In the conversion of piperidine to pyridine, palladium on carbon (Pd/C) is used. But it is reducing agent.

Dehydrogenation of piperidine

Clear the confusion, please.


2 Answers 2


Palladium on carbon ($\ce{Pd/C}$) is neither an oxidant nor a reductant. The reaction is the dehydrogenation of piperidine: $\ce{C5H11N -> C5H5N + 3H2}$ and here, $\ce{Pd/C}$ is merely a catalyst.

Even in standard hydrogenation reactions using $\ce{H2(g) + Pd/C}$, the $\ce{Pd/C}$ is not a reducing agent. The $\ce{H2}$ is the reducing agent and Pd/C is again only a catalyst.


(The above answer brings up hydrogenation using Pd/C, be weary of the answer because they're not addressing the question.)

The above answer is partly true, but on the whole incorrect and would fail in a test scenario.

Yes, Pd/C is a catalyst therefore in a full reaction cycle the Pd/C should return to its original oxidation state.

However, Pd/C can only accept the N-H bond via oxidative addition onto Pd metal. By this definition, because the Pd is oxidized while adding the N-H bond across Pd, then Pd must begin the cycle as a "reductant" (i.e. the Pd metal initially reduces the N-H bond). The organic compound must be the "oxidant" because it oxidizes the Pd (i.e. oxidative-addition to Pd).

Where the first answer fails to be correct is: if Pd/C has been previously oxidized, it cannot perform the first step in the catalytic cycle and cannot be catalytic. It is more correct to say that Pd/C is a reducing agent (reductant, what have you).

Where the first answer fails more is: most Pd/C reductions in a lab setting are NOT catalytic and require a molar equivalent or even excess of Pd/C for a reduction, which goes against the definition of a catalyst.

  • 1
    $\begingroup$ I don't agree with some of your points: (1) I already addressed OP's dehydrogenation reaction in my very first sentence. (2) Even if the Pd is initially oxidised, I don't think I would say that it acts as a "reducing agent" in the grand scheme of things. There's plenty of Pd(0) catalysis out there with the whole ox. add. / ligand exchange / red. elim. procedure, which (at least in terms of the ox. states involved) is similar. Do people call Pd(0) a reducing agent in those? I would consider that to be extra detail about the catalytic cycle, not a definition of the fundamental role of Pd. $\endgroup$ Aug 27, 2020 at 8:06
  • $\begingroup$ Am open to being convinced otherwise, though. I am also mildly surprised when you say that hydrogenation is often not catalytic – it's been a long time since I did anything, but I used to do Lindlars with 0.5 mol% Pd/CaCO3. Anecdotally, the amount required depended on the supplier. $\endgroup$ Aug 27, 2020 at 8:16

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