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Espectroscopia e Determinação Estrutural · Reading each technique

Espectroscopia no infravermelho

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O que você vai aprender

  • Explain which vibrations absorb infrared light and estimate a stretching wavenumber from bond strength and reduced mass.
  • Assign the diagnostic O–H, N–H, C–H, C≡N and C≡C bands from their position, width and intensity.
  • Place a C=O band by carbonyl class and correct it for conjugation and ring strain.
  • Use diagnostic bands, out-of-plane bending bands and the fingerprint region to tell isomers apart and to follow a reaction.

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What makes an infrared band

A bond absorbs infrared light when its vibration changes the molecule's dipole moment. N₂ and O₂ do not absorb at all, the symmetric stretch of CO₂ is silent, and the C≡C stretch of a symmetric internal alkyne such as hex-3-yne is absent or very weak. The larger the dipole change, the stronger the band: a C=O stretch is one of the most intense bands in any spectrum, while a C=C stretch is weak to medium. Positions are given in wavenumbers, ν̃ = 1/λ, in cm⁻¹; a band at 1715 cm⁻¹ corresponds to λ = 10⁴/1715 ≈ 5.83 µm. Treating a bond as a spring gives Hooke's law, ν̃ ≈ 4.12 √(k/μ), with k in dyn/cm and the reduced mass μ = m₁m₂/(m₁ + m₂) in atomic mass units. Stiffer bonds (higher bond order) and lighter atoms vibrate at higher wavenumber, which is why the spectrum divides into regions: X–H stretches ≈2500–4000, triple bonds ≈2000–2300, double bonds ≈1500–1850, and single bonds to C plus bending vibrations below ≈1500 cm⁻¹. Replacing H by D nearly doubles μ for a C–H bond, so a C–D stretch appears at about 0.73 times the C–H wavenumber.

O–H, N–H and C–H stretches

A hydrogen-bonded alcohol O–H gives a strong, broad band at ≈3200–3550 cm⁻¹; in dilute CCl₄ solution, where molecules are no longer hydrogen-bonded to each other, it sharpens and moves to ≈3600–3650 cm⁻¹. A hydrogen bond inside one molecule does not change on dilution. A carboxylic acid O–H is much broader still, ≈2500–3300 cm⁻¹, because the acid exists as a strongly hydrogen-bonded dimer, and it sits on top of the C–H bands. N–H bands are weaker and sharper than O–H: a primary amine (RNH₂) shows two bands between ≈3300 and ≈3500 cm⁻¹ (symmetric and asymmetric stretches), a secondary amine one, and a tertiary amine none. Primary amides show two N–H bands near ≈3350 and ≈3180 cm⁻¹. C–H stretches depend on hybridisation: sp³ C–H just below 3000 (≈2850–2960), sp² C–H just above 3000 (≈3010–3100), and sp C–H as a strong, sharp band at ≈3300 cm⁻¹. An aldehyde adds two weak C–H bands near ≈2820 and ≈2720 cm⁻¹.

Triple bonds

The region between ≈2000 and ≈2300 cm⁻¹ is almost empty, so any band there is informative. A nitrile C≡N gives a sharp, medium-to-strong band at ≈2210–2260 cm⁻¹ (≈2250 for alkyl nitriles, a little lower when conjugated). A terminal alkyne gives a weak C≡C band at ≈2100–2140 cm⁻¹ together with the sharp ≡C–H band at ≈3300 cm⁻¹; an internal alkyne shows a weak band near ≈2190–2260 cm⁻¹, or none when it is symmetric. A doublet near ≈2350 cm⁻¹ that changes from run to run is atmospheric CO₂ in the beam path, not part of the sample.

The carbonyl band by class

Saturated, acyclic carbonyl compounds absorb strongly at characteristic positions: acid chloride ≈1800, anhydride two bands ≈1820 and ≈1760 (coupled symmetric and asymmetric stretches), ester ≈1735–1750, aldehyde ≈1725, ketone ≈1715, carboxylic acid dimer ≈1710 and amide ≈1650–1690 cm⁻¹. Two effects compete. An electronegative substituent withdraws electrons inductively and stiffens the C=O, raising ν̃ (chloride most). A lone pair on the attached atom donates into the carbonyl, giving the C=O partial single-bond character and lowering ν̃ (nitrogen most, which is why amides are lowest). A carboxylate ion has two equivalent C–O bonds of order ≈1.5, so its C=O band is replaced by a strong asymmetric stretch at ≈1550–1610 cm⁻¹ and a symmetric one near ≈1400 cm⁻¹.

