Organic Chemistry · Aldehydes and Ketones: Nucleophilic Addition Reactions
Spectroscopy of Aldehydes and Ketones
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In 30 seconds
Spectroscopy is the art of reading molecular structure from how a molecule interacts with light or fragments in a mass spectrometer. The carbonyl group C=O is one of the most distinctive structural "reporters" in organic chemistry: it absorbs infrared light strongly, its carbon and nearby protons appear at extreme shifts in NMR spectra, and its mass-spectral fragmentation is highly predictable. Because both families contain the carbonyl group, they share most spectral features; the aldehyde's extra C-H bond provides the clearest distinction. This topic covers what to look for in IR, 1H NMR, 13C NMR, and mass spectrometry, and how to combine the evidence to identify an unknown carbonyl compound.
Why this matters
Structure determination is a daily task in research, pharmaceutical quality control, and forensic chemistry, and spectroscopy is the fastest way to do it. Aldehydes and ketones are everywhere in biology and industry: the aldehyde retinal captures light in our eyes, acetone is a common solvent and biological "ketone body," and many drugs and flavors are carbonyl compounds. Confirming "this is an aldehyde, not a ketone" from spectra is a classic exam skill and a genuine workplace skill.
The college version
Core Concepts
Infrared spectroscopy: the carbonyl stretch
The C=O double bond absorbs infrared light strongly near 1715–1725 cm⁻¹. Saturated aliphatic ketones absorb near 1715 cm⁻¹; aldehydes absorb slightly higher, near 1725 cm⁻¹. Two structural changes shift this band predictably: conjugation with a C=C double bond or aromatic ring lowers the frequency by about 30–40 cm⁻¹ (to roughly 1685 cm⁻¹ for a conjugated ketone), and ring strain raises it. Aldehydes also show a distinctive pair of weak C-H stretching bands near 2720 and 2820 cm⁻¹ — a "fingerprint" ketones lack.
1H NMR: the aldehyde proton
The proton bonded to the carbonyl carbon of an aldehyde appears far downfield, at about δ 9–10 ppm — a region where almost no other protons resonate. The α-protons (adjacent to the carbonyl) appear at δ 2.1–2.5 ppm in both families, and the aldehyde proton couples to them with a small coupling constant (1–3 Hz), so it often appears as a triplet or doublet. A ketone shows no proton at δ 9–10; its only telltale is the α-proton signal.
13C NMR: the carbonyl carbon
The carbonyl carbon is the most downfield carbon in virtually any organic molecule, appearing at δ 190–220 ppm. Aldehyde carbonyl carbons resonate near δ 200 and ketones slightly further downfield, near δ 205–210. For comparison, acids appear near δ 178–180 and esters near 165–175, so the shift immediately distinguishes a ketone or aldehyde from its oxidized or derivatized relatives.
Mass spectrometry: molecular ion and fragmentation
The molecular ion M+ gives the molecular mass and, with high-resolution mass spectrometry, the formula. Two fragmentation pathways dominate for ketones. In α-cleavage, the bond between the carbonyl carbon and an adjacent carbon breaks, ejecting the largest alkyl radical and leaving an acylium ion RCO+. In the McLafferty rearrangement MS rearrangement in which a γ-hydrogen transfers to the carbonyl oxygen, giving an enol radical cation. Full entry →, a ketone with a hydrogen on a carbon three bonds away (a γ-hydrogen) transfers that hydrogen through a six-membered-ring transition state, giving an enol radical cation.
Combining the evidence
No single spectrum is conclusive. A practical workflow: use IR to confirm the carbonyl, 1H NMR to decide aldehyde versus ketone and count protons, 13C NMR to confirm the shift, and MS to fix the formula and support the skeleton. IR answers the functional-group question fastest, but 1H NMR usually settles aldehyde versus ketone.
Common Confusions
| Do Not Confuse | With | Difference |
|---|---|---|
| Aldehyde IR band (~1725 cm⁻¹) | Ketone IR band (~1715 cm⁻¹) | The difference is small; confirm with the aldehyde C-H bands (2720/2820 cm⁻¹) and the δ 9–10 NMR proton. |
| C=O stretch (~1715 cm⁻¹, strong) | C=C stretch (~1650 cm⁻¹, weak) | Carbonyl bands are intense and higher in wavenumber. |
| ppm (NMR shift) | cm⁻¹ (IR wavenumber) | NMR shifts are relative (ppm); IR frequencies are absolute (cm⁻¹). |
| Aldehyde proton at δ 9–10 | Acid proton at δ 10–12 | The acid also shows a broad IR O-H band and a 13C shift near 180, not 200. |
| Conjugation lowering the C=O frequency | Conjugation raising it | Conjugation spreads electron density and lowers the stretch; ring strain raises it. |
Eli explains
The same idea, in plain words
Explain it like I’m 10
Every molecule has a fingerprint, just like people do. The carbonyl group leaves a strong mark in the infrared spectrum, and the aldehyde's "nose" hydrogen shouts from far downfield in the NMR spectrum. Collecting several kinds of spectra is like a detective gathering clues: each clue is weak alone, but together they identify the molecule.
