I am wondering how does PCC actually work in not oxidizing aldehyde. Like the reaction mechanism I saw online, it seemed as if not having a acidic solvent is what made aldehyde not tranform to carboxylic acid when PCC is present. So the why can't you just use chromium containing compounds without the acidic solvent and then you will be able to isolate aldehyde.

  • $\begingroup$ The crux lies in the half-equation for the reduction of PCC. It must have a source of $\ce{H+}$ ions (or $\ce{OH-}$ ions, for basic conditions) for any reaction to take place in the first place. $\endgroup$
    – Linus Choy
    Feb 4, 2017 at 5:34

1 Answer 1


The "trick" is in the enviroment, not in the PCC itself.

Alcohols tend to be oxidized to carboxylic acids when they are in aqueous enviroment.

PCC was invented as a workaround to this problem: it works as a reactant in anhydrous enviroment, hence stopping the reaction at the aldheyde stage.

An alternative way to see PCC is to follow its preparation: if you add $\ce{CrO3, HCl}$ and pyridine, PCC is formed. If you add $\ce{CrO3}$, acid and water, chromic acid is formed.

So, you see that the reagents are similar: the point is in finding at which stage the difference is.

Mechanistically speaking, the difference occurs as follows: in aqueous enviroment, water can form a geminal diol intermediate by attacking the aldheyde. This step adds the second oxygen to the aldheyde that, with further reaction with chromic acid, yelds a carboxylic acid.

When water is not present, this step doesn't occur, and the reaction stops.

  • $\begingroup$ Anhydrous PCC oxidations can also form esters as by-products with primary alcohols. Unreacted alcohol forms a hemiacetal with the aldehyde previously produced. As far as PCC is concerned, the hemiacetal is a secondary alcohol. It gets oxidized to an ester. Use dilute suspensions of PCC in dichloromethane; use excess reagent; add alcohol slowly to the oxidant. These steps minimize ester formation. $\endgroup$
    – user55119
    Nov 24, 2017 at 20:18
  • $\begingroup$ Good answer. Here's a useful page for anyone who wants more on how this works. $\endgroup$
    – harry
    Apr 27, 2021 at 2:32

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