Suzuki Reaction

by James

Description: The Suzuki reaction is a palladium-catalyzed carbon-carbon bond forming reaction between an alkenyl halide and an alkenyl boronic acid (or boronic ester).

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Notes: Although Br is shown here, other leaving groups can be used such as Cl, I, and OTf. Furthermore note that the configuration of the alkene is preserved (trans boronic acid leads to trans product). Pd(PPh3)4 is palladium [in the “zero” oxidation state] attached to four triphenylphosphine (PPh3) “ligands” [a “ligand” is the chemistry term for any group attached to a central metal atom]

Examples: 

2-examples

Notes: Lots of details here. First of all, note that the Suzuki reaction is stereospecific. That is, the double bond geometry remains the same (is “retained”). A trans boronic ester (or acid) will give a trans product and a cis boronic acid/ester will give a cis product (examples 3 and 4). Note that a boronic acid (Example 1) or a boronic ester (Example 2) can both be used to provide the same product. Either hydroxide as base (example 1) or alkoxide base (example 2) can be used. Example 4 shows the use of a “triflate” leaving group (OTf) which is a superior leaving group to Br-. The final example shows how a Z vinyl iodide combines with an E boronic ester to give a (Z, E) product. Again, note how the stereochemistry of the alkene is preserved for both components. This is true for all Suzuki reactions.

Mechanism: As with the Heck reaction, a disclaimer is in order. The background to fully understand how and why these reactions work is not provided in a typical introductory organic chemistry class. Because so much important material is quickly glossed over, resorting to memorization rather than truly understanding each reaction is likely what you will be expected to do. That is why I (JA) strongly disagree with the decision to include this material in introductory organic chemistry courses. Nevertheless, because it is touched on in an increasing number of courses, this material is provided as a courtesy.

The structure of triphenylphosphine [PPh3] is shown below. Note that other phosphines such as tricyclohexylphosphine [PCy3] or tri-o-tolyl-triphenylphosphine [PAr3] can also be used – for our purposes, the identity of the phosphine doesn’t matter.
3-mech-a-phosphines

In the first step, the Pd loses two equivalents of PPh3 to give Pd(PPh3)2 . This frees up space on the Pd to react with the aryl/alkenyl halide in an oxidative addition reaction [Step 1]. Note that now the oxidation state of Pd is (II). Next [although this might be omitted in your textbook] the halide group on the Pd is exchanged for base [OH in this case] in a “ligand exchange” reaction [Step 2]. The OH group on Pd is going to coordinate to boron in the next step, “transmetallation”[Step 3] where Pd forms a new bond to carbon and the Pd-B bond is broken, with B(OH4) as a leaving group. Now that there are two Pd-C bonds, the Pd can undergo “reductive elimination” [Step 4] forming the new C-C bond and regenerating Pd(PPh3)2 , the active catalyst in this reaction, which can go on to react with another equivalent of the aryl halide.

3-mech-b

Notes: It’s really beyond the scope of introductory organic chemistry to explain the key trends behind the different mechanistic steps, such as “ligand exchange”, “oxidative addition”, “reductive elimination”, and “transmetallation”. If you are curious, one online resource that does a great job of this is Michael Evans, “The Organometallic Reader“.

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