有机光谱化学分析 (32).pdf

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1、P2：Welcome to todays class-Organic Spectroscopic Analysis.From the previous lectures,we have known that IR spectroscopy is useful for structure analysis.In this lecture,I will talk about the IR absorption of functional groups.大家好，欢迎学习今天的课程有机波谱分析。从前面的课程中，我们知道红外光谱对于结构分析很有用。在这一节课中，我将讲述官能团的红外吸收。P3：Infra

2、red Absorption of Organic Compounds can be divided into four areas.The First IR Region is from 3700 to 2500 the reciprocal of centimeter.The Second IR Region is from 2500 to 1900 cm-1.The Third IR Region is from 1900 to 1500 cm-1.The Fourth IR Region is from 1500 to 600 cm-1.Its also called the Fing

3、er-Print Region that is individual to each molecule.有机化合物的红外吸收可分为四个峰区。第一个 IR 峰区是从 3700-2500 cm-1 第二 IR 区域是从 2500 到 1900 cm-1。第三 IR 区域为 1900 至 1500 cm-1。第四 IR 区域为 1500 至 600 cm-1。也称为指纹图谱区，每一个分子都不同。P5：Both of O-H,N-H occur around 3300 cm-1,but they look different.Alcohol O-H,broad with rounded tip.Pri

4、mary amine(RNH2),broad with two sharp spikes.Secondary amine(R2NH),broad with one sharp spike.No signal for a tertiary amine(R3N).O-H,N-H 都出现在 3300 cm-1左右，但看上去却有所不同。醇 O-H，宽峰。伯胺（R-NH2），宽且有两个尖峰。仲胺（R2-NH），宽且尖锐。叔胺（R3N）无信号。P6：Both O-H and N-H bonds can form hydrogen bonding.H-bonds vary in strength and l

5、ength.They cause a weakening of X-H bonds.The range of strengths leads to a range of absorptions.O-H 和 N-H 键均可形成氢键。氢键的强度和长度各不相同。它们导致 X-H 键减弱。不同的强度范围导致吸收范围宽。P7：However,in the structure of BHT,the steric bulk of the two tert-butyl groups prevents two molecules of BHT from hydrogen bonding with each ot

6、her and we get a sharp peak in the IR spectrum.但是，在 BHT 的结构中，两个叔丁基的空间位阻阻止了两个 BHT 分子彼此之间的氢键结合，在 IR 光谱中出现了一个尖峰。P8：O-H Stretch of a Carboxylic Acid,this O-H absorbs broadly,from 2500-3500 cm-1,due to the strong hydrogen bonding.羧酸的 O-H 伸缩振动，由于强的氢键作用，O-H 吸收谱带宽从2500-3500 cm-1。P9：And for Carbon-Hydrogen S

7、tretching,bonds with more s character absorb at a higher frequency.sp3 C-H bond is just below 3000 cm-1,as see(to the right)sp2 C-H bond is just above 3000 cm-1(to the left)sp C-H bond is at 3300 cm-1 对于碳氢伸缩振动，具有更多 s 特性的键会吸收频率更高。sp3 C-H 键正好在 3000 cm-1以下,如（右侧）sp2 C-H 键刚好在 3000 cm-1以上（左侧）sp C-H 键在 330

8、0 cm-1左右。P10：Stronger bonds absorb at higher frequencies:Conjugation lowers the frequency:更强的键吸收频率更高：共轭作用降低了吸收频率：P11：C-H stretch in alkane is just below 3000 cm-1.烷烃中 C-H 的伸缩振动刚好在 3000 cm-1以下。P12：C-H stretch in alkene is just above 3000 cm-1.C-H 在烯烃中的伸缩刚好在 3000 cm-1以上。P13：The C=O bond of simple keto

9、nes,aldehydes,and carboxylic acids absorb around 1710 cm-1.Usually,its the strongest IR signal.Carboxylic acids also have O-H group.Aldehydes have two C-H signals around 2700 and 2800 cm-1.简单的酮，醛和羧酸的 C=O 键吸收峰约 1710 cm-1。通常，这是最强的红外信号。羧酸也将具有 O-H 基团。醛在 2700 和 2800 cm-1 附近有两个 C-H 信号。P16：The strengths an

10、d the limitations of IR are the follows:The functional group is usually indicated.The absence of a signal is definite proof that the functional group is absent.IR 的优点和局限性如下：通常标明官能团。信号的缺失是功能基团缺失的明确证明。Correspondence with a known samples IR spectrum confirms the identity of the compound.IR alone cannot

11、 determine a structure.Some signals may be ambiguous.与已知样品的红外光谱相符，证实了该化合物的身份。单独的 IR 不能确定结构。有些信号可能含糊不清。P17：Lets look at a question that needs to be solved by IR.We have the IR spectra for an organic liquid that was found on the clothes of a criminal defendant.让我们看一道 IR 的题目。我们在刑事被告衣服上发现的有机液体的红外光谱。P18：

12、The crime scene was heavily contaminated with a mixture of hydrocarbons,but there were no ketones,aldehydes,acids or alcohols present.犯罪现场被碳水化合物混合物严重污染，但没有酮，醛，酸或醇类存在。P19：Is the defendants clothing contaminated with a hydrocarbon?Would this spectral evidence put the suspect at the crime scene?Please justify your answer with IR functional group information.被告的衣服是否被碳氢化合物污染？谱图会证明嫌疑人在犯罪现场吗？请用 IR 官能团信息证明你的答案合理并解决问题。Please solve the problem and leave your answer in the discussion section after this class.Thats all for todays class.Thank you for listening.下课后可在讨论版块提交你的答案。今天的课程就到这里！谢谢收听！