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Spectroscopic Techniques Questions and Answers

31.

In the proton spectrum of ethanol the methyl signal appears as a triplet because it has

Answer: C

The two methylene protons split it into three lines of intensity 1:2:1, and the methylene signal is in turn split into a quartet by the three methyl protons. The hydroxyl proton often appears as a singlet because it exchanges rapidly.

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32.

The molecular ion peak in a mass spectrum gives the compound's

Answer: D

Correct answer: molecular mass

The molecular ion is the intact molecule minus one electron, so its mass-to-charge ratio gives the molecular mass directly. High resolution instruments measure it precisely enough to distinguish formulae of the same nominal mass and so give the molecular formula. It says nothing about boiling point, and the number of double bonds follows only after the formula is known, from the degrees of unsaturation.

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Answer: A

Correct answer: greatest intensity

The base peak is the tallest peak in the spectrum, and all other intensities are quoted as a percentage of it. It corresponds to the most stable fragment ion, which is often not the molecular ion at all - in many compounds the molecular ion fragments so readily that it is weak or entirely absent, which is why soft ionisation methods were developed.

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34.

A compound containing one chlorine atom shows a peak at M and another at M plus 2 in a ratio of about

Answer: B

Correct answer: 3 to 1

Chlorine occurs as two isotopes in a natural ratio of about 3 to 1, chlorine-35 to chlorine-37, so a molecule containing one chlorine gives a peak at M and another two units higher at about a third the height. The pattern is diagnostic: seeing an M plus 2 peak at 33 per cent of M is strong evidence for a single chlorine.

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35.

A compound containing one bromine atom shows M and M plus 2 peaks in a ratio of about

Answer: C

Correct answer: 1 to 1

Bromine’s two isotopes, bromine-79 and bromine-81, are almost equally abundant, so a single bromine gives M and M plus 2 peaks of nearly equal height. The contrast with chlorine’s 3 to 1 makes the two immediately distinguishable, and the presence of both elements gives a characteristic multi-peak cluster whose pattern reveals how many of each atom are present.

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Answer: D

Correct answer: free atoms in a flame or furnace

The sample is atomised - sprayed into a flame or heated in a graphite furnace - so that the metal is present as free, ground-state atoms in the vapour. Those atoms absorb at their own sharp resonance lines, and the attenuation of the beam measures their number. The technique determines the element regardless of the compound it arrived in, since the atomisation destroys all chemical form.

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Answer: A

Correct answer: the sharp resonance lines of the element sought

Atomic absorption lines are extremely narrow, so a continuous source filtered by an ordinary monochromator would deliver mostly light the atoms cannot absorb, and the sensitivity would be hopeless. A hollow cathode lamp whose cathode is made of the element emits exactly the lines that element absorbs, and narrower than the absorption profile. The price is that a separate lamp is needed for each element.

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Answer: B

Correct answer: emitted by excited atoms in the flame

In flame emission the flame both atomises the sample and excites a fraction of the atoms thermally; the light they emit as they relax is measured, and no lamp is required. Atomic absorption measures instead the light removed from an external beam by the far larger population of ground-state atoms, which is why it is generally the more sensitive of the two for elements of high excitation energy.

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Answer: C

Correct answer: select a narrow band of wavelengths

The monochromator - a prism or, now, a diffraction grating with entrance and exit slits - disperses the light and passes only a narrow band to the sample or detector. Its bandwidth matters: too wide a band violates the monochromatic assumption behind Beer’s law and flattens sharp peaks. The detector converts light to signal, and the amplification and temperature control are separate functions.

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Answer: D

Correct answer: compensating for source and detector drift

A double beam instrument splits the light between the sample and a reference and compares them continuously, so any drift in lamp output or detector response affects both alike and cancels in the ratio. A single beam instrument must be re-zeroed against a blank at each wavelength and is vulnerable to drift between that and the measurement. Both use a monochromator and both may report either quantity.

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