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Espectroscopia e Determinação Estrutural · Reading each technique

Espectrometria de massas

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O que você vai aprender

  • Choose an ionization method and identify the molecular ion or molecular-related ion it produces, using the nitrogen rule as a check.
  • Read M+1, M+2 and M+4 isotope patterns to count carbons and detect chlorine, bromine and sulfur.
  • Calculate monoisotopic masses and ppm errors and choose a molecular formula from high-resolution data.
  • Predict the main fragment ions from α-cleavage, benzylic cleavage, the McLafferty rearrangement and simple neutral losses, and use them to identify a structure.

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Making ions: hard and soft ionization

A mass spectrometer measures ions by their mass-to-charge ratio, m/z. In electron ionization (EI), a 70 eV electron beam knocks one electron out of the gas-phase molecule, giving the molecular ion M⁺•, a radical cation with one unpaired electron. Its m/z equals the molecular mass, because the lost electron weighs only 0.00055 u. EI leaves the ion with several electronvolts of excess energy, so it breaks into fragments; the tallest peak in the spectrum is the base peak, set to 100%. Soft methods transfer little energy. Chemical ionization (CI) protonates the molecule with a reagent-gas ion such as CH₅⁺ and gives [M+H]⁺. Electrospray ionization (ESI) moves ions that already exist in solution into the gas phase: [M+H]⁺ and [M+Na]⁺ (22 above [M+H]⁺) in positive mode, [M−H]⁻ in negative mode. Large molecules such as proteins pick up many protons in ESI, so a 12 000 Da protein appears as a series of peaks [M+zH]^z⁺ near m/z 1000. Soft ionization is the tool of choice when the EI molecular ion is weak or absent.

Finding the molecular ion and the nitrogen rule

The molecular ion is the highest-mass peak of the spectrum's main cluster, but only if it is present at all; tertiary alcohols and highly branched alkanes often show none. Molecular-ion intensity is high for aromatic compounds (naphthalene's M⁺• is its base peak), moderate for straight-chain alkanes and ketones, and low or zero for alcohols. A candidate molecular ion must also give sensible losses: M−15 (CH₃) and M−18 (H₂O) are common, while a loss of 3 to 14 mass units almost never occurs. The nitrogen rule checks the nominal mass: a molecule made of C, H, O, S and halogens, or containing an even number of nitrogens, has an even nominal mass, while one with an odd number of nitrogens has an odd nominal mass. Pyridine, C₅H₅N, has M = 79; benzene has M = 78. So for a compound without nitrogen, any odd-m/z peak is a fragment, never M⁺•.

Isotope patterns

Every element contributes its heavier isotopes. ¹³C is about 1.1% of carbon, so the M+1 peak is about 1.1% of M for each carbon atom: an M+1 of 7.7% points to seven carbons. Chlorine is ³⁵Cl : ³⁷Cl ≈ 3 : 1 and bromine ⁷⁹Br : ⁸¹Br ≈ 1 : 1, both two mass units apart, so one Cl gives an M : M+2 pair of about 3 : 1 and one Br a pair of about 1 : 1. Several halogens multiply out like a binomial: Cl₂ gives M : M+2 : M+4 ≈ 9 : 6 : 1, Br₂ 1 : 2 : 1, BrCl 3 : 4 : 1 and Cl₃ 27 : 27 : 9 : 1. Sulfur gives a small M+2 of about 4.4% from ³⁴S, well above the ≈0.2% from oxygen or two ¹³C atoms. Fluorine, phosphorus and iodine have one stable isotope and give no M+2 peak; iodine shows itself instead by an I⁺ ion at m/z 127 or a loss of 127.

Exact mass and molecular formulas

Nominal masses are whole numbers, but exact isotope masses are not: ¹H 1.007825, ¹²C 12.000000 by definition, ¹⁴N 14.003074, ¹⁶O 15.994915, ³²S 31.972071, ³⁵Cl 34.968853 and ⁷⁹Br 78.918338. The monoisotopic mass of a formula adds the masses of the most abundant isotope of each atom, so CO (27.9949), N₂ (28.0061) and C₂H₄ (28.0313) all have nominal mass 28 yet are easy to tell apart with a high-resolution instrument. Caffeine (shown), C₈H₁₀N₄O₂, has monoisotopic mass 8 × 12 + 10 × 1.007825 + 4 × 14.003074 + 2 × 15.994915 = 194.0804. The fit of a measured mass is reported in parts per million: error (ppm) = (measured − calculated)/calculated × 10⁶; a good formula usually fits within about 5 ppm. Compare like with like: an [M+H]⁺ ion is M + 1.007276 (a proton), and M⁺• is M − 0.000549 (an electron). A formula must also make chemical sense: its degree of unsaturation, DBE = (2C + 2 + N − H − X)/2, has to be a whole number of zero or more.

