《有机化学波谱》PPT课件.ppt

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1、2023/7/18,第十一章 波谱,2023/7/18,11.1 Infrared(IR)Spectroscopy:an instrumental method for detecting functional groups,1.Hydrocarbons（1）C-H bond strength和bond stiffness都有spsp2sp3的顺序，其中bond stretching有：sp杂化的C-H stretching peaks 在3300cm-1附近，如1-己炔的C-H在3320cm-1；sp2杂化的C-H stretching peaks 在3000-3100cm-1区间，如1-己

2、烯端基C=C-H在3080cm-1，而甲苯苯环C-H在3090cm-1；sp3杂化的C-H stretching peaks 在2800-3000cm-1区间，如辛烷、甲基苯、1-己炔、1-己烯的甲基、亚甲基的C-H。,2023/7/18,(2)烃的C-C的bond stretchings也能给出IR吸收peaks,C-C单键,一般很弱的peaks，在assigning structures中用处不大；C=C在1620-1680 cm-1，苯环C-C在1450-1600 cm-1处有特征尖峰;CC在2100-2260 cm-1;C=C和CC的stretching peaks不总是强，对称取代者

3、则无吸收；,2023/7/18,（3）烯烃C-H bending vibration在600-1000 cm-1之间，吸收峰具体位置可用于判断substitution pattern of the double bond and its configuration（当无释电子或吸电子取代（烃基除外）时，上述位置可靠性很高；否则会超出上述范围）。,2023/7/18,2.Carbonyl functional groups在1630-1780 cm-1之间（1）酰胺的羰基在1630-1690 cm-1之间；（2）酮的羰基在1680-1750 cm-1之间；（3）醛的羰基在1690-1740 cm

4、-1之间；（4）酸的羰基在1710-1780 cm-1之间；（5）酯的羰基在1735-1750 cm-1之间。,2023/7/18,3.Hydroxyl groups of alcohols and phenols 如果用不能形成氢键的溶剂稀释成很稀溶液，醇和酚O-H尖锐的stretching peaks出现在3590-3650 cm-1之间；醇和酚的浓溶液因氢键在3200-3550 cm-1之间出现宽吸收带。4.Carboxylic acid group 如果羰基和羟基的stretching absorptions 在IR谱中都出现，可判断有羧基（当然羰基和羟基也可能单独存在于分子中）。,2

5、023/7/18,5.Amines 很稀的1和2胺在3300-3500 cm-1之间有尖锐的N-H 其中RNH2在3300-3500 cm-1之间有对称和不对称2个尖锐stretching peaks;R2NH在3300-3500 cm-1之间只有1个stretching peak；氢键使得1和2胺的stretching peaks变宽；酰胺会给出羰基及类似的N-H absorption peaks,2023/7/18,11.2 Nuclear Magnetic Resonance(NMR)and Mass Spectrometry:Tools for Structure Determinat

6、ion,1.1H-NMR 质子的磁矩在无外加磁场时随机分布取向；在有外加磁场时，因质子的自旋(m=+1/2,m=-1/2)，质子磁矩与外加磁场同向（自旋态，低能态）或反向（自旋态，高能态）排列。由低能态向高能态的反转需要能量，在NMR光谱仪这一能量来自射频范围的电磁辐射electromagnetic radiation in the RF(radio frequency)region。实际上，在质子周围存在着电子，电子（bond of a C-H group）运动产生小磁场，称作诱发磁场（induced field）并与外加磁场方向相反，这意味着质子真正感受到的磁场比外加磁场稍低，即电子对质子

7、具有屏蔽作用。高屏蔽的质子与弱屏蔽的质子当然不会在相同的频率下吸收能量。去屏蔽质子将在高频（低场）吸收能量。,2023/7/18,离域的电子既可能屏蔽也可能去屏蔽邻近的质子。由于去屏蔽效应，苯的H(absorb at)7.27，1，4-二甲基苯的芳H(absorb at)7.05.而乙炔和其它端炔的质子由于离域的电子的屏蔽效应在高场吸收。如果仅考虑C在不同杂化态的相对电负性，与C相连的质子应当有如下顺序：(low field strength)spsp2sp3(high field strength)而实际端炔在2.0和3.0之间吸收,端炔向高场的位移是由于端炔三键电子的屏蔽效应：(low f

