# Regioselectivity in Carbometallation

I am a student of organic chemistry and I frequently watch online open course lectures by various professors around the globe. Of these, in https://www.youtube.com/watch?v=OYAMls5x4EI, at about 27:10, I had a question, as mentioned in the image below.

Mechanism

What factors determine where the palladium will attach?

My initial thoughts were to find out the electron-rich regions of the $\pi$ bond, and the palladium atom in the $\mathrm{+II}$ oxidation state will act as an electrophile. But I wasn't able to figure it out exactly.

I understand that the ratio of products shown in the figure can be controlled by using various ligands, especially based on their steric effects, in this case.

• @EashaanGodbole One thing that is important to note is that the hydroxyl group acts as a hydride source. This is only possible for the above product (see my answer). – logical x 2 Nov 1 '16 at 21:31

Your mechanism is missing an important step: between the oxidative addition of palladium into the $\ce{C-Br}$ bond and the alkene insertion there should be an alkene annealing forming a palladium-alkene complex. This is displayed in the mechanism scheme of a generic Heck reaction below, which I sheepishly borrowed from Wikipedia.

Scheme 1: Catalytic cycle of a Heck reaction. Image taken from Wikipedia, where a full list of authors is available. Note that the β hydrode elimination in step C should be syn and not anti or gauche as displayed.

The annealing step and the π complex are important because they influence the Heck reaction’s selectivity. Due to the avoidance of steric stress, both residues will adopt an anti-type conformation as depicted in structure 3. On the (syn-selective) insertion of the alkene into the $\ce{C-Pd}$ bond, this geometry is kept resulting in the palladium catalyst ending up on the higher-substituted alkene carbon. For the inverse insertion, both organic residues — in your case two phenyl groups — would have to occupy the limited space on one side of the catalyst.

The subsequent β hydride elimination can theoreticly occur in both directions in your molecule. However, eliminating towards the hydroxy group allows keto-enol tautomerism with further stabilisation. Since β hydride elimination is a reversible process, the reaction will select the ketone over the allylic alcohol that would be formed upon opposite β hydride elimination.

The alkene insertion into the $\ce{Pd-C}$ bond, is not reversible. Two $\ce{C-C}$ single bonds are formed at the expense of a palladium-π interaction resulting in a notably downhill process.

Just have a look at the reductive elimination step of this Heck reaction:

In case of the second reaction the resulting allylalcohol is far less stable than the enol tautomerizes to the ketone. The other thing are sterics aspects that make the migratory insertion step far less favourable for the second product as well.

• No. The selectivity is determined when the $\ce{C-C}$ bond is formed. Ease or unease of the following beta hydride elimination is irrelevant for the proceeding of the reaction; if the $\ce{C-C}$ bond formed a branched compound, the final product will be branched. – Jan Nov 1 '16 at 22:25
• @Jan You sound so definite in what you are saying. Then could you please provide proof of what you are saying, i.e. that the C-C bond formation in a Heck reaction is irreversible? – logical x 2 Nov 1 '16 at 22:37
• @Jan Looking at e.g. the pinacol-pinacolon rearrangement, benzoin condensation and so on it seems entirely possible that a C-C bond formation is reversible, and I am actually pretty sure that all steps of the Heck reaction are reversible... – logical x 2 Nov 1 '16 at 22:40
• @EashaanGodbole I really don't think so. Typically, all steps of cross coupling reactions are reversible. In fact, the outcome of the Heck reaction is in most cases controlled by the final beta hydride elimination (so the reductive elimination). So I dont see the problem with my 'product controlled reaction' argument. But sure, the formation of the first Intermediate after carbometalation will also be far less favorable for the second reaction. However, this has nothing to do with kinetic or thermodynamic control its only thermodynamic control. – logical x 2 Nov 2 '16 at 8:12
• @EashaanGodbole Looking at both arguments either arguing about the disfavoured carbometalation and/or the disfavoured beta hydride elimination: Even if the reaction would be allowed to react for 10 billion years the first product would still be the only product, as the other product is just too unstable to be formed anyway. It is just a thermodynamic argument. By the way, the lectures of Dyker are really nice : ) – logical x 2 Nov 2 '16 at 8:15