A-LevelChemistryHydrocarbons9701/41 · Oct/Nov 2017 · Q625 Marks

(a) 4-nitromethylbenzene can be prepared via an electrophilic substitution reaction as shown. [Figure showing reaction of methylbenzene to 4-nitromethylbenzene via intermediate T]

(a)(i)This reaction also forms an isomer of 4-nitromethylbenzene as a by-product. Draw the structure of this by-product.[1]
(a)(ii)Write an equation for the reaction between HNO₃ and H₂SO₄ that forms the electrophile for this reaction.[1]
(a)(iii)Describe how the structure and bonding of the six-membered ring in intermediate T differs from that in methylbenzene.[3]
(b)(i)Give the systematic name of compound W.[1]
(b)(ii)Suggest the reagents and conditions for steps 1–5. step 1 step 2 step 3 step 4 step 5[6]
(c)Suggest how the basicity of benzocaine would compare to that of ethylamine. Explain your answer.[2]
(d)(i)Predict the number of peaks that would be seen in the carbon-13 NMR spectrum.[1]
(d)(ii)Benzocaine was dissolved in CDCl₃ and the proton NMR spectrum of this solution was recorded. Suggest why CDCl₃ and not CHCl₃ is used as the solvent when obtaining a proton NMR spectrum.[1]
(d)(iii)Use the Data Booklet and the spectrum in (d)(ii) to complete the table for the proton NMR spectrum of benzocaine. The actual chemical shifts, δ, for the four absorptions have been added.[4]
(d)(iv)Explain the splitting pattern for the absorption at δ1.2ppm.[1]
(d)(v)The proton NMR spectrum of benzocaine dissolved in D₂O was recorded. Suggest how this spectrum would differ from the spectrum in (d)(ii). Explain your answer.[1]
(e)(i)Suggest the reagents used for step 1.[1]
(e)(ii)Suggest structures for compounds R and S and draw them in the boxes.[2]

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 (Oct/Nov 2017). The mark or mark type is shown in brackets.

[1] (a)(i) [Diagram showing 3-nitromethylbenzene (or 2-nitromethylbenzene)]
[1] (a)(ii) HNO3 + 2H2SO4 → H3O+ + NO2+ + 2HSO4-
[1] (a)(iii) Point 1: bonds/electrons are partially delocalised in T or delocalised/π system/π bonding extends over only five carbons
[1] (a)(iii) Point 2: four π-electrons in the (delocalised system of T) or methylbenzene has (two) more π-electrons /(two) more delocalised electrons
[1] (a)(iii) Point 3: contains a carbon that is sp³ hybridised in T or (all the) carbons are sp² hybridised in methylbenzene
[1] (a)(iii) Point 4: one carbon has a bond angle of 109.5Y/tetrahedral (in T) or (C-C) bond strengths/lengths are not all the same or not all the bond angles are 120Y(in T)
[1] (b)(i) 4-aminobenzoic acid
[1] (b)(ii) step 1 Sn + HCl
[1] (b)(ii) step 1 concentrated/reflux/heat
[1] (b)(ii) step 2 CH3COCl
[1] (b)(ii) step 3 KMnO4/manganate(VII)/MnO4- (acidified/alkaline) and heat
[1] (b)(ii) step 4 aqueous HCl and heat
[1] (b)(ii) step 5 ethanol, H2SO4, concentrated/reflux/heat
[1] (c) (benzocaine) is less (basic than ethylamine) AND lone pair (on N) is less available to accept a proton/H+
[1] (c) since (lone pair on N) is delocalised over the ring or phenyl ring is electron withdrawing group
[1] (d)(i) 7 peaks
[1] (d)(ii) CDCl3 will produce no signal in the spectrum or CHCl3 would produce a signal/would be detected
[1] (d)(iii) 1.2: CH(3), 3 H atoms, triplet
[1] (d)(iii) 3.5: CH(2)O, 2 H atoms, quartet
[1] (d)(iii) 5.5: NH2, 2 H atoms, singlet (broad)
[1] (d)(iii) 7.1–7.4: H attached to aromatic/benzene ring, 4 H atoms, multiplet
[1] (d)(iv) neighbouring/adjacent carbon atom has two protons/H (attached to it) or there is an adjacent CH2(O) group
[1] (d)(v) peak at 5.5/NH2 peak will disappear and NH2/protons exchange/swap with deuterium
[1] (e)(i) NaNO2 + HCl or HNO2
[1] (e)(ii) [Structure of diazonium salt R: phenyl-N⁺≡N Cl⁻ with -CO₂C₂H₅ at para position]
[1] (e)(ii) [Structure of azo dye S: phenyl-N=N-phenyl with -CO₂C₂H₅ and -OH groups]