Arrow Lab

Espectroscopia e Determinação Estrutural · Solve the structure

Problemas de espectros combinados

Uma lição resolvida gratuita e 40 problemas de prática em que você mesmo desenha cada seta curva.

Travou em problemas de espectros combinados? Escaneie o problema.

Tire uma foto da reação da sua lista ou do livro e desenhe o mecanismo com cada seta corrigida. 1 scan grátis agora, sem cadastro; mais 1 com uma conta gratuita.

Escaneie sua lição de graça

O que você vai aprender

  • Turn a molecular formula and mass spectrum into a budget: degree of unsaturation, nitrogen count, halogen isotope pattern and exact-mass formula.
  • Use IR bands, present and absent, to decide which functional groups the unknown can contain.
  • Build fragments from ¹H shifts, integrals and splitting, and test them against the number and position of ¹³C signals.
  • Use characteristic mass-spectral fragments (acylium, tropylium, α-cleavage, McLafferty) to confirm how the fragments are joined.
  • Choose the one candidate that fits every datum, explain why each rival fails, and pick the measurement that separates candidates that remain.

Lição gratuita

O texto da lição resolvida. No Arrow Lab, cada etapa vem com as estruturas e as setas desenhadas.

Ler a aula gratuita (7 partes)

Start with the formula and the molecular ion

The formula sets the budget before any spectrum is read. The degree of unsaturation, DBE = (2C + 2 + N − H − X)/2, counts rings plus π bonds; oxygen and sulfur do not enter it, and a benzene ring uses four. 4′-Bromoacetanilide (shown), C₈H₈BrNO, has DBE = (16 + 2 + 1 − 8 − 1)/2 = 5: the ring (4) plus the C=O (1). The nitrogen rule checks the molecular ion: a compound of C, H, O, halogens and S with an even number of nitrogens (including none) has an even nominal mass, and an odd number of nitrogens gives an odd mass. Isotopes give the halogen count: one Cl gives M and M+2 in a ≈3 : 1 ratio, one Br a ≈1 : 1 pair, two Br ≈1 : 2 : 1. The M+1 peak is about 1.1 % of M per carbon atom. High-resolution MS separates formulas with the same nominal mass, because ¹H (1.00783), ¹⁴N (14.00307) and ¹⁶O (15.99491) are not whole numbers: C₆H₁₀O₂ is 114.0681 but C₇H₁₄O is 114.1045.

IR: which groups are present, and which are absent

Read the IR spectrum as a checklist. A broad O–H at ≈3200–3550 cm⁻¹ means an alcohol or phenol; a much broader band from ≈2500 to 3300 cm⁻¹ with a C=O near 1710 cm⁻¹ means a carboxylic acid. One N–H band means a secondary amine or amide, two mean NH₂. A sharp ≡C–H at ≈3300 cm⁻¹ and a weak band at ≈2100–2140 cm⁻¹ mean a terminal alkyne; a sharp band at ≈2210–2260 cm⁻¹ a nitrile. The C=O position sorts the carbonyl class: acid chloride ≈1800, γ-lactone ≈1770, saturated ester ≈1735–1750, aldehyde ≈1725 (with C–H bands near 2820 and 2720), ketone ≈1715, amide ≈1650–1690 cm⁻¹, each lowered by about 20–40 cm⁻¹ by conjugation. An aromatic nitro group gives two strong bands near 1520 and 1345 cm⁻¹. Absences are just as firm: no band between 1650 and 1850 cm⁻¹ rules out aldehydes, ketones, esters, acids and acid chlorides; only some amides and strongly hydrogen-bonded C=O groups absorb lower.

¹H NMR: the pieces and their neighbours

Integrals scaled to the formula give the number of hydrogens in each signal. Shifts say what each group is attached to: CH₃ on carbon δ ≈0.9–1.3, CH₃C=O or ArCH₃ ≈2.1–2.6, OCH₃ ≈3.3–3.9, ester OCH₂ ≈4.1, alkene H ≈4.6–6.5, aromatic H ≈6.5–8.2, CHO ≈9.5–10.0, CO₂H ≈10–12. Splitting connects neighbours: a 2H quartet with a 3H triplet is an ethyl group, a 1H septet with a 6H doublet an isopropyl group, and a singlet sits next to an atom with no hydrogens. A para-disubstituted ring with two different groups gives two 2H doublets (J ≈ 8–9 Hz); a monosubstituted ring gives 5H. O–H and N–H protons usually exchange fast, appear as broad singlets that do not split their neighbours, and disappear after a D₂O shake. In CDCl₃ a small singlet at δ 7.26 is residual CHCl₃ from the solvent, not part of the sample.

