HOCH2CH(R)CH2NH{C6H4}OCH3 is refluxed with (C3H3N2)2CO, using CH3CN solvent for two hours,
forming (C3H3N2)C(=O)OCH2CH(R)CH2NH{C6H4}OCH3. This was then heated to 150degC for 2 hours under vacuum pressure (0.05 mbar) to form R{C3H4N}NH{C6H4}OCH3. (C3H3N2) is an imidazole ring (nitrogens in 1- and 3- positions) which is connected on one of its nitrogen atoms.--{C3H4N}-- is a 1,3-substituted azetidine ring. (C3H3N2)2CO is also known as 1,1'-Carbonyldiimidazole. The variable R group ends in the 3-position on the azetidine ring. --{C6H4}OCH3 is a para-methoxyphenyl protecting group. This can usually be removed by hydrolyzing off the ester methoxy group with 20%HCl, then oxidizing. The --CH2NH{C6H4}OH turns into
--CH2N={C6H4}=O, and this resulting imine easily hydrolyzes off with more acid, leaving --CH2NH2 and benzoquinone
O={C6H4}=O. R. M. de Figueiredo, R. Fröhlich, M. Christmann, J. Org. Chem., 2006, 71, 4147-4154
I had a thought about reacting glycerol with limited bromoethane, then distilling off the glycerol mono-ethyl ether.
66% would be 1- or 3- substituted, while 33% would be desirable 2-substituted. Then react the distillate with
acetyl chloride, putting acetyl esters on the remaining hydroxyl groups. The 1,3-acetyl,2-ethoxy propane could react with acetamide CH3CONH2 (the reaction would take days) to give low yields of 1N-acetyl,3-ethoxy-azetidine.
The azetidine derivitive could be distilled out, since it would be the most volatile of any of the reaction products. I would think 10% yield could be reasonably expected. based on the initial glycerol.
The ethoxy-ester could be hydrolyzed to 1N-acetyl,3-hydroxyl-azetidine, then the acetyl hydrolyzed off in conc NH4OH solution, leaving plain 3-hydroxy-azetidine. Any comments or opinions about this idea are welcome.