Polymerisation (A Level)

• A condensation link is an ester or an amide made by losing H₂O or HCl; the repeat unit has open bonds at both ends; and hydrolysis adds water back across every link, whichever small molecule the link lost when it formed.
• Two groups per monomer; one small molecule per link; open bonds at both ends. In the chain, condensation; off the chain, addition. Acid frees the acid, alkali frees the amine.

• 35.1.1 I can describe — describe the formation of polyesters: (a) the reaction between a diol and a dicarboxylic acid or dioyl chloride (b) the reaction of a hydroxycarboxylic acid
• 35.1.2 I can describe — describe the formation of polyamides: (a) the reaction between a diamine and a dicarboxylic acid or dioyl chloride (b) the reaction of an aminocarboxylic acid (c) the reaction between amino acids
• 35.1.3 I can deduce — deduce the repeat unit of a condensation polymer obtained from a given monomer or pair of monomers
• 35.1.4 I can identify — identify the monomer(s) present in a given section of a condensation polymer molecule
• 35.2.1 I can predict — predict the type of polymerisation reaction for a given monomer or pair of monomers
• 35.2.2 I can deduce — deduce the type of polymerisation reaction which produces a given section of a polymer molecule
• 35.3.1 I can — recognise that poly(alkenes) are chemically inert and can therefore be difficult to biodegrade
• 35.3.2 I can — recognise that some polymers can be degraded by the action of light
• 35.3.3 I can — recognise that polyesters and polyamides are biodegradable by acidic and alkaline hydrolysis

Units and significant figures are marked. The syllabus states that failure to quote units, the inclusion of units in quantities defined as ratios, and answers given to an inappropriate number of significant figures are all liable to be penalised. Give a calculated answer to the same number of significant figures as the least precise data, or one more; keep full precision in the working and round only at the end. A fifth of the qualification is experimental: Papers 3 and 5 test AO3 only, and their questions may be set in contexts outside the syllabus content, so the practical work in this chapter is set out as procedure, recording and evaluation rather than as theory.

• “The repeat unit of Terylene is HO–CH₂CH₂–O–CO–C₆H₄–COOH.” Correct The repeat unit has open bonds at both ends, not –OH and –COOH: [–O–CH₂CH₂–O–CO–C₆H₄–CO–]ₙ. The reacting groups have become links. And count the small molecules per link: two different monomers make two links per unit, so 2n H₂O.
• “Nylon-6,6 and nylon-6 have the same repeat unit.” Correct Nylon-6 comes from one monomer: [–NH–(CH₂)₅–CO–]ₙ, one amide link and six carbons per unit. Nylon-6,6 comes from two: [–NH–(CH₂)₆–NH–CO–(CH₂)₄–CO–]ₙ, two links and twelve carbons.
• “It contains ester groups, so it is a polyester.” Correct Ask where the ester is. In the backbone: condensation. Hanging off an all-carbon backbone, as in poly(ethenyl ethanoate): addition, because the ester was already in the monomer.
• “The dioyl chloride route also eliminates water.” Correct A –COCl loses Cl, not OH, so each link eliminates HCl, and it happens at room temperature. The repeat unit is the same as from the diacid.
• “Acid hydrolysis of nylon gives the diamine and the diacid.” Correct In acid the amine comes out protonated, as its ammonium salt; in alkali the acid comes out as its carboxylate salt. Only the diol of a polyester is free in both media.
• “Monomers are found by cutting the polymer into equal lengths.” Correct Cut at each ester or amide link, between the carbonyl carbon and the O or N, and count carbons between links. A section that starts mid-unit makes the first fragment look shorter than it is.
• “Terylene is made by addition polymerisation.” Repair Condensation: a diol and a dicarboxylic acid (or dioyl chloride) form ester links, eliminating water (or HCl).
• A repeat unit drawn with –COOH and –OH still on the ends. Repair The reacting groups have become the links; the ends of a repeat unit are open bonds, not intact functional groups.
• “One molecule of water is lost per repeat unit of nylon-6,6.” Repair Two: one at each of the two amide links formed per unit. Nylon-6 and poly(lactic acid) lose one per unit.
• Nylon-6,6 drawn as –NH–(CH₂)₆–CO– repeating. Repair That is a single-monomer polyamide (the unit of nylon-7). Nylon-6,6 alternates –NH–(CH₂)₆–NH– with –CO–(CH₂)₄–CO–.
• “The dioyl chloride route gives off water.” Repair A –COCl loses Cl, so each link eliminates HCl, at room temperature. Same repeat unit, different small molecule.
• “The monomers of a polyester are found by cutting the C–C bonds.” Repair Cut the ester link between the carbonyl carbon and the oxygen, then add OH to the carbonyl and H to the oxygen.
• “The section starts –CH₂–CO–NH–, so one monomer is a two-carbon acid.” Repair A section can start mid-unit. Count the carbons between two links, not from the edge of the drawing.
• “Poly(ethenyl ethanoate) is a polyester.” Repair Its ester groups are side-chains; the backbone is all carbon, so it is an addition polymer of CH₂=CH–O–CO–CH₃.
• “Ethanoic acid and ethanol polymerise to a polyester.” Repair Each has only one reacting group, so one ester forms and the chain stops; a polymer needs two groups per monomer.
• “Poly(alkene)s biodegrade slowly because they are large.” Repair Because they are chemically inert: strong, non-polar C–C and C–H bonds with nothing for water, acid, alkali or enzymes to attack.
• “Polyesters are biodegradable because bacteria eat them.” Repair Because their ester links are hydrolysed, by acid, alkali or enzymes, breaking the chain into small soluble molecules.
• Acid hydrolysis of a polyamide giving the free diamine. Repair In acid the amine is protonated and appears as its ammonium salt; the free diamine is the product of alkaline hydrolysis.
• Alkaline hydrolysis of a polyester giving the free dicarboxylic acid. Repair In alkali the acid appears as its carboxylate salt; the diol is free in either medium.

