Answers to questions on Drug detection and analysis

1. (a) Morphine will show a strong broad absorption between 3200 and 3600 cm^-1 due to the two –OH groups which will be absent in heroin. Heroin will show two sharp absorptions in the 1700-1750 cm^-1 region due to the C=O absorption in the two ester groups.

(b) Aspirin will show two absorptions in the 1700-1750 cm^-1 region as it has two groups that contain C=O whereas ibuprofen will only have one.

2. (a) An ideal solution is one that obeys Raoult’s law.

(b) The mole fraction of a component is equal to the number of moles of the component in a mixture divided by the total number of moles of all the components in the mixture.

(c) When a mixture boils the vapour will be richer in the more volatile component present in the mixture. If the vapour is then condensed the new liquid formed will be richer in the more volatile component. By repeatedly boiling and condensing the mixture the vapour will eventually just consist of the more volatile component, which will condense to give the pure component – a process known as fractional distillation.

3. (a) 14.03 g mol^-1 (one –H has been substituted by a –CH₃ group).

(b) The urine is injected into a gas-liquid chromatography machine and carried by an inert gas through the heated liquid stationary phase (e.g. a long-chain alkane) coated onto a solid support in a long thin capillary tube contained inside an oven. This process separates the components. As the different components are eluted at the end of the column they are passed into a mass spectrometer. The mass spectrum for each component is compared with the spectra of known compounds and a printout obtained of all the separate compounds and their concentrations in the urine is produced.

4. (a) Ephedrone contains a carbonyl (ketone) functional group and ephedrine contains a hydroxyl functional group.

(b) Ephedrine contains two chiral carbon atoms so can form stereoisomers which are not mirror images of each other whereas both methamphetamine and ephedrine only contain one chiral carbon atom.

(c) The –CH₃ group attached to the N atom will give a singlet whereas the –CH₃ group attached to the carbon atom bonded to a hydrogen atom will be split into a doublet. This is because the small magnetic field created by the hydrogen nucleus can either be lined up with the external magnetic field or against it.

(d) Methamphetamine will give 6 signals in the ratio 1:1:2:3:3:5. Ephedrone will give 5 signals in the ratio 1:1:3:3:5 and ephedrine will give 7 signals in the ratio 1:1:1,1:3:3:5. So, in fact, they can be distinguished just from the number of signals they show.

(e) Ephedrone will have an sharp strong absorption at 1700-1750 cm^-1 due to the C=O group which will be absent in the other two compounds and ephedrine will have a strong broad absorption at 3200-3600 cm^-1 due to the OH group which will be absent in the other two compounds. All three will show a difference in the fingerprint region.

5. (a) The green colour is due to the formation of chromium(III) ions.

Cr₂O₇^2-(aq) + 14H⁺(aq) + 6e⁻ → 2Cr³⁺(aq) + 7H₂O(l)
C₂H₅OH(aq) + H₂O(l) → CH₃COOH(aq) + 4H⁺(aq) + 4e⁻

Overall equation:

2Cr₂O₇^2-(aq) + 3C₂H₅OH(aq) + 16H⁺(aq) → 4Cr³⁺(aq) + 3CH₃COOH(aq) + 11H₂O(l)

(b) O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l)

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