Polymers
11.8What the syllabus asks of you
- Define polymers as large molecules built up from many smaller molecules called monomers
- Identify the repeat units and / or linkages in addition polymers and in condensation polymers
- Deduce the structure or repeat unit of an addition polymer from a given alkene and vice versa
- Deduce the structure or repeat unit of a condensation polymer from given monomers and vice versa, limited to: (a) polyamides from a dicarboxylic acid and a diamine (b) polyesters from a dicarboxylic acid and a diol
- Describe the differences between addition and condensation polymerisation
- State that plastics are made from polymers
- Describe how the properties of plastics have implications for their disposal
- Describe the environmental challenges caused by plastics, limited to: (a) disposal in land fill sites (b) accumulation in oceans (c) formation of toxic gases from burning
- Describe and draw the structure of: (a) nylon, a polyamide O O O O O C C N N C C N N C H H H H (b) PET, a polyester O O O O C C O O C C O O The full name for PET, polyethylene terephthalate, is not required
- State that PET can be converted back into monomers and re-polymerised
- Describe proteins as natural polyamides and that they are formed from amino acid monomers with the general structure: H R O N C C H H O H where R represents different types of side-chain
- Describe and draw the structure of proteins as: H O H O N C N C N C H O
Read the notes first
Chapter 11: Polymers
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Every question on Polymers, untimed, marked as you go.
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49 questions
1May / June 2026 · Paper 11 · Q36Show mark scheme →
Anone / none
Bnone / water
Cwater / none
Dwater / water
Mark scheme
2Oct / Nov 2025 · Paper 11 · Q36Show mark scheme →
A-C(=O)-(CH2)4-C(=O)-N(H)-(CH2)6-N(H)-
B-C(=O)-(CH2)4-N(H)-C(=O)-(CH2)6-N(H)-
C-C(=O)-(CH2)6-C(=O)-N(H)-(CH2)4-N(H)-
D-C(=O)-(CH2)6-N(H)-C(=O)-(CH2)4-N(H)-
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3Oct / Nov 2025 · Paper 12 · Q36Show mark scheme →
Amonomer: CH2=CH(C2H5); type of polymerisation: addition
Bmonomer: CH2=CH(C2H5); type of polymerisation: condensation
Cmonomer: CH3-CH2-CH=CH2; type of polymerisation: addition
Dmonomer: CH3-CH2-CH=CH2; type of polymerisation: condensation
Mark scheme
4May / June 2025 · Paper 11 · Q37Show mark scheme →
AH2NCH2CH2NH2 and HOOCCH2COOH
BH2NCH2CH2NH2 and HOOCCH2CH2COOH
CH2NCH2NH2 and HOOCCH2CH2COOH
DH2NCH2NH2 and HOOCCH2COOH
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5Oct / Nov 2024 · Paper 11 · Q36Show mark scheme →
A2
B3
C4
D5
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6May / June 2024 · Paper 12 · Q37Show mark scheme →
AHO-C(=O)-[shaded box]-NH2 and HO-C(=O)-[black box]-NH2
BHO-C(=O)-[black box]-C(=O)-OH and NH2-[shaded box]-C(=O)-OH
CNH2-[black box]-C(=O)-OH and NH2-[shaded box]-NH2
DNH2-[shaded box]-NH2 and HO-C(=O)-[black box]-C(=O)-OH
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7Oct / Nov 2023 · Paper 12 · Q37Show mark scheme →
AThe repeat unit of X is [CH(CH3)CH(CH3)] and Y is a polyamide.
BThe repeat unit of X is [CH(CH3)CH(CH3)] and Y is a polyester.
CThe repeat unit of X is [CH2CH(C2H5)] and Y is a polyamide.
DThe repeat unit of X is [CH2CH(C2H5)] and Y is a polyester.
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8Oct / Nov 2022 · Paper 11 · Q39Show mark scheme →
AEvery carbon atom in the chain has one –CH3 and one hydrogen atom as side groups.
BEvery carbon atom in the chain is joined to a CH3–CH– side group.
CEvery carbon atom in the chain is joined to either two –CH3 or to two hydrogen atoms as side groups.
