1、第一章第一章 紫外紫外- -可见光谱可见光谱一基本原理二仪器装置三实验技术四紫外吸收与分子结构的关系五紫外吸收光谱的应用一、基本原理一、基本原理电磁波可见光光谱练习1返回白光的色散紫外吸收光谱的形成返回分子在入射光的作用下发生了价电子的跃迁，吸收了特定波长的光波形成。 E = hE = h = = h hc c/ / h h为普朗克常数为普朗克常数 6.636.631010-34-34 J J. .s ssamplereferencedetectorI0I0I0Ilog(I0/I) = A200700, nmmonochromator/beam splitter opticsUV-VIS so
2、urces郎伯-比耳定律A吸光度T透射率摩尔吸光系数（Lmol-1cm-1 ）c溶液的浓度（mol /L）l光在溶液中经过的距离 （比色池厚度，cm） I0入射光强度I1透过光强度 返回A = lg (I0/I1) = lg (1/T) = . c . lmax 最大吸收波长（一般有多个，数目等于最大吸收峰数目）max 最大吸收时的摩尔吸光系数（每个max 都有一个对应的max）紫外光紫外光远紫外(10200nm)近紫外(200400nm) 空气中的O2, N2, CO2, 潮气有强吸收玻璃对波长 4nE带带（ethylenic band,乙烯型谱带乙烯型谱带）：苯环的* 跃迁产生，当共轭系统
3、有极性基团取代时， E带相当于K带，吸收强度大，lg 4nB带带（benzenoid band, 苯型谱带苯型谱带）：苯环的*跃迁产生，中等强度吸收峰，lg 3，特征是峰形有精细结构nR带带（德语radikalartig，基团型基团型）：未共用电子的n*跃迁产生，特征是吸收强度弱，lg 2返回德语 konjugieren konjugierte 英语 conjugateconjugated1.1.生色团（发色团）生色团（发色团） 使分子在紫外使分子在紫外- -可见光区产生吸收带的基团，一般可见光区产生吸收带的基团，一般为带为带电子基团电子基团(C=C,C C,(C=C,C C,苯环苯环,C=O
4、,N=N,NO,C=O,N=N,NO2 2) )p 发色团间不共轭：吸收光谱包括发色团各自的吸收带p 发色团间形成共轭：原来各自发色团的吸收带消失,产生新的吸收谱带(波长和吸收强度比原来明显加大)光谱术语2.2.助色团助色团p 本身在紫外区和可见区不产生吸收，当连接到发色团后，使发本身在紫外区和可见区不产生吸收，当连接到发色团后，使发色团的吸收波长移向长波色团的吸收波长移向长波, , 同时使吸收强度增加。同时使吸收强度增加。p 助色团一般为带有孤对电子的原子或原子团，如助色团一般为带有孤对电子的原子或原子团，如-OH,-OR,-NHR, -OH,-OR,-NHR, -SR, -Cl, -Br,
5、 I-SR, -Cl, -Br, I等等例例:254nm(B带带)270nm(B带带)OH a a) ) 助色团与助色团与键相连时键相连时, ,与与键形成键形成p-p-共轭或共轭或- -超共轭超共轭, ,使使键电子容易被激发键电子容易被激发, ,发生红移发生红移. . CCClHHH b) b) 助色团与羰基相连时助色团与羰基相连时, ,使羰基的使羰基的n-n-* *跃迁吸收带向短波跃迁吸收带向短波方向移动方向移动, , 即蓝移。其中带有孤对电子的原子或原子团蓝即蓝移。其中带有孤对电子的原子或原子团蓝移更加明显。移更加明显。 H-C-H H-C-H CHCH3 3-C-H -C-H CHCH3
6、 3-C-C-CHCH3 3 CHCH3 3-C-C-OHOHO OO OO OO Omax/nm 310 290 279 204 5 17 16 41 溶剂溶剂 异戊烷 己烷 己烷 乙醇OCRYOYCnnY= -NH2,-OH,-OR 等助色基团 K 带红移 R 带蓝移 p共轭分子轨道能级图共轭分子轨道能级图nCOCOYYn 123n常见生色团和助色团返回发色团举例跃迁max, nm溶剂C=C乙烯 _ *17115,000己烷CC1-己炔 _ *18010,000己烷C=O乙醛n _ * _ *2901801510,000己烷己烷N=OCH3NO2n _ * _ *275200175,000
7、乙醇乙醇C-XCH3BrCH3In _ *n _ *205255200360己烷己烷红移红移：吸收带向长波方向移动蓝移蓝移：吸收带向短波方向移动/nmA红移红移蓝移蓝移增色效应增色效应减色效应减色效应max3. 红移和蓝移红移和蓝移4. 增色效应和减色效应增色效应和减色效应增色效应增色效应：使吸收带的吸收强度增加的效应 减色效应减色效应：使吸收带的吸收强度降低的效应5. 末端吸收末端吸收吸收峰随着波长变短而强度增强,直至仪器测量的极限,而不显示峰型(这主要是因为其最大吸收在短波长处),这种极限处吸收称为末端吸收。 返回1213241、吸收峰、吸收峰 2、谷、谷 3、肩峰、肩峰 4、末端吸收、末
8、端吸收A影响紫外吸收的因素n共轭体系的形成使吸收红移 n取代基效应n超共轭效应 ：烷基与共轭体系相连可产生少量红移 n空间效应：空间位阻，构型，跨环效应n跃迁类型n外部因素：溶剂效应 ，温度，pH值影响返回共轭效应EE*max 223nm( 22600)CH3CH=CHCH=CH2max(K) 234nm( 14000)max(K) 244nm( 5000)(CH3)2C=CH-C-CH3OC-HO返回取代基的影响 当共轭双键的两端有容易使电子流动的基团当共轭双键的两端有容易使电子流动的基团(给电子给电子基或吸电子基基或吸电子基)时，极化现象显著增加。时，极化现象显著增加。给电子基：给电子基：
