# Catalytic Hydrogenation of Alkenes

#### CuriousBanker

Summary
Why can the Catalyst break the H2 bonds?
Hello.

My question is this: so the final product is the alkene adds two hydrogens across the double bond to become an alkane. At room temperature, the alkene is not strong enough to Break the H-H bond. So palladium adsorbs the hydrogens, and then the catalyst comes from above and attaches itself to the two hydrogens. Why is the catalyst strong enough to break the H-H bond, and the alkene is not? If the catalyst is strong enough to break the H-H bond, and attach itself to the two hydrogens, why is the catalyst not so strongly bonded to the Hydrogens as to not let the alkene steal them from it?

#### TeethWhitener

Gold Member
and then the catalyst comes from above and attaches itself to the two hydrogens.
I was with you until this part. What catalyst are you referring to? Palladium is generally a good enough catalyst on its own.

#### CuriousBanker

I was with you until this part. What catalyst are you referring to? Palladium is generally a good enough catalyst on its own.
Sorry, typo. I meant the alkene comes from above and grabs the hydrogens

#### TeethWhitener

Gold Member
This can be treated at a number of different levels, but the main issue that I’m seeing from your post is that you’re mixing up kinetics and thermodynamics.

At the most basic level, the addition of H2 to an alkene to form an alkane (e.g., ethylene to ethane):
$$C_2H_4 + H_2 \longrightarrow C_2H_6$$
is generally thermodynamically favorable (that is, the reaction will release energy if it takes place). The reason we don’t observe it happening at room temperature is that the activation energy is too high (which is another way of saying that the reaction rate is very slow).

However, most of the activation energy goes into breaking the H2 bond. The activation energy of the reaction:
$$C_2H_4 + 2H \longrightarrow C_2H_6$$
is much lower (the reaction happens much faster).

$$H_2\longrightarrow 2H$$
via a mechanism that is straightforward but a little advanced. I can discuss that in more detail if you have questions. But the upshot is that the palladium lowers the overall energy barrier of the alkene hydrogenation reaction.

The reason the palladium doesn’t hold on strongly to the hydrogen is that the surface palladium hydride is not particularly thermodynamically or kinetically stable, so it is energetically more favorable (and rather rapid) for the hydrogens to move onto the alkene.

#### CuriousBanker

Thanks for that great explanation. I think most of what I'm missing is probably in the step you left out, so if you have the time/desire I would love to see how the palladium catalyzes the reaction

#### TeethWhitener

Gold Member
The basic picture is that H2 approaches the metal perpendicular to the surface. Because of orbital symmetry, the overlap between the out of plane metal orbitals is strongest with the sigma antibonding orbital in the hydrogen. This backdonation of the filled metal $d_{z^2}$ orbitals into the empty $\sigma_u^*$ H2 orbital weakens and lengthens the H-H bond to the point where it is easy to break thermally. What happens next depends on the metal, but in the case of palladium, H atoms are quite soluble, and they begin to diffuse into the palladium crystal lattice.

When an alkene is introduced to this system, the interaction of the out of plane metal orbitals and the $\pi$ and $\pi^*$ orbitals causes the alkene to bend so that it's no longer planar. In atomic orbital terms, this mixes significant s orbital character into the p orbitals on each of the C's participating in the $\pi$ bond. At this point, you basically have a radical(ish) looking alkene molecule sitting on the surface, with significant s character on each of the carbon atoms. This orbital reconfiguration allows for good overlap with other species having significant s character....for example, hydrogen atoms, which--diffusing into and out of the metal surface--attach stepwise to the metal-bound alkene. Once the alkene has been hydrogenated to an alkane, the binding to the metal surface weakens dramatically (the molecule no longer has any sites that are easy to chemically bind to), and the alkane desorbs from the metal thermally.

This is a rough picture, dependent on the identity of the metal (how soluble H is in the metal), and still subject to a decent amount of research today, but I think what I presented covers the basics of most metal-catalyzed hydrogenations.

#### CuriousBanker

Nice. Don't really understand much of that, I suck lokl

#### TeethWhitener

Gold Member
Is there something in particular you have questions about?

#### CuriousBanker

Well the main issue is that I am self studying chemistry solely for pleasure and have only read one gen chem book and now im 1/3 of the way through an orgo book. and i devote like 6 hours a week to this. So just overall my level of knowledge is low. I don't understand what you mean by "out of plane metal orbitals" or " backdonation of the filled metal dz2dz2 " im assuming im just not advanced enough to get this

#### TeethWhitener

Gold Member
Well, that’s what PF is for. If you’ve been through gen chem, you’ve most likely encountered atomic and molecular orbitals. Organic and organometallic chemistry rely heavily on how molecular orbitals interact to determine reactivity patterns. Feel free to ask specific questions as you have them.

#### CuriousBanker

I really appreciate your help. I’m having trouble figuring out exactly what to ask unfortunately. What do you mean by out of plane metal orbitals? And why do those match up with anti bonding orbitals of H, meanwhile the alkenes don’t match up with the anti bonding orbitals of H?

