Benzene can be used to make benzoic acid in the two-step process shown in Fig. 7.1. benzene methylbenzene benzoic acid step 1 step 2 CH3 COOH Fig. 7.1

1. (a)Give the reagents and conditions for step 1 and step 2. step 1 .......................................................................................................................... step 2 ..........................................................................................................................[2]
2. (b)Methylbenzene and benzoic acid each have five different peaks in the carbon (13C) NMR spectrum. Table 7.1 hybridisation environment of carbon example chemical of the carbon atom shift range atom /ppm sp3 alkyl CH3-, -CH2-, -CH<, >C< 0-50 sp3 next to alkene/arene -C-C=C, -C-Ar 25-50 sp3 next to carbonyl/carboxyl C-COR, C-O2R 30-65 sp3 next to halogen C-X 30-60 sp3 next to oxygen C-O 50-70 sp2 alkene or arene >C=C<, C-C 110-160 sp2 carboxyl R-COOH, R-COOR 160-185 sp2 carbonyl R-CHO, R-CO-R 190-220 sp nitrile R-C≡N 100-125 Use Table 7.1 to complete the two sentences to suggest descriptions of these two spectra. The carbon (13C) NMR spectrum of methylbenzene: • has ............................ peak(s) in the chemical shift range of ............................ and • has ............................ peak(s) in the chemical shift range of ............................ The carbon (13C) NMR spectrum of benzoic acid: • has ............................ peak(s) in the chemical shift range of ............................ and • has ............................ peak(s) in the chemical shift range of ............................[2]
3. (c)(i)When treated with Cl2 under suitable conditions, methylbenzene forms compound J. When treated with Cl₂ under different conditions with different reagents, methylbenzene forms compound K. Suggest and draw structures of compounds J and K in the boxes. The molecular formula of each compound is given. JC7H7Cl methylbenzene KC₂H₂CI CH3 State the reagents and conditions required to form each product. to form compound J .......................................................................................................... to form compound K ..........................................................................................................[4]
4. (c)(ii)When treated with a chlorine-containing reagent under suitable conditions, benzoic acid forms compound L. When treated with a different chlorine-containing reagent under different conditions, benzoic acid forms compound M. Suggest and draw structures of compounds L and M in the boxes. The molecular formula of each product is given. LC7H5ClO2 benzoic acid MC7H5ClO COOH State the reagents and conditions to form compound M from benzoic acid.[3]

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