DEvery carbon atom in the chain is joined to hydrogen atoms only as side groups.
Mark scheme
9Oct / Nov 2022 · Paper 11 · Q40Show mark scheme →
Amonomers: HO–C(=O)–[grey square]–NH2; polymer: –C(=O)–[grey square]–C(=O)–N(H)–[grey square]–N(H)–
Bmonomers: H–C(H)=C(H)–C(H)=C(H)–H; polymer: –C(H)–C(H)–C(H)–C(H)–C(H)–C(H)– with H substituents
Cmonomers: HO–C(=O)–[grey square]–C(=O)–OH and H2N–[white square]–NH2; polymer: –O–C(=O)–[grey square]–C(=O)–O–N(H)–[white square]–N(H)–
Dmonomers: HO–C(=O)–[grey square]–C(=O)–OH and HO–[white square]–OH; polymer: –C(=O)–[grey square]–C(=O)–O–[white square]–O–
Mark scheme
10Oct / Nov 2022 · Paper 12 · Q39Show mark scheme →
Anylon: C, H; poly(ethene): C, H, O; Terylene: C, H, N, O
Bnylon: C, H, N, O; poly(ethene): C, H; Terylene: C, H, N, O
Cnylon: C, H, O; poly(ethene): C, H, N; Terylene: C, H, O
Dnylon: C, H, N, O; poly(ethene): C, H; Terylene: C, H, O
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11Oct / Nov 2022 · Paper 12 · Q40Show mark scheme →
A1 only
B1 and 2
C1 and 3
D2 and 3
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12May / June 2022 · Paper 11 · Q39Show mark scheme →
A1 addition, 2 condensation, 3 amide, 4 ester, 5 amino acids
B1 addition, 2 condensation, 3 amide, 4 ester, 5 simple sugars
C1 addition, 2 condensation, 3 ester, 4 amide, 5 amino acids
D1 condensation, 2 addition, 3 ester, 4 amide, 5 simple sugars
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13May / June 2022 · Paper 12 · Q39Show mark scheme →
AComplex carbohydrates, such as starch, are hydrolysed to give simple sugars.
BFats have the same amide linkages as Terylene.
CProteins and nylon are polymers formed from the same monomers but with different linkages.
DProteins are natural polymers and are also called polysaccharides.
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14May / June 2022 · Paper 12 · Q40Also set in May / June 2022 Paper 11, May / June 2017 Paper 11Show mark scheme →
AH-C(CH3)=C(Cl)-Br
BCH3-C(Cl)=C(H)-Br
CCH3-C(Br)=C(H)-Cl
DCl-C(H)=C(CH3)-Br
Mark scheme
15Oct / Nov 2021 · Paper 12 · Q40Also set in Oct / Nov 2021 Paper 11Show mark scheme →
A-C(=O)-(CH2)4-C(=O)-O-(CH2)6-O-
B-C(=O)-(CH2)6-C(=O)-N(H)-(CH2)6-N(H)-
C-C(=O)-(CH2)4-C(=O)-N(H)-(CH2)6-N(H)-
D-C(CH3)(H)-C(C3H7)(H)-C(CH3)(H)-C(C3H7)(H)-
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16May / June 2021 · Paper 11 · Q40Show mark scheme →
Aa complex carbohydrate
Ba fat
Ca protein
Dan addition polymer
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17May / June 2021 · Paper 12 · Q39Also set in May / June 2021 Paper 11Show mark scheme →
ANylon and Terylene are both polyesters.
BProteins and nylon have the same monomer units.
CProteins have the same amide linkages as nylon.
DTerylene and fats are esters but with different linkages.
