有机化学 英文课件 chapter5
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高中化学竞赛——重排反应
第一节 从碳原子到碳原子的重排一Wangner-Meerwein重排
一、Wangner-Meerwein重排
醇或卤代烃在酸催化下进行亲核取代或消除反应时 发生的重排
烯烃进行亲电加成时发生的重排
R2
R1 C
R4 C
R5
H+(-H2O)
R3 OH
R1 R4 R2 C C R5
R3
重排
R1 R2
C
R4 H2O(-H+)
第一节 从碳原子到碳原子的重排
四 Favorski重排
O E tO N a O - C lH
*
C l
* C l
O E tO N a
*
O E t
O
*
O
O
* C O E t+ C O E t *
5 0 % 5 0 %
O
O
Cl C CH3
EtONa
O EtO
EtONa
C 所连接的取代基越少越稳定
COOEt CH2
H N O 2
O H
O
第一节 从碳原子到碳原子的重排
三、二苯基乙二酮 —— 二苯乙醇酸型重排
三、二苯基乙二酮 ——二苯乙醇酸型重排
机理：
OO K O H
A r CCA r
△
O HO A r CCO K
A r
OO Ar C C
OO
OH Ar C C OH
OO Ar C C OH
Ar
Ar
Ar
OHO Ar C C O
(1)催化剂:质子酸 H+ ,H2SO4 , HCl, H3PO4 非质子酸PCl5, SOCl2, TsCl, AlCl3
有机化学英文原版ppt
Number of Covalent Bonds to an Atom Atoms with one, two, or three valence electrons form one, two, or three bonds Atoms with four or more valence electrons form as many bonds as they need electrons to fill the s and p levels of their valence shells to reach a stable octet
1.1 Atomic Structure
Structure of an atom
Positively charged nucleus (very dense, protons and neutrons) and smal (10-15 m) Negatively charged electrons are in a cloud (10-10 m) around nucleus Diameter is about 2 × 10-10 m (200 picometers (pm)) [the unit angstrom (Å) is 10-10 m = 100 pm]
1.5 The Nature of the Chemical Bond
Atoms form bonds because the compound that results is more stable than the separate atoms Ionic bonds in salts form as a result of electron transfers Organic compounds have covalent bonds from sharing electrons (G. N. Lewis, 1916) Lewis structures shown valence electrons of an atom as dots Hydrogen has one dot, representing its 1s electron Carbon has four dots (2s2 2p2) Stable molecule results at completed shell, octet (eight dots) for main-group atoms (two for hydrogen)
有机化学英文课件
+ C H 3 C H 2 -O -H + C H 3 C H 2 -O -H
SN 2
+ C H 3 C H 2 -O -C H 2 C H 3 + O -H
H
H
H
A new oxonium ion
11-11
Preparation of Ethers
Step 3: proton transfer to solvent completes the reaction
+
CH 3 CH 2 -O -CH 2 C H 3 + O -H
H
H
proton tran sf er
+ CH 3 CH 2 -O -CH 2 C H 3 + H O -H
