chapter 5 stereochemistry （立体 化学） : chiral molecules （手性分 子） 5.1 isomerism :...
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Chapter 5 Stereochemistry （立体化学） : Chiral Molecules （手
性分子）
5.1 Isomerism: Constitutional isomers and Stereoisomers
( 构形异构与立体异构）
Subdivison of isomers(Different compounds with same molecular formula)
Constitutional isomers Stereoisomers(Isomers that have the same connectivity but that differ in the arrangement of their atoms in space1. CH3CH2CH2CH3 (CH3)2CHCH3
2. CH3CH2CH2Cl (CH3)2CHCl
3. CH3CH2OH CH3OCH3
Enantiomers (¶ÔÓ³Òì¹¹£© Diastereomers (·ÇÁ¢Ìå¶ÔÓ³Òì¹¹)(Stereoisomers that are mirror images of each other,
but it is notsuperposable »¥ÎªÊµÎïÓë¾µÏó¹Øϵ£¬µ«²»Öصþ)
(Stereoisomers that are not mirror images of each other)
Stereoisomers(Isomers that have the same connectivity but that differ in the arrangement of their atoms in space
Enantiomers (¶ÔÓ³Òì¹¹£© Diastereomers (·ÇÁ¢Ìå¶ÔÓ³Òì¹¹)
(Stereoisomers that are mirror images of each other,but it is notsuperposable »¥ÎªÊµÎïÓë¾µÏó¹Øϵ£¬µ«²»Öصþ)
(Stereoisomers that are not mirror images of each other)
CH3
OHH3CH2C
H
CH3
HOCH2CH3
H
2-butanol
CH3
H
CH3
H
CH3
H
H
CH3
Cis- trans-
H
Cl
H
Cl
H
Cl
Cl
H
5.2 Enantiomers and chiral molecules
• A Chiral molecule is defined as one that is not superposable on its mirror image.
• The chiral molecule and its mirror image are enantiomers, and the relationship between the chiral molecule and its mirror image is defined as enantiomeric ( 对映异构） .
• The word chiral comes from the Greek word Cheir, meaning ‘hand’. That means, relationship between left hand and right hand.
5.2 Chiral molecules ( 手性分子 )
• 1.(Chiral) 手性—— 物质分子与其镜象相似而不重合的特征。
Left hand ( 左手 ) 和 right hand( 右手 ) 相似而不重合
Because models I and II are nonsuperposable mirror images of each other, the molecules that they represen
t are enantiomers.
Problem 5.1 Classify the following objects as to whether they are chiral ( 手性） or achiral （非手
性）• (a) Screw ( b) plain spoon (c ) Fork
• (d) Cup (e) Foot (f) Ear
• (g) Shoe (h) Spiral staircase （螺旋梯）
Problem 5.2
• (a) If models are available, construct the 2-butanols represented in Fig. 5.3 and demonstrate for yourself that they are not mutually superposable.
• (b) Make similar models of 2-propanol (CH3CHOHCH3). Are they superposable?
• (c ) Is 2-propanol chiral?• (d) Would you expect to find enantiomeric forms
of 2-propanol?
CH2CH3
H
H3C
HO
CH2CH3
HOH
CH3
2-Butanol
(nonsuperposable) (chiral)
CH3
H
H3C
HO
CH3
HOH
CH3
2-Propanol
(superposable) (achiral)
one tetrahedral atom with four different groups attached to carbon, it is
called chiral carbon （手性碳）• (a) CH3CHClCH3 (b) CH3CHBrCH2CH3
• (c ) CH3CHOHCHOHCH3 (d) (CH3)2CHOH
• (e) BrCHClI
• (f) CH3CH2CHClCH2CH3
5.3 Historical origin of stereochemistry
• The following information was available to van’t Hoff and le Bel.
• 1. Only one compound for CH3X is ever found
• 2. Only one compound for CH2X2 or CH3XY is ever found.