9、未共用电子对的流动性很大，能够形成未共用电子对的流动性很大，能够形成p- 共轭，降低能量，共轭，降低能量， max红移，红移， max增加。如：增加。如：-N(C2H5)2， -N(CH3)2，-NH2，-OH，-OCH3 ，-NHCOCH3吸电子基：吸电子基：易吸引电子而使电子容易流动的基团，如：易吸引电子而使电子容易流动的基团，如：-NO2, -COR等。产生等。产生 电子的永久性转移，电子的永久性转移， max红移。红移。 电子流动性增加，吸收强度增加。电子流动性增加，吸收强度增加。给电子基与吸电子基同时存在：给电子基与吸电子基同时存在：产生分子内电荷转移产生分子内电荷转移吸收，吸收，
10、max红移，红移， max增加。增加。返回空间位阻 0 10o 90 o （共轭最差） 180 omax 466nm 370nm 490nm NO2NO2CH3NO2C2H5NO2t C4H9t C4H9max 8900 6070 5300 640 共轭逐渐变差共轭逐渐变差邻位效应邻位效应空间位阻（K带） 构型影响max (nm) 228 296 224 280max 16400 29000 24000 10500 返回溶剂：乙醇E带 K 带 B带被淹没E带 K 带 跨环效应 OOmax 300.5nm 280nmmax 292 150 返回电子云部分重叠电子云部分重叠溶剂效应1-己烷 2-9
11、5%乙醇 3-水 练习2n丙酮丙酮溶剂效应使精细结构消失返回水环己烷蒸气态温度的影响温度降低减小了振动和转动对吸收带的影响，呈现电子跃迁的精细结构返回pH值影响苯酚的紫外光谱 苯胺的紫外光谱 p 在不同的pH 值介质中生成相应阳离子阳离子或阴离子阴离子, 吸收波长发生变化。max/nm (max)230(8600) 203(7500) 280(1470) 254(160)211(6200) 236(9400) 270(1450) 287(2600)原因原因成盐后氮原子的未成键电子消失, 氨基的助色作用也随之消失，变为与苯的吸收类似。酚羟基含有两对孤对电子, 当形成酚盐负离子后氧原子孤电子对增加
12、到3对, 使p-共轭作用进一步增强。NH2NH3+H+OH-OHO_H+OH-E2带B带返回返回可见光颜色与波长的关系 紫光 400435nm 蓝光 450480nm 青光 480490nm 蓝光绿 490500nm 绿光 500560nm 黄光绿 560580nm 黄光 580595nm 橙光 595605nm 红光 605700nm 可见光谱与颜色蓝光波长：蓝光波长：450480 nm 化合物呈什么颜色？化合物呈什么颜色？分别为橙黄色和红色分别为橙黄色和红色返回二、仪器装置二、仪器装置主要包括光源、分光系统、吸收池、检测系统和记录系统光源、分光系统、吸收池、检测系统和记录系统等五个部分。
13、返回三、实验技术三、实验技术n分光光度计的校正用已知最大吸收波长的标准样品校正仪器，如：蒽醌 max = 323nm（EtOH），水杨醛 326nm（EtOH）n溶剂的选择对200-400 nm的紫外光无吸收与样品不发生化学作用常用的溶剂：己烷、环己烷、乙腈、甲醇、乙醇、异丙醇、乙醚、二氧六环等返回练习3溶剂的选择返回化合物低于溶剂化合物低于溶剂最低波长极限最低波长极限的吸收将受严重干扰！的吸收将受严重干扰！四、紫外吸收与分子结构四、紫外吸收与分子结构的关系的关系n饱和烃化合物 （烷烃和卤代烷烃的紫外吸收波长短，可用于紫外吸收测试溶剂）n简单的不饱和化合物 n共轭系统的紫外吸收光谱 n芳环化合
14、物的紫外吸收光谱 返回卤代烃（教材表1-4）化合物溶剂max（nm）maxCF4蒸气105.2CH3F蒸气 173160153169370CHCl3蒸气 175175.5950CH3Br蒸气 204175200CH2Br2异辛烷 200.51981050970CHBr3异辛烷223.41980CH3I蒸气异辛烷257257.5230370CHI3异辛烷349.4307.2274.921408301310返回简单的不饱和化合物n简单烯烃、炔烃n位于真空紫外区，助色基团的存在可以使波长红移n简单醛、酮nn*跃迁在紫外区，为弱吸收（比较特征）返回简单醛酮n化合物溶剂maxmax甲醛蒸气3041817
15、518000乙醛蒸气3105丙酮蒸气28912.5182100002-戊酮己烷278154-甲基-2-戊酮异辛烷28320环戊酮异辛烷30018环己酮异辛烷29115环辛酮异辛烷291144lg*返回n4lg*共轭系统的紫外吸收光谱n共轭烯烃共轭烯烃 n，不饱和醛和酮不饱和醛和酮n， 不饱和羧酸和酯不饱和羧酸和酯返回共轭烯烃K带吸收位置的计算波长增加因素波长增加因素max(nm)1. 开链或非骈环共轭双烯基本值 217双键上烷基取代增加值 +5环外双烯 +52. 同环共轭双烯或共轭多烯骈环异环共轭双烯基本值 214同环共轭双烯 253延长一个双键增加值 +30烷基或环残基取代 +5环外双键 +
16、5助色基团OAc 0OR +6SR +30Cl、Br +5NR2 +60此公式最多可适用于此公式最多可适用于共轭四烯共轭四烯，对于共轭五烯及以上的多烯，用，对于共轭五烯及以上的多烯，用Fieser-KuhnFieser-Kuhn公式（不作要求）。公式（不作要求）。计算举例4个环残基取代 +54 计算值 237 nm（238 nm） （1）共轭双烯基本值 2174个环残基或烷基取代 +54 环外双键 + 5 计算值 242 nm （243 nm）（2）非骈环双烯基本值 217 5个烷基取代 +55 3个环外双键 +53 延长2个双键 +302计算值 353 nm（355 nm）AcO（3）同环共
17、轭双烯基本值 253 （4）骈环异环异环共轭双烯 基本值 2143个环残基取代 +531个环外双键 + 5计算值 234 nm（235 nm） 返回骈环同环同环共轭双烯结构？，不饱和醛、酮 CCCOCCCCCO，不饱和醛、酮计算举例（1）六元环，-不饱和酮 基本值 215 2个取代 122 1个环外双键 5 计算值 244nm (251nm） O（2）六元环，-不饱和酮 基本值 215 2个烷基取代 122 O1个烷基取代 10 2个环外双键 52 计算值 259nm（258nm） O（3）直链，-不饱和酮 基本值 215延长1个共轭双键 30 1个烷基取代 181个烷基取代 18 计算值 2