H
11-12ቤተ መጻሕፍቲ ባይዱ
Preparation of Ethers
Acid-catalyzed addition of alcohols to alkenes
Acid-catalyzed dehydration of alcohols
• diethyl ether and several other ethers are made on an industrial scale this way
• a specific example of an SN2 reaction in which a poor leaving group (OH-) is converted to a better one (H2O)
CH 3 CH3C=CH2 + H
+ O CH3
H
CH3 C H 3 C+C H 3 +
Chapter 5 固液萃取
Chapter 5
5-1 5-2 算 5-3 5-4
固液萃取
概述 浸出过程计
提取设备 超临界流体
§5-1 概述
定义 提取方法 提取阶段 扩散速率 工艺流程 工艺参数
一、定义
萃取：是利用各种物质在选定溶剂中溶解 度的不同，已分离混合物中组分的一种方 法。 固-液提取：用萃取剂分离固体混合物中组 分的方法。（提取或浸提） 液-液萃取：用萃取剂分离液体混合物中组 分的方法。（萃取）
固定环则可抑制返混。
转盘塔结构简单，生产能力大，传质效率高，操作弹性大， 故在工业中应用较广泛。
图4. 51转盘萃取塔
（四）离心萃取器
利用离心力使两相快速混合和分离的萃取设备。有多种型式。
一种称为POD离心萃取器如图4.52所示 ，属卧式微分接触设 备。在外壳内有一螺旋形转子，转速高达2000~5000r/min。 轻相由外圈引入，重相由中心引入。在离心力作用下，重 相由中心向外流，轻相由外圈向中部流，两相成逆向流动。 最终，重相由螺旋最外层流出，轻相从中部流出。
若以重液为分散相时，则应将降液管改为升液管，安装在筛板上方。 在筛板塔内分散相的液体经多次分散和凝聚，而且筛板的存在又抑 制了塔内的轴向混合，故其效率高，应用广泛。
脉冲筛板塔
脉冲筛板塔的基本结构
与普通筛板相同，但 没有溢流管，如图 4.50(2)所示。 工作原理
工作原理
操作时，轻、重液相均穿过筛板面作逆流
普通筛板塔 脉冲筛板塔 往复筛板塔
普通筛板塔
普通筛板塔图4.50(1)内配 有若干层加工有许多小
孔和一个溢流管(也称为
降液管)的筛板。筛孔直
径一般为3~9mm，孔距
为孔径的3~4倍，板间距
为150~600mm。
5-1 5-2 算 5-3 5-4
固液萃取
概述 浸出过程计
提取设备 超临界流体
§5-1 概述
定义 提取方法 提取阶段 扩散速率 工艺流程 工艺参数
一、定义
萃取：是利用各种物质在选定溶剂中溶解 度的不同，已分离混合物中组分的一种方 法。 固-液提取：用萃取剂分离固体混合物中组 分的方法。（提取或浸提） 液-液萃取：用萃取剂分离液体混合物中组 分的方法。（萃取）
固定环则可抑制返混。
转盘塔结构简单，生产能力大，传质效率高，操作弹性大， 故在工业中应用较广泛。
图4. 51转盘萃取塔
（四）离心萃取器
利用离心力使两相快速混合和分离的萃取设备。有多种型式。
一种称为POD离心萃取器如图4.52所示 ，属卧式微分接触设 备。在外壳内有一螺旋形转子，转速高达2000~5000r/min。 轻相由外圈引入，重相由中心引入。在离心力作用下，重 相由中心向外流，轻相由外圈向中部流，两相成逆向流动。 最终，重相由螺旋最外层流出，轻相从中部流出。
若以重液为分散相时，则应将降液管改为升液管，安装在筛板上方。 在筛板塔内分散相的液体经多次分散和凝聚，而且筛板的存在又抑 制了塔内的轴向混合，故其效率高，应用广泛。
脉冲筛板塔
脉冲筛板塔的基本结构
与普通筛板相同，但 没有溢流管，如图 4.50(2)所示。 工作原理
工作原理
操作时，轻、重液相均穿过筛板面作逆流
普通筛板塔 脉冲筛板塔 往复筛板塔
普通筛板塔
普通筛板塔图4.50(1)内配 有若干层加工有许多小
孔和一个溢流管(也称为
降液管)的筛板。筛孔直
径一般为3~9mm，孔距
为孔径的3~4倍，板间距
为150~600mm。
Chapter 5.重氮化与偶联反应
氟硼酸苯重氮盐在二甲亚砜中与亚硝酸钠作用可使芳香 化合物芳基化，是比巴赫曼反应制备取代联苯更好的方 法。这个反应条件温和、产率高、应用范围广，固体化 合物同样可以反应。
2. 保留氮的反应 2.1 还原反应——苯肼
Na2SO3
Ar
N
N Cl
or NaHSO3 or SnCl2 / HCl or Na2S2O3 / NaOH
1.4 被含硫基团取代 a：重氮盐和一些含硫化合物反应，可以在苯核上引入含硫基