• 3. Two enantiomeric compounds for CHXYZ are found.
5.4 Tests for Chirality: Planes of Symmetry( 对称面）
All molecules with a plane of symmetry are achiral
All molecules with no plane of symmetry are chiral
三、对称因素和手性分子判据 • 1. 对称因素• ◎ 对称面 σ：可把分子分割成两部分的
平面，一部分正好是另一部分的镜象，这个平面称为对称面。 （图）
• 例 1 ： 1 ， 1- 二氯乙烷
C
CH3
Cl
ClH
例 2 ：（ E ） -1 ， 2- 二氯乙烯
C CCl
Cl
H
H
Problem 5.7 Write three-dimensional formulas and designate a plane of symmetry for all of the ach
iral molecules in Problem 5.4.a 1-Chloropropane
CH3
CH2ClH
H
b 2-BromobutaneCH3
CH2CH3Br
H
a plane of symmetry (achiral) no plane of symmetry (chiral)
c 1-Chloro-2-methylpropaneCH3
CH2ClH3C
H
d 2-Chloro-2-methylpropaneCH3
CH3Cl
H3C
a plane of symmetry (achiral) two planes of symmetry (achiral)
f 1-ChloropentaneCl
CH2CH2CH2CH3H
H
h 3-ChloropentaneCH2CH3
CH2CH3Cl
H
a plane of symmetry (achiral) a plane of symmetry (achiral)
5.5 Nomenclature of enantiomers: The (R-S) system
2-Butanol
CH3
CH2CH3OH
H
CH3
H3CH2COH
H
(2S)-2-butanol (2R)-2-butanol
OH CH2CH3 CH3 HC*
Groups containing double or triple bonds
CH2CH3
CH=CH2OH
H
CH2CH3
H2C=HCOH
H
(3S) -1-penten-3-ol (3R)-1-penten-3-ol
OH CH=CH2 CH2CH3 HC*
1-Penten-3-ol
Problem 5.11 Assign ( R) or (S) designations to each of the
following compounds
Br CH=CH2
CH3
CH2CH3
(a)
F CH=CH2
H
(b)
H3C C
H
(c)
CH
(3R) (3R) (3R)
Sample problem; Consider the following pair of structures and tell whether they represent enantiomers or two molecules of the same compound
in different orientations.
H Cl
CH3
Br
A
Br CH3
Cl
H
B
H Cl
CH3
Br
A
Br CH3
Cl
H
B
H3C H
Cl
Br
B
H Cl
CH3
Br
B
Method 1.
H Cl
CH3
Br
A
Br CH3
Cl
H
B
H Cl
CH3
Br
B
Method 2.
H Cl
CH3
Br
A
H Cl
CH3
Br
B
Method 3.
R R
Problem 5.12 Tell whether the two structures in each pair represent enantiomers or two molecules of the same compound in differ
ent orientations.
