A hydrocarbon is known to be either compound D or compound E. [Figure 7.1] D CH2CH3 CH2CH3 E CH2CH3 CH2CH3 Fig. 7.1

1. (a)Give the systematic name of E.[1]
2. (b)(i)The proton (¹H) NMR spectra of D and E are compared. They are very similar. The proton (¹H) NMR spectrum of D is shown in Fig. 7.2. [Figure 7.2] Suggest a suitable solvent for obtaining the spectrum in Fig. 7.2.[1]
3. (b)(ii)The proton (¹H) NMR spectrum of E is obtained twice, once before and once after shaking with D₂O. Describe any differences between these two spectra. Explain your answer.[1]
4. (b)(iii)Complete Table 7.1 for the proton (¹H) NMR spectrum of D. Table 7.1 chemical shift δ/ppm number of ¹H atoms responsible for the peak group responsible for the peak splitting pattern 1.3 2.7 7.1 Table 7.2 environment of proton example chemical shift range δ/ppm alkane -CH3, -CH2-, >CH– 0.9-1.7 alkyl next to C=O CH3-C=O, -CH2–C=O, >CH–C=O 2.2-3.0 alkyl next to aromatic ring CH3-Ar, -CH2-Ar, >CH-Ar 2.3-3.0 alkyl next to electronegative atom CH3-O, -CH2-O, –CH2–Cl 3.2-4.0 attached to alkene =CHR 4.5-6.0 attached to aromatic ring H-Ar 6.0-9.0 aldehyde HCOR 9.3-10.5 alcohol ROH 0.5-6.0 phenol Ar-OH 4.5-7.0 carboxylic acid RCOOH 9.0-13.0 alkyl amine R-NH- 1.0-5.0 aryl amine Ar-NH2 3.0-6.0 amide RCONHR 5.0-12.0[3]
5. (b)(iv)Compounds D and E can be distinguished by carbon-13 NMR spectroscopy. State the number of peaks in each spectrum. The carbon-13 NMR spectrum of D has ................................ peaks. The carbon-13 NMR spectrum of E has ................................ peaks.[1]
6. (c)(i)Compound D can be oxidised to compound F by alkaline KMnO4 followed by dilute acid. [Figure 7.3] D CH2CH3 CH2CH3 F COOH COOH Fig. 7.3 Write an equation for this reaction using molecular formulae for D and F. The products of this reaction are F, water and carbon dioxide. Use [O] to represent one atom of oxygen from the oxidising agent.[1]
7. (c)(ii)F reacts with an excess of SOCl2 to form compound G. The molecular formula of G is C8H4Cl2O2- G reacts with ethane-1,2-diol, HOCH2CH2OH, to form a mixture of products that includes compounds J, molecular formula C10H8O4, and K, molecular formula C18H12Cl2O6. Draw the structures of compounds G, J and K in Fig. 7.4. [Figure 7.4][3]

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