团，得到相应的硫酚、硫醚等化合物。
例如：
ArN2
2ArN2
HS
S2-
ArSH
Ar-S-Ar
N2
2N2
ArN 2
RS
ArSR
N2
若重氮盐与黄原酸钾酯作用，生成的中间体经加热分解或水 解即得硫酚(Leukar反应) P119
例如：
c. 被氰基取代
例如：
• 氰基可以水解成羧酸，所以可以通过重氮盐在芳环 上引入羧基：
d. 希曼反应——（芳香族氟化物的制备） 必须将氟硼 酸加到重氮盐溶液中，使生成重氮盐氟硼酸沉淀，经 分离并干燥后再小心加热，即逐渐分解而得相应的芳 香族氟化物：
• 近来报导用重氮氟磷酸盐代替重氮氟硼酸盐，产率高:
O3S
N2
30~40oC时仍稳定
•(Ⅲ) Structure and Property of Diazo Compound •重氮化合物的结构：[ArN+N]X- 或 ArN2+XCharacter:P104-105 1.Stability:正电荷分散程度越大， 重氮盐稳定性越大。 2.重氮盐基带有正电荷，强吸 电子性，相当于两个硝基。 3. 氮原子上的正电荷主要在重 氮基末端的氮原子上，显示亲 电特性。 4.重氮盐有如铵盐性质，其氯 化物和硫酸盐一般可溶于水 。
第 5 章 酸碱和酸碱反应 (2)
前言（酸碱理论的发展） 1．日常生活中，人们几乎天天与酸碱打交道。 2．Arrhenius 的电离学说于 19 世纪 80 年代提出，使人们对酸和碱的认识产生了飞跃， 以电离理论为基础去定义酸和碱，使人们对酸和碱的本质有了极为深刻的了解。是 酸碱理论发展的重要里程碑。至今仍被广泛应用。 不足点：仅局限在水溶液中，对近年来兴起的大量非水溶剂体系乃至无溶剂体系无 能为力；仅限于某些只含氢原子或氢氧根的物质； 因此要扩大酸碱的物种范围和应用范围，重要的理论有将要讲的三种。
（1）HB + H2O
B- + H3O+
K
θ a
= [C(B− )/mol ⋅ dm −3 ] ⋅ [C(H`3O + )/mol ⋅ dm −3 ] [C(HB)/mol ⋅ dm −3 ]
若 HB 的起始浓度为 C0（HB）
K
θ a
=
[C(H`3O + )/mol ⋅ dm −3 ]2
[C0 (HB)/mol ⋅ dm −3 ] − [C(H`3O + )/mol ⋅ dm −3 ]
Chemistry
西北大学精品课程·重点课程·学科核心课程 ---- 普通化学
第5章 酸、碱和酸碱反应
Chapter 5 Acid, Base & Acid-base GENERAL
Reaction
教学要求：
1、 理解布朗斯特酸碱、路易斯酸碱和软硬酸碱理论的意义和要点； 2、 掌握一元弱酸、弱碱的 pH 的计算公式； 3、 掌握缓冲溶液 pH 的计算 4、盐的水解及其溶液 pH 的计算。
即
C(H 3O + )/mol ⋅ dm −3
=
K
θ a
[C
（1）HB + H2O
B- + H3O+
K
θ a
= [C(B− )/mol ⋅ dm −3 ] ⋅ [C(H`3O + )/mol ⋅ dm −3 ] [C(HB)/mol ⋅ dm −3 ]
若 HB 的起始浓度为 C0（HB）
K
θ a
=
[C(H`3O + )/mol ⋅ dm −3 ]2
[C0 (HB)/mol ⋅ dm −3 ] − [C(H`3O + )/mol ⋅ dm −3 ]
Chemistry
西北大学精品课程·重点课程·学科核心课程 ---- 普通化学
第5章 酸、碱和酸碱反应
Chapter 5 Acid, Base & Acid-base GENERAL
Reaction
教学要求：
1、 理解布朗斯特酸碱、路易斯酸碱和软硬酸碱理论的意义和要点； 2、 掌握一元弱酸、弱碱的 pH 的计算公式； 3、 掌握缓冲溶液 pH 的计算 4、盐的水解及其溶液 pH 的计算。
即
C(H 3O + )/mol ⋅ dm −3
=
K
θ a
[C
Chapter 5 Clayden Organics 大学有机化学
Electrophile accept electrons into empty low-energy orbitals represented by one of the following:
Curly arrows represent reaction mechanisms
Nucleophiles
Nucleophiles are either (i) negatively charged, (ii) neutral species with lone pair(s) of electrons, (iii) carbon with lone pairs of electrons
Bond polarity
Polarity can arise from σ bonds too.