¹³C NMR: count the carbons and use symmetry

Each set of symmetry-equivalent carbons gives one signal, so the signal count tests a candidate's symmetry directly. 1,4-Dichlorobenzene (shown, left) gives two signals and 1,2-dichlorobenzene (right) three, although both have six ring carbons. Shift ranges: C=O of ketones ≈205–215 (≈195–200 when conjugated with a ring or C=C) and aldehydes ≈190–205; acids, esters and amides ≈165–185; aromatic and alkene carbons ≈110–160; nitrile ≈115–125; alkyne ≈65–90; C–O ≈50–80 (OCH₃ ≈49–60); alkyl carbons ≈10–45. The solvent CDCl₃ appears as a three-line signal at δ 77.2. Two carbons can be non-equivalent even when they look alike: the two methyl groups of an isopropyl group next to a stereocentre are diastereotopic and give two signals, and the two N-methyl groups of an amide are different because rotation about the C–N bond is slow on the NMR time scale.

Mass-spectral fragments that confirm the connectivity

Electron ionisation breaks a molecule at predictable places. Carbonyl compounds cleave next to the C=O (α-cleavage) to give acylium ions: CH₃CO⁺ at m/z 43, C₂H₅CO⁺ at 57 and PhCO⁺ at 105, which then loses CO to give C₆H₅⁺ at 77. A benzyl group gives C₇H₇⁺ at m/z 91. A tert-butyl group gives (CH₃)₃C⁺ at m/z 57. A carbonyl compound with a hydrogen on the γ-carbon undergoes the McLafferty rearrangement: the γ-H moves to oxygen through a six-membered transition state and the bond between the α- and β-carbons breaks, expelling an alkene. 1-Phenylbutan-1-one (shown), m/z 148, loses ethene (28) to give m/z 120. Alcohols, especially tertiary ones, break a C–C bond on the carbon that bears the OH (α-cleavage) to give resonance-stabilised oxonium ions, so their molecular ion is weak or absent.

Test every candidate against every datum

Work in a fixed order: formula and DBE, then IR for the functional groups, then ¹H integrals, shifts and splitting to assemble fragments, then ¹³C signal count and shifts to check symmetry and carbon types, then MS fragments to confirm how the fragments join. Write the candidates, then strike out each one with a datum it cannot explain. Isomers often agree on several measurements (4-methoxyacetophenone and methyl 4-methylbenzoate both have formula C₉H₁₀O₂, a para ring and two 3H singlets), so the deciding datum is the one where they differ: here the ketone C=O at δ ≈197 and ≈1675 cm⁻¹ against the ester at δ ≈167 and ≈1720 cm⁻¹. When two candidates survive, choose the next measurement that gives them different answers rather than one they share.

What to carry forward

Budget first (DBE, nitrogen rule, isotope pattern), list the groups the IR allows, assemble fragments from the ¹H spectrum, check symmetry and carbon types against the ¹³C count, and confirm the joins with MS fragments. The answer is the single structure that no datum contradicts.

Problemas de prática (40)

Praticar é grátis com uma conta.

Preveja o produto 3

Formula and mass budget

  • Isotope pattern of a bromochloro compoundPredict the molecular-ion isotope cluster of a compound containing one bromine and one chlorine.

Aromatic unknowns

  • McLafferty ion of 1-phenylbutan-1-onePredict the m/z of the McLafferty rearrangement ion of an aryl ketone that has a γ-hydrogen.

Check, compare and plan

  • Predict the ¹H spectrum of a para amino esterPredict the full ¹H NMR spectrum of a para-disubstituted aromatic ester from shift and splitting rules.

Classificar 2

Formula and mass budget

  • Where does the fifth degree of unsaturation go?Allocate the degrees of unsaturation of a formula between a benzene ring and the functional group the IR and ¹³C data reveal.
  • Which formula fits an odd molecular ion?Use the nitrogen rule and an integer DBE to reject a molecular formula proposed for an odd-mass molecular ion.

Comparar 2

Check, compare and plan

  • Rank isomers by ¹³C signal countRank substituted benzenes by the number of ¹³C signals that their symmetry predicts.
  • Rank proton shifts from upfield to downfieldRank the chemical shifts of named protons in common compounds from the effects of electronegative atoms, C=O groups, π bonds and ring currents.

Calcular 4

Formula and mass budget

  • Degree of unsaturation with N and BrCompute the degree of unsaturation of a formula that contains nitrogen and a halogen.
  • Degree of unsaturation of C₉H₁₀O₂Compute the degree of unsaturation of an oxygen-containing formula as the first step of a structure problem.
  • Exact mass of C₈H₈O₂Calculate the monoisotopic mass that high-resolution MS would report for a molecular formula.
  • Count carbons from the M+1 peakEstimate the number of carbon atoms from the M+1 to M intensity ratio.

Encontre o erro 2

Check, compare and plan

  • Find the datum that rules out a proposalIdentify which observation contradicts a proposed regioisomer when every other measurement fits both isomers.
  • A small singlet at δ 7.26Recognise the residual solvent signal in CDCl₃ instead of assigning it to the sample.