(a) Br F
Cl
H
Br Cl
F
H
and
(s) Enantiomers (R)
(b) F CH3
Cl
H
H Cl
CH3
F
and
H OH
CH2CH3
CH3
OH CH2CH3
CH3
H
and(C)
( R ) Same (R)
(2S)-Butanol Enantiomers (2R)-Butanol
‘ 十字式’的 R/S 的命名Br F
Cl
H
FBr
H
Cl
(S) (S)
OHH
CH3
CH2CH3
(S)
H OH
CH3
CH2CH3
(s)
Ðýת·½ÏòÓë R/S Ïà·´
Ðýת·½ÏòÓë R/S Ò»ÖÂ
a c
b
d
(R)
(S)
d b
a
c
Ðýת·½ÏòÓë R/S Ïà·´
Ðýת·½ÏòÓë R/S Ò»ÖÂ
a c
d
b
( S )
b d
a
c
(R)
C* a b c d
CH3
H OH
H OH
CH3
S
R
(2S,3R)-2,3-butanediol
2C* OH CHOHCH3CH3 H
CH3
HO H
H OH
CH3
R
R
(2R,3R)-2,3-butanediol
CH3
H OH
HO H
CH3
S
S
(2S,3S)-2,3-butanediol
Problem; Assign （ R) or (S) designations to each of the following compounds;
HBr
H Br
CH3
CH2CH3
CH3
H Br
H Br
CH2CH3
HBr
Br H
CH3
CH2CH3
CH3
H Br
Br H
CH2CH3
(?)-2,3-Dibromopentane
(?)-2,3-Dibromopentane
5.6 Properties of enantiomers; Optical Activity ( 光学活
性）The molecules of enantiomers are not superposable one on the other, and on this basis alone, we have concluded that enantiomers are different compounds. Do enantiomers resemble constitutional isomers and diastereomers in having different melting and boing points? The answer is no. Enantiomers have identical melting and boiling points. Enantiomers have identical (refraction( 折射率） , solubilities （溶解度） , infrared spectra （红外光谱） and rate
s of reaction （反应速率 )
Table 5.1 Physical properties of ( R )- and ( S )-
2-butanol
Physical property ( R )-2-butanol (S )-2- butanol
Boiling point ( 1 atm) 99.5 oC 99.5 oC
Density (g mL-1 at 20 oC ) 0.808 0.808Index of refraction (20 oC) 1.397 1.397
5.6 A . Plane-polarized light ( 偏振光）
One easily observable way in which enantiomers differ is in their behavior toward plan-polarized light. When a beam of plane-polarized light passes throught an enantiomer, the plane of polarization rotates. Because of their effect on plane-polarized light, separate enantiomers are said to
be optically active compounds.
物质的旋光性 (The matter rotation for plane-polarized light)
• 一、平面偏振光 (plane-polarized light) 和旋光性(rotation)
• The Characteristic of electric wave of light ( 光波的特点 ) ：
• ① 振动方向与前进方向垂直 (vertical) ；• ② 在垂直于前进方向的任何可能的平面上振动。• Nicol 棱晶 (Nicol polarizer) ：只允许与晶轴平行
的平面上振动的光线透过。• 这种只在一个平面上振动的光称为平面偏振光或
简称偏振光 (plane-polarized light) 。
使偏光使振动平面旋转的物质——旋光性物质（光学活性物质 (optically active compounds）使偏光向右旋——右旋体“ +”(Levorotatory)使偏光向左旋——左旋体“ -” (Dextrorotatory)例：肌肉乳酸 (CH3CHOHCOOH) +3.8° 发酵乳酸 (CH3CHOHCOOH) -3.8°
Fig. 5.8 The oscillating ( 振荡） electric and magnetic fields of a beam of ordinary light in one plane
Fig. 5.9 Oscillation of the electrical field of ordinary light occurs in all possible planes Perpendicular ( 垂直） to the direction of propagation （传播） .