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18May / June 2021 · Paper 12 · Q40Show mark scheme →
Aa polyester
Ba protein
Cnylon
Dstarch
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19Oct / Nov 2020 · Paper 11 · Q39Show mark scheme →
Apolyamide
Bpolyester
Cpoly(ethene)
Dpolysaccharide
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20Oct / Nov 2020 · Paper 12 · Q39Show mark scheme →
Anylon
Bpoly(ethene)
Cstarch
DTerylene
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21Oct / Nov 2020 · Paper 12 · Q40Also set in Oct / Nov 2020 Paper 11Show mark scheme →
Amonomer: CH2=CH(C2H5); type of polymerisation: addition
Bmonomer: CH2=CH(C2H5); type of polymerisation: condensation
Cmonomer: CH3-CH2-CH=CH(CH3); type of polymerisation: addition
Dmonomer: CH3-CH2-CH=CH(CH3); type of polymerisation: condensation
Mark scheme
22May / June 2020 · Paper 11 · Q38Show mark scheme →
Aaddition
Bcondensation
Ccracking
Desterification
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23May / June 2020 · Paper 11 · Q39Show mark scheme →
A1, 2 and 3
B1 and 3 only
C1 and 4 only
D2, 3 and 4
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24May / June 2020 · Paper 11 · Q40Show mark scheme →
ANylon and Terylene are produced by addition polymerisation.
BNylon and Terylene both contain the amide linkages.
CSimple sugars are produced by hydrolysing proteins.
DStarch contains the elements carbon, hydrogen and oxygen.
Mark scheme
25May / June 2020 · Paper 12 · Q39Show mark scheme →
APolymer chain with repeating unit -C(H)(H)-C(F)(Cl)-
BPolymer chain with repeating unit -C(Cl)(H)-C(F)(H)-
CPolymer chain with repeating unit -C(H)(Cl)-C(F)(H)-
DPolymer chain with repeating unit -C(F)(H)-C(Cl)(H)-
Mark scheme
26May / June 2020 · Paper 12 · Q40Show mark scheme →
ANylon and Terylene are produced by addition polymerisation.
BNylon and Terylene both contain the amide linkages.
CSimple sugars are produced by hydrolysing proteins.
DStarch contains the elements carbon, hydrogen and oxygen.
Mark scheme
27Oct / Nov 2019 · Paper 11 · Q39Show mark scheme →
AH2C=CH-CH2-CH2-CH3 (structure A)
BH2C=CH-CH2-CH3 (structure B)
CH2C=CH2 (structure C)
DH3C-CH=CH-CH3 (structure D)
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28Oct / Nov 2019 · Paper 11 · Q40Show mark scheme →
AFats and nylons all contain the C(=O)-O linkage.
BMonomers used in condensation polymerisation must contain both –CO2H and –OH groups.
CPoly(ethene) will decolourise bromine.
DProteins with the C(=O)-N(H) linkage are biodegradable as they can be hydrolysed.
Mark scheme
29Oct / Nov 2019 · Paper 12 · Q36Show mark scheme →
Apolyalkene
Bpolyamide
Cpolyester
Dprotein
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30May / June 2019 · Paper 11 · Q39Show mark scheme →
Afat
Bprotein
Cstarch
DTerylene
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31May / June 2019 · Paper 11 · Q40Show mark scheme →
Aaddition; partial structure with alternating shaded and H substituents
Baddition; partial structure with shaded substituents on alternating carbons
Ccondensation; partial structure with shaded substituents on alternating carbons
Dcondensation; partial structure with shaded substituents on every carbon
Mark scheme
32May / June 2019 · Paper 12 · Q39Show mark scheme →
AH-C(H)(H)-C(Cl)(H)-H
BCl-C(H)(H)-C(Cl)(H)-H
CH-C(H)=C(Cl)-H
DCl-C(H)=C(H)-Cl
Mark scheme
33May / June 2019 · Paper 12 · Q40Show mark scheme →
Aan acid and an alcohol | addition
Ban acid and an alcohol | condensation
Can alkene | addition
Dan alkene | condensation
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34Oct / Nov 2018 · Paper 11 · Q39Show mark scheme →
ANylon contains oxygen atoms but not nitrogen atoms.
BProteins contain both nitrogen atoms and oxygen atoms.
CTerylene contains nitrogen atoms.
DThe polymer used to make clingfilm contains oxygen atoms.
