DP Chemistry (last assessment 2024)

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Question 19M.3.sl.TZ2.14b(ii)

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Date May 2019 Marks available [Maximum mark: 2] Reference code 19M.3.sl.TZ2.14b(ii)
Level sl Paper 3 Time zone TZ2
Command term Explain Question number b(ii) Adapted from N/A
b(ii).
[Maximum mark: 2]
19M.3.sl.TZ2.14b(ii)

Medicines and drugs are tested for effectiveness and safety.

(b(ii))

Explain why diamorphine (heroin) is more potent than morphine using section 37 of the data booklet.

[2]

Markscheme

morphine has «two» hydroxyl groups AND diamorphine has «two» ester/ethanoate/acetate groups
OR
molecule of diamorphine is less polar than morphine
OR
groups in morphine are replaced with less polar/non-polar groups in diamorphine  [✔]

«less polar molecules» cross the blood–brain barrier faster/more easily
OR
diamorphine is more soluble in non-polar environment of CNS/central nervous system than morphine [✔]

 

Note: Accept “alcohol/hydroxy” for “hydroxyl” but not ”hydroxide”.

Accept “fats” for “lipid”.

Accept “heroin” for “diamorphine”.

Examiners report

This question was reasonably well answered with many students receiving at least one of the two marks.