Fig. 5.10 The plane of oscillation of the electrical field of plane-polarized light. In this example the plane of polarization is vertical (´¹Ö±£©
5.6B The polarimeter （旋光仪 或偏振仪） The device that is used for measuring the effect of plane-polarized light on optically active compounds is a polarimeter
Fig 5.11 The principal working parts of a polarimeter and the measurement of optical rotation
Fig 5.11 a polarimeter
5.6 C Specific Rotation ( 比旋光度） [α]
The number of degrees that the plane of polarization is rotated as the light passes through a solution of an enantiomer depends on the number of chiral molecules that it encounters. This, of course, depends on the length of the tube and the concentration of the enantiomer. In order to place measured rotations on a standard basis, chemists calculate a qua
ntity called the specific rotation [α]
Specific Rotation ( 比旋光度 [α])
c . l
Where the specific rotation ±ÈÐý¹â¶È
the observed rotation ¹Û²ìµÄÐý¹â¶È
c the concentration of the solution in grams per milliliterof solution ( Ũ¶È, g / ml)
l the length of the tube in decimeters (1 dm = 10 cm)£¨Ê¢Òº¹ÜµÄ³¤¶È£©
The specific rotations of ( R )-2-butanol and ( S )-2-butanol
HO H
CH2CH3
CH3
H OH
CH2CH3
CH3
( R )-2-butanol ( S )-2-butanol
D
25-13.52 o
D
25+13.52 o
a sodium lamp ( = 599.6 nm)
No necessary correlation exits between the ( R ) and ( S )
designation and the direction of rotation of plane-polarized light
HOH2C H
CH2CH3
CH3
ClH2C H
CH2CH3
CH3
( R )-(+)-2-methyl-1-butanol ( R )-(-)-1-Chloro-2-methylbutanol
D
25 -13.52 o D
25 +13.52 o= =
5.7 the origin of optical activity
CH3
CH3
OHH
H3C
HO
H3C
H
CH3
CH2CH3
OHH
H3CH2C
HO
H3C
H
No net rotation
Rotation(R)-2-butanol (S)-2-butanol
2-propanol
Achiral molecule
Chiral molecule
5.7A Racemic forms ( 外消旋体 )
An equimolar mixture of two enantiomers is called a racemic form ( 左旋体和右旋体
的等量混合物称为外消旋体 )
CH2CH3
OH
CH3
H
( R)-2-butanol
H3CH2C
HO
H3C
H
( S)-2-butanol50% 50%
( + )- 2-butanol-
Racemic forms ( （ ± ）外消旋体 )
• An equimolar mixture of two enantiomers
• ( 左旋体 (levorotatory (-) ) 和右旋体（ dextrorotatory (+) ) 的等量混合物。（ ± ） )
The characteristic of racemic forms; ( 外消旋体的特点 ) ：
1 ） Chemical property is at same ( 化性基本相同 ) ； 2 ） (No rotation ) 没有旋光性； 3 ） (Biological activity is different for enantiomers) 左旋
体和右旋体的相应生理作用不同。 （氯霉素的左旋体有杀菌作用，它的右旋体无作用）
Meso-Compounds ( 内消旋体 )
The characteristic of meso-Compounds;
• 1) 旋光在分子的内部抵消（ Achiral molecule) ；• 2 ） (there is a plane of symmetry) 分子有对称
面，没有旋光活性 (no rotation) ；• 3 ） (No enantiomers) 无对映异构体；
OHH
OHH
COOH
COOH
m
5.7B Enantiomeric purity, optical purity, and enantiomeric exc
ess
Percent enantiomeric purity
moles of one enantiomer - moles of other enantiomer
moles of both enantiomers100
Percent optical purity specific rotation of the pure enantiomers
100observed specific rotation
Enantiomeric excess ( 对映体过量百分率 ) （ %e.e ）：
[R]-[S]%e.e = ×100% = %R-%S [R]+[S]
[R] ：过量对映体的量[S] ：为其对映体的量
Problem 5.13 What relative molar proportions of ( S )- (+)-2-butanol
and ( R)- (-)-2-butanol would give a specific rotation, [α], equal to
+6.76o ? (S)-(+)-2-butanol%? And ( R )%?
