Flashcards · 9701 · A Level Topic 37

Analytical techniques (A Level) flashcards

17 cards on A Level Topic 37, subtopics 37.1 to 37.4. Pick a subtopic or study them all.

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Stationary phase and mobile phase: Stationary phase: does not move. Mobile phase: carries the sample through. A component with greater affinity for the stationary phase moves more slowly.
Rf value: Rf = distance moved by the spot ÷ distance moved by the solvent front. Between 0 and 1; constant for a substance only under identical conditions.
TLC phases, and which components have low Rf: Stationary: silica or alumina on a plate (polar). Mobile: a solvent rising by capillary action. More polar components are held more strongly by the polar stationary phase, so have lower Rf.
Practical points in TLC: Draw the baseline in pencil (ink would run); keep the spot above the solvent level; locate colourless spots with UV light or a locating agent (e.g. ninhydrin for amino acids); run a known reference alongside.

GLC: stationary and mobile phases: Stationary: a high-boiling liquid on an inert solid support in a heated column. Mobile: an inert carrier gas, such as nitrogen or helium.
Retention time, and what affects it: The time from injection to detection. Components with lower boiling points and lower solubility in the stationary phase have shorter retention times.
Peak areas 15.0, 25.0 and 20.0 for A, B and C. Percentage of A?: Total area = 60.0, so A = (15.0 ÷ 60.0) × 100 = 25.0%. Use peak area, not height.
Limitations of GLC: The sample must be volatile and thermally stable; two compounds can share a retention time, so a known reference is needed for identification.

What does the number of peaks in a ¹³C NMR spectrum show?: The number of chemically different carbon environments (allowing for symmetry). There is no splitting in ¹³C NMR at this level.
Number of ¹³C NMR peaks: propan-1-ol vs propan-2-ol: Propan-1-ol: 3 (all carbons different). Propan-2-ol: 2 (the two CH₃ groups are equivalent).
Why is TMS used as the reference in NMR?: It is defined as δ = 0; it is inert, volatile and non-toxic, and gives a single sharp peak (all its H and C environments are equivalent) away from most others.
Number of ¹³C environments in benzene and in ethanol: Benzene: 1 (all six carbons equivalent). Ethanol: 2.

Three pieces of information from a ¹H NMR spectrum: Chemical shift (δ): the type of proton environment. Relative peak area (integration): the ratio of H atoms in each environment. Splitting: the number of H on adjacent carbons.
The n + 1 rule: A peak is split into n + 1 lines, where n is the number of H atoms on adjacent carbons: 0 singlet, 1 doublet, 2 triplet, 3 quartet.
Predict the ¹H NMR spectrum of ethanol (CH₃CH₂OH): CH₃: triplet, 3H. CH₂: quartet, 2H. OH: singlet, 1H (broad).
How is an O–H or N–H peak identified?: Add D₂O and re-run the spectrum: the O–H/N–H hydrogen exchanges for deuterium, so that peak disappears. These protons do not split or get split.
Why are deuterated solvents such as CDCl₃ used?: Ordinary solvents contain H atoms, which would give large peaks and obscure the sample spectrum.