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Cho đến nay các phương pháp phổ học được lựa chọn là: hồng ngoại, tử ngoại, khối lượng và cộng hưởng hạt nhân. Nhiệm vụ của phổ tử ngoại là phát hiện các hệ thống tiếp cách (conjugated), bởi vì sự kích thích các điện tử từ trạng thái cơ bản lên trạng thái kích thích của những hệ thống như thế mới gây nên sự hấp thu trong vùng tử ngoại. Nhưng sự phát triển hiện nay của các kĩ thuật cộng hưởng từ hạt nhân đã đạt đến mức có thể giảm tối đa vai trò của phổ tử ngoại. Hơn nữa, nhiều dao động của nhiều liên kết hoá học không hoạt động trong phổ hồng ngoại, mà hoạt động trong phổ Raman, do đó phổ học Raman được chọn bổ sung cho việc nghiên cứu dao động của các phân tử hữu cơ
Problem C5H10O MW 86 The band at 1716 indicates a carbonyl, probably a ketone The bands at 3000-2850 indicate C-H alkane stretches Problem 1, NMR intrepreted Problem2 C7H14O MW 114 The band at 1718 indicates a carbonyl, probably a ketone The bands at 3000-2850 indicate C-H alkane stretches Prob NMR, interpreted Problem C4H10O MW 74 The broad band at 3339 indicates O-H stretch, probably an alcohol The bands at 3000-2850 indicate C-H alkane stretches The band at 1041 is C-O stretch, consistent with an alcohol Prob NMR, interpreted Problem C6H14O MW 102 The broad band at 3350 indicates O-H stretch, probably an alcohol The bands at 3000-2850 indicate C-H alkane stretches The bands from 1320-1000 indicate C-O stretch, consistent with an alcohol Prob NMR, interpreted Note that the structure has a chiral center and the mixture is racemic Methylene protons adjacent to a chiral center may not be identical Problem C4H8O2 MW 88 Prob 5, IR answer The band at 1743 indicates a carbonyl, probably a saturated aliphatic ester The bands at 3000-2850 indicate C-H alkane stretches The bands in the region 1320-1000 could be due to C-O stretch, consistent with an ester Prob NMR, interpreted Problem C5H10O2 MW 102 Prob 6, IR answer The band at 1740 indicates a carbonyl, probably a saturated aliphatic ester The bands at 3000-2850 indicate C-H alkane stretches The bands in the region 1320-1000 could be due to C-O stretch, consistent with an ester Prob NMR, interpreted Problem C5H10O MW 86 Prob 7, IR answer The band at 1728 indicates a carbonyl, probably an aldehyde; an aldehyde is also suggested by the band at 2719 which is likely the C-H stretch of the H-C=O group The bands at 3000-2850 indicate C-H alkane stretches Prob NMR, interpreted Problem C8H8O MW 120 Prob 8, IR answer The band at 1703 indicates a carbonyl; that the carbonyl is an aldehyde is suggested by the bands at 2828 and 2733 (C-H stretch of the H-C=O group) The bands at 3000-2850 indicate C-H alkane stretches The band (unmarked on the graph below) just to the left of 3000 indicates aromatic C-H stretch Aromatics also show bands in the regions 1600-1585 and 1500-1400 (C-C in-ring stretch), and 900-675 (C-H out-of-plane) Prob NMR, interpreted Note: There are aromatic protons (B, 7-8 ppm), indicating a disubstituted aromatic ring You are not expected to assign the individual peaks in the aromatic region to specific hydrogens on the aromatic ring The aromatic protons show a pattern characteristic of para substitution Problem C9H10O2 MW 150 Prob 9, IR answer The band at 1697 indicates an alpha, beta-unsaturated carbonyl; that the carbonyl is an aldehyde is suggested by the bands at 2828 and 2739 (C-H stretch of the H-C=O group) The bands at 3000-2850 indicate C-H alkane stretches The band (unmarked on the graph below) just to the left of 3000 indicates aromatic C-H stretch Aromatics also show bands in the regions 1600-1585 and 1500-1400 (C-C in-ring