Infrared Spectroscopy Questions and Answers

Answer: A

Resolution is inversely proportional to the maximum optical path difference, so a longer mirror travel resolves finer structure. The aperture affects throughput, averaging affects noise, and detector speed affects how fast the mirror may move. That resolution depends on a mechanical travel rather than an optical slit is the structural difference from a dispersive instrument and is why the two are specified in quite different terms. A limit that survives making every external component ideal is an intrinsic one, and worth knowing for that reason.

Answer: B

Averaging n scans improves the signal-to-noise ratio by the square root of n, since the signal adds coherently and the random noise does not. Every scan covers the same range, the travel is fixed, and detector warming is not a mechanism. That four times as many scans buys only twice the improvement is the practical consequence, and it sets a limit on how much patience is worth spending. A single measurement chosen to respond to the one asymmetry beats several that respond to shared behaviour.

Answer: C

The improvement goes as the square root of the number of scans:

\[ \sqrt{n} = 4 \Rightarrow n = 16 \]

giving sixteen. Four scans would buy only a doubling. Because the return diminishes as the square root, doubling the improvement again would need sixty-four scans and sixteen times the time - which is why averaging is a remedy for a modest noise problem rather than a large one.

Answer: D

Where photons are scarce and the detector contributes most of the noise, measuring every wavelength all the time is worth a great deal - which is exactly the infrared situation and not the ultraviolet one. Sample preparation, band width and grating manufacture are not decisive. A technique whose advantage depends on which noise source dominates will win in one spectral region and not another, and that is precisely what happened. The quantity that varies and the quantity that does not are worth naming before any substitution is attempted.

Answer: A

The single-beam spectrum of the empty instrument contains the source's output, the detector's response, and the absorption of atmospheric water and carbon dioxide, all of which divide out when the sample spectrum is ratioed against it. Wavelength calibration comes from the laser, dark current is handled separately, and mirror travel is monitored continuously. Measuring everything except the sample, and dividing, is the same principle as a blank in photometry.

Answer: B

Water vapour and carbon dioxide both absorb strongly in the mid infrared and both vary with the room's conditions, so a background recorded minutes earlier no longer matches. Nitrogen, oxygen and the noble gases are infrared inactive. The remedy is purging with dry nitrogen or recording the background immediately before the sample, and recognising the characteristic sharp structure near 3700 and 2350 prevents it being mistaken for the sample's. The quantity that varies and the quantity that does not are worth naming before any substitution is attempted.

Answer: C

A very broad hydroxyl stretch together with a carbonyl is the signature of a carboxylic acid, whose hydrogen-bonded dimer gives an exceptionally wide band from about 3300 down to 2500. A ketone would show the carbonyl alone, an alcohol the hydroxyl alone, and an aromatic hydrocarbon neither. Reading two bands together rather than one at a time is what distinguishes a functional group from a mere pair of features. Naming the cost of a choice alongside its benefit is what turns a fact into a design decision.

Answer: D

A band near 2250 is the carbon-nitrogen triple bond of a nitrile, and the absence of any carbonyl rules out the other obvious assignment. Alkenes absorb near 1650 and alcohols near 3300. The triple bond region is nearly empty in most compounds, which makes any band there immediately informative - a nitrile, an alkyne or an isocyanate, and the three are distinguished by their exact position and intensity. The shared properties explain why these belong to one family, and the differences explain why more than one exists.

Answer: A

Two carbonyl bands separated by about sixty reciprocal centimetres, with no hydroxyl stretch, is the classic anhydride pattern arising from symmetric and asymmetric stretching of the coupled pair. An acid would show the broad hydroxyl, an ester one carbonyl near 1735, and an amide one near 1650. A band count as well as a band position is what makes this assignment unambiguous, and it is why the number of features is always worth noting explicitly. Working one numerical case through both cases side by side fixes the distinction better than any rule of thumb.

Answer: B

An ester contains a carbon-oxygen single bond as well as the double, and its stretch gives a strong band between 1100 and 1300. Methyl absorptions lie near 2900 and 1450, an overtone would be far weaker and near 3400, and there is no hydrogen bonding in a simple ester. Looking for the second band that a functional group implies, rather than stopping at the obvious one, is what turns a tentative assignment into a confident one. Checking the units of each quantity catches a wrong formula before the arithmetic begins.