Infrared spectroscopy of organic compounds explained: HSC Chemistry Module 8

A 4 mark answer needs the diagnostic absorption ranges for each compound and the discrimination logic.

Ethanol $CH_3CH_2OH$.

• Broad O-H stretch 3200 to 3550 cm$^{-1}$ (hydrogen-bonded alcohol).
• C-O stretch around 1050 cm$^{-1}$.
• No C=O peak.

Ethanal $CH_3CHO$.

• Strong C=O stretch around 1720 to 1740 cm$^{-1}$ (aldehyde, slightly higher than ketones).
• Two aldehyde C-H stretches around 2720 and 2820 cm$^{-1}$ (a doublet, the "Fermi doublet").
• No broad O-H peak.

Ethanoic acid $CH_3COOH$.

• Very broad O-H stretch 2500 to 3300 cm$^{-1}$ (carboxylic acid, much broader and lower frequency than alcohol O-H).
• Strong C=O stretch around 1710 cm$^{-1}$.
• C-O stretch around 1250 cm$^{-1}$.

Discrimination. Look for the C=O peak first (1700 to 1750 cm$^{-1}$). If absent, the compound is the alcohol. If present, look at the O-H region: a very broad band stretching down to 2500 cm$^{-1}$ confirms the acid; absence of broad O-H plus the aldehyde C-H doublet confirms the aldehyde.

Markers reward (1) correct C=O range, (2) the O-H range and shape for alcohol vs acid, (3) one feature unique to the aldehyde, (4) a clear discrimination logic.

The IR spectrum below about 1500 cm$^{-1}$ is the fingerprint region. It contains many overlapping bands from C-C stretches, C-O stretches and bending vibrations of the whole molecular skeleton. The exact pattern of peaks in this region is so sensitive to molecular structure that two different compounds almost never have the same fingerprint region, even if their functional groups are the same. A match between an unknown spectrum and a reference spectrum in this region is therefore as good as a fingerprint match.

Optical isomers (enantiomers) have identical connectivity and identical bond strengths; they differ only in the spatial arrangement at a chiral centre. IR detects bond vibrations, which depend on connectivity and force constants, not on three-dimensional arrangement. Enantiomers therefore give superimposable IR spectra and cannot be distinguished by IR alone. To resolve enantiomers, use chiral chromatography, polarimetry, or NMR with a chiral shift reagent.

Markers reward (1) the fingerprint region defined, (2) the matching-to-reference logic, (3) the reason IR is blind to chirality.

A 3-mark identify needs the correct range AND shape descriptor for each bond.

(a) Alcohol O-H

3200 to 3550 $cm^{-1}$, broad and strong (hydrogen-bonded).

(b) Carbonyl C=O

1700 to 1750 $cm^{-1}$, sharp and strong.

(c) Alkene C=C

1620 to 1680 $cm^{-1}$, medium intensity.

Marking criteria: 1 mark per bond for the correct range with the correct shape/intensity descriptor; a range alone without shape/intensity earns half credit under most marking guidelines.

(a) Functional group class. Carboxylic acid.

(b) Justification. The peak at 1705 $cm^{-1}$ falls in the carbonyl $C=O$ range (1700 to 1750 $cm^{-1}$), confirming a carbonyl group is present. The broad absorption from 2600 to 3300 $cm^{-1}$ is far broader than an alcohol O-H (which is centred around 3300 to 3400 and does not extend below about 3200) and matches the very broad, hydrogen-bonded O-H of a carboxylic acid (2500 to 3300 $cm^{-1}$). The combination of a carbonyl peak WITH this specific broad low-wavenumber O-H is diagnostic of -COOH; an aldehyde or ketone would show the carbonyl peak with no broad O-H at all.

Marking criteria: 1 mark for identifying carboxylic acid, 1 mark for correctly reading the carbonyl peak, 1 mark for correctly reading the broad acid O-H range, 1 mark for explicitly contrasting with what an aldehyde/ketone or alcohol spectrum would look like instead.

(a) Degree of unsaturation.

One degree of unsaturation implies one ring or one pi bond (here, one $C=O$ double bond, consistent with the carbonyl peak observed and no other unsaturation).

(b) Functional group class. The sharp strong peak at 1715 $cm^{-1}$ sits within the carbonyl range and specifically within the ketone sub-range (1705 to 1725 $cm^{-1}$), slightly lower than the aldehyde sub-range (1720 to 1740 $cm^{-1}$). More decisively, there is no absorption above 3000 $cm^{-1}$, so the aldehyde's diagnostic Fermi doublet near 2720 and 2820 $cm^{-1}$, and any O-H, are both absent. This combination (carbonyl present, no O-H, no aldehyde doublet) identifies a ketone.

(c) Structure and name. With formula $C_4H_8O$ and one ketone $C=O$, a consistent structure is $CH_3COCH_2CH_3$, butan-2-one.

Marking criteria: 1 mark for the correct degree-of-unsaturation calculation, 1 mark for correctly interpreting it as one ring/pi bond, 1 mark for identifying ketone using the exact wavenumber sub-range, 1 mark for using the ABSENCE of the aldehyde doublet as confirming evidence, 1 mark for a valid structure and correct IUPAC name.

(a) Identifying the two peaks using the chart's ranges.

• 3320 $cm^{-1}$ falls in the terminal alkyne C-H range (around 3300 $cm^{-1}$, sharp and strong), not the broader alcohol O-H range (3200 to 3550, broad) or amine N-H range (3300 to 3500, medium), because this peak is described as sharp rather than broad or medium.
• 2230 $cm^{-1}$ falls in the nitrile $C\equiv N$ range (2200 to 2260 $cm^{-1}$, sharp, medium intensity) shown on the chart, rather than the alkyne $C\equiv C$ range (2100 to 2260, weak), because it is described as medium rather than weak intensity.

