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第14章 芳香族化合物(Aromatic compounds) 1)芳香族化合物

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第14章 芳香族化合物(Aromatic compounds) 1)芳香族化合物

有機化合物被人為地分成兩大類;即:脂肪族 (aliphatic)和芳 香族 (aromatic)

最簡單的aromatic 化合物:苯(benzene,C 6 H 6 )是由Michael Faraday在1825年所合成。

其分子式是由Eilhardt Mitscherlich在1834年所確定.

C 6 H 5 CO 2 H + CaO heat

C 6 H 6 + CaCO 3

(2)

1865年, Kekulé將苯的結構定義為:

(3)

有別於其它有機化合物(aliphatic),芳香族化合物有其特殊的性質。

(4)

2) 苯類衍生物的化學命名

a) 以苯環結構為母體

(5)

若有三個以上取代基,則使編號總和為最小 b) 以苯環單取代物結構為母體

(6)

母體取代基的編號為1 c) 以苯環為取代基

(7)

練習(page 655, 14.16)

CO2H

NO2 3-Nitrobenzoic acid

CH3

p-Bromotoluene Br

o-Dibromobenzene Br

Br

m-Dinitrobenzene NO2

NO2

3,5-Dinitrophenol NO2

NO2 HO

p-Nitrobenzoic acid O2N

CO2H

3-Chloro-1-ethoxybenzene

Cl OEt

p-Chlorobenzenesulfonic acid SO3H

Cl

Methyl p-toluenesulfonate S

H3C

O O OCH3

Benzyl bromide CH2Br

p-Nitroaniline NO2

H2N

(8)

o-Xylene CH

3

CH

3

p-Cresol OH

H

3

C

p-Bromoacetophenone Br

O

CH

3

OH

3-Phenylcyclohexanol

OH

2-Methyl-3-phenyl-1-butanol

Cl

OCH

3

o-Chloroanisole

(9)

3) 苯的特殊化學穩定性,及穩定性的解釋 a)苯有別於烯烴的特殊穩定性:

Br 2 , CCl 4

Br Br

Br 2 , CCl 4

no reaction

KMnO 4 , NaHSO 3

OH OH H2O

KMnO 4 , NaHSO 3

H2O

no reaction

CH 3 OH

H 3 O + H 3 O +

no reaction

(10)

苯在Lewis acid的催化作用下與Br 2 發生取代反應而並非加成反應而

且僅生成一種取代產物(implying苯環上的氫均為相等):

(11)

Kekule結構可以解釋在環上的六個氫均為相等的氫;但不能夠合理解釋 1)為什 麼只有一種1,2-dibromobenzene存在而不是兩種,2)苯環的六個鍵長(1.39 Å;

< 1.47 Å) 及鍵角(120o)都相等 3)為什麼cyclooctatraene則不具有苯的穩定 性

(12)

b)從能量的角度來觀察苯的穩定性:resonance energy

(13)

c)用共振(resonance)理論解釋苯的穩定性

I與II單獨均不能反應苯的真實結構。其真正的結構應為I和II的共振混合體 (III):

Delocalized

(14)

d)用近代分子軌道理論解釋苯的穩定性

There are six π molecular orbitals for benzene

(15)

Huckel's rule:

Cyclooctatetraene has two nonbonding orbitals each with one electron

Polygon-and-circle method: 將polygon的一角置於圓圈的底部

(16)

Huckel's rule:若有一個成平面結構的環狀化合物,其

delocalized的π電子數為4n + 2 (n = 0,1, 2, 3…),這一化合物 則具有aromatic的性質

The [14]and [18]annulenes are aromatic (4n+2, where n= 3,4) The [16] annulene is not aromatic

The [10]annulenes below should be aromatic but none of them can be planar

分別解釋

(17)

解釋為什麼此化合物屬於aromatic類

(18)
(19)

Exercise in page 635

(20)

4)具有芳香性的離子(Aromatic ions) 解釋:

H H H

H

PKa = 16 PKa = 36

(21)

課堂練習(page 638)

14.4

14.5

不是aromatic, 不符合4n + 2 rule.

不是aromatic, 不符合4n + 2 rule.

(22)

14.7

H H

Br 2 , CCl 4

Br Br

heat

Br

Br -

ionic compound

(23)

5)Aromatic, Antiaromatic, and nonaromatic之定義

當我們說一個化合物為aromatic,即代表此化合物的π電子為delocalized.

(24)

比較成環前後能量的大小

H 2 C CH CH 2 + energy decrease + H 2

aromatic

(25)

6)其它類型的aromatic化合物

Polycyclic benzenoid aromatic compounds

(26)

fullerene

(27)

heteroaromatic compounds

Pyridine has an

sp

2 hybridized nitrogen

The

p

orbital on nitrogen is part of the aromatic π system of the ring

The nitrogen lone pair is in an

sp

2 orbital orthogonal to the

p

orbitals of the ring;

(28)

The nitrogen in pyrrole is

sp

2

hybridized and the lone pair resides in the p orbital

In furan and thiophene an electron pair on the heteroatom is also in a

p

orbital which is part of the aromatic system

(29)
(30)
(31)
(32)

根據4n + 2 規則, 判斷下列化合物是否具有芳香性 (aromatic)

II

III

IV I

N V

VI

參考文獻

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