[Optimist]

My starting material having three Methyl ester and isonitrile group. I want to deprotect methyl ester using IM LiOH in methanol. But I have seen in the literature that deprotection of methyl ester using LiOH required longer reaction time (3days). Also, during workup it must be quenched with dilute acid or NH4Cl to get required acid but I want to avoid acidic workup due instability of isonitrile group in acidic condition.

1M NaOH in Methanol are also used for the deprotection of methyl ester. As, NaOH is stronger base compared to LiOH. Does using NaOH will significantly reduce the reaction time and how can I do the work up in case of NaOH?

[rolnor]

NH4Cl is not acidic?

[orgopete]

I don't think the acid should be a problem. If you are using NaOH or LiOH, you will be generating a salt. I see two options. You could just neutralize the base, 10 mL of 1N NaOH will require 10 mL of 1N acid. The other is to use an excess of a volatile (weak) acid and evaporate off the excess. However, you still must remove the salt of the acid. Without the compound being hydrolyzed, I don't know if this would be a problem or not. You could also use a carboxylic acid ion exchange column. The sodium or lithium ions would be retained on the column.