Deprotection of carboxybenzyl and methyl ester can easily be achieved by using ammonium formate followed by addition of 10% Pd/C and LiOH/MeOH respectively.
But how we can achieve deprotection of carboxybenzyl group in the presence of methyl ester?
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
Just leave out the last step, i.e., LiOH/MeOH and you should be just fine, the Cbz group gets removed under reductive conditions which leave the methyl ester unaffected. If you lose the methyl ester, you could just add it again afterwards.
answered Oct 15, 2016 at 15:17
-
$\begingroup$ Are you sure that reduction using Pd/C will not cleave the methyl group. As, I know t-Butyl ester is stable against reducing agent like Pd/C but I am not sure about Methyl group. any insight or reference ? $\endgroup$– KhanCommented Oct 16, 2016 at 3:44
-
1$\begingroup$ I would at least not suspect esters to be very labile in Pd/C hydrogenation. While this is in principle possible, the carbamate of the Cbz will be cleaved quicker. So, I don't really see that to be a significant problem. Here are two references that look pretty similar to your problem: T. Maki, T. Tsuritani, T. Yasukata, Org. Lett., 2014, 16, 1868-1871. G. S. Vanier, Synlett, 2007, 131-135. (see organic-chemistry.org/protectivegroups/amino/cbz-amino.htm) One other approach would be to use another reducing reagent for the deprotection that leaves the ester unaffected (NaBH3CN?) $\endgroup$ Commented Oct 16, 2016 at 9:52