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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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