Mark scheme
35Oct / Nov 2018 · Paper 11 · Q40Show mark scheme →
AH-C(H)(C6H5)-C(H)(H)-H
BH-C(H)=C(C6H5)-H
CH-C(H)(C6H5)-C(H)(H)-C(H)(C6H5)-H
DC6H5-C(H)=C(H)-C6H5
Mark scheme
36Oct / Nov 2018 · Paper 12 · Q39Show mark scheme →
A1 and 3
B1 and 4
C2 and 3
D2 and 4
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37Oct / Nov 2018 · Paper 12 · Q40Show mark scheme →
AP: fat, Q: nylon
BP: fat, Q: Terylene
CP: nylon, Q: protein
DP: protein, Q: Terylene
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38May / June 2018 · Paper 11 · Q40Show mark scheme →
AC(=O)-[shaded box]-C(=O)-O-[white box]-O-C(=O)-[shaded box]-C(=O)-O-[white box]-O
BC(=O)-[shaded box]-C(=O)-N(H)-[black box]-N(H)-C(=O)-[crosshatched box]-C(=O)-N(H)-[grey box]-N(H)
CC(=O)-[shaded box]-C(=O)-N(H)-[white box]-N(H)-C(=O)-[shaded box]-C(=O)-N(H)-[white box]-N(H)
DC(=O)-[shaded box]-C(=O)-O-[black box]-O-C(=O)-[crosshatched box]-C(=O)-O-[grey box]-O
Mark scheme
39May / June 2018 · Paper 12 · Q40Show mark scheme →
AHO2C–(CH2)6–CO2H and H2N–(CH2)6–NH2
BHO2C–(CH2)4–CO2H and H2N–(CH2)4–NH2
CHO2C–(CH2)4–CO2H and H2N–(CH2)6–NH2
DHO2C–(CH2)6–CO2H and H2N–(CH2)4–NH2
Mark scheme
40Oct / Nov 2017 · Paper 11 · Q40Show mark scheme →
Amonomer: CH2=CH(C2H5); type of polymerisation: addition
Bmonomer: CH2=CH(C2H5); type of polymerisation: condensation
Cmonomer: CH3-CH=CH2; type of polymerisation: addition
Dmonomer: CH3-CH=CH2; type of polymerisation: condensation
Mark scheme
41Oct / Nov 2017 · Paper 12 · Q40Show mark scheme →
ANylon and Terylene are produced by addition polymerisation.
BNylon and Terylene both contain the amide linkage.
CSimple sugars can be produced by hydrolysing proteins.
DStarch contains the elements carbon, hydrogen and oxygen.
Mark scheme
42May / June 2017 · Paper 12 · Q39Also set in May / June 2017 Paper 11Show mark scheme →
Aprotein
Bpoly(ethene)
Cpoly(propene)
Dstarch
Mark scheme
43May / June 2017 · Paper 12 · Q40Also set in May / June 2017 Paper 11Show mark scheme →
ANylon and Terylene are both polyesters.
BProteins and nylon have the same monomer units.
CProteins have the same amide linkages as nylon.
DTerylene and fats are esters but with different linkages.
Mark scheme
44Oct / Nov 2016 · Paper 11 · Q40Show mark scheme →
Aa fat
Ba nylon
Ca polyester
Da protein
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45Oct / Nov 2016 · Paper 12 · Q33Show mark scheme →
ACH₃CH=CH₂
BCH₃CH₂CH=CH₂
CCH₃CH₂CH=CHCH₃
DCH₃CH=CHCH₃
Mark scheme
46Oct / Nov 2016 · Paper 12 · Q40Show mark scheme →
Aan addition polymer
Ba carbohydrate
Ca polyamide
Da polyester
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47May / June 2016 · Paper 11 · Q35Show mark scheme →
Acarbohydrates
Bfats
Cproteins
Dsugars
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48May / June 2016 · Paper 12 · Q37Also set in May / June 2016 Paper 11Show mark scheme →
A1 and 3
B1 and 4
C2 and 3
D2 and 4
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49May / June 2016 · Paper 12 · Q38Show mark scheme →
AThe monomer is [structure: Cl-C(H)=C(Cl)-H].
BThe monomer is [structure: H-C(Cl)=C(Cl)-H].
CThe polymer is formed by a condensation reaction.
DThe polymer has a lower melting point than the monomer.
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