# Why are alkenes used to make polymers?

Jun 4, 2016

Well, it looks to me like since alkenes have to break one $\pi$ bond and make one $\sigma$ bond to add a monomer to the chain, that could be why they work so well. It's a thermodynamically favorable trade-off.

One popular method used to make controlled-length polymers, first introduced in 1955, is the Zeigler-Natta catalyst (which has since been updated to incorporate ${\text{MgCl}}_{2}$ to further enhance the catalytic activity).

This is commonly used as ${\text{TiCl}}_{4}$, dissolved in hydrocarbon solvents in the presence of "Al"("C"_2"H"_5)_3 (Inorganic Chemistry, Miessler et al., Ch. 14.4.1, pg. 570).

So, we have this happening to form a titanium alkyl complex first:

Next, this activated complex can react with alkenes. Let's use ethene. The ethene ligand first binds dihapto (${\eta}^{2} -$, i.e. by two atoms) to titanium.

Then, a 1,2-insertion occurs (keep your eyes peeled for this one!), changing ethene into a monohapto ligand (${\eta}^{1} -$, i.e. by one atom):

In these two steps, note that we have broken one $\pi$ and one $\sigma$ bond, and made two $\sigma$ bonds.

Breaking net-weaker bonds and forming net-stronger bonds is thermodynamically favorable. Hence, alkenes are a good choice for polymerization.

SIDENOTE: The above process is cyclic until it is terminated.

It allows for further polymerization by coordinating more alkenes and performing more 1,2-insertions to lengthen the alkyl chain:

and so on.

This is known as the Cossee-Arlman Mechanism, and experiments by Robert H. Grubbs have supported this mechanism as "the likely pathway for polymerization in most cases" (Inorganic Chemistry, Miessler et al., Ch. 14.4.1, pg. 571).