附件2 粒大浮 教 案 2003~~2004学年第Ⅰ学期 院(系、所、部)化学与环境学院有机化学研究所 教研室有机化学 课程名称有机化学(双语教学 授课对象化学教育 授课教师杨定乔 职称职务教授 教材名称 Organic Chemistry 2003年09月01日
附件 2 教 案 2003~~ 2004 学年 第 I 学期 院（系、所、部）化学与环境学院有机化学研究所 教 研 室 有机化学 课 程 名 称 有机化学(双语教学) 授 课 对 象 化学教育 授 课 教 师 杨定乔 职 称 职 务 教授 教 材 名 称 Organic Chemistry 2003 年 09 月 01 日
有机化学课程教案 授课题目(教学章节或主题):第六章.对映异构|授课类型|理论课 Enantiomers 授课时间第7周第23-28节 教学目标或要求:了解物质的旋光性,对映异构体基本理论。掌握对映异构体。 教学内容(包括基本内容、重点、难点) 对映异构体 本章的重点是对所学的各种概念的理解和应用,在多做练习的基础上加深对基本内容及有关立 体化学知识的理解。包括RS命名法、各种表示构型的方法及相互间的转换、对称元素及其操作、 反应过程中的立体化学问题及对映异构体、非对映异构体、外消旋体及内消旋体等概念。难点主要 体现在对立体化学的理解上。如表示构型的方法及相互间的转换和反应过程中的立体化学问题 Chirality and Stereogenic Centers The word chirality refers to the property of handness. You right and left hands are very similar, yet they are not identical. They are related to each other as mirror images, and as such they can not be superimposed on top of each other. Molecules can also be chiral if they contain one or more chiral centers For the purposes of introductory organic chemistry, a chiral center can be defined as an sp hydridized carbon which is bonded to four different groups. A more contemporary term for chiral center"(or chiral carbon) is stereogenic center (or stereocenter), and the terms will be used interchangeably. Mirror PE 11 Hc H A Padr of Enantiomen(Minor Images
有机化学 课程教案 授课题目（教学章节或主题）：第六章．对映异构 （Enantiomers） 授课类型 理论课 授课时间 第 7 周第 23-28 节 教学目标或要求：了解物质的旋光性，对映异构体基本理论。掌握对映异构体。 教学内容（包括基本内容、重点、难点）： 对映异构体 本章的重点是对所学的各种概念的理解和应用,在多做练习的基础上加深对基本内容及有关立 体化学知识的理解。包括 R/S 命名法、各种表示构型的方法及相互间的转换、对称元素及其操作、 反应过程中的立体化学问题及对映异构体、非对映异构体、外消旋体及内消旋体等概念。难点主要 体现在对立体化学的理解上。如表示构型的方法及相互间的转换和反应过程中的立体化学问题。 Chirality and Stereogenic Centers The word "chirality" refers to the property of "handness". You right and left hands are very similar, yet they are not identical. They are related to each other as mirror images, and as such they can not be superimposed on top of each other. Molecules can also be chiral if they contain one or more chiral centers. For the purposes of introductory organic chemistry, a chiral center can be defined as an sp3 hydridized carbon which is bonded to four different groups. A more contemporary term for "chiral center" (or chiral carbon) is stereogenic center (or stereocenter), and the terms will be used interchangeably
Chirality, resulting from the presence of four different groups around a carbon results from an asymmetry in the molecule. This can be seen by examining the molecules shown below; when two identical groups are on one carbon, there is an internal mirror plane passing through the molecule; when four groups are present, there is no internal mirror plane (no symmetry) and hence the carbon is chiral Plane of Symmetry No Plane of Symmetry Mismatch H2 Consider 2-bromobutane HcH The carbon which is bound to the bromine (carbon #2)is a stereogenic center it is sp hydridized and it is bonded to four different groups 1. a hydroge 2. a bromine 3. a methyl group, and 4. an ethyl group All of the other carbons in this molecule are not stereogenic centers, since they are all bonded to at least two hydrogens. 2-Chloro-3-bromobutane has two stereogenic centers
Chirality, resulting from the presence of four different groups around a carbon results from an asymmetry in the molecule. This can be seen by examining the molecules shown below; when two identical groups are on one carbon, there is an internal mirror plane passing through the molecule; when four groups are present, there is no internal mirror plane (no symmetry) and hence the carbon is chiral. Consider 2-bromobutane: The carbon which is bound to the bromine (carbon #2) is a stereogenic center; it is sp3 hydridized and it is bonded to four different groups: 1. a hydrogen, 2. a bromine, 3. a methyl group, and 4. an ethyl group. All of the other carbons in this molecule are not stereogenic centers, since they are all bonded to at least two hydrogens. 2-Chloro-3-bromobutane has two stereogenic centers:
Carbon #2 in this molecule is sp hydridized and and is attached to four different groups 2. a chlorin 3. a methyl group 4. a-CHbrCh3 Likewise, carbon #3 is also sp hydridized and is attached to four different groups hydroge 2. a brom ine 3. a methyl group, and 4. a-CHCICH; group The remaining carbons are both sp hydridized, but each is attached to three hydrogens,and neither is a stereogenic center. Examine the three molecules shown be low and attempt to identify any stereogenic centers. You may click on the image to check your answers Assigning Absolute Configuration: The Cahn-Ingold-Prelog R/S System The rules for assignment of priorities in order to assign absolute configuration are based on the same set of rules which have been used previously for assigning E and Z stereochemistry. The procedure is fairly straightforward for simple compounds; first, you assign priorities to the groups attached around the chiral center. Next, you rotate the molecule so that the lowest priority group is pointing towards the back(away from you). Finally, you examine the remaining I group priorities and determine if they are now arranged so that the priority
