2
$\begingroup$

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.

$\endgroup$
0

2 Answers 2

9
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

(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.

$\endgroup$
2
  • 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. $\endgroup$ Commented 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. $\endgroup$ Commented Aug 27, 2020 at 8:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.