Química Orgânica 1 · Molecular evidence and structure
Dos espectros combinados à estrutura
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- C₅H₈O with a strained carbonyl
- C₇H₇Br with a 2H singlet
- C₄H₈O₂ with a very broad O–H band
- A sharp band at 2250 cm⁻¹
- IR bands of pentan-2-one
- C₄H₈O₂: quartet, singlet, triplet
- C₃H₇Cl: doublet and septet
- C₈H₁₀: an aromatic ring and an ethyl group
- C₄H₁₀O with no O–H band
Ver todos os 34 problemas de dos espectros combinados à estrutura →
O que você vai aprender
- Convert a molecular formula into degrees of unsaturation and use it to bound the rings and π bonds a structure may contain.
- Assign the functional group from diagnostic IR bands, treating an absent band as evidence too.
- Assemble structural fragments from ¹H NMR shift, integration and splitting, supported by ¹³C signal counts and mass-spectral peaks.
- Choose between isomeric candidates by finding the one observation that only the correct structure explains.
Lição gratuita
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Ler a aula gratuita (5 partes)
Unsaturation first
Before reading any spectrum, turn the formula into degrees of unsaturation (DoU): DoU = (2C + 2 + N − H − X)/2, where X counts halogens and oxygen is ignored. For C₄H₈O this gives (10 − 8)/2 = 1: the structure contains exactly one ring or one π bond. Butanal (left) spends it on a C=O; but-3-en-1-ol (right) spends it on a C=C and carries an O–H instead. Same formula, different functional groups, so the formula alone cannot decide; it tells you how many rings and π bonds the spectra must account for. A DoU of 4 or more in a compound with six or more carbons should make you look for a benzene ring (one ring plus three C=C).
IR narrows the class
IR answers “which functional group?” fastest. Ethyl acetate shows a strong C=O stretch at 1740 cm⁻¹ and strong C–O stretches near 1240 and 1050 cm⁻¹, with nothing in the 3200–3550 cm⁻¹ O–H region. Learn the landmarks: acyclic ketones near 1715; aldehydes near 1725 with a C–H pair at about 2720 and 2820; carboxylic acids near 1710 with a very broad O–H spanning 2500–3300; alcohols a rounded band at 3200–3550; nitriles a sharp band near 2250; C=C near 1640–1680. Treat an absent band as hard evidence: no band near 3300 rules out O–H and N–H, and no band near 1700 rules out a carbonyl.
¹H NMR builds fragments
Each ¹H signal reports one set of equivalent hydrogens through its chemical shift (environment), integration (how many) and splitting (n + 1 lines for n neighbouring hydrogens on adjacent carbons). For ethyl acetate: 4.12 ppm, quartet, 2H is a CH₂ bonded to the ester oxygen and next to a CH₃; 2.04 ppm, singlet, 3H is the CH₃ on the carbonyl carbon, which has no hydrogen-bearing neighbour; 1.26 ppm, triplet, 3H is a CH₃ next to a CH₂. A quartet and triplet in a 2H : 3H ratio is the signature of an ethyl group. Write each signal as a fragment, then join the fragments so the formula is used exactly once.
One datum decides
When several candidates survive the formula and the IR, look for the single observation only one of them can produce. 1-Propanol (left) and 2-propanol (right) are both C₃H₈O alcohols with an O–H band near 3350 cm⁻¹. 2-Propanol has one CH carrying the oxygen and six equivalent methyl hydrogens, so it shows a 1H septet near 4.0 ppm and a 6H doublet near 1.2 ppm. 1-Propanol instead shows a 2H triplet near 3.6 ppm, a 2H sextet near 1.6 ppm and a 3H triplet near 0.9 ppm. Their ¹³C spectra differ too: two signals for 2-propanol, three for 1-propanol. Name the deciding datum explicitly in every answer.
What to carry forward
Work in a fixed order: formula → degrees of unsaturation; IR → functional group, using absent bands as evidence; ¹H NMR → fragments from shift, integration and splitting; then assemble the fragments and test every candidate against every datum until only one survives.
Problemas de prática (34)
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Preveja o produto 5
Formula and IR first
- IR bands of pentan-2-onePredict which IR absorptions an acyclic ketone shows and which it lacks.
Assemble from ¹H NMR
- Count ¹H signals in 2,2-dimethylpropan-1-olPredict the number of ¹H NMR signals from sets of equivalent hydrogens.
Monte a estrutura a partir do RMN de ¹H
- Padrão de RMN de ¹H do anel aromático do 4-clorotoluenoPreveja o padrão aromático de RMN de ¹H de um benzeno para-dissubstituído.
Decide between isomers
- C₅H₁₀O ketone with two signalsUse symmetry and splitting to choose among C₅H₁₀O ketone isomers.
- C₇H₈O with no O–HPlace the oxygen of an aromatic C₇H₈O isomer from IR and ¹H NMR.
Classificar 2
Formula and IR first
- C₄H₈O₂ with a very broad O–H bandClassify a C₄H₈O₂ compound from the shape and position of its O–H and C=O bands.
