Topic 19 · page 8 of 13

Addition polymers

In addition polymerisation the C=C bond of the monomer opens and nothing is lost. Poly(propenamide) and poly(ethenol) follow the same pattern as poly(ethene). Draw the two chain carbons in brackets, with the side group attached.

10 min read19.8(ii)19.9

Read first: Polyamides and proteins

Addition polymerisation 19.8(ii)

You met addition polymers in Topic 5. A monomer with a C=C double bond opens that bond, and the monomer units join one after another into a long saturated chain.

Poly(propenamide) is made from propenamide, CH₂=CHCONH₂. The side group is the amide group $\ce{-CONH2}$ :

n\,\ce{CH2=CHCONH2} \rightarrow \text{poly(propenamide)}

Poly(ethenol) has the repeat unit –[CH₂–CH(OH)]– and the same pattern, with –OH as the side group. The name comes from the monomer, ethenol, CH₂=CHOH.

Addition polymerisation: propenamide, CH2=CHCONH2, opens its C=C bond to give poly(propenamide) with repeat unit CH2-CH(CONH2). Ethenol, CH2=CHOH, would give poly(ethenol) with repeat unit CH2-CH(OH). Only the two carbon atoms of the chain are inside the brackets. — Each monomer opens its C=C bond. The repeat unit has the two chain carbons inside the brackets, and the side group stays on the carbon it was on.

Drawing the repeat unit 19.9

MethodRepeat unit of an addition polymer

Draw the monomer with the C=C in the middle and groups above and below.
Change the C=C into a single bond and add an extension bond at each end.
Put the two chain carbons inside the brackets, with their side groups, and n outside.

For poly(propenamide) the repeat unit is $\ce{-[CH2-CH(CONH2)]-}$ . For poly(ethenol) it is $\ce{-[CH2-CH(OH)]-}$ .

To go backwards, from repeat unit to monomer, take the two chain carbons, remove the extension bonds, and put the double bond back between them. The relative formula mass of the repeat unit equals that of the monomer: propenamide, C₃H₅NO, has $M_\text{r}$ = 71.0, and ethenol, C₂H₄O, has $M_\text{r}$ = 44.0.

Addition or condensation?

Look at the backbone of the repeat unit.

	Addition polymer	Condensation polymer
backbone	all carbon, saturated	contains amide (or ester) links
small molecule lost	none	yes: H₂O or HCl
monomer	alkene, one C=C	two functional groups, one at each end
example	poly(propenamide), poly(ethenol)	nylon-6,6, protein

Worked exampleFrom repeat unit to monomer and formula mass

A chain of poly(ethenol) has 250 repeat units. Give the monomer and the relative molecular mass of the chain.

Repeat unit: –[CH₂–CH(OH)]–. Put the double bond back: CH₂=CHOH, which is ethenol.
$M_\text{r}$ of one repeat unit = 2 × 12.0 + 4 × 1.0 + 16.0 = 44.0
$M_\text{r}$ of the chain = 250 × 44.0 = 11 000

Check yourself

Give the repeat unit of poly(propenamide).
Name the monomer that gives –[CH₂–CH(OH)]– and state what is formed from it in addition polymerisation.
Explain why no small molecule is lost in addition polymerisation.
How can you tell from the repeat unit whether a polymer is an addition or a condensation polymer?
Write the empirical formula of poly(propenamide).
A sample of poly(propenamide) has 600 repeat units per chain. Calculate its $M_\text{r}$ .

Answers cover these until you have tried the questions

$\ce{-[CH2-CH(CONH2)]-}$ , with extension bonds through the brackets.
Ethenol, CH₂=CHOH. It gives poly(ethenol).
The monomer's C=C opens to make the bonds in the chain; every atom of the monomer ends up in the polymer, and nothing is left over.
An addition polymer has an all-carbon, saturated backbone. A condensation polymer has links such as amide (–CO–NH–) or ester (–CO–O–) in the backbone.
The empirical formula is the same as the monomer's formula, C₃H₅NO.
$M_\text{r}$ (repeat unit) = 3 × 12.0 + 5 × 1.0 + 14.0 + 16.0 = 71.0. Chain: 600 × 71.0 = 42 600.