Percent optical purity specific rotation of the pure enantiomers +13.52o
100observed specific rotation +6.76o
50 %
Answer
Percent optical purity
50 %
(That means that sample contains 50% of the ( S) enantiomer and 50% of the racemic form(25%R + 25%S)The tatal percentage of (S) enantiomer in the sample is 75%, the percentage of ( R ) enantiomer is 25%
5.8 The synthesis of enantiomers
The hydrogenation of ketone
CH3CH2CCH3
O
+ H-HNi
CH3CH2CHCH3
OH
H3CH2C
OH
H
CH3
( +_ )
H3C
CH2CH3
HO
H
(achiral) (achiral) (chiral) (50%R+50%S)
50% R-(-)-2-butanol 50% S-(+)-2-butanol
Fig 5.14. Shows why a racemic form of 2-butanol is obtaned
+Ni H3CH2C
OH
H
CH3 H3C
CH2CH3
HO
H
(achiral)50% R-(-)-2-butanol 50% S-(+)-2-butanol
O
H3CH2C CH3
H
H
H
H
( racemic form)
1) Compounds contain two different chiral carbon ( 含两个不相同手性碳原子的化合物 ) HOOCCHCHCOOH
OHCl
**
OHH
ClH
COOH
COOH
HOH
HCl
COOH
COOH
OHH
HCl
COOH
COOH
HOH
ClH
COOH
COOH
[¦Á] -7.1¡ã +7.1¡ã +9.3¡ã-9.3¡ã
¢ñ ¢ò ¢ó ¢ô
2n
5.9 Molecules with more than one stereocenter
OHH
ClH
COOH
COOH
HOH
HCl
COOH
COOH
OHH
HCl
COOH
COOH
HOH
ClH
COOH
COOH
¢ñ ¢ò ¢ó ¢ôⅠandⅡor III and IV are enantiomers ( 对映体 ) ；ⅠandⅢ or Ⅳ are diasteromers( 为非对映体 ) ； (Ⅱ 与Ⅲ或Ⅳ为非对映体 ) ；2.The charecteristic of diasteromers ( 非对映异构体的特点 ) ：1 ） Physical property , specific rotation is different ( 物性不同、比旋光度不同 ) ；2 ） Chemical property is similar( 化性相似，但速度有差异 )
二、 At the same two chiral carbon (含两个相同手性碳原子的化合物 )1. 光学异构体数目 HOOCCHCHCOOH
OHOH
**
OHH
HOH
COOH
COOH
HOH
OHH
COOH
COOH
OHH
OHH
COOH
COOH
HOH
HOH
COOH
COOH
¢ñ ¢ò ¢ó ¢ô
ⅠandⅡ are enantiomers ( 对映异构体）；Ⅲ and Ⅳ are at the same compounds) ；Ⅰ and Ⅲ are diastereomers ；Ⅱ andⅢ are distereomers ( 非对映体) ；Ⅲ is a meso compound ( 内消旋体 )
(2n-1);
5.9A Meso Compounds ( 内消旋化合物）
A structures with two stereocenters will not always have four possible steroisomers. Sometimes there are only three (2n-1). This happens because some molecules with stereo
centers are, overall, achiral
2,3-Dibromobutane
CH3
CHBr
CHBr
CH3
*
*
CH3
Br H
Br H
CH3
CH3
H Br
H Br
CH3
CH3
Br H
H Br
CH3
CH3
H Br
Br H
CH3
The plane of symmetry
Meso compounds
Enantiomers
Problem 5.17 Write three-dimensional formulas for all of the steroisomers of each of the following compounds. In answer to parts (a)-(e) label pairs of each enantiomers and mes
o compounds.• ( a) CH3CHClCHClCH3
• ( b) CH3CHBrCHClCH3
• ( c) CH3CHBrCHBrCH2Br
• ( d) CH3BrCHBrCHBrCH2Br
• ( e) CH3CHClCHClCHClCH3
5.10 Naming compounds with more than one stereocente
r
CH3
Br H
Br H
CH3
CH3
H Br
H Br
CH3
CH3
Br H
H Br
CH3
CH3
H Br
Br H
CH3
The plane of symmetry
Meso compounds
Enantiomers
H Br
Br H
CH3
CH3
S
S (2S,3S)- (2R,3R)-
2S
2R
5.11 Fischer projection formulas
（费歇尔投影式 ）
CH3
H Br
H Br
CH3
CH3
Br H
H Br
CH3
CH3
H Br
Br H
CH3
Meso compounds
Enantiomers
H Br
Br H
CH3
CH3 (2S,3S)- (2R,3R)-
Br H
H Br
CH3
CH3
H Br
H Br
CH3
CH3
Fischer projection formula
The rotation 180o in plane forms at the same structure
CH3
H Br
Br H
CH3
CH3
Br H
H Br
CH3
CH3
Br H
H Br
CH3
Enantiomers
CH3
Br H
H Br
CH3
rotate 180o
in plane
A
Same structure
AB
Not the same
Not the same
B
don't allowed to flip them over
5.12 Stereoisomerism of cyclic compounds
H3C
HCH3
H H
H3CH
CH3
trans-1,2-dimethylcyclopentane (Enantiomers)
S S
H3C
HH
CH3
Plane of symmetry (meso compounds)
cis-1,2-dimethylcyclopentane
S R
Problem 5.21 Write structural formulas for all of the stereoisomers of 1,3-dimethylcyclopentane. Label pairs of enantiomers and meso commpounds if they exist.
