The questions are on the amines and polymers test. Take it under real conditions before reading the mark scheme.
Quick answer: Amines are organic bases because the lone pair on nitrogen accepts a proton, and amino acids contain both an amine and an acid group so they form zwitterions. This mark scheme gives the model answers and the marks for the 8 questions of the Chemistry 11-12 test on amines and polymers, from basic strength to choosing the reagent, with the equivalent answers to accept.
Below you find the model answers with the marking points. Firstly, award each mark only for a correct statement or a correct step. Then allow error carried forward in the calculations.
Level: Chemistry grades 11-12 (A Level · IB · Class 11-12) · Curriculum: A Level 3.3.11-3.3.13
Question 1 Basic strength (3 marks)
a. Ethylamine (smaller pKb).
b. In ethylamine the alkyl group donates electron density to nitrogen, so the lone pair is more available to accept H⁺. However, in phenylamine the lone pair is delocalised into the ring, so it is less available.
Marking: 1 for ethylamine; 1 for donation by the alkyl group; 1 for delocalisation in phenylamine.
Question 2 pH of dimethylamine (3 marks)
[OH⁻] = √(5.4 × 10⁻⁴ × 0.0400) = √(2.16 × 10⁻⁵) = 4.65 × 10⁻³ mol/dm³. pOH = −log(4.65 × 10⁻³) = 2.33. pH = 14.00 − 2.33 = 11.67.
Marking: 1 for [OH⁻]; 1 for pOH; 1 for pH.
Question 3 Serine melting point (2 marks)
a. HOCH₂CH(NH₃⁺)COO⁻, the zwitterion.
b. It is an ionic lattice with strong electrostatic attraction, so you need a lot of energy to melt it.
Marking: 1 for the zwitterion; 1 for ionic attraction.
Question 4 Poly(lactic acid) (3 marks)
a. An ester link forms and the molecule loses water.
b. Repeat unit –OCH(CH₃)CO– is C₃H₄O₂: 36.0 + 4.0 + 32.0 = 72.0. Then the chain: 450 × 72.0 = 32 400.
Marking: 1 for ester and water; 1 for 72.0; 1 for 32 400.
Question 5 Tripeptide (3 marks)
a. X and Y are peptide (amide) links, –CO–NH–.
b. Two links form, so the molecule loses two water molecules: 75 + 89 + 105 − 2 × 18 = 269 − 36 = 233.
Marking: 1 for peptide link; 1 for subtracting 2 × 18; 1 for 233.
Question 6 Nylon or polypropene (2 marks)
Nylon-6,6 has polar amide links that water and acid hydrolyse, so the chain breaks into monomer salts. In contrast, poly(propene) has only non-polar C–C and C–H bonds that water and acid cannot attack.
Marking: 1 for hydrolysis of amide links; 1 for non-polar carbon chain.
Question 7 Hydrogen bonds in DNA (2 marks)
Since A = T = 15, A–T pairs = 15. G + C = 80 − 30 = 50, so G–C pairs = 25. hydrogen bonds = 15 × 2 + 25 × 3 = 30 + 75 = 105.
Marking: 1 for 15 and 25 pairs; 1 for 105.
Question 8 Choosing the reagent (2 marks)
Excess ammonia in ethanol, heated in a sealed tube. If ammonia is limited, the propylamine that forms acts as a nucleophile and reacts further with 1-bromopropane to give secondary and tertiary amines.
Marking: 1 for excess ammonia with ethanol; 1 for further substitution.
