When answering this question it should be assumed that together all the hydrogen atoms in a benzene ring result in a single unsplit peak at δ = 7.2 in a proton (¹H) NMR spectrum. The structures of five isomeric ketones, P, Q, R, S and T are given. P C₆H₅COCH(CH₃)₂ Q C₆H₅COCH₂CH₂CH₃ R C₆H₅CH₂COCH₂CH₃ S C₆H₅CH₂CH₂COCH₃ T C₆H₅CH(CH₃)COCH₃

1. (a)Identify all the chiral carbon atoms on the structures above. Label each chiral carbon atom with an asterisk (*). [1][1]
2. (b)The proton (¹H) NMR spectrum of one of the five isomers, P, Q, R, S or T, is shown in Fig. 8.1. [Figure 8.1: ¹H NMR spectrum] δ/ppm Fig. 8.1
3. (b)(i)Identify which of the compounds P, Q, R, S or T gives this spectrum. Draw the displayed formula of the compound you have identified. Identify the protons responsible for the peaks at δ = 3.7, δ = 2.5 and δ = 1.0 on the structure you have drawn. [2][2]
4. (b)(ii)Name the splitting pattern of the peak at δ = 3.7. Explain why it has this splitting pattern. [1][1]
5. (c)Choose from the letters P, Q, R, S and T to identify:
6. (c)(i)the two compounds that each have a doublet peak in the proton (¹H) NMR spectrum [1][1]
7. (c)(ii)the compound with only three peaks in its proton (¹H) NMR spectrum. [1][1]
8. (d)Suggest a suitable solvent that should be used for obtaining the spectrum shown in Fig. 8.1. [1][1]
9. (e)The proton (¹H) NMR spectrum of compound T is compared in the presence of D₂O and in the absence of D₂O. Describe any difference between the two spectra. Explain your answer. [1][1]
10. (f)Complete Table 8.1 below to give the number of peaks in the carbon-13 NMR spectrum of each compound. Table 8.1 compound | number of peaks | compound | number of peaks [Figure 8.2: Structures of Ethylbenzene, Isopropylbenzene, Toluene, t-Butylbenzene] | | | [2][2]

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