• Read the command word before you decide how much to write. This syllabus has twenty-two of them: analyse, calculate, compare, consider, contrast, deduce, define, demonstrate, describe, determine, discuss, evaluate, examine, explain, give, identify, justify, predict, show (that), sketch, state and suggest. Comment, estimate, name and outline are not among them: where a question wants something named it says identify, which the syllabus glosses as “name/select/recognise”. State and give want a fact and nothing more. Describe wants the points or the features. Explain wants the reasons and the relationships — a describe-level answer to an explain question is incomplete however well written it is. Deduce and determine want a conclusion reached from the information given, with the reasoning visible.
• Interleave with the chapters that use this one. Topic 36 (organic synthesis) puts polymers inside multi-step routes: when you reach it, re-answer “which monomers, which link, which small molecule?” for every polymer a route produces. Topic 34’s amide and peptide chemistry and topic 18’s ester hydrolysis are the same reactions as section I: revisit that section when you revise either. Recalling a topic inside a new context is worth more than another pass over this chapter on its own; at A Level, Paper 4 assumes the whole of the AS content, so nothing here is ever finished with.

• Cambridge International AS & A Level Chemistry 9701 has five components. Topic 35 is A Level content, so it is examined in Papers 4 and 5. A Level content: examined in Paper 4 (A Level structured, which also requires the AS content) and, as practical context, Paper 5. AS Level candidates take Papers 1, 2 and 3; A Level candidates take all five, either staged over two years (Papers 1–3 in year one, Papers 4 and 5 in year two) or together in one series. Examinations are available in the June and November series, and in March in India.
• Across both the AS Level and the A Level the assessment objectives are weighted AO1 40% (knowledge and understanding), AO2 40% (handling, applying and evaluating information) and AO3 20% (experimental skills and investigations). AS candidates are graded a–e; A Level candidates A*–E.
• There is no multiple-choice paper on A Level content. A Paper 4 question on this topic asks you to deduce a repeat unit from given monomers, identify the monomers in a given section, predict or deduce the type of polymerisation, describe a formation with its small molecule, or explain why one polymer biodegrades and another does not. Because topic 36 synthesis questions may use any reaction in the syllabus, polymerisation can also appear inside them.
• The numbers are small: the formula and Mᵣ of a repeat unit (the monomers minus the small molecules), the mass of water or HCl eliminated, and the number of repeat units in a chain from its Mᵣ. Every Aᵣ comes from the Periodic Table in the Data section; nothing has to be recalled.
• The interfacial preparation of nylon-6,6 is an observation-and-hazard context. The alkaline hydrolysis of a polyester, followed by loss of mass, is a natural Paper 5 plan: variables, a control, the balance precision, drying to constant mass and the percentage error of a small difference.
• Read the command word before you decide how much to write. This syllabus has twenty-two of them: analyse, calculate, compare, consider, contrast, deduce, define, demonstrate, describe, determine, discuss, evaluate, examine, explain, give, identify, justify, predict, show (that), sketch, state and suggest. Comment, estimate, name and outline are not among them: where a question wants something named it says identify, which the syllabus glosses as “name/select/recognise”. State and give want a fact and nothing more. Describe wants the points or the features. Explain wants the reasons and the relationships — a describe-level answer to an explain question is incomplete however well written it is. Deduce and determine want a conclusion reached from the information given, with the reasoning visible.