18、81nm（281nm） 1). 1). 溶剂极性对溶剂极性对，不饱和醛、酮不饱和醛、酮* *跃迁谱带的影响跃迁谱带的影响基态基态激发态激发态溶剂极性溶剂极性 * *跃迁的吸收跃迁的吸收谱带发生谱带发生红移红移例如例如: : 溶剂由环己烷溶剂由环己烷变为乙醇，变为乙醇，红移红移溶剂效应溶剂溶剂己烷己烷乙醚乙醚乙醇乙醇甲醇甲醇水水介电常数介电常数2.04.325.83181max/nm (max)* *229.5 (12600)230 (12600)237 (12600)238 (10700)244.5 (10000)nn* *327(97.5)326(96)315(78)312(74)305(6
19、0)异亚丙基丙酮异亚丙基丙酮CH3COCH=C(CH3)2吸收带与溶剂极性的关系吸收带与溶剂极性的关系2).2).溶剂极性对溶剂极性对，不饱和醛、酮不饱和醛、酮nn* *跃迁谱带的影响跃迁谱带的影响基态基态基发态基发态溶剂极性溶剂极性 nn* *跃迁的吸收跃迁的吸收谱带发生蓝移谱带发生蓝移例如例如: : 环己烷改环己烷改乙醇乙醇: : 蓝移蓝移7nm, 7nm, 水水: : 蓝移蓝移8nm8nm溶剂溶剂己烷己烷乙醚乙醚乙醇乙醇甲醇甲醇水水介电常数介电常数2.04.325.83181max/nm (max)* *229.5 (12600)230 (12600)237 (12600)238 (10
20、700)244.5 (10000)nn* *327(97.5)326(96)315(78)312(74)305(60)异亚丙基丙酮异亚丙基丙酮CH3COCH=C(CH3)2吸收带与溶剂极性的关系吸收带与溶剂极性的关系，不饱和醛、酮紫外K吸收带max溶剂校正 返回思考如何校正？（P17,表1-9）预测预测max = 计算计算 max （+ or -？）？） 溶剂甲醇氯仿二氧六环乙醚己烷环己烷水(nm) 0+1+5+7+11+11-8减号！减号！,-不饱和羧酸和酯计算规则CCCCCO计算举例CH3-CH=CH-CH=CH-COOH单取代羧酸基准值 208延长一个共轭双键 30烷基取代 18 计算值
21、 256nm （254nm） 返回芳环化合物的紫外吸收光谱溶剂：异辛烷 溶剂：庚烷 基团吸电子能力越强，影响越大芳环化合物的紫外吸收光谱K带B带R带芳环化合物的紫外吸收光谱返回五、紫外吸收光谱的应用五、紫外吸收光谱的应用n化合物的鉴定 n异构体的确定 （ 互变异构）n成分分析（定量分析，与色谱连用，如液相色谱）n位阻作用的测定 n氢键强度的测定 n紫外光谱法在工业生产中的应用n纯度检查(定量):如乙醇中少量苯的检查 化合物的鉴定推测化合物分子骨架推测化合物分子骨架（鉴定共轭体系、羰基的存在）： 200-800nm 没有吸收，说明分子中不存在共轭结构 (-C=C-C=C-， -C=C-C=O,苯
22、环等），可能为饱和化合物。 200-250nm 有强吸收峰，为发色团的K带，分子中有上述共轭结构单元。同样在260，300，330nm处有高强度K吸收带，则表示有三、四和五个共轭体系存在。 250-300nm 有中等强度的吸收峰（=2001000），则表示有B带吸收，体系中可能有苯环存在。如果苯环上有共轭的生色基团存在时，则可以大于10000。 270-350nm 有弱吸收峰，为R带， :14N, 2H, 10BEven atomic mass & odd numberI = +1, 0 & -1spinning charged nucleus creates a magnetic field
23、Similar to magnetic field created by electric current flowing in a coilMagnetic momentdetermines the direction of the magnetic field around a current-carrying wire and vice-versa related to the relative sensitive of the NMR signalmagnetic moment (m) is created along axis of the nuclear spinwhere: p
24、angular momentum g gyromagnetic ratio (different value for each type of nucleus)magnetic moment is quantized (m)m = I, I-1, I-2, , -Ifor common nuclei of interest: m = + & -IIhm mm m g g 2pg gm m Bo= g g h / 4 In the absence of external field,each nuclei is energetically degenerateAdd a strong exter
25、nal field (Bo) and the nuclear magnetic moment: aligns with (low energy) against (high-energy) Magnetic moment are no longer equivalentMagnetic moments are oriented in 2I + 1 directions in magnetic field)1( II)1(cos IIm oBImEm m For I = 1/2The energy levels are more complicated for I 1/2Magnetic mom