(b) Compound class. A terminal alkyne C-H together with a nitrile $C\equiv N$, and the absence of any carbonyl peak, is consistent with a terminal alkynenitrile, an organic compound containing both a terminal $C\equiv CH$ group and a $-C\equiv N$ group, with no oxygen-containing functional group present.

Marking criteria: 1 mark for correctly reading each range from the chart, 1 mark for using the sharp/medium intensity descriptor (not just the wavenumber) to choose between overlapping ranges for each peak, 1 mark per correct functional-group identification (2 marks), 1 mark for a compound class consistent with BOTH peaks and the absence of a carbonyl.

Bond vibration frequency (and hence wavenumber, since wavenumber is proportional to frequency) follows $\nu = \frac{1}{2\pi}\sqrt{k/\mu}$, where $k$ is the bond force constant (stiffness) and $\mu$ is the reduced mass of the two vibrating atoms.

C=O versus C-O. A double bond has a higher force constant $k$ than a single bond between the same two atoms, because it involves an additional pi bond that resists stretching more strongly. Since both involve the same atoms (so the same $\mu$), the higher $k$ of the double bond directly gives a higher vibration frequency and wavenumber for $C=O$ (1700 to 1750) than for $C-O$ (1000 to 1300).

O-H versus C=O. The reduced mass of an O-H bond, $\mu = \frac{m_O m_H}{m_O + m_H}$, is very small because hydrogen's mass (1) is tiny compared with oxygen's (16), so $\mu \approx m_H = 1$. The reduced mass of $C=O$ is much larger, roughly $\frac{12 \times 16}{12+16} \approx 6.9$. Because $\nu \propto 1/\sqrt{\mu}$, the much smaller reduced mass of O-H drives its frequency (and wavenumber) higher than $C=O$, even though $C=O$ has a stiffer (double) bond, because the mass effect dominates for the very light hydrogen atom.

Marking criteria: 1 mark for correctly stating the vibration equation and identifying $k$ and $\mu$, 2 marks for the $C=O$ vs $C-O$ comparison via force constant with the same atoms, 2 marks for the O-H vs C=O comparison via reduced mass (explicitly noting hydrogen's very small mass), 1 mark for a coherent final ranking that matches the numeric evidence.

This is a 7-mark ANALYSE and EVALUATE: markers reward full use of the given data plus a genuine evaluation of what IR alone cannot tell you.

Band 6 PLAN.

• Degree of unsaturation from $C_3H_6O_2$: $\text{DoU} = \frac{2(3)+2-6}{2} = 1$, consistent with a single carbonyl and no ring.
• Peak at 1740 $cm^{-1}$: at the high end of the carbonyl range, specifically matching the ester sub-range (1735 to 1750 $cm^{-1}$), which is higher than aldehyde (1720 to 1740), ketone (1705 to 1725) and acid (1700 to 1725).
• No broad O-H rules out a carboxylic acid (which would show both a carbonyl AND a very broad O-H down to 2500 $cm^{-1}$) despite matching the molecular formula $C_3H_6O_2$ (propanoic acid is also $C_3H_6O_2$).
• Conclude: Y is an ester. With formula $C_3H_6O_2$, the ester $-COO-$ group accounts for $CO_2$, leaving $C_2H_4$ split across the two alkyl fragments either side of the ester oxygen: methyl ethanoate, $CH_3COOCH_3$, fits exactly.
• Evaluation: IR confirms the FUNCTIONAL GROUP (ester) and the exact wavenumber narrows it away from acid, aldehyde or ketone, but IR alone cannot distinguish which specific ester isomer of $C_3H_6O_2$ is present, since methyl ethanoate is the only reasonable isomer here, but for larger esters more than one arrangement of the two alkyl groups is often possible and IR gives no atom connectivity beyond the functional group. Mass spectrometry would resolve this: the molecular ion peak confirms the exact molar mass, and characteristic fragment losses (e.g. loss of $OCH_3$, 31 mass units, or loss of $COOCH_3$) reveal which side of the ester oxygen each alkyl group sits on. $^1H$ NMR would add the number and splitting of proton environments, confirming the exact carbon skeleton.

Model paragraph (excerpt). The IR evidence for Y is internally consistent with an ester: a single carbonyl-range peak at exactly 1740 $cm^{-1}$ sits at the ester end of the carbonyl range, and the complete absence of any O-H absorption rules out the carboxylic acid isomer that shares the same molecular formula. However, IR spectroscopy is fundamentally a functional-group technique; it cannot by itself distinguish between esters that share a carbonyl environment but differ in exactly how the two alkyl fragments are arranged, because ester connectivity beyond the $C=O$ and $C-O$ stretches falls into the unassignable fingerprint region. Confirming the specific structure of Y as methyl ethanoate rather than any other conceivable $C_3H_6O_2$ ester therefore requires a complementary technique: mass spectrometry to fix the exact molecular mass and reveal fragment losses either side of the ester linkage, and $^1H$ NMR to confirm the number of distinct proton environments and their integration ratios.

Marker's note: top-band answers (1) compute and use the degree of unsaturation, (2) use the EXACT wavenumber sub-range (not just "it's a carbonyl") to choose ester specifically, (3) explicitly rule out the isomeric carboxylic acid using the absence of broad O-H, (4) propose a structure consistent with the formula, and (5) give a genuine evaluation naming a SPECIFIC complementary technique and what new information it adds, rather than just asserting "more tests are needed."