Fragmentation by simple cleavage

A radical cation breaks where the products are most stable. In α-cleavage the bond next to the carbon bearing a heteroatom or carbonyl breaks, a radical leaves, and the charge stays as a cation stabilised by a lone pair: ketones give acylium ions R–C≡O⁺ (pentan-3-one gives m/z 57), amines give iminium ions (propan-1-amine gives CH₂=NH₂⁺ at m/z 30), and alcohols and ethers give oxonium ions (primary alcohols give CH₂=OH⁺ at m/z 31). When an atom can lose more than one group, the largest alkyl radical is usually lost preferentially. Alkylbenzenes cleave at the benzylic bond to give C₇H₇⁺ at m/z 91, which rearranges to the aromatic tropylium ion. Alkyl halides lose X• to give the carbocation, and alcohols often lose H₂O (M−18). Simple cleavage of an odd-electron ion gives an even-electron ion plus a radical, so in a nitrogen-free compound these ions appear at odd m/z. These fragmentations involve radical cations, which this course describes in words rather than with curved arrows.

Rearrangements: McLafferty and retro-Diels–Alder

In the McLafferty rearrangement, a carbonyl radical cation with a hydrogen on the γ-carbon transfers that hydrogen to the carbonyl oxygen through a six-membered transition state and splits the α–β bond, expelling a neutral alkene. The ion that remains is an enol radical cation, still odd-electron, so it appears at even m/z in a nitrogen-free compound: when the α-carbon is an unbranched CH₂, methyl ketones give m/z 58, aldehydes 44, carboxylic acids 60 and methyl esters 74. A carbonyl compound with no γ-hydrogen cannot do it. Cyclohexene radical cations undergo a retro-Diels–Alder fragmentation that expels ethylene and leaves the buta-1,3-diene radical cation at m/z 54. Losses of a neutral molecule from M⁺• (H₂O 18, CO 28, C₂H₄ 28, HCl 36) also leave odd-electron ions, and an ambiguous loss of 28 is settled by exact mass: CO is 27.9949 and C₂H₄ is 28.0313.

What to carry forward

Find the molecular ion first, by soft ionization if needed, and check it with the nitrogen rule and sensible losses. Read the M+1 and M+2 peaks for carbon count and Cl, Br or S, then fix the formula with exact mass and ppm error. Finally explain the large fragments: α-cleavage to acylium, iminium and oxonium ions, benzylic cleavage to m/z 91, and McLafferty rearrangement to even-mass ions when a γ-hydrogen is present.

Problemas de prática (40)

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Preveja o produto 6

Isotope patterns

  • The CCl₃⁺ fragment of tetrachloromethanePredict the four-peak isotope cluster of an ion that contains three chlorine atoms.

Fragmentation pathways

  • α-Cleavage of a symmetric ketonePredict the acylium ion formed by α-cleavage of a dialkyl ketone.
  • Base peak of triethylaminePredict the m/z of the iminium ion formed by α-cleavage of a tertiary amine.
  • Which radical does a tertiary alcohol lose?Predict the dominant α-cleavage ion when a carbinol carbon carries alkyl groups of different size.
  • The McLafferty ion of hexan-2-onePredict the m/z and identity of the McLafferty rearrangement ion of a methyl ketone.
  • Retro-Diels–Alder in the mass spectrometerPredict the ion formed by retro-Diels–Alder fragmentation of a cyclohexene radical cation.

Classificar 5

Ionization and the molecular ion

  • What ion does electron ionization make?Identify the species formed when a molecule is ionized by a 70 eV electron beam and where it appears on the m/z scale.
  • Which molecular ion has odd mass?Apply the nitrogen rule to predict which compound has a molecular ion at odd m/z.
  • Ionizing an intact proteinRecognize which ionization method can bring an intact, involatile protein into the gas phase as ions.

Fragmentation pathways

  • Which ketone cannot do a McLafferty rearrangement?Identify whether a carbonyl compound has the γ-hydrogen required for a McLafferty rearrangement.
  • Spot the odd-electron fragmentUse the parity of m/z to tell an odd-electron rearrangement ion from even-electron cleavage ions in a nitrogen-free compound.