Carbon #2 in this molecule is sp3 hydridized and and is attached to four different groups: 1. a hydrogen, 2. a chlorine, 3. a methyl group, and 4. a -CHBrCH3 group. Likewise, carbon #3 is also sp3 hydridized and is attached to four different groups: 1. a hydrogen, 2. a bromine, 3. a methyl group, and 4. a -CHClCH3 group. The remaining carbons are both sp3 hydridized, but each is attached to three hydrogens, and neither is a stereogenic center. Examine the three molecules shown below and attempt to identify any stereogenic centers. You may click on the image to check your answers. Assigning Absolute Configuration: The Cahn-Ingold-Prelog R/S System The rules for assignment of priorities in order to assign absolute configuration are based on the same set of rules which have been used previously for assigning E and Z stereochemistry. The procedure is fairly straightforward for simple compounds; first, you assign priorities to the groups attached around the chiral center. Next, you rotate the molecule so that the lowest priority group is pointing towards the back (away from you). Finally, you examine the remaining group priorities and determine if they are now arranged so that the priority
decreases clockwise (R, for rectus) or counterclockwise (S, for sinister) These rules are restated below, with exampl Exam ine the four atoms directly attached to the chiral center in question. Assign priorities in order of decreasing atomic number: the atom with the highest atom ic num ber is #1. the next is #2.etc H: Atomic #=1 o: Atomic #=8 F: Atomic #=9 S: Atomic # 16 Cl: Atomic # 17 Br: Atomic # 35 2. If a decision regarding priority cannot be reached using Rule #1, compare the atom ic num bers of the econd atoms in each substituent then the third. etc. until a difference is found Oxyge OCH2CH3 H lowest Atomic Carbon ttached to(Cl,, H attached to(C, H, H) C(C,H, H) C(Cl C, H) CH3CH2CH 3. Multiple bonds count twice(or three times) when examining substituents C(C H, H) C(Cl C, Once the priorities have been assigned, rotate the molecule in space so that the lowest priority group is pointing back. Connect the three remaining groups n order of decreasing priority and exam ine the direction of the resu lting rotation Rotation which is clockwise is termed r(rectus; right)and rotation which is counterclockwise is termed s(sinister: left)
decreases clockwise (R, for rectus) or counterclockwise (S, for sinister). These rules are restated below, with examples: 1. Examine the four atoms directly attached to the chiral center in question. Assign priorities in order of decreasing atomic number; the atom with the highest atomic number is #1, the next is #2, etc. 2. If a decision regarding priority cannot be reached using Rule #1, compare the atomic numbers of the second atoms in each substituent, then the third, etc., until a difference is found. 3. Multiple bonds count twice (or three times) when examining substituents. 4. Once the priorities have been assigned, rotate the molecule in space so tha t the lowest priority group is pointing back. Connect the three remaining groups in order of decreasing priority and examine the direction of the resulting rotation. Rotation which is clockwise is termed R (rectus; right) and rotation which is counterclockwise is termed S (sinister; left)
lOCH.CH C(C H, H) (C1 C, H) H3CH2 CH2 1 OCH2 CH3 C(C, H C(Clc.H CH3CH2CH2' Optical Activity which is capable of rotating the plane of polarized light. In a pair of optically active enantiomers, each enantiomer will rotate the plane of polarized light in equal and opposite directions; the enantiomers can therefore be referred to as (+)and (-) enantiomers, depending upon the direction of the observed rotation. A racemic mixture, in which there are equal concentrations of both enantiomers, will display no net optical rotation. 米 obseryed rotation sample tube The angle of rotation is measured in a device known as a polarimeter. The substance in question is typically dissolved in a solvent at a known concentration (c in grams per mL) and placed in an analysis tube of a known length (1 given in decimeters; one dm =10 cm). The rotation observed in the polarimeter is denoted with the Greek letter a, and the specific rotation for a molecule (denoted as [a]) is given by the equation shown below a]=/e
