2-methylbut-1-ene reacts with acidified manganate(VII) ions, under specific conditions, to produce two organic compounds X and Y. X immediately reacts with the acidified manganate(VII) ions to form carbon dioxide and water. Y has the structural formula CH₃CH₂COCH₃. [Figure showing reaction scheme: 2-methylbut-1-ene --(acidified manganate(VII) ions)--> X + Y (CH₃CH₂COCH₃). X further reacts to CO₂ + H₂O]

1. (a)Draw the skeletal formula of 2-methylbut-1-ene.[1]
2. (b)(i)State the specific conditions required for the acidified manganate(VII) ions to react with 2-methylbut-1-ene in this way.[1]
3. (b)(ii)Name the type of reaction occurring to the functional group in 2-methylbut-1-ene in the reaction in (b)(i).[1]
4. (c)Draw the structural formula of X.[1]
5. (d)Describe a chemical test and the expected observation(s) to confirm the presence of the carbonyl functional group in Y.[2]
6. (e)The infra-red spectrum of 2-methylbut-1-ene is shown. [Figure with IR spectrum]. Predict two main differences that would be seen between the spectra of Y, CH₃CH₂COCH₃, and of 2-methylbut-1-ene. Give reasons for your predictions. Your answer should refer only to the region of each spectrum above 1500 cm⁻¹.[2]
7. (f)(i)Write an equation to show this reaction. Use [H] to represent an atom of hydrogen from the reducing agent.[1]
8. (f)(ii)Name the organic product formed in this reaction.[1]
9. (g)(i)State the molecular formula of W.[1]
10. (g)(ii)Draw a possible structure of W.[1]

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