Which would be the best reagent to use to determine whether an unknown substance was 2-methylpropan-1-ol or 2-methylpropan-2-ol?
- Bromine water
- Potassium nitrate solution
- Sodium carbonate solution
- Acidified potassium permanganate solution
CHEMISTRY, M7 2020 HSC 29
| The flow chart shows reactions involving five different organic compounds, |  |
to |  |
| Draw the structure of each compound | |
to | |
in the corresponding space provided. (5 marks) |
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CHEMISTRY, M8 2022 HSC 27
A bottle labelled 'propanol' contains one of two isomers of propanol.
- Draw the TWO isomers of propanol. (2 marks)
- Describe how \( \ce{^13C NMR}\) spectroscopy might be used to identify which isomer is in the bottle. (2 marks)
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- Each isomer produces a different product when oxidised.
- Write equations to represent the oxidation reactions of the two isomers. Include reaction conditions. (3 marks)
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a. Isomer 1:
Isomer 2:
b. Identifying isomers with \( \ce{^13C NMR} \) spectroscopy:
→ this can be used to identify the isomers in the bottle because they show a different number of signals which helps deduce the carbon environment.
→ Propan-1-ol contains 3 \( \ce{C}\) environments so it would have 3 peaks on a \( \ce{^13C NMR}\) spectrum whereas propan-2-ol only contains 2 \( \ce{C}\) environments (due to symmetry), so it would only have 2 signals on a \( \ce{^13C NMR}\) spectrum.
c.
Show Worked Solution
a. Isomer 1:
Isomer 2:
b. Identifying isomers with \( \ce{^13C NMR} \) spectroscopy:
→ this can be used to identify the isomers in the bottle because they show a different number of signals which helps deduce the carbon environment.
→ Propan-1-ol contains 3 \( \ce{C}\) environments so it would have 3 peaks on a \( \ce{^13C NMR}\) spectrum whereas propan-2-ol only contains 2 \( \ce{C}\) environments (due to symmetry), so it would only have 2 signals on a \( \ce{^13C NMR}\) spectrum.
c.
♦ Mean mark (c) 52%.
CHEMISTRY, M6 2021 HSC 35
A manufacturer requires that its product contains at least 85% v/v ethanol.
The concentration of ethanol in water can be determined by a back titration. Ethanol is first oxidised to ethanoic acid using an excess of acidified potassium dichromate solution.
\(\ce{3C2H5OH($aq$) + 2Cr2O7^2-($aq$) + 16H^+($aq$) ->3CH3COOH($aq$) + 4Cr^3+($aq$) + 11H2O($l$)}\)
The remaining dichromate ions are reacted with excess iodide ions to produce iodine \(\ce{(I2)}\)
\(\ce{Cr2O7^2-($aq$) + 14H^+($aq$) + 61^-($aq$) -> 2Cr^3+($aq$) + 7H2O($l$) + 3I2($aq$)}\)
The iodine produced is then titrated with sodium thiosulfate \(\ce{(Na2S2O3)}\).
\(\ce{I2($aq$) + 2S2O3^2-($aq$) -> 2I^-($aq$) + S4O6^2-($aq$)}\)
A 25.0 mL sample of the manufacturer's product was diluted with distilled water to 1.00 L. A 25.0 mL aliquot of the diluted solution was added to 20.0 mL of 0.500 mol L¯1 acidified potassium dichromate solution in a conical flask. Potassium iodide (5.0 g) was added and the solution titrated with 0.900 mol L¯1 sodium thiosulfate. This was repeated three times.
The following results were obtained.
The density of ethanol is 0.789 g mL¯1.
Does the sample meet the manufacturer's requirements? Support your answer with calculations. (7 marks)
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CHEMISTRY, M7 2021 HSC 26
A sequence of chemical reactions, starting with 2-methylprop-1-ene, is shown in the flow chart.
- Complete the flow chart by drawing structural formulae for compounds `text{A}`, `text{B}`, `text{C}`, and `text{D}`. (4 marks)
- Reflux is used in the synthesis of methyl 2-methylpropanoate.
- Provide TWO reasons for using this technique. (2 marks)
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a. Compound A:
Compound B:
Compound C:
Compound D:
b. Reasons for reflux technique:
→ Reflux heats the reaction mixture which increases the average kinetic energy, and thus increases the reaction rate.
→ Heating causes the volatile substances to form vapour molecules. Refluxing uses a condenser to cool the vapour molecules into liquids, and thus retains the substances.
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a. Compound A:
Compound B:
Compound C:
Compound D:
b. Reasons for reflux technique:
→ Reflux heats the reaction mixture which increases the average kinetic energy, and thus increases the reaction rate.
→ Heating causes the volatile substances to form vapour molecules. Refluxing uses a condenser to cool the vapour molecules into liquids, and thus retains the substances.
♦ Mean mark (b) 46%.
CHEMISTRY, M7 2021 HSC 13 MC
A chemist synthesises a substance using the following pathway.
\[\ce{ X ->[{hydration}] {Y} ->[{oxidation}] Z}\]
What are compounds `text{X, Y, Z}` ?
