SolutionsProtection of amines
Exercise 1
A molecule has a primary and a secondary amine. With one equivalent of (Boc)2O at 0 °C, which is protected first? How would you then remove the Boc without affecting a methyl ester?
The primary amine, being less hindered, reacts faster with the bulky dicarbonate. The Boc is removed with anhydrous acid (trifluoroacetic acid in dichloromethane, or HCl in dioxane or ethyl acetate): in the absence of water, the methyl ester is not hydrolysed.
Exercise 2
Boc-Ala-OH and H-Phe-OBn are used in the synthesis of a dipeptide. Once the peptide bond is formed, what conditions would give H-Ala-Phe-OBn, Boc-Ala-Phe-OH and the free dipeptide?
H-Ala-Phe-OBn: trifluoroacetic acid (removes the Boc, leaves the benzyl ester). Boc-Ala-Phe-OH: H2, Pd/C (hydrogenolysis of the benzyl ester; the Boc survives). Free dipeptide: both operations, in either order. It is an example of orthogonal protection of the amino and carboxyl groups.