8、ield strength)sp2spsp3(high field strength),2023/7/18,对映异构式的H原子有相同的化学位移只给出单一1H-NMR信号(Enantiotopic hydrogen atoms have the same chemical shift and give only one 1H-NMR signal)，如CH3-CH2-Br的-CH2-,(当溶于手性溶剂时，则化学位移不同)。如果用基团Q置换2个H中的1个，得到的是非对映体，这2个H被称为是非对映的(diastereotopic),如氯乙烯的=CH2其2个H就是非对映的，因此氯乙烯给出3个H信号：,

9、2023/7/18,对于2-丁醇的-CH2-,也是非对映异构的，应具不同化学位移，显示2个1H-NMR信号：,2023/7/18,信号的裂分源于自旋-自旋耦合,耦合效应可通过成键电子传递，但如果耦合质子被3个以上的键分隔一般观察不到耦合(Signal splitting arises from a phenomenon known as spin-spin coupling.Spin-spin coupling effects are transferred primarily through the bonding electrons and are not usually observed

10、 if the coupled protons are separated by more than three bonds.Note:Long-range coupling can be observed over more than three bond lengths in some conformationally inflexible molecules and systems where bonds are involved.):如(CH3)3COCH3为2个单峰。,2023/7/18,对于化学等效(chemically equivalent,homotopic)或对映异构的质子没

11、有峰的裂分。即具有相同化学位移的质子之间不会出现信号裂分。如乙烷只有1个单峰。同样对于CH3OCH2CN为2个单峰。裂分时峰之间间隔称为耦合常数（用hertz表示），由于耦合常数完全是内在的力，因此其大小与施加磁场大小无关(The separation of these peaks in frequency units is called the coupling constant and is abbreviated Jab.Coupling constants are generally reported in hertz.Because coupling is caused entire

12、ly by internal forces,the magnitudes of coupling constants are not dependent on the magnitude of the applied field.For example,coupling constants measured(in hertz)on an instrument operating at 60MHz will be the same as those measured on an instrument operating at 300MHz or any other magnetic field

13、strength.)。,2023/7/18,1个H自旋耦合能使邻近H裂分成2重峰（1:1）;2个H自旋耦合能使邻近H裂分成3重峰（1:2:1）;3个H自旋耦合能使邻近H裂分成4重峰（1:3:3:1）;,2023/7/18,质子NMR还有其它特点，使得在判断某一化合物结构时帮助不大(Proton NMR spectra have other features,however,that are not all helpful when we try to determine the structure of a compound.)。如：1、信号重叠（Signals may overlap）Thi

14、s happens when the chemical shifts of the signals are very nearly the same.In the 60-MHz spectrum of ethyl chloroacetate we see that the singlet of the CH2Cl group falls directly on top of one of the outermost peaks of the ethyl quartet.Using NMR spectrometers with higher magnetic field strength(cor

15、responding to 1H resonance frequencies of 300,500,or 600MHz)often allows separation of signals that would overlap at lower magnetic field strengths.,2023/7/18,2、非邻近原子的质子也能发生自旋-自旋耦合(Spin-spin couplings between the protons of nonadjacent atoms may occur.)This long-range coupling happens frequently in

16、compounds when-bonded atoms intervene between the atoms bearing the coupled protons and in some cyclic molecules that are rigid.,2023/7/18,3、芳基的裂分图形有时难以分析(The splitting patterns of aromatic groups can be difficult to analyze.)A monosubstituted benzene ring(a phenyl group)has three different kinds of

17、 protons:The chemical shifts of these protons may be so similar that the phenyl group gives a signal that resembles a singlet.Or the chemical shifts may be different,and because of long-range coupling,the phenyl group signal appears as a very complicated multiplet.,2023/7/18,在室温附近由C-C单键连接的基团快速旋转，所得H

18、的NMR反映的是某1个H所处的平均环境(At temperatures near room temperature,groups connected by carbon-carbon single bonds rotate very rapidly(unless rotation is prevented by some structural constraint,e.g.,a rigid ring system).Because of this,when we determine spectra of compounds with single bonds that allow rotati

19、on,the spectra that we obtain often reflect the individual hydrogen atoms in their average environment-that is,in an environment that is an average of all the environments that the protons have as a result of the group rotations.)。如溴乙烷的甲基给出的是1个信号(3重峰)。,2023/7/18,乙醇的羟基H和亚甲基H虽然邻近，但一般观察不到耦合引起的裂分(In ord