#### TeethWhitener

Gold Member
What do you mean by out of plane metal orbitals?
The surface of a solid (say, palladium) forms a plane. The orbitals that the electrons of an individual palladium atom occupy have certain spatial distributions, and in the bulk of a solid, these atomic orbitals blend with those of their neighbors to determine the overall electronic structure of the solid. However, at the surface, you can (roughly) imagine that there are orbitals sticking out of the plane of the surface, that would have bound other Pd atoms in the bulk, are now “dangling bonds,” partially filled orbitals that are fairly reactive.

And why do those match up with anti bonding orbitals of H, meanwhile the alkenes don’t match up with the anti bonding orbitals of H?
It’s easiest to explain if you know what the shapes of the orbitals are. In the least sterically hindered configuration (perpendicular), the interaction of H2 with an alkene pi bond is basically zero, as different phases of the orbitals cancel each other out. Additionally, aligning the sigma (anti-)bonding orbital of H2 with the pi (anti-)bonding orbital of an alkene doesn’t do any good: the two bonding orbitals are both full and the two antibonding orbitals are both empty. However, for the interaction of Pd with H2 in a perpendicular configuration, a (partially) filled Pd orbital interacts with an empty H2 antibonding orbital, which means some of the electrons from the Pd orbital can move onto the H2 antibonding orbital, which weakens the H2 bond (since it increases the antibonding character of the molecule).

#### CuriousBanker

The surface of a solid (say, palladium) forms a plane. The orbitals that the electrons of an individual palladium atom occupy have certain spatial distributions, and in the bulk of a solid, these atomic orbitals blend with those of their neighbors to determine the overall electronic structure of the solid. However, at the surface, you can (roughly) imagine that there are orbitals sticking out of the plane of the surface, that would have bound other Pd atoms in the bulk, are now “dangling bonds,” partially filled orbitals that are fairly reactive.
I get the surface statement now, sorry, don't know why it was not clear to me the first time. However, why do these partially filled orbitals interact with H2 anti-bonding orbitals? Since this is an unstable configuration, why doesn't the H2 just ignore the palladium and stay as H2?

It’s easiest to explain if you know what the shapes of the orbitals are. In the least sterically hindered configuration (perpendicular), the interaction of H2 with an alkene pi bond is basically zero, as different phases of the orbitals cancel each other out. Additionally, aligning the sigma (anti-)bonding orbital of H2 with the pi (anti-)bonding orbital of an alkene doesn’t do any good: the two bonding orbitals are both full and the two antibonding orbitals are both empty. However, for the interaction of Pd with H2 in a perpendicular configuration, a (partially) filled Pd orbital interacts with an empty H2 antibonding orbital, which means some of the electrons from the Pd orbital can move onto the H2 antibonding orbital, which weakens the H2 bond (since it increases the antibonding character of the molecule).
Unfortunately I still don't quite understand how phase works (I'm guessing/hoping it'll make more sense after I learn QM?)

However, overall, you took a question I didn't think I would understand at all, and now I think I kind of understand it. So thank you.

One more question about this second quote though. I get why the Pd weakens the H2 bond...but, why does it weaken it enough to actually bond with the hydrogen? Like, after the H2 bond breaks, why doesn't it reform rather than making the unstable PdH bond?

#### TeethWhitener

Gold Member
Since this is an unstable configuration,
What makes you think this? The perpendicular binding configuration is actually the most stable configuration of H2 with a Pd surface. It’s strongly chemisorbed, whereas other configurations are mostly just physisorbed.
Like, after the H2 bond breaks, why doesn't it reform rather than making the unstable PdH bond?
Again, the Pd hydride isn’t unstable; it’s merely less stable than the ultimate reaction product (an alkane).

#### CuriousBanker

You’re saying it’s stable? In your first reply in this thread you said “The reason the palladium doesn’t hold on strongly to the hydrogen is that the surface palladium hydride is not particularly thermodynamically or kinetically stable,”

#### CuriousBanker

Also, how does the Alkene react with the hydrogen that’s bonded to the palladium? The reason it didn’t interact with the H2 is because the bonding orbitals were full and anti bonding orbitals were empty. But what about in the case of the palladium hydrogen bond? Isn’t that the same deal? Between Pd and H there is one anti bonding orbital and one bonding orbital. There are two electrons in the bonding and none in the anti bonding. So how does it interact with the alkene to make an alkane, when it’s in the same situation as before ?

#### CuriousBanker

“When an alkene is introduced to this system, the interaction of the out of plane metal orbitals and the π and π∗ orbitals causes the alkene to bend so that it's no longer planar. “

How does the alkene interact with the out of plane metal orbitals?