Comparar 2

Ionization and the molecular ion

  • How strong is the molecular ion?Rank compound classes by the relative intensity of their EI molecular-ion peaks.

Isotope patterns

  • Rank by the size of the M+2 peakRank compounds by the M+2/M intensity ratio expected from the isotopes they contain.

Calcular 6

Ionization and the molecular ion

  • Deconvolute two charge states of a proteinFind the charge state and neutral mass of a protein from two adjacent peaks of an electrospray charge-state series.

Isotope patterns

  • Count carbons from the M+1 peakEstimate the number of carbon atoms from the M+1/M intensity ratio.
  • The M+2 peak of dibromomethaneCompute the relative M+2 intensity for a molecule with two bromine atoms from the isotope abundance ratio.

Exact mass and formulas

  • Degree of unsaturation from an HRMS formulaCompute the degree of unsaturation of a nitrogen-containing formula obtained from exact mass.
  • Monoisotopic mass of ibuprofenCalculate a monoisotopic mass to four decimal places from isotope masses.
  • Mass error of a protonated amino acidCalculate the signed ppm error between a measured and a calculated ion mass.

Encontre o erro 3

Ionization and the molecular ion

  • A molecular ion at m/z 87?Use the nitrogen rule to reject an odd-mass peak as the molecular ion of a nitrogen-free compound.

Isotope patterns

  • No M+2 peak, so no halogen?Recognize iodine from the I⁺ ion and a loss of 127 even though it gives no M+2 peak.

Exact mass and formulas

  • A steady −4 ppm error on EI molecular ionsIdentify the electron-mass correction needed when comparing a radical-cation m/z with a neutral formula mass.

Interprete as evidências 12

Ionization and the molecular ion

  • Two peaks 22 apart in an electrospray spectrumAssign protonated and sodiated molecule ions in a positive-mode electrospray spectrum and recover the neutral mass.

Isotope patterns

  • Three peaks in a 3 : 4 : 1 ratioDecode a three-peak M, M+2, M+4 cluster into its halogen content by combining isotope ratios.
  • A 3 : 1 pair at m/z 112 and 114Recognize the 3 : 1 M : M+2 pattern of one chlorine atom and match it to a structure.
  • A small but clear M+2 peakUse the size of a small M+2 peak to recognize one sulfur atom.

Exact mass and formulas

  • Pick the formula from an exact massChoose the molecular formula whose calculated monoisotopic mass lies within the stated ppm window of a measured mass.
  • Three gases of nominal mass 28Use exact masses to choose between formulas that share a nominal mass.
  • An odd molecular mass measured exactlyCombine the nitrogen rule, a ppm window and a whole-number DBE to select a molecular formula.

Fragmentation pathways

  • Which butanol gives an oxonium ion at m/z 45?Match an oxonium ion formed by α-cleavage to the alcohol that can produce it.
  • Peaks at 105 and 77 from methyl benzoateAssign the acylium ion formed when an aromatic ester loses its alkoxy radical.

Spectrum to structure

  • A 1 : 1 pair at m/z 122 and 124Combine a halogen isotope pattern with the mass of the halogen-loss fragment to identify an alkyl halide.
  • Odd molecular ion, base peak at m/z 30Use the nitrogen rule and the mass of the α-cleavage iminium ion to choose among amine isomers.
  • Which C₅H₁₀O carbonyl compound?Identify a carbonyl isomer from its α-cleavage ions and the absence of a McLafferty ion.

Trabalhe de trás para frente 1

  • Work back from m/z 105 and 77Work backwards from the observed neutral losses to the C₈H₈O isomer that produced them.

Planeje uma síntese 2

Ionization and the molecular ion

  • No molecular ion for a tertiary alcoholChoose an ionization strategy that reveals the molecular mass when the EI molecular ion is absent.

Spectrum to structure

  • Is the loss of 28 CO or ethylene?Choose the measurement that identifies an ambiguous neutral loss.

Explique por quê 3

Exact mass and formulas

  • Why exact mass gives a formulaExplain why an exact mass can distinguish molecular formulas that share the same nominal mass.

Fragmentation pathways

  • Why m/z 91 dominates for ethylbenzeneExplain why alkylbenzenes give a strong C₇H₇⁺ ion at m/z 91.

Spectrum to structure

  • Why methyl esters show m/z 74Explain the origin of the characteristic even-mass m/z 74 ion of long-chain methyl esters.

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