DP Chemistry (last assessment 2024)

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Question 23M.3.HL.TZ1.21

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Date May 2023 Marks available [Maximum mark: 5] Reference code 23M.3.HL.TZ1.21
Level HL Paper 3 Time zone TZ1
Command term Deduce, Explain, State Question number 21 Adapted from N/A
21.
[Maximum mark: 5]
23M.3.HL.TZ1.21

Analytical chemistry is very important in drug detection.

(a)

An example of a steroid is testosterone.

(a.i)

State the technique used to separate steroids, such as testosterone, from biological fluids.

[1]

Markscheme

chromatography ✓

Once separation has been completed, the components can be identified using mass spectrometry. The following mass spectrum is of testosterone:

[Source: SDBS, National Institute of Advanced Industrial Science and Technology.]

(a.ii)

Explain how the fragments at m/z 288 and 273 can be used to show that it is testosterone, C19H28O2, Mr 288.

 

m/z 288:

 

 


m/z 273:

 

 

 

[2]

Markscheme

m/z 288:
molecular ion/M+/C19H28O2 + ✓

m/z 273:
C18H25O2 +
OR
loss of methyl/CH3 group" or "(M − CH3)+ ✓

(b)

One breathalyser technique is to measure the change in colour when the dichromate ion is reduced to the chromium (III) ion:

Orange                                         Green

Cr2O72− (aq) + 14H+ (aq) + 6e−   →   2Cr3+ (aq) + 7H2O (l)

Deduce the half-equation for the oxidation of ethanol and the overall redox equation occurring in the breathalyser.

 

Half-equation for oxidation of ethanol: 

 

 

Overall balanced redox equation:

 

 

[2]

Markscheme

ALTERNATIVE 1:
C2H5OH (g) → C2H4O (aq) + 2H+ + 2e− ✓

3C2H5OH (g) + Cr2O72− (aq) + 8H+ (aq) → 3C2H4O (aq) + 2Cr3+ (aq) + 7H2O (l) ✓

ALTERNATIVE 2:
C2H5OH (g) + H2O (l) → CH3COOH (aq) + 4H+ + 4e– ✓

3C2H5OH (g) + 2Cr2O72– (aq) + 16H+ (aq) → 3CH3COOH (aq) + 4Cr3+ (aq) + 11H2O (l)