Polymerisation: Question 6

Worked solution

Step 1: Locate the double bond

But-2-ene is $\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3$. The $\text{C=C}$ double bond sits between the two middle carbons (C2 and C3), each of which already carries a methyl group and a hydrogen atom. The two end carbons (the methyl groups) are not part of the double bond and take no further part in the polymerisation.

Step 2: Open the double bond to form the repeat unit

In addition polymerisation only the $\pi$-bond of the $\text{C=C}$ breaks; the two backbone carbons keep every substituent they already had and instead form new single bonds to the neighbouring repeat units:

$\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3 \;\longrightarrow\; -\text{CH}(\text{CH}_3)-\text{CH}(\text{CH}_3)-$

Each backbone carbon in the repeat unit therefore carries one methyl branch and one hydrogen atom. The repeat unit is:

$\left(-\text{CH}(\text{CH}_3)-\text{CH}(\text{CH}_3)-\right)_n$

Why the other options are wrong

• B is the repeat unit of but-1-ene, $\text{CH}_2=\text{CHCH}_2\text{CH}_3$, which has a terminal double bond and would give one unbranched $-\text{CH}_2-$ carbon and one carbon bearing a whole ethyl side chain, but-2-ene’s double bond is internal, not terminal.
• C wrongly turns the entire four-carbon monomer into an unbranched backbone, as though but-2-ene reacted like two ethene units end to end; it discards the two methyl branches that but-2-ene actually has.
• D is the repeat unit of 2-methylpropene, $\text{CH}_2=\text{C}(\text{CH}_3)_2$, which places both methyl groups on the same backbone carbon rather than one methyl group on each of the two backbone carbons.

Final answer

A, $\left(-\text{CH}(\text{CH}_3)-\text{CH}(\text{CH}_3)-\right)_n$.