# How do alkenes react to form polymers in polymerisation?

Feb 4, 2017

Typically ethylene inserts into a ""^(+)M-R bond, where the $R$ group starts as methyl.......

#### Explanation:

The typical olefin polymerization reaction starts with a metal dimethyl species, ${L}_{2} M {\left(C {H}_{3}\right)}_{2}$, where $\left(i\right)$ $M$ is a metal such as titanium or zirconium, and $\left(i i\right)$ $L$ is some ligand that stabilizes and solubilizes the metal centre.

Addition of a Lewis acid abstracts one of the methyl groups to yield a species conceived to be ${L}_{2} {M}^{+} \left(C {H}_{3}\right)$:

${L}_{2} M {\left(C {H}_{3}\right)}_{2} + \text{aluminum reagent"rarrL_2M^(+)(CH_3) + "methylated aluminium reagent}$

Ethylene can bind to the cationic metal centre:

L_2M^(+)(CH_3)+H_2"C=C"H_2rarrL_2M^(+)(CH_3)(eta^2-H_2"C=C"H_2)

The methyl group bound to the metal centre can migrate to the ${\eta}^{2} - \text{bound}$ olefin to give a propyl group. And such insertion generates an open coordination site on the metal centre, to which another ethylene can bind, and this reinserts into the $M - \text{alkyl}$ residue. Long chain alkyl residues, polyethylenes, can be built up this way.

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