Primeiro a fórmula e o IV
- Classifique uma amina C₄H₁₁N pelas bandas de N–HClassifique uma amina como primária, secundária ou terciária pelo número de bandas de estiramento N–H.
Comparar 5
Decide between isomers
- Which C₃H₈O alcohol shows a septet?Compare 1-propanol and 2-propanol and name the ¹H signal that decides between them.
- A C₄H₁₀O alcohol with two ¹³C signalsCompare ¹³C signal counts of three C₄H₁₀O alcohols to pick the most symmetric one.
- C₄H₈O₂: acid or ester?Separate carboxylic acid and ester isomers of C₄H₈O₂ using IR and ¹H NMR.
Decida entre isômeros
- Isômeros de C₃H₆Cl₂: dois sinais de ¹³C, tripleto e quintetoEscolha entre os quatro dicloropropanos combinando o número de sinais de ¹³C e o desdobramento no RMN de ¹H.
- C₄H₆: but-1-ino ou but-2-ino pelo IV e pelo RMN de ¹HDistinga alcinos C₄H₆ terminal e interno de um dieno usando IV e RMN de ¹H.
Calcular 4
Formula and IR first
- C₅H₈O with a strained carbonylCompute degrees of unsaturation for C₅H₈O and use IR to decide how they are spent.
- C₇H₇Br with a 2H singletUse the degree-of-unsaturation count for C₇H₇Br to expect a benzene ring, then place the bromine from IR and ¹H NMR.
- Degrees of unsaturation in C₈H₈O₂Compute degrees of unsaturation for a formula containing oxygen.
Primeiro a fórmula e o IV
- Graus de insaturação de C₅H₉N com banda em 2250 cm⁻¹Calcule os graus de insaturação de uma fórmula com nitrogênio e atribua-os com o IV.
Encontre o erro 3
Decide between isomers
- A pentanone misassignedDiagnose why pentan-2-one cannot produce a spectrum of only a quartet and a triplet.
- An equal-height pair at m/z 136 and 138Diagnose a formula error by reading the M and M+2 isotope pattern of bromine.
Decida entre isômeros
- Dicloroetano a partir de um agrupamento em m/z 98/100/102Conte os cloros a partir de um agrupamento M/M+2/M+4 e posicione-os com RMN de ¹H.
Interprete as evidências 11
Formula and IR first
- A sharp band at 2250 cm⁻¹Identify the functional group responsible for a sharp 2250 cm⁻¹ band in C₄H₇N.
- An M+2 peak one third of MIdentify a halogen from the M : M+2 intensity ratio.
Assemble from ¹H NMR
- C₄H₈O₂: quartet, singlet, tripletAssemble an ester from a quartet at 4.12 ppm, a singlet at 2.04 ppm and a triplet at 1.26 ppm.
- C₃H₇Cl: doublet and septetRecognise an isopropyl group from a 6H doublet and a 1H septet.
- C₈H₁₀: an aromatic ring and an ethyl groupDistinguish ethylbenzene from the xylenes using integration and splitting.
- C₄H₁₀O with no O–H bandCombine formula, IR and ¹H NMR evidence to identify a symmetric ether.
Decide between isomers
- M⁺ 58, base peak 43, one singletCombine mass-spectral, IR and ¹H NMR evidence to identify a C₃H₆O isomer.
- C₃H₇Br: which bromopropane?Tell the two bromopropanes apart from ¹H NMR signal count and splitting.
Monte a estrutura a partir do RMN de ¹H
- C₄H₉Cl com um único simpleto de 9H: cloreto de terc-butilaIdentifique um isômero de C₄H₉Cl cujos nove hidrogênios são todos equivalentes.
- C₅H₁₃N sem banda de N–H: N,N-dietilmetilaminaMonte uma amina terciária a partir da regra do nitrogênio, da ausência de banda de N–H e do desdobramento no RMN de ¹H.
Decida entre isômeros
- Desconhecido C₈H₉Br completo: 1-bromo-4-etilbenzenoCombine EM, graus de insaturação e RMN de ¹H e de ¹³C para posicionar substituintes num anel benzênico.
Trabalhe de trás para frente 1
- Associando espectros ao 1,3,5-trimetilbenzeno (mesitileno)Preveja os dados de RMN de ¹H, de ¹³C e de EM de um areno altamente simétrico e associe-os a um conjunto de dados.
Planeje uma síntese 1
- Distinguindo o-, m- e p-diclorobenzeno pelo RMN de ¹³CEscolha a única medida que distingue os três isômeros do diclorobenzeno.
Explique por quê 2
Assemble from ¹H NMR
- Why ethanol’s CH₂ appears at 3.7 ppmExplain the downfield shift of hydrogens on a carbon bonded to oxygen.
Monte a estrutura a partir do RMN de ¹H
- Por que o C–H de um alcino terminal aparece perto de 2 ppmExplique, pela anisotropia magnética, o deslocamento para campo alto do próton de um alcino terminal.
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