A hydrocarbon is known to be either compound D or compound E. [Figure 7.1] (a) Give the systematic name of E.

1. (a)Give the systematic name of E.[1]
2. (b)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][1]
3. (b)(i)Suggest a suitable solvent for obtaining the spectrum in Fig. 7.2.[1]
4. (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]
5. (b)(iii)Complete Table 7.1 for the proton (1H) NMR spectrum of D. [Table 7.1][3]
6. (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]
7. (c)Compound D can be oxidised to compound F by alkaline KMnO4 followed by dilute acid. [Figure 7.3][1]
8. (c)(i)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]
9. (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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