stretch), and 900-675 (C-H outof-plane) Prob NMR, interpreted Problem 10 C3H6O2 MW 74 Prob 10, IR answer The band at 1716 indicates a carbonyl The wide band from 3300-2500 is characteristic of the O-H stretch of carboxylic acids The bands at 3000-2850 indicate C-H alkane stretches The bands in the region 1320-1000 indicate the C-O stretch of carboxylic acids Prob 10 NMR, interpreted Spectroscopy Problem 2: C5H10O Spectra from A Spectrum of Spectra, CD Version by Richard A Tomasi Spectroscopy Problem 3: C10H14 Spectra from A Spectrum of Spectra, CD Version by Richard A Tomasi Spectroscopy Problem 4: C10H14O Spectra from A Spectrum of Spectra, CD Version by Richard A Tomasi Spectroscopy Problem 5: C6H12O2 Spectra from A Spectrum of Spectra, CD Version by Richard A Tomasi Spectroscopy Problem 6: C8H8O2 Spectroscopy Problem 7: C6H8O2 Spectroscopy Problem 8: C8H10O Answers to Spectroscopy Problems C5H10O: 2-pentanone Step 1: the obvious stuff • • • Four distinct 1H NMR signals, integrating 2:3:2:3, triplet, singlet, multiplet, triplet Five distinct 13C NMR signals, one of which is a carbonyl, either ketone or aldehyde Other signals all aliphatic IR shows ketone-type carbonyl, but no OH, no alkene Step 2: look closer at 1H signals • • • 3H singlet at ~2.12 ppm almost certainly CH3C(O)– 3H triplet at ~0.95 ppm probably CH3CH2– 2H triplet at ~2.4 ppm probably –CH2CH2C(O)– Step 3: propose a structure • You now have all of the carbons and hydrogens figured out Structure is therefore: O H3C CH3 C5H10O: pentanal Step 1: the obvious stuff • • • • Five distinct 1H NMR signals, integration 1:2:2:2:3, singlet, triplet, multiplet, multiplet, triplet H signal at ~9.75 ppm says aldehyde (sharp singlet) Five 13C NMR signals, one definitely ketone or aldehyde carbonyl, four aliphatic IR confirms ketone or aldehyde carbonyl at 1726 cm-1, no OH (slight blob at 3400-3500 cm-1 is an overtone of the carbonyl plus a trace of water) Step 2: look closer at 1H and IR • • • IR bands at 2719 and ~2800 cm-1 confirm aldehyde (C(O)–H stretch) 2H triplet at ~2.4 ppm consistent with –CH2CH2CHO Note triplet is quite broad due to small coupling to aldehyde H as well as larger coupling to CH2 neighbor 3H triplet at 0.9 ppm certainly due to –CH2CH3 Step 3: propose a structure • There is really only one possibility: O H CH 3 C10H14: 2-butylbenzene Step 1: the obvious stuff • • • • Five distinct 1H NMR signals, integrating 5.6:1:2:3:3, multiplet, multiplet, multiplet, doublet, triplet 5H signal cluster at ~7.2 ppm says aromatic, probably monosubstituted benzene Eight distinct 13C NMR signals, four definitely aliphatic, four probably aromatic, one of which is extremely weak and further downfield from others IR shows aromatic C-H stretches, C=C stretch cluster of weak signals ~1700-1950 cm-1, no other obvious functionality Step 2: look closer at 1H signals • • • • 3H triplet at ~0.8 ppm almost certainly CH3CH2– 3H doublet at ~1.2 ppm almost certainly CH3CH– 1H multiplet at ~2.55 consistent with CH3CH(Ph)CH2– 2H multiplet at ~1.6 ppm consistent with CH3CH2CH– (doublet of quartets, overlapped) Step 3: look closer at 13C signals • • A monosubstituted benzene ring would have four different types of C atoms, in a ratio of 1:2:2:1 Three of these atoms have hydrogens and would be expected to yield stronger signals due to NOE They would also be expected to have similar chemical shifts The fourth carries an alkyl substituent and would be shifted downfield This is what is observed Four aliphatic C atoms is consistent with a butyl chain Signal at 42 ppm consistent with connection to phenyl ring Step 4: propose a structure • H strongly suggests a 2-butyl