摘要:
Abstract3‐Benzoyl‐thiocarbazic acid O‐methyl ester 7a and 3‐benzoyl‐dithiocarbazic acid methyl ester 7b are methylated, in presence of alkali, at the SH group of 1a, the ene‐thiol tautomer of 1, to give the 3‐benzoyl‐isothiocarbazic acid O, S‐dimethyl ester 8a and the 3‐benzoyl‐isodithiocarbazic acid dimethyl ester 8b resp., which clearly differ from the N‐methylated compounds 11a and 12a or 11b and 12b prepared from the two N‐methyl benzohydrazides 9 and 10 resp. (melting points, thin‐layer chromatography and NMR. spectra).The previously reported [1] ring closure of 3‐(ω‐chloroalkanoyl)‐thiocarbazic acid O‐alkyl esters and ‐dithiocarbazic acid alkyl esters can be interpretated as an intramolecular auto‐alkylation of CONHNHCSYR: the 3‐(chloroacetyl)‐compounds 2 are S‐alkylated (enethiol form) to the six‐membered thiadiazinones 4, but the 3‐(3‐chloropropionyl)‐compounds 3 are N‐alkylated to the pyrazolidinones 6, the five‐membered ring being preferred in the latter case to the seven‐membered ring 5 which would be formed by S‐alkylation. Hence the position of alkylating attack depends on the size of the ring to be formed.As a consequence, starting from 3‐(4‐chlorobutyryl) compounds 13, an alternative between the two N‐alkylation products 14 (five‐membered ring) and 15 (six‐membered ring) would be expected. On the contrary however, a combination of 1,3,4‐thiadiazole ring closure and chlorine elimination predominates, giving the 5‐(3‐hydroxypropyl)‐1,3,4‐thiadiazoles 16. This reaction may proceed via the butyrolactone intermediates 18 and could therefore be interpreted as O‐alkylation followed by rearrangement.