26、ents are oriented in one of two directions in magnetic field (for I =1/2)Difference in energy between the two states is given by: E = g g h Bo / 2 where:Bo external magnetic fieldh Plancks constantg gyromagnetic ratio Frequency of absorption: = g g Bo / 2 Transition from the low energy to high energ
27、y spin state occurs through an absorption of a photon of radio-frequency (RF) energyRFClassical DescriptionSpinning particle precesses around an applied magnetic fieldA Spinning Gyroscopein a Gravity Field)1(cos IIm Classical DescriptionAngular velocity of this motion is given by:w wo = g gBowhere t
28、he frequency of precession or frequency is: = g gBo/2 Same as quantum mechanical descriptionClassical DescriptionNet MagnetizationNuclei either align with or against external magnetic field along the z-axis.Since more nuclei align with field, net magnetization (Mo, MZ) exists parallel to external ma
29、gnetic field.Net Magnetization along +Z, since higher population aligned with Bo.Magnetization in X,Y plane (MX,MY) averages to zero.MozxyyxzBoBoNet Magnetization in a Magnetic FieldMoyxzxyzBoBoBo 0 E = h a ab bBoClassic View:- Nuclei either align with or against external magnetic field along the z-
30、axis.- Since more nuclei align with field, net magnetization (Mo) exists parallel to external magnetic fieldQuantum Description:- Nuclei either populate low energy (a a, aligned with field) or high energy (b b, aligned against field) - Net population in a a energy level.- Absorption of radio- freque
31、ncy promotes nuclear spins from a a b b.MoyxzxyzBoBoWe have a net magnetization precessing about Bo at a frequency of wo with a net population difference between aligned and unaligned spins. Now What?Perturbed the spin population or perform spin gymnasticsBasic principal of NMR experimentsThe Basic
32、1D NMR ExperimentExperimental details will effect the NMR spectra and the corresponding interpretation B1 off(or off-resonance)MozxB1zxMxyyyw w1w w1Right-hand ruleresonant condition: frequency (w1) of B1 matches Larmor frequency (wo)energy is absorbed and population of a and b states are perturbed.
33、Mo now precesses about B1 (similar to Bo) for as long as the B1 field is applied.Again, keep in mind that individual spins flipped up or down(a single quanta), but Mo can have a continuous variation.Classical DescriptionNMR PulseApplying the B1 field for a specified duration (Pulse length or width)
