Butanedioic acid can be made in a three-step synthesis using ethane as the starting material. The reaction scheme shows: Ethane (CH₃-CH₃) undergoing Step 1 to form Compound A (Br-CH₂-CH₂-Br). Compound A undergoes Step 2 to form Compound B. Compound B undergoes Step 3 to form Butanedioic acid (HOOC-CH₂-CH₂-COOH).

1. (a)Describe the reagents and conditions needed for step 1.[1]
2. (b)Name the other product also formed in step 1 which is an isomer of A.[1]
3. (c)Give the structural formula of B.[1]
4. (d)State the reagents and conditions needed for step 2.[2]
5. (e)(i)Identify the oxidising agent that could be used to oxidise ethanedioic acid.[1]
6. (e)(ii)State the product(s) of the reaction in (e)(i).[1]
7. (f)Compare and explain the relative acidities of hexanoic acid, hexan-1-ol and phenol.[3]
8. (g)(i)Identify the protons in the methylpropanedioic acid molecule that are responsible for each area of the proton NMR spectrum. δ 12.7 δ 3.3 δ 1.1[2]
9. (g)(ii)Name the splitting pattern shown at δ 3.3 and explain how it arises.[1]
10. (g)(iii)The carbon-13 NMR spectra of butanedioic acid, HO2CCH2CH2CO2H, and methylpropanedioic acid, HO₂CCH(CH3)CO2H are different. State the number of peaks • in the carbon-13 NMR spectrum of butanedioic acid • in the carbon-13 NMR spectrum of methylpropanedioic acid.[1]

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