trans-1,3-dimethylcyclopentane (Enantiomers)
Plane of symmetry (meso compounds)cis-1,3-dimethylcyclopentane
H CH3
HH3C HCH3
HH3C
H3C CH3
HH
R R
5.12A Cyclohexane derivatives
H3C
H
CH3
H
H3C
H
H
CH3
CH3
H
H
CH3
H3C
H
H
CH3
cis-1,4-dimethylcyclohexane trans-1,4-dimethylcyclohexane
There is a plane of symmetry, both compounds are achiral
1) 1,4-Dimethylcyclohexanes
Fig 5.17 cis-1,3-Dimethylcyclohexane
H3C
H
cis-1,3-dimethylcyclohexane
It has a plane of symmetry, it is achiral
CH3
H
CH3
CH3
Fig 5.18 trans-1,3-Dimethylcyclohexane
H3C
H
CH3
H
trans-1,3-dimethylcyclohexane
It has not a plane of symmetry, it is chiral (Enantiomers)
H
CH3
CH3
CH3
H3C
CH3
H3C
H
Fig 5.19 trans-1,2-Dimethylcyclohexane
trans-1,2-dimethylcyclohexane
It has not a plane of symmetry, it is chiral (Enantiomers)
H
CH3
H
CH3
H
H
H3C
H3C
CH3CH3
H3CH3C
Fig 5.20 cis-1,2-Dimethylcyclohexane
cis-1,2-dimethylcyclohexane
It has a plane of symmetry, it is achiral (meso-compounds)
H
CH3
CH3
H
Plane of symmetry
interconvertible
H3C
CH3 CH3
CH3
a b
equivalent
Problem 5.22 Write formulas for all of the isomers of each of the following. Designate pairs of enantiomers and ach
iral compounds where they exist.
• ( a) 1-Bromo-2-chlorocyclohexane
• ( b) 1-Bromo-3-chlorocyclohexane
• ( c) 1-Bromo-4-chlorocyclohexane
( a) 1-Bromo-2-chlorocyclohexane
cis-1-Bromo-2-chlorocyclohexane
It has not plane of symmetry, it is chiral for each Both compounds are Enantiomers
Br
H
H
Cl Cl
H
H
Br
Br
Cl Cl
Br
R
S
trans-1-Bromo-2-chlorocyclohexane
trans-1-Bromo-2-chlorocyclohexane
It has not plane of symmetry, it is chiral for each Both compounds are Enantiomers
H
Br
H
Cl Cl
H
Br
H
BrCl Cl
Br
S
S
( c) 1-Bromo-4-chlorocyclohexane
(cis or trans)-1-Bromo-4-chlorocyclohexane
It has plane of symmetry, it is achiral for each Both compounds are diasteromers
Br
BrCl
Cl
Br
Cl
BrCl
5.13 Relating configurations through reactions in which no bonds to the stereocenter are broke
nIf a reaction takes places in a way so that no bonds to the stereocenter are broken, the product will of necessity have the same general configuration of groups around the stereocenter as the reactant. Such a reaction is said to proceed with r
etention of configuration( 构形保持） .
