Infrared Spectroscopy Questions and Answers

Answer: A

A hydrogen-bonded hydroxyl gives a broad band centred around 3300, and its breadth is as diagnostic as its position. The 1700 region belongs to carbonyl, 2200 to triple bonds, and 1100 to carbon-oxygen single bond stretches. Reading a band's shape alongside its position is what distinguishes a hydroxyl from an amine, which absorbs nearby but sharply.

Answer: B

Each hydroxyl group experiences a slightly different degree of hydrogen bonding, and each degree corresponds to a slightly different frequency, so the observed band is the sum of many. The force constant is high, not low, which is why the band sits at 3300 in the first place. Overlap and resolution are not the cause. That a band's width can report on intermolecular interaction is a second channel of information beyond its position. Naming the cost of a choice alongside its benefit is what turns a fact into a design decision.

Answer: C

Dilution breaks up the hydrogen-bonded network, so free hydroxyl groups predominate and give a sharp band near 3600 rather than a broad one near 3300. Hydrogen bonding lengthens and weakens the bond, which is why the bonded band sits lower. Observing that change as concentration falls is the standard way to confirm that a broad band really is a hydroxyl and that hydrogen bonding is responsible. Naming the cost of a choice alongside its benefit is what turns a fact into a design decision.

Answer: D

With two hydrogens on the nitrogen a primary amine has a symmetric and an asymmetric stretch and shows two bands near 3300 and 3400; a secondary amine has one hydrogen and shows one. Breadth and position differ far less than the count. Counting bands rather than reading their position is what settles this distinction, and it is a good example of the group frequency region answering a structural question directly. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.

Answer: A

Sharing electron density with the adjacent pi system gives the carbon-oxygen bond some single bond character, lowering its force constant and hence its frequency by twenty or thirty reciprocal centimetres. Reduced mass is unchanged, polarity affects intensity, and no energy is absorbed elsewhere. That a shift of a few tens of wavenumbers reports on the electronic environment is what makes the carbonyl band so informative beyond merely announcing that a carbonyl is present.

Answer: B

The nitrogen's lone pair delocalises into the carbonyl, reducing its double bond character and dropping the band to about 1650 against a ketone's 1715. Hydrogen bonding contributes a little more, reduced mass is barely affected, and no ring is required. Ordering the carbonyl derivatives by how much the adjacent atom donates or withdraws - anhydride highest, amide lowest - turns a list of numbers into a pattern. A limit that survives making every external component ideal is an intrinsic one, and worth knowing for that reason.

Answer: C

An anhydride has two carbonyls sharing an oxygen, each withdrawing from the other, which raises the force constant and puts the bands near 1800 and 1760. An amide is lowest at about 1650, with ketones and acids between. The two bands an anhydride shows are themselves diagnostic. Ordering the family by electronic effect rather than memorising eight numbers is what makes the pattern recoverable when one value is forgotten. The quantity that varies and the quantity that does not are worth naming before any substitution is attempted.

Answer: D

Silicate glass absorbs heavily across the mid infrared, so cells are made from sodium chloride, potassium bromide or similar salts, which are transparent. Scattering and reactivity are not the issue, and thin glass is easy to make. The consequence is that infrared cells dissolve in water and must be handled dry, which shapes the whole practice of sample preparation in a way no other spectroscopy demands. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Answer: A

A mull is the solid ground to a fine powder and suspended in a drop of mineral oil, which matches the refractive index well enough to suppress scattering. A pressed disc is the potassium bromide pellet technique, and dissolution and evaporation are other methods. The drawback is that the oil is a hydrocarbon and contributes its own carbon-hydrogen bands, which must be recognised and discounted. A limit that survives making every external component ideal is an intrinsic one, and worth knowing for that reason.

Answer: B

Mineral oil is a hydrocarbon, so it contributes strong bands near 2900 and 1450 that conceal any of the sample's own in those regions. It does not dissolve most solids, absorbs only in specific regions, and is inert towards the plates. The standard remedy is to run the sample a second time in a fluorinated oil, which is transparent where Nujol is not - covering the whole spectrum between the two runs. A single measurement chosen to respond to the one asymmetry beats several that respond to shared behaviour.