A section of an addition polymer chain is represented by the repeat unit (-CH2-CH(CH(CH3)2)-)n Which alkene was used as the monomer to form this polymer?

AS Chemistry 9701: Identifying the Monomer from a Repeat Unit

Step 1: Undo the addition polymerisation

In addition polymerisation, the $\text{C=C}$ double bond of the monomer opens up so that each carbon forms a new single bond to the next monomer unit, with no atoms lost. To find the monomer, reverse this process: take one repeat unit and turn the two backbone single bonds (the ones joining it to its neighbours) back into a double bond.

The given repeat unit is:

$-\text{CH}_2-\text{CH}(\text{CH}(\text{CH}_3)_2)-$

Reconnecting the backbone carbons with a double bond gives:

$\text{CH}_2=\text{CH}-\text{CH}(\text{CH}_3)_2$

Step 2: Check the branching carefully

The backbone carbon that is not $-\text{CH}_2-$ carries exactly one branch, $-\text{CH}(\text{CH}_3)_2$ (an isopropyl group), plus one hydrogen atom. This matches a monomer with a single isopropyl branch on the carbon next to the terminal $\text{CH}_2=$ . This is 3-methylbut-1-ene.

Why the other options are wrong

B (pent-1-ene) has a straight, unbranched chain, so its repeat unit would be $-\text{CH}_2-\text{CH}(\text{CH}_2\text{CH}_2\text{CH}_3)-$ , a straight-chain propyl branch, not the branched isopropyl group shown.
C (2-methylbut-2-ene) is an internal alkene: both carbons of its double bond already carry alkyl groups, so neither backbone carbon in its polymer would be a plain, unsubstituted $-\text{CH}_2-$ . This does not match the given repeat unit.
D (2-methylbut-1-ene) puts the branch on the substituted backbone carbon itself, giving repeat unit $-\text{CH}_2-\text{C}(\text{CH}_3)(\text{CH}_2\text{CH}_3)-$ . A carbon with a methyl and an ethyl group and no hydrogen, which is different from the $-\text{CH}(\text{CH}(\text{CH}_3)_2)-$ carbon (one hydrogen, one isopropyl branch) shown in the question.

Final answer

A, 3-methylbut-1-ene, $\text{CH}_2=\text{CH}-\text{CH}(\text{CH}_3)_2$ .