Heterocyclic analogs of pleiadiene

Пожарский, А. Ф.; Kashparov, I. S.

doi:10.1007/bf00480649

Cited by 6 publications

(3 citation statements)

References 2 publications

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“…Naphtho-annelated iminodihydrotetrazole 1 was synthesized from di( tert -butyl)dinitronaphthalene 3a in five steps (Scheme ). The first four steps have precedence in the series devoid of tert -butyl groups, – and these reactions are detailed in the Supporting Information. Owing to the ambident character of 1-substituted 5-aminotetrazoles, treatment of 6 with dimethyl sulfate afforded a mixture (87:13) of the yellow, 1,4,5-substituted tetrazole 1 and the deep-red, mesoionic 1,3,5-substituted tetrazole 7 .…”

Section: Resultsmentioning

confidence: 99%

A Perimidine-Derived Non-Kekulé Triplet Diradical

et al. 2008

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6,9-Di(tert-butyl)-1-methyltetrazolo[1,5-a]perimidine (1) has been synthesized from naphthalene in seven steps. The EPR spectra, recorded after irradiation of 1 in a butyronitrile matrix at 77 K (lambda = 351 nm) and in Ar and Xe matrixes at 4.6 K (lambda > or = 345 nm), showed a six-line, high-field signal (Delta m(S) = +/- 1), centered at 3350 G in butyronitrile, along with a half-field signal (Delta m(S) = +/- 2), which is characteristic for triplets. Simulation of the observed EPR spectra gave values for the zero-field splitting parameters of |D/hc|/cm(-1) = 0.0105, |E/hc|/cm(-1) = 0.0014 in butyronitrile and |D/hc|/cm(-1) = 0.0107, |E/hc|/cm(-1) = 0.0016 in Ar. These EPR parameters are consistent with the diradical 5,8-di(tert-butyl)-2-(N-methylimino)perimidine-1,3-diyl ((3)2) as source of the EPR spectra. Linearity of the Curie-Weiss plot and UB3LYP and (14/14)CASPT2 calculations of the singlet-triplet energy difference (DeltaE(ST) approximately 8-10 kcal/mol) indicate that the triplet is the ground state of 2, as predicted for such a nondisjoint diradical.

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Section: Resultsmentioning

confidence: 99%

A Perimidine-Derived Non-Kekulé Triplet Diradical

et al. 2008

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“…2-Methoxy-2-methyl-1 H-perimidzne (11). To a s o h of 2-chloro-I-methyl-lH-penmidine [7] (0.4 g, 1.89 mmol) in dry MeOH (6 ml), Na (0.43 g, 18.9 mmol) in anh. MeOH (4 ml) was added.…”

Section: Experimental Partmentioning

confidence: 99%

“…Naphthalene-I ,8-diamine, 1H-benzimidazol-2(3H)-one (3), l-methyl-lH-henzimidazol-2(3H)-one (13), l-ethyl-lH-benzimidazol-2(3H)-one (15), and 1,3-diphenylurea (4) are commercial products and were used without further purification. The following compounds were prepared according to published procedures: 1,3-dimethyl-lH-benzimidazol-2(3H)-one (14) [9], 2-methoxy-I-methyl-IH-benzimidazole (17) [lo], lH-perimidine (1) (291, 12-methyl-IH-perimidine [29], IH-perimidin-2(3H)-one (2) [7], l-methyl-1H-permidin-2(3H)-one (5) [7], 1,3-dimethyl-IH-perimidin-2(3H)-one (6) [8], 1,3-diethyl-lH-penmidin-2(3H)-one (8) [8] and 1,3-dipropylperimidin-2(3H)-one (10) [8]. Flash Vacuum Pyrolsis: the pyrolysis system using a quartz tube 50-mm long and 3-mm wide was described elsewhere [30].…”

Section: Experimental Partmentioning

confidence: 99%

The Structure of 1H‐Perimidin‐2(3H)‐one and Its Derivatives in the solid state (x‐ray crystallography and CP/MAS ¹³C‐NMR), in solution (¹³C‐NMR), and in the gas phase (mass spectrometry)

Claramunt

Dotor

Sanz

et al. 1994

Helvetica Chimica Acta

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The X-ray crystal and molecular structures of 1 -methyl-1H-perimidin-2(3H)-one (5b) and 1,3-dimethyl-IHperimidin-2(3H)-one (6) were determined. The crystals are built of piles of dimers faced head-to-head in 5b and of alternating independent head-to-tail molecules in 6, both along the 6 axis. Semiempirical calculations at the AM1 level revealed that the eclipsed conformation of the Me groups with respect to the C=O group, found in the crystals, is the most stable. The lack of planarity of the whole molecules is a consequence of the packing forces since it is not found in the calculations. A comparative NMR study was carried out in solution ('H and I3C) and in the solid state (CP/MAS I3C) for lH-perimidin-2(3H)-one (2) and 1H-benzimidazol-2(3H)-one (3) with the conclusion that in both heterocycles the 0x0 tautomer is the most abundant. The structure in the gas phase was approached by mass spectrometry. In the case of 3, the 0x0 tautomer loses CO after ionization, while the 0x0 form of 2 tautomerizes to the hydroxy form which loses H,O after a [1,3]-H shift. AM1 calculations were carried out on the ground and ionized states of 2.

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