I am currently designing a synthesis requiring the use of hydroxy protecting groups. The problem I am having is that the protecting group must be able to survive highly acidic conditions. I would have loved to use a TMBS protecting group since it can withstand LAH, lewis acids, and strong base; however, not acidic conditions. I really do not want to have to use a t-butyl ether protecting group, as its removal requires very harsh conditions. I will provide a scheme indicating all of the conditions I have proposed so far (this scheme is just a rough draft, so there are a few reagents that I have changed to be compatible with the reaction, but the general conditions should remain the same). Does anyone know any protecting groups that would match this criteria? I am only referring to replacing the TMBS protecting group on the second intermediate.

enter image description here

  • $\begingroup$ A substituted benzyl ether such as 4-nitrobenzyl, or 2,2,2-trichloroethyl ether. I wouldn't do the sulfonation step to block that site $\endgroup$ – Waylander Nov 5 at 18:44
  • $\begingroup$ Do you know of any other good blocking groups? If I could use a blocking group that didn't involve such harsh conditions, I might be able to use a silyl ether protecting group (if I altered some of the other steps of course) $\endgroup$ – ELI JONES Nov 5 at 19:19
  • $\begingroup$ Why use a blocking group? If you form methoxymethyl ethers of the hydroxyls of the diphenol then you can put in both your carbon chains by sequential directed lithiations and cut your number of steps drastically. $\endgroup$ – Waylander Nov 5 at 19:31
  • $\begingroup$ How about Br? Then you can put your second carbon chain in by a Pd catalysed coupling $\endgroup$ – Waylander Nov 5 at 19:47
  • $\begingroup$ If I first converted the hydroxy groups from the second step into -MOM groups, would that not direct the metalation to occur where the -SO3H is now since both of the -MOM groups are strong ortho-directors? $\endgroup$ – ELI JONES Nov 5 at 20:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.