chain and a linked phenyl group confirms Structure is: H3C CH3 13 C C10H14O: 4-t-butylphenol Step 1: the obvious stuff • • • • • H NMR shows four signals, integration 2:2:1:9, doublet, doublet, broad singlet, singlet Pair of 1H doublets at ~6.85 and ~7.35 ppm strongly suggest paradisubstituted phenyl ring Broad 1H signal at ~4.9 consistent with exchangeable OH Six distinct 13C NMR signals, four probably aromatic (two of which are weak and two strong) and two aliphatic (one weak and one very strong) IR definitely says OH, broad signal at 3250 cm-1 Step 2: look closer at 1H and 13C signals • • 9H 1H singlet at ~1.35 ppm combined with 13C signals at 34 and 31.5 ppm can only be a tertiary butyl group, CH3C– Aromatic 13C signals and 1H doublet pattern say para-disubstituted for sure Step 3: propose a structure • The only possibility given the formula is: CH H 3C OH CH C6H12O2: isobutyl acetate Step 1: the obvious stuff • • • H NMR shows four signals, integration 2:3:1:6, doublet, singlet, multiplet, doublet 13 C NMR shows five signals, a carbonyl, no alkenyl or aryl, one aliphatic probably attached to a heteroatom and three other aliphatics IR carbonyl at 1745 probably ester, no OH Step 2: look closer at the 1H signals • • • • 3H singlet at ~2.05 ppm almost certainly CH3C(O)– 2H doublet at ~3.85 ppm consistent with –CHCH2O– 6H doublet at ~0.95 ppm consistent with (CH3)2CH– 1H multiplet at ~1.9 ppm consistent with (CH3)2CHCH2– Step 3: propose a structure • Given that IR and 13C NMR say it is an ester, and the proton signals suggesting an isobutyl group, the structure can only be: O H 3C O CH3 CH3 C8H8O2: p-methoxybenzaldehyde Note that the molecular formula indicates degrees of unsaturation - that is, rings and/or double bonds • • • Infrared Spectrum o no OH stretches o strong C=O at 1684 - conjugated ketone or aldehyde H H NMR Spectrum O o 1-proton singlet at ~9.9 ppm - aldehyde o 2-proton doublets at 7.85 and 7.0 - para-disubstituted phenyl ring o 3-proton singlet at ~3.9 ppm - OCH3 13 C NMR Spectrum o 190.7 - conjugated aldehyde C o 164.6 - quaternary aromatic C attached to O O o 131.9, 114.3 - aromatic CH (2 each) H3C o 130.0 - quaternary aromatic C attached to C=O o 55.5 - OCH3 C6H8O2: E,E-2,4-hexadienoic acid degrees of unsaturation • • • Infrared Spectrum o broad OH stretch o strong C=O at 1693 - conjugated carbonyl o strong bands at 1637, 1612 - conjugated alkene, probably asymmetric diene H NMR Spectrum o 1-proton broad signal at ~11.8 ppm - COOH o 1-proton wide multiplet at ~7.8 ppm, 2-proton multiplet at ~6.2 ppm, 1-proton doublet at ~5.75 ppm large J suggests R-CH=CHCH=CH-COOH o 3-proton broad doublet at 1.87 ppm - CH-CH3 13 C NMR Spectrum o 172.8 - COOH o 147.3, 140.8, 129.6, 118.2 - CH=CH-CH=CH o 18.4 - CH3 The double bonds are probably both E, as suggested by the large J seen in the doublet at 1.87 and the width of the multiplet at 7.8 (i.e one large coupling constant) O OH C8H10O: 2-phenylethanol degrees of unsaturation • • • Infrared Spectrum o strong OH stretch o no C=O o C-H stretches at >3028, weak overtone bands 1800-1950, medium to weak bands 1604, 1498, 1454 indicate phenyl ring, probably monosubstituted o medium to strong bands 748, 700 also suggest monosubstituted phenyl H NMR Spectrum o 5-proton multiplet 7.15-7.35 ppm - phenyl o two 2-proton triplets, same J - X-CH2-CH2-Y o broad singlet at ~2 ppm - OH The integral is a bit high but this may simply indicate that the OH is exchanging with water in the solvent 13 C NMR Spectrum o 138.7 - quaternary aromatic o 129.1, 128.5 - aromatic CH, two each o 126.4 - aromatic CH, para to substituent o 63.5 - CH2-OH o 39.2 - CH2-phenyl OH