20、inary ethanol we observe no signal splitting arising from coupling between the hydroxyl proton and the protons of the CH2-group even though they are on adjacent atoms.)。但是，对于很纯的乙醇，羟基H裂分成3重峰，亚甲基H裂分成8重峰，显然，羟基H和亚甲基H之间发生了耦合(If we were to examine a 1H NMR spectrum of very pure ethanol,however,we would fi

21、nd that the signal from the hydroxyl proton was split into a triplet and that the signal from the protons of the CH2-group was split into a multiplet of eight peaks.Clearly,in very pure ethanol the spin of the proton of the hydroxyl group is coupled with the spins of the protons of the CH2-groups.)。

22、,2023/7/18,是否发生上述耦合决定于质子在某一特定乙醇分子上的时间。质子与带孤对电子的电负性原子相连时，能发生快速化学交换，即从1个分子交换到另1个分子。对于很纯的乙醇化学交换慢，可观察到羟基质子的裂分和其引起的裂分；对于一般的乙醇，酸性或碱性杂质催化化学交换，并且交换很快，以至于羟基质子给出的是不裂分的信号、亚甲基给出的也仅是被与甲基质子的耦合所裂分。快速的交换导致自旋去耦(Whether coupling occurs between the hydroxyl protons and the methylene protons depends on the length of ti

23、me the proton spends on a particular ethanol molecule.Protons attached to electronegative atoms with lone pairs such as oxygen(or nitrogen)can undergo rapid chemical exchange.That is,they can be transferred rapidly from one molecule to another.The chemical exchange in very pure ethanol is slow and,a

24、s a consequence,we see the signal splitting of and by the hydroxyl proton in the spectrum.In ordinary ethanol,acidic and basic impurities catalyze the chemical exchange;the exchange occurs so rapidly that the hydroxyl proton gives an unsplit signal and that of the methylene protons is split only by

25、coupling with the protons of the methyl group.We say,then,that rapid exchange causes spin decoupling.)。,2023/7/18,自旋去耦在醇、胺和羧酸1H-NMR都有，OH和NH的质子一般不被裂分(Spin decoupling is found in the 1H-NMR spectra of alcohol,amines,and carboxylic acids,and the signals of OH and NH protons are normally unsplit.)。能够进行快

26、速化学交换的质子（如与O或N相连的质子）通过放入到D2O中可很容易地检出。质子很快被D所替代，质子信号从谱图上消失(Protons that undergo rapid chemical exchange(i.e.,those attached to oxygen or nitrogen)can be easily detected by placing the compound in D2O.The protons are rapidly replaced by deuterons,and the proton signal disappears from the spectrum.)。,2

27、023/7/18,2.13C-NMR 有机物分子每一个特别C原子只给出一个13C NMR峰简化了析谱过程。1H-NMR邻近质子彼此耦合使得每个H的信号变成多重峰，但由于每100个C原子只有1个C原子是13C核（1.1%丰度），因此2个13C原子相邻的可能性只有万分之一(1.1%1.1%),基本可以排除相邻C原子互相裂分信号成为多重峰的可能性。13C的低丰度和其固有的低灵敏性还有另外1个效果：13C NMR只能用具平均信号的FT NMR获得(One aspect of 13C NMR spectra that greatly simplifies the interpretation proce

28、ss is that each unique carbon atom in an ordinary organic molecule produces only one 13C NMR peak.There is no carbon-carbon coupling that causes splitting of signals into multiple peaks.Recall that in 1H-NMR spectra,hydrogen nuclei that are near each other(within a few bonds)couple with each other a

29、nd cause the signal for each hydrogen to become a multiple of peaks.This does not occur for adjacent carbons because only one carbon atom of every 100 carbon atoms is a 13C nucleus(1.1%natural abundance).Therefore,the probability of there being two 13C atoms adjacent to each other in a molecule is o

30、nly about 1 in 10,000(1.1%1.1%),essentially eliminating the possibility of two neighboring carbon atoms splitting each others signal into a multiplet of peaks.The low natural abundance of 13C nuclei and its inherently low sensitivity also have another effect:Carbon-13 NMR spectra can be obtained onl

31、y by pulse FT NMR spectrometers,where signal averaging is possible.)。,2023/7/18,虽然无C-C裂分，连接于C的H能使13C信号裂分，这可通过仪器参数的选择对质子-碳相互作用进行去耦，所得谱图称为宽频质子去耦(broadband(BB)proton decoupled)。与1H-NMR一样，13C-NMR 在C核低电子密度时去屏蔽使得向左在低场；相对高的电子密度屏蔽效应向右在高场。,2023/7/18,2023/7/18,11.3 Mass Spectrometry,M+1元素：C,H,N;M+2元素：O,S,Br,C

