Explain why trichloroethanoic acid, CCl₃COOH, is more acidic than ethanoic acid, CH₃COOH.

1. (a)Explain why trichloroethanoic acid, CCl₃COOH, is more acidic than ethanoic acid, CH₃COOH.[1]
2. (b)(i)Acyl chlorides are formed by reacting carboxylic acids with thionyl chloride, SOCl₂. Ethanedioyl chloride, (COCl)₂, can be prepared by reacting ethanedioic acid, (COOH)₂, with an excess of SOCl₂. Write an equation for this reaction.[1]
3. (b)(ii)Samples of (COCl)₂ are reacted separately with an excess of warm acidified KMnO₄(aq) and with H₂NCH₂CH₂NH₂. The carbon-containing product from the reaction with H₂NCH₂CH₂NH₂ has the molecular formula C₄H₆N₂O₂. Complete the boxes in Fig. 9.1 to suggest the structure of the carbon-containing product in each reaction. [Figure 9.1 with two empty boxes, one for "with warm acidified KMnO₄(aq)" and one for "with H₂NCH₂CH₂NH₂"][2]
4. (b)(iii)A polyester can be synthesised from the reaction of (COCl)₂ with ethane-1,2-diol, HOCH₂CH₂OH. Draw two repeat units of the polymer formed. Any functional groups should be displayed.[2]
5. (c)(i)Compound H, C₆H₁₀O₃, reacts with alkaline I₂(aq) to form yellow precipitate J but does not react with Na₂CO₃(aq). The proton (¹H) NMR spectrum of H in CDCl₃ is shown in Fig. 9.2. [Figure 9.2 showing the NMR spectrum with peaks at ~1.15 ppm (triplet), ~2.25 ppm (singlet), ~3.60 ppm (singlet), ~3.95 ppm (quartet)] [Table 9.1 of chemical shift ranges is also provided] Identify yellow precipitate J.[1]
6. (c)(ii)Complete Table 9.2 for the proton (¹H) NMR spectrum of H, C₆H₁₀O₃. Table 9.2 chemical shift δ/ppm splitting pattern number of ¹H atoms responsible for the peak number of protons on adjacent carbon atoms 1.15 2.25 3.60 3.95[4]
7. (c)(iii)Suggest a structure for H, C₆H₁₀O₃.[1]

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