Consider as an example the reaction that takes place when (S)-
(-)-2-methyl-1-butanol is heated with concentrated hydrochloric
acid
H CH2OH
CH2CH3
CH3
+ HClheat
H CH2Cl
CH2CH3
CH3
+ H2O
Same configuration
( S )-(-)-2-Methyl-1-butanol
( S )-(+)-1-Chloro-2-methylbutane
no bonds to the stereocenter are broken
S S
Example---retention of configuration
H OH
CH2CH3
CH2Br
H OH
CH2CH3
CH3
Same configuration
( R)-1-Bromo-2-butanol ( S )-2-butanolno bonds to the stereocenter are broken
Zn, H+ (-ZnCl2)
retention of configuration, because the -CH2Br changes to -CH3 ( R change S, -CH2Br has a higher priority than -CH2CH3)
5.13A Relative and absolute configurations ( 相对构形和绝对构形）
H OH
CH2OH
CHO
HO H
CH2OH
CHO
( R)-GlyceraldehydeD-Glyceraldehyde(¸ÊÓÍÈ©£©
( S )-GlyceraldehydeL-Glyceraldehyde
H OH
CH3
COOH
( R)-(-)-Lactic acidD-(-)-Lactic acid£¨ ÈéËᣩ
OH H
CH3
COOH
( S )-(+)-Lactic acidL-(+)-Lactic acid
Tartaric acid ( 酒石酸）
HO H
CH2OH
CHO
( S )-GlyceraldehydeL-Glyceraldehyde
HO H
COOH
HOH
COOH
L-(+)-Tartaric acid
5.14 Separation of enantiomers ( 对映体的分离） ; Resolution ( 拆
分）• How are enantiomers separated?
• Enantimores have identical solubilities in ordinary solvents. You couldn’t do crystallzation for separations of racemic form.
• You have made diastereomers for each (because they have different melting points, different boing points, and different solubilities) and then resolution for each one
将外消旋体的两个对映体分开使之成为纯净的左旋体或右消
旋体 ----- 称为拆分 (resolution) 。 1. 化学法（拆分外消旋（ ± ） Tartaric acid 酒石
酸）
2. Biological 生化法
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HCl
HCl
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5.15 Compounds with stereocenters other than carbon
• Any tetrahedral atom with four different groups attached to it is a stereocenter.
• Listed here are general formulas of compounds whose molecules contain stereocenters other than carbon. Silicon and germanium are in the same group of the periodic table as carbon. They form tetrahedral compounds as carbon does.
The molecules are chiral and the enantiomers can be separate
d
CR4 R2
R1
R3
SiR4 R2
R1
R3
GeR4 R2
R1
R3
N+R4 R2
R1
R3
Stereocenter;
Chiral carbon Chiral silicon Chiral germanium Chiral nitrogen
5.16 Chiral molecules that do not posses a tetrahedral atom with four different groups( 丙二烯型的手性分
子）H
H3C
C
CH3
H H
H3C
H
CH3
H
H3C
C
CH3
H H
H3C
C
CH3
H
Enantiomers
Fig 5.21 Enantiomeric forms of 1,3-dichloroallene
H
Cl
C
Cl
H H
Cl
C
Cl
H
Both compounds are enantiomers
C
H
HH
H
C
H3C
HH
H
C
H3C
H3CH
H
achiral achiral achiral
Diphenyl compounds( 联苯型化合物） -----Chiral molecules
O2N
COOHHOOC
NO2
NO2
COOHCOOH
NO2
Chiral Enantiomers Chiral
NO2
NO2
COOH
COOH
Achiral There are two planes of symmetry
Chiral Nitrogen compounds
N+
H
Ph
H
COOCH2CH3
N
H
H3CH2COOC Ph
H+
Chiral ChiralEnantimoers
Homeworks
• 5.26 5.28 5.30
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