Polymers Worked ExamplesIGCSE Chemistry 0620

Plastics are made from polymers. Most, like poly(ethene), are unreactive and non-biodegradable: microorganisms cannot break them down. Once thrown away they stay in the environment for a very long time. The answer is B.

A is a correct statement about all polymers, but it does not explain a disposal problem. C describes why the plastic is easy to shape (and can be recycled by melting), not a problem. D is true but has nothing to do with disposal.

(a) Read the landfill bar: $45\%$.

(b) Read the oceans and rivers bar: $8\%$.

(c)(i) Plastics are non-biodegradable, so they do not decompose in landfill: the sites fill up and more land is used up for waste.

(ii) Plastics accumulate in the oceans because they are not broken down; sea animals and birds swallow them or become trapped, and are harmed or killed.

(iii) Burning plastics forms toxic gases — for example, incomplete combustion makes carbon monoxide. (Burning also releases carbon dioxide, a greenhouse gas.)

(a) Addition polymerisation needs a monomer with a $\text{C}=\text{C}$ double bond — an alkene (the name ends in -ene). The only alkene in the list is propene, $\text{CH}_2=\text{CHCH}_3$. Propane (ends in -ane) is saturated and cannot polymerise.

(b) Put the monomer's name in brackets after poly: poly(propene).

(c) Only the two carbons of the double bond go into the chain. The $\text{CH}_3$ group is a side group hanging off the chain. Draw both carbons with single bonds, continuation bonds out of both sides, brackets and $n$.

(d) The double bond opens (breaks to a single bond); each carbon then forms a new single bond to a neighbouring monomer. Only one product forms — the polymer.

Method: keep the two $\text{C}=\text{C}$ carbons as the backbone, change the double bond to a single bond, keep every group attached to them in the same place, and add a continuation bond on each side.

(a) F has one H and one $\text{CH}_3$ on each carbon, so both backbone carbons carry a $\text{CH}_3$ side group. The backbone is still only two carbons long.

(b) In G one carbon carries two H atoms and the other carries one H and the $\text{CH}_2\text{OH}$ group. The $\text{CH}_2\text{OH}$ stays as a side group, bonded to the backbone through its carbon.

(a) In an addition polymer the repeat unit is always two backbone carbons (one monomer). Here every carbon has one H and one Cl, so the repeat unit is $-\text{CHCl}-\text{CHCl}-$.

(b) Each monomer adds two carbons to the chain. The section has $6$ backbone carbons, so it was made from $6 \div 2 = 3$ monomer molecules.

(c) Put the double bond back between the two carbons of the repeat unit and remove the continuation bonds. The monomer has one H and one Cl on each carbon.

PET is a polyester: it is made by condensation polymerisation of a dicarboxylic acid and a diol, and its monomers are joined by ester linkages, $-\text{COO}-$. Row C is correct.

A gives the wrong linkage for PET — amide linkages are in nylon. B is wrong because nylon is made by condensation, not addition (water is lost at each linkage). D is wrong because proteins are natural polyamides formed by condensation of amino acids.

A fact that is often tested alongside this: PET can be converted back into its monomers and re-polymerised, which allows it to be recycled.

(a) Two different monomers join and a small molecule is lost at each link: this is condensation polymerisation. The polymer is a polyester.

(b) The $-\text{OH}$ from a $-\text{COOH}$ group and the $\text{H}$ from an alcohol $-\text{OH}$ group join to form water, $\text{H}_2\text{O}$.

(c) The acid's carbonyl carbon bonds to the diol's oxygen, making an ester linkage $-\text{C}(=\text{O})-\text{O}-$ at each end of the blue block. The repeat unit contains one acid block and one diol block, with continuation bonds at both ends.

Check: the repeat unit contains two ester linkages, so two water molecules are lost for every repeat unit formed.

(a) Each block is joined to the next through $-\text{C}(=\text{O})-\text{N}(\text{H})-$: this is the amide linkage.

(b) The pattern repeats after one blue block and one red block: $-\text{NH}-$blue$-\text{NH}-\text{CO}-$red$-\text{CO}-$. The section shows two of these.

(c) Split every amide link and add back water ($\text{H}$ to N, $\text{OH}$ to C). The blue block had an $-\text{NH}_2$ at each end: a diamine. The red block had a $-\text{COOH}$ at each end: a dicarboxylic acid.

An amide linkage, $-\text{CO}-\text{NH}-$, forms between a $-\text{COOH}$ group and an $-\text{NH}_2$ group. So a polyamide needs a dicarboxylic acid and a diamine.

This is condensation polymerisation, so a small molecule — water — is lost at every linkage. The answer is B.

A makes a polyester (ester links). C has the right monomers but condensation always gives a second product. D has no carboxylic acid group, so no amide link can form.

(a) Each unit is joined to the next through $-\text{C}(=\text{O})-\text{N}(\text{H})-$, the amide linkage. Proteins are natural polyamides.

(b) Split the chain into amino acid units, each $-\text{NH}-\text{CH}(\text{R})-\text{CO}-$. There are five units, with side-chains H, $\text{CH}_3$, H, $\text{CH}_2\text{OH}$, $\text{CH}_3$. Different amino acids have different side-chains, and only three different side-chains appear (H, $\text{CH}_3$, $\text{CH}_2\text{OH}$), so 3 different amino acids were used.

(c) One water molecule is eliminated for each amide linkage formed. Count the $\text{C}(=\text{O})-\text{N}$ bonds between units in the diagram: there are 4 (five units need four links; the bonds at the two ends lead to units not shown).

(d) An amino acid has an amine group, $-\text{NH}_2$, and a carboxylic acid group, $-\text{COOH}$, on the same carbon, which also carries an H and the side-chain R. Display both functional groups fully.

(a) In addition polymerisation the polymer is the only product, so the polymer's mass is simply $n$ times the monomer's. $M_r(\text{C}_3\text{H}_6) = 3(12) + 6(1) = 42$.

(b)(i) Dicarboxylic acid $\text{C}_6\text{H}_{10}\text{O}_4$: $6(12) + 10(1) + 4(16) = 146$. Diamine $\text{C}_6\text{H}_{16}\text{N}_2$: $6(12) + 16(1) + 2(14) = 116$.

(ii) One repeat unit forms from one of each monomer, with two amide linkages, so two water molecules ($M_r = 18$) are lost.

(iii) Divide the molecule's $M_r$ by the repeat unit's $M_r$ (ignoring the tiny end groups).

(iv) $0.500$ mol of each monomer makes (almost exactly) $0.500$ mol of repeat units, and each repeat unit releases $2$ water molecules.

(a) P has a backbone of carbon atoms only, with no linkage groups: it is an addition polymer. Q contains $-\text{C}(=\text{O})-\text{O}-$ ester linkages and R contains $-\text{C}(=\text{O})-\text{N}(\text{H})-$ amide linkages: both are condensation polymers.

(b) P's repeat unit is $-\text{CH}_2-\text{CHCl}-$. Put the double bond back between these two carbons: the monomer is chloroethene, $\text{CH}_2=\text{CHCl}$ (one carbon with two H, the other with one H and one Cl).

(c) Q is a polyester: a dicarboxylic acid (blue block with $-\text{COOH}$ at both ends) and a diol (red block with $-\text{OH}$ at both ends).

(d) Addition: the monomers are alkenes (contain $\text{C}=\text{C}$), usually of one kind, and the polymer is the only product. Condensation: the monomers have two functional groups each (often two different monomers), and a small molecule such as water is also formed at every linkage.