#### TeethWhitener

Gold Member
You've asked 6 questions in 16 minutes. You'll have to bear with me while I try to formulate a response.
You’re saying it’s stable?
Stability is relative.
How does the alkene interact with the out of plane metal orbitals?
The same way anything else does: by orbital overlap and either reinforcement or cancellation of appropriate wavefunctions. In the alkene's case, (if I remember correctly) the orbitals of surface palladium atoms can donate electron density into the alkene antibonding $\pi^*$ orbital. You can either think of this as weakening the $\pi$ bond or as adding s character to each of the carbons. In any case, it changes the geometry of ethylene from planar to bent (away from the surface).
Also, how does the Alkene react with the hydrogen that’s bonded to the palladium? The reason it didn’t interact with the H2 is because the bonding orbitals were full and anti bonding orbitals were empty.
But what about in the case of the palladium hydrogen bond? Isn’t that the same deal? Between Pd and H there is one anti bonding orbital and one bonding orbital.
At the temperature of the experiment, the Pd-H interaction is fairly labile; that is, the H diffuses quickly through the Pd lattice. Another way of thinking about it is that weak Pd-H bonds are formed and broken very quickly as H hops through the Pd crystal. What this means for the alkene is that an individual H atom has a certain probability of coming out of the Pd at the location of the alkene on the surface (the alkene doesn't really dissolve into the Pd to an appreciable extent, as far as I know). The bond formed between the alkene and the H is much stronger than any bond formed between Pd and H, so when the alkene-H bond forms, it's game over for that H. Once two alkene-H bonds form, you get an alkane and now even the chemisorption that was holding the Pd-alkene surface complex together is no longer present, so the newly-formed alkane desorbs thermally.

#### CuriousBanker

Haha. Sorry for the billion questions In a short time frame, I was in a waiting room with nothing to do. I appreciate you.

Ok. Let me see if I got this right: The H2 bond is strong, so the alkene can't supply enough energy to activate the reaction. However, the paladium has unpaired electrons which are able to weaken the H2 bond by jumping into the anti bonding orbital. (Question: is the H-Pd bond stronger or weaker than the H-H bond?) So the hydrogens split up and attach onto Pd. Then, other Pd atoms (not the ones attached to the H's) weaken the pi bond in the alkene, which lowers the energy of the transition state for the alkene to bond with the hydrogens, so they bond strongly.

Another question, then, What is to stop more surface Pd atoms from weakening the alkane bond and turning it back into alkene + Pd-H? I'm guessing that as soon as the alkane forms it floats away since it is a gas, before it gets the chance to get messed up again by the Pd?

#### TeethWhitener

Gold Member
The H2 bond is strong, so the alkene can't supply enough energy to activate the reaction.
I’m not sure I’d put it this way. The alkene isn’t supplying any energy. The alkene and hydrogen either react or they don’t. Under most normal conditions, the energy barrier for reaction is simply too high for the reagents to overcome.
However, the paladium has unpaired electrons
I wouldn’t really put it this way either, but maybe there is some value of viewing it this way. Solid state electronic structure generally uses a different language than molecular electronic structure, so it just strikes me as a bit odd. The important point is that the palladium can feed electron density into the H2 sigma antibond.
Question: is the H-Pd bond stronger or weaker than the H-H bond?
H-Pd, at least at STP, but not by a lot. I’m not sure for the kind of conditions you’d see in a Parr shaker. They’re probably comparable.
So the hydrogens split up and attach onto Pd.
Onto and into.
which lowers the energy of the transition state for the alkene to bond with the hydrogens, so they bond strongly.
The strength of the CH bond is a thermodynamic quantity, and is not really related to the transition state energy, which is a kinetic quantity.
Another question, then, What is to stop more surface Pd atoms from weakening the alkane bond and turning it back into alkene + Pd-H?
Nothing, the reaction is in equilibrium. (Edit: I got a bit sloppy with the language here. What I mean is that the reaction will go both ways, but that the alkane product will be most heavily favored in the reaction conditions, so the alkene concentration will drop throughout the reaction.) But at the conditions of the synthesis, the equilibrium is strongly in favor of the alkane.
I'm guessing that as soon as the alkane forms it floats away since it is a gas, before it gets the chance to get messed up again by the Pd?
The alkane interacts much more weakly than the Pd than the alkene. I’m sure this fact is probably exploited in industrial hydrogenation (e.g., removing the alkane to use Le Chatelier to shift the reaction more toward the desired product), but in the lab, I’m not sure how much of a factor it is, except for the fact that the alkene outcompetes the alkane for Pd surface binding sites.

#### CuriousBanker

Thanks I think I mostly get it now. But why does the alkane interact much more weekly? Because it doesn’t have a pi orbital?

#### TeethWhitener

Gold Member
Thanks I think I mostly get it now. But why does the alkane interact much more weekly? Because it doesn’t have a pi orbital?
Pretty much. I’m sure theoreticians can rattle off a whole number of reasons, but that’s how I think of it: a strong interaction between the Pd surface and the pi orbitals of an unsaturated organic compound.

#### CuriousBanker

You = good at splainin stuff