34、Net Magnetization precesses about B1 a defined angle (90o, 180o, etc)B1 off(or off-resonance)MozxB1zxMxyyyw w1w w1w w1 = g gB190o pulseClassic View:- Apply a radio-frequency (RF) pulse a long the y-axis- RF pulse viewed as a second field (B1), that the net magnetization (Mo) will precess about with
35、an angular velocity of w w1- precession stops when B1 turned offQuantum Description:- enough RF energy has been absorbed, such that the population in a a/b b are now equal- No net magnetization along the z-axisB1 off(or off-resonance)MozxB1zxMxyyyw w1w w1w w1 = g gB190o pulseBo 0 E = h a ab bPlease
36、Note: A whole variety of pulse widths are possible, not quantized dealing with bulk magnetizationWhat Happens Next?The B1 field is turned off and Mxy continues to precess about Bo at frequency wo. zxMxyReceiver coil (x)y NMR signalw woFID Free Induction DecayyyyMxy is precessing about z-axis in the
37、x-y plane Time (s) = g gBo/2 RF pulse along YDetect signal along XXyClassical DescriptionObserve NMR SignalRemember: a moving magnetic field perpendicular to a coil will induce a current in the coil.The induced current monitors the nuclear precession in the X,Y plane The FID reflects the change in t
38、he magnitude of Mxy as the signal is changing relative to the receiver along the y-axisAgain, the signal is precessing about Bo at its Larmor Frequency (wo).RF pulse along YDetect signal along XThe appearance of the FID depends on how the frequency of the signal differs from the Larmor FrequencyTime
39、 DomainFrequency DomainLarmor FrequencySo, the NMR signal is collected in the Time - domainBut, we prefer the frequency domain.Fourier Transform is a mathematical procedure that transforms time domain data into frequency domainAfter the NMR Signal is Generated and the B1 Field is Removed, the Net Ma
40、gnetization Will Relax Back to Equilibrium Aligned Along the Z-axisT2 relaxationTwo types of relaxation processes, one in the x,y plane and one along the z-axisPeak shape also affected by magnetic field homogeneity or shimmingNMR Relaxationa)No spontaneous reemission of photons to relax down to grou
41、nd state Probability too low cube of the frequencyb)NMR signal relaxes back to ground state through two distinct processes NMR Relaxationc) Spin-Lattice or Longitudinal Relaxation (T1)i. transfer of energy to the lattice or solvent materialii. coupling of nuclei magnetic field with magnetic fields c
42、reated by the ensemble of vibrational and rotational motion of the lattice or solvent.iii. results in a minimal temperature increase in sampleMz = M0(1-exp(-t/T1) Recycle Delay: General practice is to wait 5xT1 for the system to have fully relaxed.NMR Relaxationc) Spin-Lattice or Longitudinal Relaxa