32、l参考下述方法确定分子式:如果M+不是基峰，首先将所有峰强转换并使M+的峰强为100，然后M+是奇数还是偶数？根据N规则，如果是偶数，化合物必定含有偶数个N原子（0是偶数）；M+1的相对丰度显示C原子的数目。C数=（M+1）的相对丰度/1.1(13C对M+1峰的贡献最大，而13C的自然丰度为1.1%)；M+2峰的相对丰度可说明是否有S(4.4%),Cl(33%),Br(98%);通过确定H原子数和添加需要适当数目的O确立分子式。,2023/7/18,11.4 Ultraviolet-Visible Spectroscopy,最大吸收波长,max 摩尔吸光度(molar absorptivity

33、),(在老的文献中被称为摩尔消光系数molar extinction coefficient);A=Cl A为在某一特定波长（）的吸光度；C为样品的摩尔浓度；l为光束在样品池中的光程(cm)。,2023/7/18,如2,5-二甲基-2,4-己二烯溶于甲醇，在其最大吸收波长(242.5nm)的摩尔吸光度为13,100M-1cm-1,在化学文献中报告为 2,5-二甲基-2,4-己二烯,maxmethanol 242.5nm(=13,100)烯和非共轭二烯最大吸收一般低于200nm，空气中氧同样也吸收，因此在测试时必须采用特殊的隔绝空气的技术。分子含有共轭重键的化合物在长于200nm波长有最大吸收，

34、如1，3-丁二烯在217nm。当1分子吸收最长波长的光，1个电子从最高占据轨道(highest occupied molecular orbital,HOMO)被激发到最低空轨道(lowest unoccupied molecular orbital,LUMO)。一般，共轭重键数目越大，吸收波长越长。,2023/7/18,维生素A的前体-胡萝卜素（叶红素）(-carotene)有11个共轭双键，最大吸收在497nm(蓝绿色)，使人感觉到红橙色：,2023/7/18,番茄红素(lycopene)也含有11个共轭双键，对番茄显示红色起到了一定作用。番茄红素在505nm有最大吸收，而且吸收强度很高。

35、（1kg成熟的新鲜番茄能分离出0.02g番茄红素。）：,2023/7/18,有C=O双键的化合物在UV区也吸收光。如，丙酮在280nm有很宽的吸收对应于未共享电子对中一对电子的1个电子（非键电子或称为“n”电子）向C=O双键*的激发跃迁：,2023/7/18,对于C=O双键与C=C双键共轭的化合物有对应于n*激发和*激发的最大吸收。其中n*最大吸收出现在较长的波长处但弱得多（即摩尔吸光度较小）：,2023/7/18,第十一章 习题,1、化合物A和B具有相同的分子式C6H8。二者在铂存在下都能与2摩尔氢气反应生成环己烷。A在其宽频去耦13C-NMR谱中有3个信号，而B只有2个信号。A在256nm

36、有最大吸收，而B在波长大于200nm时没有吸收。A和B的结构式。2、化合物D、E、F有相同的分子式C5H6。在铂催化剂存在下，都能吸收3摩尔氢成为戊烷。化合物E和F在3300cm-1附近有IR吸收；而化合物D在此区域没有IR吸收。化合物D和E在230nm附近有UV-Vis最大吸收，而化合物F在200nm以上没有最大吸收。判定D、E和F的结构。,2023/7/18,第十一章 习题答案,1、化合物A和B具有相同的分子式C6H8。二者在铂存在下都能与2摩尔氢气反应生成环己烷。A在其宽频去耦13C-NMR谱中有3个信号，而B只有2个信号。A在256nm有最大吸收，而B在波长大于200nm时没有吸收。A和B的结构式。,2023/7/18,2、化合物D、E、F有相同的分子式C5H6。在铂催化剂存在下，都能吸收3摩尔氢成为戊烷。化合物E和F在3300cm-1附近有IR吸收；而化合物D在此区域没有IR吸收。化合物D和E在230nm附近有UV-Vis最大吸收，而化合物F在200nm以上没有最大吸收。判定D、E和F的结构。能吸收3摩尔氢成为戊烷证明三化合物的不饱和度为3；E和F在3300cm-1附近有IR吸收说明二者是端炔；D和E在230nm附近有UV-Vis最大吸收，F在200nm以上没有最大吸收，说明D和E为共轭结构，而F为非共轭结构。根据以上分析，可得出：,