Amines and polymers worksheet answers for A Level chemistry (free PDF download)

Problem 1 Classifying four amines

1. CH₃CH₂NH₂ is primary (one carbon group on N); (CH₃)₂NH is secondary; (CH₃)₃N is tertiary; C₆H₅NH₂ is primary (an aryl group is one carbon group).
2. First bottle: ethylamine; last bottle: phenylamine.

Problem 2 Why an amine accepts a proton

1. The lone pair on nitrogen forms a dative bond with H⁺.
2. CH₃NH₂ + H⁺ → CH₃NH₃⁺.
3. CH₃NH₂ + HCl → CH₃NH₃⁺Cl⁻, the salt is methylammonium chloride.

Problem 3 Glycine in a sports drink

1. A is the ammonium (amine) group, –NH₃⁺; B is the carboxylate group, –COO⁻.
2. C₂H₅NO₂: 2 × 12.0 + 5 × 1.0 + 14.0 + 2 × 16.0 = 24.0 + 5.0 + 14.0 + 32.0 = 75.0.
3. +1 and −1 cancel, so the overall charge is zero.

Problem 4 Addition or condensation

1. Nylon-6,6 and PET are condensation polymers; poly(propene) is an addition polymer.
2. PET forms with loss of water.
3. Nylon-6,6 has an amide link; PET has an ester link.

Problem 5 Reading base pairs

1. X is the A–T pair with 2 hydrogen bonds; Y is the G–C pair with 3 hydrogen bonds.
2. The drawing at Y shows four dotted lines instead of three.
3. The complementary strand is TACGGCAT. Pairs: A–T, T–A, G–C, C–G, C–G, G–C, T–A, A–T. A–T pairs = 4 × 2 = 8; G–C pairs = 4 × 3 = 12; total = 20 hydrogen bonds.

Problem 6 Ranking four bases

1. Strongest to weakest (smallest pKb first): dimethylamine, methylamine, ammonia, phenylamine.
2. Kb = 10⁻⁹·¹³ = 10⁻⁹ × 10⁻⁰·¹³ = 0.741 × 10⁻⁹ = 7.4 × 10⁻¹⁰ mol/dm³.
3. The lone pair on nitrogen in phenylamine is partly delocalised into the benzene ring, so it is less available to accept a proton.

Problem 7 pH of a methylamine solution

1. Kb = [CH₃NH₃⁺][OH⁻] ÷ [CH₃NH₂].
2. [OH⁻] = √(4.6 × 10⁻⁴ × 0.20) = √(9.2 × 10⁻⁵) = 9.6 × 10⁻³ mol/dm³.
3. pOH = −log(9.6 × 10⁻³) = 2.02; pH = 14.00 − 2.02 = 11.98.

Problem 8 Titrating propylamine

1. n(HCl) = 0.100 × 0.0184 = 1.84 × 10⁻³ mol.
2. The ratio is 1 : 1, so c = 1.84 × 10⁻³ ÷ 0.0250 = 0.0736 mol/dm³.
3. 0.0736 × 59.0 = 4.34 g/dm³.

Problem 9 Making ethylamine from bromoethane

1. CH₃CH₂Br + 2NH₃ → CH₃CH₂NH₂ + NH₄Br (H: 5 + 6 = 11 on the left and 5 + 2 + 4 = 11 on the right; N 2 and Br 1 on both sides).
2. Nucleophilic substitution, with NH₃ as the nucleophile.
3. The product amine is also a nucleophile and would attack more bromoethane to give secondary and tertiary amines. A large excess of NH₃ makes it far more likely that bromoethane meets NH₃, so mostly the primary amine forms; the second NH₃ also removes the HBr as NH₄Br.

Problem 10 Forms of alanine

1. Strong acid: CH₃CH(NH₃⁺)COOH, charge +1.
2. Strong alkali: CH₃CH(NH₂)COO⁻, charge −1.
3. The acidic species (+1) moves towards the cathode (negative electrode); the alkaline species (−1) moves towards the anode (positive electrode).

Problem 11 Two possible dipeptides

1. The –NH₂ of one acid reacts with the –COOH of the other, giving –CO–NH–, the peptide (amide) link, with loss of water.
2. Two dipeptides, Gly-Ala and Ala-Gly, because either amino acid can supply the –COOH group.
3. Mr = 75 + 89 − 18 = 146.

