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As a result of their different symmetry, the four structural isomers of the title compound (the C4h isomer 1 is depicted on the right) could be characterized by 1H NMR spectroscopy. With these complexes the complete chromatographic separation of the four structural isomers of tetrasubstituted metal phthalocyanines was achieved for the first time. R  2‐EtC6H12O.

The 1(4)-and 2(3)-tetraalkoxy-substituted nickel (5), copper (6), and metal-free phthalocyanines 7 and 8 were synthesized from the corresponding substituted phthalonitriles 2 and 4, respectively, and the four structural isomers formed for each phthalocyanine were separated by HPLC. In the case of 1(4)-tetraalkoxy-substituted phthalocyanines, the ratio of the four isomers was found to be different from the expected statistical distribution of D 2h :C s :C 2v :C 4h 12.5:50:25:12.5. For the 1(4)-substituted metal-free phthalocyanine 7 a very high proportion of the C 4h isomer (87 %) is formed. In the case of the 1(4)-substituted metal phthalocyanines 5 and 6 the strikingly low proportion of the D 2h isomer (found: 2 ± 4 % compared to statistical distribution: 12.5 %) is interpreted by a template mechanism (given in Scheme 2) in which strong steric hindrance of the respective neighboring groups prevent the formation of the D 2h isomer. To investigate further the mechanism of formation of phthalocyanines the synthesis of 1(4)-and 2(3)-tetraalkoxy-substituted metal phthalocyanines containing chiral alkoxy groups (13 ± 17) was studied under different reaction conditions starting from the corresponding alkoxy-substituted phthalonitriles 10 and 11. In all investigated cases the chiral alkoxy groups in the starting phthalonitrile again affect the distribution of the structural isomers of the formed phthalocyanines, leading to a higher proportion of the C 4h isomer in comparison with the 1(4)-tetraalkoxysubstituted phthalocyanines with racemic alkoxy groups.

Attempts to prepare vanous peripheral tetra-and octasubstituted (phtha1ocyaninato)iron derivatives R,PcFe [n = 4:

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