Matching Six Compounds to Their IR Spectra: 04-BS-12, May 2017, Q9

Question 9: Matching Six Compounds to Their IR Spectra

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

A — pentanoic acid

B — 2-methylhex-1-ene

C — cumene (isopropylbenzene)

D — diisopropyl ether

E — tert-butyl acetate

F — triethylcarbinol (given)

Check

Reading the drawn structures shows A is unambiguously a carboxylic acid (COOH, not CH₂OH), B's condensed formula has only six main-chain carbons (a hexene, C₇H₁₄, not a heptene), and E is drawn with an ester oxygen between the carbonyl and the tert-butyl group (an ester, not a ketone). All spectral assignments below use the structures actually drawn on the page, verified against each compound's molecular formula.

Each spectrum is assigned from its most diagnostic feature(s):

Spectrum [5] → A (pentanoic acid). An extremely broad, deep absorption spanning roughly 2500–3300 cm^-1 (the classic hydrogen-bonded, dimeric carboxylic-acid O–H, broader and lower-reaching than any alcohol O–H) plus a strong C=O stretch near 1710 cm^-1 is unique to a carboxylic acid among these six compounds.
Spectrum [2] → F (triethylcarbinol). A broad O–H centred nearer 3300–3400 cm^-1 (not extending down to 2500 the way A's dimeric acid O–H does) with no carbonyl absorption identifies the lone alcohol.
Spectrum [6] → E (tert-butyl acetate). The deepest band on the spectrum is a sharp, strong carbonyl near 1740 cm^-1 (esters sit higher than ketones/acids) with no O–H above 3100, followed by the strong ester C–O stretches near 1250 and 1170 cm^-1 and the tert-butyl C–H bend near 1370 cm^-1 — the only ester.
Spectrum [4] → C (cumene). Aromatic (sp²) C–H just above 3000 cm^-1, the weak overtone/combination bands near 1700–2000 cm^-1 characteristic of a monosubstituted/alkyl-substituted benzene ring, and no O–H/C=O, identifies the aromatic compound.
Spectrum [3] → D (diisopropyl ether). No O–H, no C=O, but a cluster of strong C–O–C stretching bands in the 1000–1150 cm^-1 fingerprint region identifies the ether by elimination.
Spectrum [1] → B (2-methylhex-1-ene). A distinct =C–H stretch just above 3000 cm^-1 (~3080), a sharp medium C=C stretch near 1650 cm^-1 and a very strong band near 890 cm^-1 (the out-of-plane wag of a 1,1-disubstituted =CH₂), with no O–H and no C=O, identifies the lone alkene.

Check

Spectra [1] and [6] are the two that are easy to swap from a caption alone, so they were read directly from the printed figure: [1] has only one band between 2000 and 1500 cm^-1, a medium peak at ~1650, plus the deep ~890 wag — no carbonyl; [6] has its deepest band at ~1740 with a strong 1250/1170 pair — an ester.

Compound	Spectrum	Diagnostic feature
A — pentanoic acid	[5]	very broad 2500–3300 O–H + C=O ~1710
B — 2-methylhex-1-ene	[1]	=C–H ~3080 + C=C ~1650 + =CH₂ wag ~890
C — cumene	[4]	aromatic C–H + overtone bands
D — diisopropyl ether	[3]	C–O–C cluster ~1000–1150, no O–H/C=O
E — tert-butyl acetate	[6]	sharp ester C=O ~1740 + C–O ~1250/1170, no O–H
F — triethylcarbinol	[2]	broad O–H ~3300–3400, no C=O

Concept & Theory

Topic. Functional-group identification from characteristic IR absorption regions.

Key relations. O–H (alcohol): broad, 3200–3550. O–H (carboxylic acid, H-bonded dimer): very broad, 2500–3300, often overlapping C–H. C=O: ketone ~1715, ester ~1735–1745, carboxylic acid ~1710–1760. C–O–C (ether): 1000–1150. C=C (alkene): ~1620–1680, sharp. Aromatic ring: weak overtones 1700–2000 + C–H just above 3000.

Why this works. Each functional group has a bond (or bond combination) with a vibrational frequency essentially unique to that group; presence/absence of the O–H and C=O regions alone is usually enough to sort compounds into acid/alcohol/ester/ketone/ether/hydrocarbon classes before finer fingerprint-region detail is needed.

Common pitfall.

Source. McMurry, Organic Chemistry, 9th ed., Ch.12 (infrared spectroscopy).

Question 9 of 13: Matching Six Compounds to Their IR Spectra

Question 9: Matching Six Compounds to Their IR Spectra