Conjugation and ring strain

Conjugating a C=O with a C=C or an aromatic ring delocalises the π electrons and lowers the band by about 20–40 cm⁻¹: cyclohex-2-en-1-one ≈1680, acetophenone ≈1685, methyl benzoate ≈1720 cm⁻¹. Conjugation on the oxygen side of an ester works the other way: in phenyl and vinyl esters the alkoxy lone pair is drawn into the ring or C=C, so it donates less to the C=O, and the band rises to ≈1760–1770 cm⁻¹. Small rings raise the carbonyl band: cyclohexanone ≈1715, cyclopentanone ≈1745 and cyclobutanone ≈1780 cm⁻¹. The same trend holds for lactones (six-membered ≈1735, five-membered ≈1770, four-membered ≈1840 cm⁻¹) and lactams (six-membered ≈1670, five-membered ≈1700, four-membered β-lactam ≈1745 cm⁻¹).

Fingerprint region and telling isomers apart

Below ≈1500 cm⁻¹ C–O stretches (≈1000–1300 cm⁻¹; aryl alkyl ethers ≈1250 and ≈1040), C–C stretches and bending vibrations overlap into a pattern unique to each compound, so an exact match to a reference spectrum confirms identity even between isomers whose functional-group bands coincide. Some bending bands are diagnostic on their own: a trans-disubstituted C=C gives a strong band at ≈965 cm⁻¹, a cis C=C a broad band near ≈700 cm⁻¹, a monosubstituted vinyl group two bands at ≈990 and ≈910 cm⁻¹, and a 1,1-disubstituted C=CH₂ one at ≈890 cm⁻¹. To identify an unknown, first compute the degree of unsaturation, DBE = (2C + 2 + N − H − X)/2, then ask which candidate bands are present and which are absent. Absence is often the strongest evidence: no broad band above 3200 cm⁻¹ rules out an O–H, and no strong band at 1650–1850 cm⁻¹ rules out a carbonyl.

What to carry forward

Check the X–H region first (broad O–H, very broad acid O–H, one or two N–H bands, C–H above or below 3000), then the triple-bond window, then the carbonyl: place it by class, lower it for conjugation, raise it for ring strain. Use what is absent as firmly as what is present, and confirm an isomer from bending bands or a fingerprint match.

Problemas de prática (40)

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Carbonyl positions

  • Benzoic acid becomes a saltPredict how deprotonating a carboxylic acid changes its infrared spectrum.

Telling isomers and reactions apart

  • Follow an alcohol oxidation by IRPredict which bands disappear and appear when a secondary alcohol is oxidised to a ketone.

Classificar 5

Bands, bonds and selection rules

  • Which diatomic gas is IR-silent?Apply the dipole-change selection rule to decide whether a stretching vibration absorbs infrared light.

O–H, N–H, C–H and triple bonds

  • Name the group behind a broad bandAssign a strong, broad band near 3350 cm⁻¹ to a hydrogen-bonded alcohol O–H.
  • Which C₆H₁₂ isomer is an alkene by IR?Classify a C₆H₁₂ isomer as an alkene from an sp² C–H stretch above 3000 cm⁻¹ and a C=C stretch.
  • Two N–H bandsClassify an amine as primary, secondary or tertiary from the number of N–H stretching bands.

Carbonyl positions

  • A carbonyl band at 1800 cm⁻¹Recognise an acid chloride from its unusually high carbonyl wavenumber.

Comparar 6

Bands, bonds and selection rules

  • Rank four stretching bandsRank stretching wavenumbers using bond order and the mass of the atoms involved.

Carbonyl positions

  • Rank carbonyl bands by classRank the C=O stretching wavenumbers of an amide, a ketone, an ester and an acid chloride.
  • Which ketone absorbs lowest?Predict which cyclic ketone has the lowest C=O wavenumber from conjugation and ring size.
  • Where the aryl group sits mattersContrast the effects of conjugation on the acyl side and on the alkoxy side of an ester carbonyl.
  • Rank lactones by C=O wavenumberApply the ring-strain trend to rank lactones by carbonyl wavenumber.
  • Rank cyclic ketones by C=O wavenumberRank cyclic ketones by combining the effects of conjugation and ring strain on the carbonyl band.