An electronegative element bonded to an electropositive element causes bond polarization. The negative end of the dipole is attracted to the electropositive end
(v) σ bonds too can act as nucleophiles
(vi) compounds with carbon-metal bonds
Electrophiles
Electrophiles are neutral or positively charged species with an empty atomic orbital (the opposite of a lone pair) or a low-energy antibonding orbital
有机化学基础英文版
3
CH3
1
Is 5-(1’-methylethyl)-2,2,4-trimethyloctane
CH3 CH3
Is 4,5-diethyl-2,2-dimethylheptane
It is NOT 3,4-diethyl-6,6-dimethylheptane!
4
H3C
H2 C
2
CH OH
CH3
4
H3C
R C O R
I
Bromide
R C O NH2
Iodide
R C O Cl
Carboxylic acid
R C O O C O
Ester
R R C N
Ketone
H R N R O O
Amide
Acid Chloride
Anhydride
Imine
Nitro groupFra bibliotekA cyclic ester is called a lactone, a cyclic amide a lactam
The ‘substitution level’ of a carbon atom in an organic compound is determined by the number of attached hydrogen atoms:
tertiary carbon (one H)
H3C H2 C H3C C H2 CH3 CH CH CH CH3 H2 C C CH3 CH3 CH3
CH3 CH C H2 CH3
is 3-methyloctane, not 5-methyloctane
2'
H3C
major chain
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3
CH2 = CH2
CH3 CH= CH 2
CH3 C= CH2
IUPAC: Common:
Ethene Ethylene
Propene Propylene
2-Methylpropene Isobutylene
5-12
Common Names
• the common names methylene, vinyl, and allyl are often used to show the presence of the following alkenyl groups
Allyl
5-13
The Cis,Trans System
Configuration
is determined by the orientation of atoms of the mainபைடு நூலகம்chain
H C CH3 CH C
CH2 CH3 H
H
1
2
3
CH3
4
C
C
H3 C
CH( CH3 ) 2
trans 2 -3-Hexene
cis -3,4-D imethyl-2-penten e
5-14
The E,Z System
• uses priority rules (Chapter 3) • if groups of higher priority are on the same side, the configuration is Z (German, zusammen) • if groups of higher priority are on opposite sides, the configuration is E (German, entgegen)
5-3
Unsaturated Hydrocarbons
Arenes:
benzene and its derivatives (Ch 21-22)
H C C H C H C C H H Benzene Alternative rep res entations for th e ph enyl grou p C6 H5 Ph-
H H H H Cyclohexen e H
H Cyclobuten e
H Cyclop entene
H Cycloh eptene
5-17
Cis,Trans Isomerism
• trans-cyclooctene is the smallest trans cyclooctene that has been prepared in pure form and is stable at room temperature • the cis isomer is 38 kJ (9.1 kcal)/mol more stable than the trans isomer • the trans isomer is chiral even though it has no chiral center
H C C H H H Ethylene (an alk e ne )
Alkyne:
contains a carbon-carbon triple bond and has the general formula CnH2n-2
H-C C-H Ace tyle ne (an alk yne )
.