43、tion (T1)i. Commonly measure T1 with an inversion recovery pulse sequence5xT1 fully relaxedNMR Relaxationd) spin-spin or transverse relaxation (T2)i. exchange of energy between excited nucleus and low energy state nucleusii. randomization of spins or magnetic moment in x,y-planeMx = My = M0 exp(-t/T
44、2)NMR Relaxationd) spin-spin or transverse relaxation (T2)iii. related to NMR peak line-width(derived from Heisenberg uncertainty principal)Please Note: Line shape is also affected by the magnetic fields homogeneity Eggh Bo / 2 NMR signal (s) depends on:1) Number of Nuclei (N) (limited to field homo
45、geneity and filling factor)2) Gyromagnetic ratio (in practice g g3)3) Inversely to temperature (T)4) External magnetic field (Bo2/3, in practice, homogeneity)5) B12 exciting field strength (RF pulse)Na a / Nb b = e E / kT EBog gs g g- Intrinsic property of nucleus can not be changed.(g (gH H/g /gC)3
46、 for 13C is 64x(g (gH H/g /gN)3 for 15N is 1000 x1H is 64x as sensitive as 13C and 1000 x as sensitive as 15N !Consider that the natural abundance of 13C is 1.1% and 15N is 0.37%relative sensitivity increases to 6,400 x and 2.7x105x !Relative sensitivity of 1H, 13C, 15N and other nuclei NMR spectra
47、depend on1H NMR spectra of caffeine8 scans 12 secs13C NMR spectra of caffeine8 scans 12 secs13C NMR spectra of caffeine10,000 scans 4.2 hoursIncrease in Magnet Strength is a Major Means to Increase SensitivityBut at a significant cost!$800,000$2,00,000$4,500,000An Increase in concentration is a very
48、 common approach to increase sensitivity30 mg1.2 mghttp:/web.uvic.ca/pmarrs/chem363/nmr%20files/363%20nmr%20signal%20to%20noise.pdfThe applied magnetic field causes an energy difference between the aligned (a) and unaligned (b) nucleiNMR signal results from the transition of spins from the a to b st
49、ateStrength of the signal depends on the population difference between the a and b spin statesThe population (N) difference can be determined from the Boltzmann distribution and the energy separation between the a and b spin states:Na a / Nb b = e E / kTBo = 0Bo 0 E = h a ab bLow energy gapSince:and
50、 = g g Bo / 2 then:The E for 1H at 400 MHz (Bo = 9.39 T) is 6 x 10-5 Kcal / molN Na a/N/Nb b .000060.000060Na a/Nb b = e(g ghBo/2 kT)Na a / Nb b = e E / kTBut, this can lead to concentration dependent changes in the NMR spectra (chemical shift & line-shape) resulting from compound property changesBe