Problem 12 Cutting nylon into monomers

1. The amide link is –CO–NH– (the C=O carbon bonded to N).
2. Hydrolysis gives H₂N(CH₂)₆NH₂ (hexane-1,6-diamine) and HOOC(CH₂)₄COOH (hexanedioic acid).
3. With aqueous HCl and heat the amide links hydrolyse and the polymer breaks into the diacid and the diammonium chloride salt, because the amine is protonated.

Problem 13 Bottles from ethane-1,2-diol

1. 166 + 62 − 2 × 18 = 228 − 36 = 192.
2. n(repeat units) = 1000 ÷ 192 = 5.208 mol; m(diol) = 5.208 × 62 = 323 g.
3. n(H₂O) = 2 × 5.208 = 10.42 mol; m = 10.42 × 18 = 188 g. Check: acid 865 g + diol 323 g − water 188 g = 1000 g.

Problem 14 Breaking down polymers

1. Hydrolysis with NaOH gives the disodium salt of the diacid and the diol; acid hydrolysis gives the free diacid and diol.
2. Poly(ethene) has a chain of strong, non-polar C–C and C–H bonds with no ester or amide link for water to attack, so hydrolysis does not occur.
3. Monomers can be recovered and reused, and the polymer breaks down in the environment more readily.

Problem 15 Why Kevlar is so strong

1. The link is an amide link, and HCl is lost.
2. Kevlar has hydrogen bonds between the N–H and C=O groups of neighbouring chains, much stronger than the weak London forces between poly(ethene) chains.
3. The benzene rings make the chain stiff and straight, so chains line up in sheets and many hydrogen bonds act along the length, which gives high tensile strength.

Problem 16 Identifying an amino acid

1. %O = 100 − (40.4 + 7.9 + 15.7) = 36.0 %.
2. Moles per 100 g: C 3.37, H 7.9, N 1.12, O 2.25. Dividing by 1.12 gives C 3.0 : H 7.05 : N 1 : O 2.0, so the empirical formula is C₃H₇NO₂ (Mr = 89).
3. The empirical mass is 89, equal to the Mr, so the molecular formula is C₃H₇NO₂: alanine (2-aminopropanoic acid).

Problem 17 Glycine across the pH range

1. A (pH below 2.34): ⁺H₃NCH₂COOH (cation, +1); B: ⁺H₃NCH₂COO⁻ (zwitterion, 0); C (pH above 9.60): H₂NCH₂COO⁻ (anion, −1).
2. pI = (2.34 + 9.60) ÷ 2 = 11.94 ÷ 2 = 5.97.
3. At pH 12, glycine is in region C, so it is an anion and moves towards the anode.

Problem 18 Yield of nylon-6,6

1. n(diamine) = 8.70 ÷ 116 = 0.0750 mol, so n(repeat units) = 0.0750 mol and m = 0.0750 × 226 = 16.95 g.
2. Yield = 14.1 ÷ 16.95 × 100 = 83.2 %.
3. n(HCl) = 2 × 0.0750 = 0.150 mol; V = 0.150 × 24.0 = 3.60 dm³.

Problem 19 Counting bonds in DNA

1. A = 30.0 % of 1200 = 360; T = A = 360; G + C = 1200 − 720 = 480, so G = C = 240.
2. A–T pairs = 360, giving 360 × 2 = 720; G–C pairs = 240, giving 240 × 3 = 720; total = 1440 hydrogen bonds.
3. Each G–C pair has three hydrogen bonds against two for A–T, so more energy is needed to separate the strands.

Problem 20 Four amines with one formula

1. Propylamine (primary, CH₃CH₂CH₂NH₂); propan-2-amine (primary, (CH₃)₂CHNH₂); N-methylethanamine (secondary, CH₃NHCH₂CH₃); N,N-dimethylmethanamine or trimethylamine (tertiary, (CH₃)₃N).
2. Trimethylamine (about 3 °C): it has no N–H bond, so its molecules cannot form hydrogen bonds with each other.
3. Propylamine (48 °C) boils higher than propan-2-amine (32 °C). Its straight chain has a larger surface area of contact, so stronger London forces act between molecules, while the branched molecule is more compact.