• the molecular formula of the reference hydrocarbon is CnH2n+2
5-8
Index of Hydrogen Deficiency
IDH =
) (H reference - Hmolecule
2
• for each atom of a Group 7 element (F, Cl, Br, I), add one H • no correction is necessary for the addition of atoms of Group 6 elements (O,S) to the reference hydrocarbon • for each atom of a Group 5 element (N, P), subtract one hydrogen
5-4
Structure of Alkenes
A
double bond consists of
• one sigma bond formed by the overlap of sp2 hybrid orbitals and one pi bond formed by the overlap of parallel 2p orbitals • the two carbon atoms of a double bond and the four atoms bonded to them lie in a plane, with bond angles of approximately 120°
Cl (a) (b)
Cl Cl (c) (d) Br
5-16
Cis,Trans Isomerism
Cycloalkenes
• in small-ring cycloalkenes, the configuration of the double bond is cis • these rings are not large enough to accommodate a trans double bond
Alk enyl Grou p CH2= Methylidene CH2=CHEth enyl CH2=CHCH2 3-Propenyl Common N ame Methylene Vin yl Example H2 C CH2 =CH CH2 =CHCH2 IUPA C N ame (Common name) Methylidenecyclopen tane (Methylenecyclopentan e) Eth enylcyclop entane (Vin ylcyclopen tane) 3-Propenylcyclopentane (Allylcyclopentane)
5-6
Cis,Trans Isomerism in Alkenes
Cis,trans
isomers: isomers that have the same connectivity but a different arrangement of their atoms in space due to the presence of either a ring or a carbon-carbon double bond
5-5
Structure of Alkenes
• it takes approximately 264 kJ (63 kcal)/mol to break the pi bond in ethylene; that is, to rotate one carbon by 90° with respect to the other so that there is no overlap between 2p orbitals on adjacent carbons
1. Number the longest chain of carbon atoms that contains the double bond in the direction that gives the carbons of the double bond the lowest numbers 2. Locate the double bond by the number of its first carbon 3. Name substituents 4. Number the carbon, locate and name substituents, locate the double bond, and name the main chain
5-9
Index of Hydrogen Deficiency
Problem: isopentyl acetate has a molecular formula of C7H14O2. Calculate its IHD O • reference hydrocarbon C7H16 O • IHD = (16-14)/2 = 1 Isopentyl acetate
Problem: calculate the IHD for niacin, molecular formula O C6H6N2O NH2 • reference hydrocarbon C6H16 • IHD = (16 - 6)/2 = 5 N
Ni aci n
5-10
IUPAC Nomenclature
higher C lower C lower lower high er higher C C higher lower
Z (zusammen)
E (entgegen)
5-15
The E,Z System
Example: name each alkene and specify its configuration by the E,Z system
H H H CH3 C C C C H3 C CH3 cis -2-Butene mp -139° C, bp 4° C H3 C H t rans- 2-Butene mp -106° C, bp 1° C
5-7
Index of Hydrogen Deficiency
Index
of hydrogen deficiency (IHD): the sum of the number of rings and pi bonds in a molecule To determine IHD, compare the number of hydrogens in an unknown compound with the number in a reference hydrocarbon of the same number of carbons and with no rings or pi bonds
CH2 = CH2
CH3 CH= CH 2
CH3 C= CH2
IUPAC: Common:
Ethene Ethylene
Propene Propylene
2-Methylpropene Isobutylene
5-12
Common Names
• the common names methylene, vinyl, and allyl are often used to show the presence of the following alkenyl groups
Allyl
5-13
The Cis,Trans System
Configuration
is determined by the orientation of atoms of the mainபைடு நூலகம்chain
H C CH3 CH C
CH2 CH3 H
H
1
2
3
CH3
4
C
C
H3 C
CH( CH3 ) 2
trans 2 -3-Hexene
cis -3,4-D imethyl-2-penten e
5-14
The E,Z System
• uses priority rules (Chapter 3) • if groups of higher priority are on the same side, the configuration is Z (German, zusammen) • if groups of higher priority are on opposite sides, the configuration is E (German, entgegen)
5-3
Unsaturated Hydrocarbons
Arenes:
benzene and its derivatives (Ch 21-22)
H C C H C H C C H H Benzene Alternative rep res entations for th e ph enyl grou p C6 H5 Ph-
H H H H Cyclohexen e H
H Cyclobuten e
H Cyclop entene
H Cycloh eptene
5-17
Cis,Trans Isomerism
• trans-cyclooctene is the smallest trans cyclooctene that has been prepared in pure form and is stable at room temperature • the cis isomer is 38 kJ (9.1 kcal)/mol more stable than the trans isomer • the trans isomer is chiral even though it has no chiral center
H C C H H H Ethylene (an alk e ne )
Alkyne:
contains a carbon-carbon triple bond and has the general formula CnH2n-2
H-C C-H Ace tyle ne (an alk yne )
.