Optical Activity A substance which is optically active is one which is capable of rotating the plane of polarized light. In a pair of optically active enantiomers, each enantiomer will rotate the plane of polarized light in equal and opposite directions; the enantiomers can therefore be referred to as (+) and (-) enantiomers, depending upon the direction of the observed rotation. A racemic mixture, in which there are equal concentrations of both enantiomers, will display no net optical rotation. The angle of rotation is measured in a device known as a polarimeter. The substance in question is typically dissolved in a solvent at a known concentration (c in grams per mL) and placed in an analysis tube of a known length (l given in decimeters; one dm = 10 cm). The rotation observed in the polarimeter is denoted with the Greek letter , and the specific rotation for a molecule (denoted as []) is given by the equation shown below: [] = /(c x l)
The specific rotation for a molecule is also dependent on the wavelength of the plane polarized light. A common light source for simple polarimeters is a lamp with enhanced output at the sodium D-line"; in this instance, the specific rotation would be shown as [a] Sample problem A pure enantiomer has an observed optical rotation of -0 82 when measured in a one dm tube at a concentration of 0.3 g/10 mL. Calculate the specific rotation for this molecule Solution The concentration of 0.3 g/10 mL is equivalent to 0.03 g/mL; c=0.03 g/mL The length of the analysis tube is one dm: 1=1.0 dm The specific rotation is therefore [a]=082)103gmLx10dm) [a]=-27 3"- dr If the specific rotation of a pure enantiomer is known, the observed rotation can also be used to calculate optical purity, or the level of contamination of one compound with its enantiomer, using the simple convention optical purity =( of one enantiomer )-(% of the other enantiomer) Sample problem The specific rotation for a pure enantiomer is known to be -39 g mL- dmr. A sample containing both enantiomers is found to have an observed rotation of 0. 62. in a one dm tube at a concentration of 3. 5 g/100 mL. What is the optical purity of the sample Solutio For this sample, the apparent specific rotation is [a]=-17 7 g"' mL- dr
The specific rotation for a molecule is also dependent on the wavelength of the plane polarized light. A common light source for simple polarimeters is a lamp with enhanced output at the "sodium D-line"; in this instance, the specific rotation would be shown as []D. Sample problem: A pure enantiomer has an observed optical rotation of -0.82o when measured in a one dm tube at a concentration of 0.3 g/10 mL. Calculate the specific rotation for this molecule. Solution: The concentration of 0.3 g/10 mL is equivalent to 0.03 g/mL; c = 0.03 g/mL The length of the analysis tube is one dm; l = 1.0 dm The specific rotation is therefore: [] = -0.82o) /(0.03 g/mL x 1.0 dm) [] = -27.3o g -1 mL-1 dm-1 If the specific rotation of a pure enantiomer is known, the observed rotation can also be used to calculate optical purity, or the level of contamination of one compound with its enantiomer, using the simple convention: optical purity = (% of one enantiomer) - (% of the other enantiomer) Sample problem: The specific rotation for a pure enantiomer is known to be -39o g -1 mL-1 dm-1. A sample containing both enantiomers is found to have an observed rotation of -0.62o in a one dm tube at a concentration of 3.5 g/100 mL. What is the optical purity of the sample? Solution: For this sample, the apparent specific rotation is: [] = -0.62o) /(0.035 g/mL x 1.0 dm) [] = -17.7o g -1 mL-1 dm-1
If the fraction of the ( enantiomer is x, then (1x gives the fraction of the (+) enantiomer. For any mixture of the two, the apparent specific rotation will be given by x(-39)+(1x)(+39)=[a]-m For this mixture x(-39)+(1x)(+39)=-17.7 (-39x) (-39x)=-17.7 78x=-56.7 X=0.73 Therefore, the mixture contains 73% of the (-)enantiomer and 27% of the (+) enantiomer 教学手段与方法:课堂讲授 思考题、讨论题、作业:( Homeworks; Page204: Additional problems525-527) 参考资料(含参考书、文献等) 1. Solomons, Organic Chemistry, fifth adition 2. Oxford; Organic Chemistry 3.北京大学,有机化学 4.南京大学,有机化学,(上,下) 5.邢其毅,有机化学,(上,下) 注:1、每项页面大小可自行添减;2一次课为一个教案;3、“重点"、“难点"、“教学手段 与方法”部分要尽量具体;4、授课类型指:理论课、讨论课、实验或实习课、练习或习题 课等
If the fraction of the (-) enantiomer is x, then (1x) gives the fraction of the (+) enantiomer. For any mixture of the two, the apparent specific rotation will be given by: x(-39o) + (1x)(+39o) = []apparent For this mixture: x(-39o) + (1x)(+39o) = -17.7o (-39x) + 39 (-39x) = -17.7 -78x = -56.7 x = 0.73 Therefore, the mixture contains 73% of the (-) enantiomer and 27% of the (+) enantiomer. 教学手段与方法：课堂讲授 思考题、讨论题、作业：(Homeworks;Page 204: Additional Problems:5.25-5.27) 参考资料（含参考书、文献等）： 1. Solomons, Organic Chemistry, fifth adition 2. Oxford; Organic Chemistry 3. 北京大学， 有机化学 4．南京大学， 有机化学，（上，下） 5．邢其毅，有机化学， （上，下） 注：1、每项页面大小可自行添减；2 一次课为一个教案；3、“重点”、“难点”、“教学手段 与方法”部分要尽量具体；4、授课类型指：理论课、讨论课、实验或实习课、练习或习题 课等
