How do alkenes react to form polymers in polymerisation?

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How do alkenes react to form polymers in polymerisation?

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Answer 1 · 2017-02-04T19:19:22

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

Explanation:

The typical olefin polymerization reaction starts with a metal dimethyl species, $L_{2} M {(C H_{3})}_{2}$ $L_2M(CH_3)_2$ , where $(i)$ $(i)$ $M$ $M$ is a metal such as titanium or zirconium, and $(i i)$ $(ii)$ $L$ $L$ 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 $L_{2} M^{+} (C H_{3})$ $L_2M^(+)(CH_3)$ :

$L_{2} M {(C H_{3})}_{2} + aluminum reagent \to L_{2} M^{+} (C H_{3}) + methylated aluminium reagent$ $L_2M(CH_3)_2 + "aluminum reagent"rarrL_2M^(+)(CH_3) + "methylated aluminium reagent"$

Ethylene can bind to the cationic metal centre:

$L_{2} M^{+} (C H_{3}) + H_{2} C=C H_{2} \to L_{2} M^{+} (C H_{3}) (η^{2} - H_{2} C=C H_{2})$ $L_2M^(+)(CH_3)+H_2"C=C"H_2rarrL_2M^(+)(CH_3)(eta^2-H_2"C=C"H_2)$

The methyl group bound to the metal centre can migrate to the $η^{2} - bound$ $eta^2-"bound"$ 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 $M - alkyl$ $M-"alkyl"$ residue. Long chain alkyl residues, polyethylenes, can be built up this way.

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How do alkenes react to form polymers in polymerisation?

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