A-LevelChemistryHalogen derivatives9701/43 · May/June 2019 · Q915 Marks

Organochlorine compounds can undergo hydrolysis. R-Cl + H2O → R−OH + HCl

(a)Organochlorine compounds can undergo hydrolysis. R-Cl + H2O → R−OH + HCl State and explain the relative rates of hydrolysis of the following compounds. CH3CH2Cl CH3COCl C6H5Cl[3]
(b)(i)Epibatidine is a naturally occurring organochlorine compound. [Figure 9(b)] (Structure of epibatidine shown.) Epibatidine is a weak base. State what is meant by the term weak base.[1]
(b)(ii)A molecule of epibatidine contains two nitrogen atoms, both of which can act as a base. Epibatidine reacts with HCl(aq). Complete the structure to suggest the product formed in this reaction. [Figure 9(b)(ii)] (Partial structure of epibatidine shown with Cl atom, N-H bond, and aromatic ring, missing the protonated nitrogen.)[1]
(c)(i)Polyamides, such as nylon-6, can be prepared from a monomer that contains both an amine and an acyl chloride functional group. nylon-6 monomer [Figure 9(c)] (Structure of H2N-(CH2)5-COCl shown.) When the nylon-6 monomer is hydrolysed, bonds are broken and formed. By considering the two steps in the mechanism of the reaction, complete the table by placing one tick (✔) in each row to indicate the types of bonds broken and formed during the mechanism. bonds broken bonds formed σ bonds only π bonds only both σ and π bonds[1]
(c)(ii)Draw two repeat units of nylon-6. The amide bond should be shown fully displayed.[2]
(d)An addition polymer made from two different alkene monomers is called a co-polymer. A section of a polyalkene co-polymer is shown. [Figure 9(d)] (Section of a polyalkene co-polymer chain shown.) Draw the structure of the two alkene monomers which produce this co-polymer.[2]
(e)Explain why polyamides normally biodegrade more readily than polyalkenes.[1]
(f)(i)The alkene phenylethene can be prepared from benzene in three steps. [Figure 9(f)] (Reaction scheme showing benzene -> H (step 1) -> phenylethene (step 2) and also H -> phenylethene (step 3).) Deduce the identity of compound H and draw its structure in the box. [Figure 9(f)(i)] (Empty box for structure of H.)[1]
(f)(ii)Suggest reagents and conditions for each of the steps 1–3. step 1 step 2 step 3[3]

This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.

Mark scheme

Marking points from the official mark scheme (May/June 2019). The mark or mark type is shown in brackets.

[1] (a) M1: CH3COCl > CH3CH2Cl > C6H5Cl
[1] (a) M2 & M3 any two from: • in C6H5Cl (no hydrolysis) C-Cl bond is part of delocalised system OR p-orbital on Cl overlaps with π system OR electrons from Cl overlap with π system • CH3COCl carbon in C-Cl bond is more electron deficient since it is also attached to an oxygen atom (ora) or C-Cl bond strength is weakest in CH3COCl (ora) • CH3CH2Cl carbon in C-Cl bond strengthened by positive inductive effect of alkyl group
[1] (b)(i) partially ionised and proton acceptor
[1] (b)(ii) [Diagram shows epibatidine protonated on the bridgehead nitrogen, forming N-H+]
[1] (c)(i) bonds broken: σ bonds only; bonds formed: both σ and π bonds (Both ticks correct)
[1] (c)(ii) [Drawing showing two repeat units of nylon-6, fully displayed amide bond. -N-(CH2)5-C(=O)-N-(CH2)5-C(=O)-]
[1] (d) CH3CCl=CH2 and C2H5CH=CHCH3 each correct structure scores one mark
[1] (e) C-C bonds are non-polar / polyalkenes cannot be hydrolysed and polyamides can be broken down by hydrolysis
[1] (f)(i) OH (structure of (1-phenylethyl)ethanol, also called 1-phenylethanol)
[1] (f)(ii) M1: step 1: CH3COCl + AlCl3 [1]
[1] (f)(ii) M2: step 2: NaBH4 / LiAlH4 [1]
[1] (f)(ii) M3: step 3: conc. H2SO4, heat [1]