Interprete as evidências 23

Aromatic unknowns

  • Identify C₈H₈O₂ with a δ 9.88 singletIdentify a para-alkoxy benzaldehyde by combining an aldehyde proton, an aryl methyl ether and para symmetry.
  • Identify C₉H₁₀O₂ with a 2H singlet at δ 5.10Place a CH₂ between a phenyl ring and an ester oxygen from its shift, its carbon at δ 66 and the benzyl ion.
  • Identify C₈H₇N with a band at 2250 cm⁻¹Identify an unconjugated nitrile from its C≡N band position, a benzylic CH₂ singlet and a high-field CH₂ carbon.
  • Identify C₇H₉N with an exchangeable 2H signalIdentify a primary benzylic amine from two N–H bands, an exchangeable 2H signal and a benzylic CH₂ carbon near δ 46.
  • Identify C₆H₆BrN from a para patternCombine a bromine doublet, an odd molecular ion, two N–H bands and para symmetry to identify a haloaniline.
  • Identify C₇H₅ClO with a chlorine patternCombine a chlorine isotope pattern, aldehyde evidence and para symmetry to distinguish a chlorobenzaldehyde from benzoyl chloride.
  • Identify C₉H₁₀O₂ with a para ring and two singletsDecide between a para aryl ketone and a para aryl ester with the same ¹H pattern using the carbonyl shift, IR band and fragment ion.
  • Identify C₇H₇NO₂ with no O–H, N–H or C=ORecognise an aromatic nitro compound from its paired N–O bands, an odd molecular ion and para symmetry, rejecting isomers that hide the same atoms in amine, acid or amide groups.
  • Identify C₈H₁₀O with a quartet and a doubletIdentify a secondary benzylic alcohol from a CH quartet coupled to a CH₃ doublet and an O–H band.
  • Identify C₉H₁₀O from an aryl ketone patternIdentify an aryl ethyl ketone from a conjugated C=O band, an ethyl quartet near δ 3.0 and the benzoyl ion.

Aliphatic unknowns

  • Identify C₄H₆O₂ with three CH₂ groupsUse the absence of alkene protons to choose a five-membered lactone over a vinyl ester that absorbs in the same IR region.
  • Identify C₆H₁₀O from four carbon signalsUse a ¹³C signal count to recognise a symmetric cyclic ketone among C₆H₁₀O isomers.
  • Identify C₆H₁₂O₂ with a ketone and an alcoholAssemble a β-hydroxy ketone from singlet-only ¹H data, a broad O–H band and two α-cleavage oxonium/acylium ions.
  • Identify C₇H₁₂O₄ from a symmetric diesterRecognise a symmetric diester from doubled integrals and a lone CH₂ singlet between two carbonyls.
  • Identify C₄H₉NO with three methyl singletsIdentify a tertiary amide whose two N-methyl groups give separate signals because of restricted C–N rotation.
  • Identify C₅H₁₀O₂ with two ethyl groupsPlace two ethyl groups on opposite sides of an ester linkage from their shifts and an acylium fragment.
  • Identify C₄H₈O₂ from a full data setIdentify a branched carboxylic acid from its O–H band, septet–doublet pattern and three ¹³C signals.
  • Identify C₅H₈O₂ from uncoupled alkene protonsUse the size of the alkene coupling constant to tell a 1,1-disubstituted acrylate from its trans-crotonate isomer.
  • Identify C₅H₁₂O with no O–H bandIdentify a methyl ether from the absence of O–H, an OCH₃ singlet and a tert-butyl singlet.
  • Identify C₅H₈O with a terminal alkyneRecognise a tertiary propargylic alcohol from paired O–H/≡C–H bands, a 6H singlet and alkyne carbon shifts.
  • Identify C₆H₁₂O from two singletsRecognise a tert-butyl methyl ketone from a 9H singlet, a 3H singlet and α-cleavage ions.
  • Identify C₄H₉Br from a 1H CHBr signalCombine a bromine isotope doublet with ¹H integration of the CHBr signal to identify a secondary alkyl bromide.

Formula and mass budget

  • Pick the formula from an exact massAssign a molecular formula from a high-resolution mass and confirm it against IR evidence.

Trabalhe de trás para frente 1

  • Which ketone gave this alcohol?Work back from the ¹³C symmetry of a reduction product to the carbonyl compound it came from.

Planeje uma síntese 1

  • Choose the experiment that separates two isomersChoose the measurement that gives different answers for two candidate isomers rather than one they share.

Explique por quê 2

Check, compare and plan

  • Why does the O–H not split the CH?Explain why an alcohol O–H appears as a broad singlet and does not split the neighbouring C–H in CDCl₃.
  • Where is the molecular ion?Explain why a tertiary alcohol shows almost no molecular ion in an EI spectrum and identify its α-cleavage ions.

Travou em problemas de espectros combinados? Escaneie o problema.

Tire uma foto da reação da sua lista ou do livro e desenhe o mecanismo com cada seta corrigida. 1 scan grátis agora, sem cadastro; mais 1 com uma conta gratuita.

Escaneie sua lição de graça

Antes deste tópico