This question is about compounds with the molecular formula C6H12O2.
Hexanoic acid, C5H11COOH, is a weak acid.
i) Write the equation for the acid dissociation constant, Ka, of hexanoic acid.
(1)
ii) Calculate the pH of a 0.100 mol dm–3 solution of hexanoic acid. [pKa of hexanoic acid = 4.88]
(4)
iii) The solubility of hexanoic acid in water is 1.08 g per 100 g of water. The isomer of hexanoic acid, butyl ethanoate, CH3CO2C4H9, has a solubility of 0.68 g per 100 g of water.
Explain the differences in these data in terms of the hydrogen bonding between hexanoic acid and water, and between butyl ethanoate and water.
(3)
i) Compound A is thought to be another isomer of hexanoic acid.
10 g of compound A is found to contain 6.21 g of carbon and 1.03 g of hydrogen, with the remainder being oxygen.
Use the data to calculate the empirical formula of compound A.
You must show all your working.
(3)
ii) State how you might use your answer to (b)(i) and a mass spectrum of compound A to prove that compound A is an isomer of hexanoic acid.
(1)
iii) A series of tests was performed on compound A.
| Observation | |
1 | addition of phosphorus(V) chloride | misty fumes |
2 | addition of 2,4-dinitrophenylhydrazine | orange precipitate |
3 | addition of Benedict’s or Fehling’s reagent | solution remains blue with no precipitate |
4 | addition of acidified potassium dichromate(VI) | solution remains orange |
5 | test using polarimetry | plane of plane polarised light is rotated |
Deduce the structure for compound A. Justify your answer by using all the test results.
(6)
Compound B, another isomer with the molecular formula C6H12O2, contains a ring of six carbon atoms.
The carbon-13 NMR spectrum has only two peaks, one of which is at 69 ppm.
Draw the structure of compound B.
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