(a) The table shows three pairs of monomers that are capable of polymerisation. Complete the table by identifying each type of polymerisation. [Table: Monomer pairs: HOCH2CH2OH and HO2CCH2CO2H; a dicarboxylic acid and a diol, both aromatic; CH3CHCF2 and CH3CHCH2]

1. (a)Complete the table by identifying each type of polymerisation. [Table: Monomer pairs: HOCH2CH2OH and HO2CCH2CO2H; 1,4-benzenedicarboxylic acid and 1,4-benzenediol; CH3CHCF2 and CH3CHCH2][1]
2. (b)(i)2-aminopropanoic acid, CH3CH(NH2)CO2H, can polymerise under suitable conditions. No other monomer is involved in this reaction. Draw a section of the polymer chain formed including three monomer residues. Clearly identify one repeat unit on your diagram.[3]
3. (b)(ii)2-aminopropanoic acid, CH3CH(NH2)CO2H, exists as two stereoisomers. Draw three-dimensional diagrams to show the two stereoisomers of 2-aminopropanoic acid. State the type of stereoisomerism shown.[2]
4. (c)(i)The skeletal formula of compound W is shown. [Figure shows the skeletal formula of compound W, an epoxide] When W is mixed with a second compound, called a hardener, a polymerisation reaction occurs, producing a non-solvent-based adhesive. Give the name of this type of non-solvent-based adhesive.[1]
5. (c)(ii)The hardener is a diamine. A diamine has an alkyl chain with two amine groups which are not bonded to the same carbon atom. Draw the structural formula of a compound that would make a suitable hardener.[1]

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