A Sequential Method to Prepare Polymorphs and Solvatomorphs of [Fe(1,3‐bpp)₂](ClO₄)₂⋅nH₂O (n=0, 1, 2) with Varying Spin‐Crossover Behaviour

Abstract

Two polymorphs of the spin crossover (SCO) compound [Fe(1,3-bpp)2 ](ClO4 )2 (1 and 2; 1,3-bpp=2-(pyrazol-1-yl)-6-(pyrazol-3-yl)pyridine) were prepared using a novel, stepwise procedure. Crystals of 1 deposit from dry solvents, while 2 is obtained from a solid-state procedure, by sequentially removing lattice H2 O molecules from the solvatomorph [Fe(1,3-bpp)2 ](ClO4 )2 ⋅2 H2 O (2⋅2 H2 O), using single-crystal-to-single-crystal (SCSC) transformations. Hydrate 2⋅2 H2 O is obtained through the same reaction as 1, now with 2.5 % of water added. Compounds 2 and 2⋅2 H2 O are unstable in the atmosphere and absorb or lose one equivalent of water, respectively, to both yield the stable solvatomorph [Fe(1,3-bpp)2 ](ClO4 )2 ⋅H2 O (2⋅H2 O), also following SCSC processes. The four derivatives have been characterised by single-crystal X-ray diffraction (SCXRD). Furthermore, the homogeneity of the various compounds as well as their SCSC interconversions have been confirmed by powder X-ray diffraction (PXRD). Polymorphs 1 and 2 exhibit abrupt SCO behaviour near room temperature with T1/2↑ =279/316 K and T1/2↓ =276/314 K (near 40 K of shift) and different cooperativity.

A probe of steric ligand substituent effects on the spin crossover of Fe(ii) complexes

“…The structure of complex 1 has been already described in a previous publication. 41 This compound can be obtained following two different procedures as two polymorphs, 1a and 1b, showing two different organizations of the Fe(II) complex cations closely related to these observed for compounds 2 and 3, respectively (see below). The structure of the solvatomorph 2b is briefly described in the ESI (Table S1 and Fig.…”

“…The ligands met1,3bpp and dimet1,3bpp were prepared in three steps using a procedure analogous to the synthesis of 1,3bpp. 41 Thus, the appropriate substituted pyrazole ring was first coupled through one N atom to 2-acetyl-6-bromopyridine. The pyrazole with only one substituent couples through the less crowded nitrogen atom, leading ultimately to the ligand met1,3bpp with the substituent on position 3.…”

“…The complex [Fe(1,3bpp) 2 ](ClO 4 ) 2 (1) was prepared as published by our group. 41 It could be obtained following two distinct procedures as two different polymorphs, 1a and 1b (see below). Complexes [Fe(met1,3bpp) 2 ](ClO 4 ) 2 (2) and [Fe(dimet1,3bpp) 2 ](ClO 4 ) 2 (3) were obtained by direct reaction of the hydrated Fe(ClO 4 ) 2 salt with the met1,3bpp and dimet1,3bpp ligands, respectively, in the presence of a catalytic amount of ascorbic acid to prevent the oxidation of Fe(II) to Fe(III).…”

“…40 We have now designed a ligand system to probe these two not ligand field related opposite effects within an analogous series of Fe(II) complexes. Thus, three ligands have been prepared (Scheme 1); 2-(pyrazol-1-yl)-6-(1H-pyrazol-3-yl)pyridine (1,3bpp), 41 2-(3-methylpyrazol-1-yl)-6-(1H-pyrazol-3-yl)pyridine (met1,3bpp) and 2-(3,5-dimethylpyrazol-1-yl)-6-(1H-pyrazol-3-yl) pyridine (dimet1,3bpp), showing zero, one and two methyl substituents on the pyrazol-1-yl rings. The corresponding homoleptic Fe(II) complexes [Fe(1,3bpp) 2 ](ClO 4 ) 2 (previously published, 41 1), [Fe(met1,3bpp) 2 ](ClO 4 ) 2 (2) and [Fe(dimet1,3bpp) 2 ](ClO 4 ) 2 (3) were prepared and their magnetic properties determined in the solid state and in solution.…”

“…Thus, three ligands have been prepared (Scheme 1); 2-(pyrazol-1-yl)-6-(1H-pyrazol-3-yl)pyridine (1,3bpp), 41 2-(3-methylpyrazol-1-yl)-6-(1H-pyrazol-3-yl)pyridine (met1,3bpp) and 2-(3,5-dimethylpyrazol-1-yl)-6-(1H-pyrazol-3-yl) pyridine (dimet1,3bpp), showing zero, one and two methyl substituents on the pyrazol-1-yl rings. The corresponding homoleptic Fe(II) complexes [Fe(1,3bpp) 2 ](ClO 4 ) 2 (previously published, 41 1), [Fe(met1,3bpp) 2 ](ClO 4 ) 2 (2) and [Fe(dimet1,3bpp) 2 ](ClO 4 ) 2 (3) were prepared and their magnetic properties determined in the solid state and in solution. The results show that the methyl in position 3 favours the HS state compared to the unsubstituted system, and that the same substituent on position 5 favours the LS state, with a stronger incidence.…”