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Base removal of an Fmoc group: the mechanism, step by step

The base takes a hydrogen the fluorene ring makes unusually acidic, the carbamate is pushed out, and what is left gives off carbon dioxide.

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. The base takes the acidic hydrogen

    The base removes the hydrogen from the carbon between the two rings. That carbon can hold the electrons unusually well, which is what makes this protecting group come off in base.

    Deprotonation at the fluorene 9-position. The resulting anion is aromatic across the five-membered ring, and that stabilisation is what makes the proton acidic enough to remove with a mild amine.

  2. The carbamate is pushed out

    The electrons left behind drop down to make a double bond, and that pushes the whole carbamate group off.

    E1cb elimination giving dibenzofulvene and the carbamate. The alkene formed is conjugated across both rings, which is what makes this step fast.

  3. Carbon dioxide leaves and the amine is free

    The carbamate falls apart: carbon dioxide comes off, and the nitrogen left behind picks up a hydrogen from whatever is around.

    Loss of carbon dioxide gives the amide anion, which is protonated at once by the solvent. The gas escaping is what makes the deprotection irreversible.

What this drawing assumes

Read these before you quote the mechanism
  • Drawn through the carbanion. The hydrogen is acidic because the anion it leaves behind is aromatic in the five-membered ring, which is the entire reason this group comes off in base and the Boc group does not.
  • The dibenzofulvene is usually trapped by the amine used as the base; here it is simply left.

Other pathways this class runs by

  • Nothing else removes it usefully: the point of Fmoc is that it is stable to the acid that removes a Boc group, so the two can be used on the same molecule.

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.