Worked examples
An unknown has the formula C3H6O. Its IR shows a strong band at 1725 cm⁻¹ and weak bands near 2720 and 2820 cm⁻¹. Its 1H NMR shows a triplet (1H) at δ 9.80, a multiplet (2H) at δ 2.4, and a triplet (3H) at δ 1.1.
First, the unsaturation count:
DU = 2C + 2 - H2 = 2(3) + 2 - 62 = 1
One degree of unsaturation plus a strong 1725 cm⁻¹ band means a carbonyl, not a C=C bond. The 2720/2820 cm⁻¹ bands and the proton at δ 9.80 identify an aldehyde, and the 1:2:3 integration matches an ethyl group next to the aldehyde carbon. The compound is propanal, CH3CH2CHO. Had it been acetone, the IR band would sit near 1715 cm⁻¹ and the NMR would show a 6H singlet near δ 2.1 with no δ 9–10 proton.
Three bottles — cyclohexanone, cyclohexene, and cyclohexanol — have lost their labels; IR identifies each. Cyclohexanone shows a strong C=O band near 1715 cm⁻¹ and no O-H band. Cyclohexene shows a weak C=C stretch near 1650 cm⁻¹ (a nearly symmetrical double bond absorbs weakly) and no carbonyl band. Cyclohexanol shows a broad O-H band near 3300 cm⁻¹ plus a C-O band near 1060 cm⁻¹, and no carbonyl band. Three IR clues — carbonyl, alkene, hydroxyl — separate the isomers in seconds.
A ketone gives a molecular ion at m/z 100 (formula C6H12O), a base peak at m/z 58, and a strong fragment at m/z 43. The m/z 58 fragment is the enol radical cation CH3C(OH)=CH2+• from the McLafferty rearrangement, which requires a γ-hydrogen — so the carbonyl sits near the end of the chain. The m/z 43 fragment is the acylium ion CH3CO+. Together these identify 2-hexanone, CH3COCH2CH2CH2CH3; a 3-hexanone would give different fragments, showing how fragmentation reveals the carbonyl's position.
Key takeaways
- IR: strong C=O stretch at 1715–1725 cm⁻¹; aldehydes near 1725, ketones near 1715; conjugation lowers it to about 1685 cm⁻¹.
- Aldehyde IR "doublet" at about 2720 and 2820 cm⁻¹ is unique to aldehydes.
- 1H NMR: aldehyde proton at δ 9–10 ppm (unique); α-protons at δ 2.1–2.5; ketones show no δ 9–10 signal.
- 13C NMR: carbonyl carbon at δ 190–220; aldehydes near 200, ketones near 205–210.
- MS: α-cleavage gives an acylium ion RCO+; McLafferty rearrangement (requires a γ-hydrogen) gives an enol radical cation, e.g., m/z 58 for 2-hexanone.
- A broad IR O-H band at 2500–3300 cm⁻¹ rules out a simple aldehyde or ketone.
- Never rely on one spectrum alone: IR finds the carbonyl, NMR distinguishes aldehyde from ketone, MS gives the formula.
Check yourself
5 review questions from the chapter. Try each one, then open the answer.
What IR absorption identifies the carbonyl group, and how does the frequency differ between a saturated and a conjugated ketone?
Show answer
The C=O stretch near 1715–1725 cm⁻¹; conjugation lowers it by about 30–40 cm⁻¹ to roughly 1685 cm⁻¹.
Which single NMR signal unambiguously identifies an aldehyde, and where does it appear?
Show answer
The aldehyde proton at δ 9–10 ppm in 1H NMR; ketones have no such signal.
Where does the carbonyl carbon appear in 13C NMR for aldehydes versus ketones?
Show answer
Aldehydes near δ 200; ketones near δ 205–210, both within 190–220 ppm.
What two fragmentation pathways dominate ketone mass spectra, and what does the McLafferty rearrangement require?
Show answer
α-cleavage (giving RCO+) and the McLafferty rearrangement (giving an enol radical cation); the latter requires a γ-hydrogen.
How would you use IR alone to rule out a carboxylic acid when you see a strong carbonyl band?
Show answer
Look for the broad O-H stretch at 2500–3300 cm⁻¹; its absence rules out an acid, and a 13C shift near 180 rather than 200 would confirm one if present.
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- wavenumber (cm⁻¹)
- The unit of IR frequency, proportional to energy.
- C=O stretch
- The carbonyl vibration that absorbs IR light strongly.
- chemical shift (δ)
- Position of an NMR signal in ppm, reflecting the local magnetic environment.
- α-proton
- A proton on the carbon directly attached to the carbonyl carbon.
- α-cleavage
- MS fragmentation in which the bond next to the carbonyl breaks, giving RCO+.
- McLafferty rearrangement
- MS rearrangement in which a γ-hydrogen transfers to the carbonyl oxygen, giving an enol radical cation.
- n→π* transition
- Low-energy electronic transition of the carbonyl oxygen lone pair into the π* orbital.
- n→π transition
- Low-energy electronic transition of the carbonyl oxygen lone pair into the π orbital.
Sources & references
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