• the molecular formula of the reference hydrocarbon is CnH2n+2
5-8
Index of Hydrogen Deficiency
IDH =
) (H reference - Hmolecule
2
• for each atom of a Group 7 element (F, Cl, Br, I), add one H • no correction is necessary for the addition of atoms of Group 6 elements (O,S) to the reference hydrocarbon • for each atom of a Group 5 element (N, P), subtract one hydrogen
5-4
Structure of Alkenes
A
double bond consists of
• one sigma bond formed by the overlap of sp2 hybrid orbitals and one pi bond formed by the overlap of parallel 2p orbitals • the two carbon atoms of a double bond and the four atoms bonded to them lie in a plane, with bond angles of approximately 120°
Cl (a) (b)
Cl Cl (c) (d) Br
5-16
Cis,Trans Isomerism
Cycloalkenes
• in small-ring cycloalkenes, the configuration of the double bond is cis • these rings are not large enough to accommodate a trans double bond
Alk enyl Grou p CH2= Methylidene CH2=CHEth enyl CH2=CHCH2 3-Propenyl Common N ame Methylene Vin yl Example H2 C CH2 =CH CH2 =CHCH2 IUPA C N ame (Common name) Methylidenecyclopen tane (Methylenecyclopentan e) Eth enylcyclop entane (Vin ylcyclopen tane) 3-Propenylcyclopentane (Allylcyclopentane)
5-6
Cis,Trans Isomerism in Alkenes
Cis,trans
isomers: isomers that have the same connectivity but a different arrangement of their atoms in space due to the presence of either a ring or a carbon-carbon double bond
5-5
Structure of Alkenes
• it takes approximately 264 kJ (63 kcal)/mol to break the pi bond in ethylene; that is, to rotate one carbon by 90° with respect to the other so that there is no overlap between 2p orbitals on adjacent carbons
1. Number the longest chain of carbon atoms that contains the double bond in the direction that gives the carbons of the double bond the lowest numbers 2. Locate the double bond by the number of its first carbon 3. Name substituents 4. Number the carbon, locate and name substituents, locate the double bond, and name the main chain
5-9
Index of Hydrogen Deficiency
Problem: isopentyl acetate has a molecular formula of C7H14O2. Calculate its IHD O • reference hydrocarbon C7H16 O • IHD = (16-14)/2 = 1 Isopentyl acetate
Problem: calculate the IHD for niacin, molecular formula O C6H6N2O NH2 • reference hydrocarbon C6H16 • IHD = (16 - 6)/2 = 5 N
Ni aci n
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IUPAC Nomenclature
higher C lower C lower lower high er higher C C higher lower
Z (zusammen)
E (entgegen)
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The E,Z System
Example: name each alkene and specify its configuration by the E,Z system
H H H CH3 C C C C H3 C CH3 cis -2-Butene mp -139° C, bp 4° C H3 C H t rans- 2-Butene mp -106° C, bp 1° C
5-7
Index of Hydrogen Deficiency
Index
of hydrogen deficiency (IHD): the sum of the number of rings and pi bonds in a molecule To determine IHD, compare the number of hydrogens in an unknown compound with the number in a reference hydrocarbon of the same number of carbons and with no rings or pi bonds