Infrared Spectroscopy Questions and Answers

Answer: C

The number of adjacent hydrogens on the ring determines the pattern, so a monosubstituted, an ortho, a meta and a para compound each give a distinguishable set of bands. Ring count, heteroatoms and molecular weight are not read there. That a bending region can settle a substitution pattern is what keeps the technique useful alongside nuclear magnetic resonance for aromatic compounds.

Answer: D

Both aromatic and alkene hydrogens stretch just above 3000, while saturated ones fall just below, so a band above 3000 indicates unsaturation of some kind and the ring modes near 1600 and 1500 then distinguish an aromatic ring. A band near 1650 belongs to an alkene, below 3000 to saturated carbon, and 3300 to hydroxyl. That one band narrows the possibilities and a second settles them is the ordinary way an infrared spectrum is read.

Answer: A

Only triple bonds and cumulated double bonds have force constants high enough, with atoms light enough, to reach that region, and they are uncommon. Detector, atmosphere and source are all adequate there apart from the carbon dioxide band at 2350. That the region's emptiness makes any band in it immediately significant is a good example of a negative expectation carrying information.

Answer: B

A band at exactly 2350 that changes between the background and the sample scan is carbon dioxide in the beam path rather than anything in the sample. A nitrile would sit near 2250 and would not vary, an alkyne near 2120, and a carbonyl overtone near 3400. That a band's variability between scans identifies it as atmospheric is a diagnostic no band position alone could provide.

Answer: C

The hydrogen is drawn towards the acceptor atom, which lengthens and weakens the original bond, lowering its force constant and hence its frequency. Reduced mass and dipole moment are essentially unchanged, the latter affecting intensity rather than position. That an intermolecular interaction shows up as an intramolecular frequency shift is what makes infrared so sensitive to association, and it is why concentration studies are so informative.

Answer: D

An internal hydrogen bond cannot be broken by dilution, so the band stays low and broad where an intermolecular one would sharpen and rise. It does not disappear or split. That dilution distinguishes the two kinds of hydrogen bonding is a clean experiment: change one variable and see whether the observation responds, which is the general shape of every assignment in this topic.

Answer: A

The first two are correct and the third is the reverse of the technique's chief advantage: a solid or liquid is simply pressed against the crystal with no preparation at all. That is precisely why it displaced mulls and pellets in routine work. A statement that inverts a technique's main selling point is worth doubting on that ground before any detail is checked.

Answer: B

The first three are correct and the fourth is not, so three match. The region near 1450 holds methylene and methyl bending modes, while a carbonyl sits between 1650 and 1800 depending on its neighbours. Placing a band four hundred wavenumbers from where it belongs would misidentify almost any spectrum, which is why the carbonyl region's boundaries are worth knowing precisely.

Answer: C

A broad hydroxyl stretch with a strong carbon-oxygen single bond band near 1050 and no carbonyl at all points to an alcohol. An acid would show a carbonyl near 1710 as well, a ketone would show the carbonyl and no hydroxyl, and a nitrile would show a band near 2250. Reading what is absent alongside what is present is what makes this assignment secure rather than merely plausible.

Answer: D

The requirement is a changing dipole moment, so that the oscillating electric field has something to couple to. Magnetic moments matter in magnetic resonance, quadrupole moments in specialised nuclear techniques, and angular momentum in rotational selection rules. One sentence accounts for why symmetric molecules are transparent, why carbon dioxide shows only two of its four modes, and why Raman is complementary.