link the tetrameric units to give an infinite two‐dimensional network. Iodo(β‐naphtyl)tellurium(II) is less stable than the phenyl derivative. Solutions of this compound decompose under formation of elemental tellurium and (β‐naphtyl)2TeI2. (β‐Naphtyl)2TeI2 crystallises in the monoclinic space group C 2/c (a = 21.198(6) Å, b = 5.8921(8) Å, c = 16.651(5) Å, β = 114.77(2)°). The tellurium atom is situated on a two‐fold crystallographic axis and Te–I and Te–C bond lengths of 2.899(1) and 2.108(7) Å have been determined.

Synthesis and X-ray investigation of novel Fe and Mn phenyltellurenyl-halide complexes: (CO)3FeBr2(PhTeBr), (η5-C5H5)Fe(CO)2(PhTeI2) and CpMn(CO)2(PhTeI)

“…Complex 3 is the first example of PhTeI ligand bonded exclusively with metal atom in contrast with a large number of charge transfer (CT) complexes of aryltellurenyl halides with phosphines and other P-, N-, O-, S-, Se-donor ligands. It is noteworthy that in (Ph 3 P)PhTeI [3] which is the typical example of such CT-complexes, the phosphine ligand is in trans position to Te-I bond so that P-Te-I angle is 179.45(5)°and Te-I distance (3.0930(9) Å) is lengthened on 0.32 Å as compared to the sum of Fe and Te covalent radii [4] To investigate the effect of substitution of iodine atoms in 1 by more electronegative and smaller bromine atoms we prepared the bromine analog of 1, complex (CO) 3 ) and slightly shifted into the high-frequency area due to the higher electronegativity of bromine. XRD investigation of the yellow crystals of 4 (see Fig.…”

“…Normally, such bonds are formed due to the additional p-inter- Recently we reported the oxidative addition of PhTeI 3 to Fe(CO) 5 to yield an unusual complex (CO) 3 FeI 2 (PhTeI) 3 (1) which was the first example of a transition metal complex with unstable aryltellurenyl halide ligand PhTeI. Stabilization of PhTeI in 1 was achieved due to the formation of a Fe-Te bond (Te-Fe (2.5451(6) Å) which is reduced on 0.15 Å in comparison with the sum of Fe and Te covalent radii [4] Te-I antibonding orbital (Scheme 1).…”

“…Stabilization of PhTeI in 1 was achieved due to the formation of a Fe-Te bond (Te-Fe (2.5451(6) Å) which is reduced on 0.15 Å in comparison with the sum of Fe and Te covalent radii [4] Te-I antibonding orbital (Scheme 1). As part of our ongoing interest in unusual reactivity and molecular structure of metal-aryltellur-halogenide complexes and our interest in complexes with PhTeI coordinated exclusively with a metal center without any additional secondary intramolecular bonds, we have looked into the chemistry of CpMn(CO) 2 (PhTeI) (3). Absence of M-I bonds in 3 would suggest that additional secondary intramolecular TeÁ Á ÁI interactions are not possible.…”

Synthesis and characterization of the first aryltellurium(II) halide complex stabilized by a Te–Te bond from a tellurium ether

“…We have obtained PhTeI in a tetrameric form by the reaction between diphenylditelluride and equivalent amount of iodine [2]. The reaction between [b-naphtyl-Te] 2 and an equivalent amount of iodine affords [(b-naphtyl) 2 TeI 2 ] and elemental tellurium [2].…”

“…We have obtained PhTeI in a tetrameric form by the reaction between diphenylditelluride and equivalent amount of iodine [2]. The reaction between [b-naphtyl-Te] 2 and an equivalent amount of iodine affords [(b-naphtyl) 2 TeI 2 ] and elemental tellurium [2]. When the same reaction is carried out using [(2,4,6-Ph 3 C 6 H 2 )Te] 2 , the compound (2,4,6-Ph 3 C 6 H 2 )TeI 3 [3] is formed as a result of a disproportion reaction in inert atmosphere or an oxo-bridged tellurium(IV) compound in the presence of air [4].…”