Calcular 4

Bands, bonds and selection rules

  • Where does a C–D stretch appear?Predict the wavenumber shift on deuterium substitution from the ratio of reduced masses.
  • Estimate a C=O stretch with Hooke's lawEstimate a stretching wavenumber from a force constant and the reduced mass of the two atoms.
  • Convert a ketone band to wavelengthConvert a band position from wavenumber to wavelength in micrometres.

Telling isomers and reactions apart

  • Degree of unsaturation before the IRCompute the degree of unsaturation of a formula as the first constraint on an IR assignment.

Encontre o erro 2

Bands, bonds and selection rules

  • A band near 2350 cm⁻¹Recognise atmospheric carbon dioxide as an artefact rather than a sample band.

Telling isomers and reactions apart

  • Find the flaw in an acid assignmentDiagnose an incorrect carboxylic acid assignment from a missing O–H band and a high carbonyl position.

Interprete as evidências 13

Bands, bonds and selection rules

  • Which bond absorbs at ≈1650 cm⁻¹?Map a band position onto the X–H, triple-bond, double-bond and single-bond regions of the spectrum.

O–H, N–H, C–H and triple bonds

  • Aldehyde or ketone?Use the aldehyde C–H stretching pair near 2720 and 2820 cm⁻¹ to separate an aldehyde from a ketone.
  • Which C₄H₈O₂ isomer is it?Identify a carboxylic acid from its very broad O–H band combined with a carbonyl band.
  • Which C₄H₇N isomer absorbs at 2250 cm⁻¹?Distinguish a nitrile from an alkyne and an imine by the triple-bond band and the X–H region.
  • Which C₄H₉NO isomer is it?Combine the N–H band count with the carbonyl position to identify a primary amide.
  • A sharp band at 3300 cm⁻¹Recognise a terminal alkyne from its sharp ≡C–H band and weak C≡C band.

Carbonyl positions

  • Two carbonyl bands, 60 cm⁻¹ apartIdentify an acid anhydride from its pair of coupled carbonyl bands.
  • An amide that absorbs at 1745 cm⁻¹Assign a lactam's ring size from its carbonyl wavenumber.

Telling isomers and reactions apart

  • Which unsaturated alcohol is it?Use out-of-plane C–H bending bands to tell a terminal alkene from a trans-disubstituted alkene.
  • Which C₃H₆O isomer is it?Choose among C₃H₆O isomers using the presence and absence of carbonyl, aldehyde C–H and O–H bands.
  • Which C₇H₈O isomer has no O–H?Separate an aryl ether from isomeric alcohols and phenols using the O–H and C–O regions.
  • Which C₈H₈O carbonyl compound is it?Combine the carbonyl position with the aldehyde C–H test to identify an aryl ketone among isomers.
  • cis or trans hex-2-ene?Assign alkene geometry from the out-of-plane C–H bending band.

Trabalhe de trás para frente 1

  • Which reagent made this product?Work back from a product's IR spectrum to the reagent that transformed a ketone.

Planeje uma síntese 2

Bands, bonds and selection rules

  • Confirm a ketone isomer by IRChoose the part of an IR spectrum that distinguishes two isomers with the same functional groups.

Telling isomers and reactions apart

  • Is the alkyne all used up?Choose the IR band that shows a starting material has been fully consumed.

Explique por quê 5

Bands, bonds and selection rules

  • Strong C=O, weak C=CRelate infrared band intensity to the dipole change of a vibration.
  • The missing alkyne bandExplain why the C≡C stretch of a symmetric internal alkyne is absent from its infrared spectrum.

O–H, N–H, C–H and triple bonds

  • Dilution sharpens an O–H bandExplain how intermolecular hydrogen bonding lowers and broadens an O–H stretching band.
  • A band that ignores dilutionUse the effect of dilution to distinguish intramolecular from intermolecular hydrogen bonding.

Carbonyl positions

  • Why amides absorb low and acid chlorides highExplain the carbonyl positions of amides and acid chlorides through resonance donation and inductive withdrawal.

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