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Acid removal of a Boc group: the mechanism, step by step

Acid protonates the carbamate, the tert-butyl group leaves as a cation because it can, and what is left gives off carbon dioxide to free the amine.

rule-derived3 stepsNot a record of what was observed

The pathway runs in 3 steps. What sits between them is a real structure: it is what the reaction passes through, and on a bad day it is what you isolate instead of the product.

Below is the sequence as it is taught. On the platform it is drawn on whatever structures you supply, with the curly arrows resolved to the actual atoms involved rather than to a general case with R groups on it.

The steps

  1. Acid protonates the carbamate

    The acid gives a hydrogen to the oxygen of the C=O group.

    Protonation at the carbonyl oxygen, which makes the bond to the tert-butyl group far easier to break.

  2. The tert-butyl group leaves as a cation

    The bond to the tert-butyl group breaks, both electrons staying behind. That group can carry a positive charge comfortably, which is why this protecting group comes off in acid.

    Heterolysis of the alkyl-oxygen bond to give the tertiary carbocation and the carbamic acid. The stability of that cation is the whole basis of the method.

  3. Carbon dioxide leaves and the amine is free

    What is left falls apart: carbon dioxide comes off as a gas and the nitrogen takes back its hydrogen.

    Loss of carbon dioxide from the carbamic acid. The gas escaping is what drives the deprotection to completion.

What this drawing assumes

Read these before you quote the mechanism
  • The cation is drawn as free. It is stabilised by its three methyl groups, which is the whole reason this group comes off in acid and the benzyl one does not.
  • The carbamic acid is drawn as an intermediate; it loses carbon dioxide quickly and is rarely seen.

Other pathways this class runs by

  • The cation is usually trapped by a scavenger in the flask, which is why these deprotections are run with one: otherwise it alkylates whatever else is present.

A class is not one pathway. Which of these runs depends on the substrate, the solvent and what else is in the flask, and the drawing above does not decide that for you.

Run it on your own structures

Enter the reactants and the product you expect. If the transformation is one the engine can perform and this pathway reaches that product from those structures, it is drawn on them — with the arrows on the right atoms and the intermediates you would actually pass through. If it does not reach the product, you are told that instead of being shown a drawing that does not apply.