I am quoting my Clayden et al 2nd edition:

Some catalysts are particularly selective towards certain classes of compound – for example, Pt, Rh and Ru will selectively hydrogenate aromatic rings in the presence of benzylic C-O bonds, while with Pd catalysts the benzylic C-O bonds are reduced faster.

My question is, why $\ce{Pt, Ru, Rh, Pd}$ all being transition metals still show different chemoselectivity in hydrogenation reactions?

1st edit: (After reading ringo's comment I think I should add something more) Well, I first thought that the number of $\ce{d}$ electrons is the cause. But, both $\ce{Pd}$ and $\ce{Pt}$ have same number of $\ce{d}$ electrons. Yet, they show different chemoselectivity. I think the difference is because $\ce{Pt}$ has an $\ce{f}$ orbital filled up.

And, I guess as $\ce{Ru, Rh, Pd}$ belong to same period, they show similar chemoselectivity.

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    $\begingroup$ Well, Palladium and Platinum have same number of $d$ electrons, as they both belong to group 10, but they show different chemoselectivity. $\endgroup$ Commented Jun 20, 2017 at 5:54
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    $\begingroup$ @Ringo And talking of valence electrons, all transition metals have valence shell configuration $ns^2$. $\endgroup$ Commented Jun 20, 2017 at 5:56
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    $\begingroup$ It’s because each element is subtly different. All four will interact with both functional groups in some way but some interactions will be stronger due to the intrinsic electronic properties. And the stronger, i.e. better interactions will drive the reaction. Sometimes even, seemingly random things happen, like my colleague who managed to completely reduce an N-protected 3-acyltetramic acid to the fully saturated compound leaving only the amide carbonyl functionality. That was funny. $\endgroup$
    – Jan
    Commented Jun 22, 2017 at 12:11
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    $\begingroup$ The actual problem is that I’m just the organic chemist who uses the catalysts in published ways and that I have little clues as to what has been done or not. It could well be that someone has already checked out that question and it is published somewhere. $\endgroup$
    – Jan
    Commented Jun 22, 2017 at 13:34
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    $\begingroup$ One important piece that's missing here is what ligands each metal is coordinated to (or if it's the plain metal). $\endgroup$
    – Zhe
    Commented Jun 22, 2017 at 15:27

1 Answer 1


Unfortunately, this is extremely complicated and the topic of active research in both computational and inorganic chemistry and chemical engineering. There is no simple answer to your question. We can discuss various forces that go to determining chemoselectivity though.

1) Crystal structure: You said that you are primarily focused on bare metal catalysis, so I will restrict myself to those systems. Not all surfaces of even the same material behave the same. These are commonly described by Miller indices (1), which describe the surface by how a plane cuts through the material's unit cell. The reactivity and selectivity are heavily determined by the exact surface structure (think dangling bond sites, binding sites, multicenter reactivity, all of which are influenced by the surface atomic structure both local and in its immediate environs). Different materials will have different crystal structures based on both electronic effects (like how many valence electrons there are) and size effects (how big atoms are, although there is relatively less variation on this in the 2nd and 3rd rows).

2) Chemistry of binding sites: Some materials are more likely to bind molecules than others. The Pt-H bond is extremely strong, much more than a Pd-H bond (these trends down a period are often explained in terms of the diffuseness of the 5d orbitals vs 4d which lead to better overlap with the main group orbitals of hydrogen and other light elements). This leads to greater coverage of a Pt surface with hydride equivalents than a Pd surface (generally, caveats apply). On the other hand, Pd tends to bind oxygen atoms more strongly than Pt does (one can think about hard-soft acid base theory here to explain that, but the situation is probably more subtle and complicated than that). One can easily imagine a benzylic C-O oxygen bind to the Pd surface which holds the site for hydrogenation close the the surface until a hydride equivalent can diffuse over to react (hydrides on surfaces are quite mobile and diffusion is often quite rapid, of course, hydride diffusion rates are also strongly determined by surface structure and material properties (because almost nothing can be simple)). On the other hand, $\pi$-system can interact strongly with metal surface (especially of soft (again in the HSAB sense) metal). So, for Pt, the aromatic system is brought close to the surface, allowing hydrides to migrate over to it.

So we can hand-wavily discuss why Pd is more likely to hydrogenate benzylic C-O bonds and Pt is more likely to hydrogenate aromatic systems as we did above, but the nitty-gritty detail is quite complicated. This is a massive area of research. Inorganic chemistry is its own field not because everything is so similar, but because metals have hugely varying behavior. The fact that these are "all being transition metals" should NOT imply that their reactivity is similar in the slightest. Being in the same period as a transition metal tells you as much about being in the same period in the main group. C and O and N have totally different chemistry, as do Rh and Ru and Pd.

P.S. f orbitals do not figure into this.


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