Structure chemistry Chapter 03 Lecture 5 molecular spectrum
3 化学热力学基础
●在恒压过程中 U Q p p ex V
U 2 U1 Q p p ex V2 V1
Q p ( U 2 p 2 V2 ) U1 p1 V1
U 2 U1 Q p p 2 V2 p1 V1
焓： H U pV 焓变： Qp H 2 H1 H
2 2 2 θ -1 2 2 2 θ -1
kPa 下进行 时，习惯上可不再予以注明。 (3)焓变值与一定的反应式相对应。 N2(g) ＋ 3H2(g) ＝ 2NH3(g) Hθ ＝-92.20 KJ.mol-1 (4)在相同条件下，正向反应和逆向反应的反应热绝 2NH3(g) ＝ N2(g) ＋ 3H2(g) Hθ ＝+92.20 KJ.mol-1 对值相等，符号相反。
Inorganic Chemistry
第三章 化学热力学基础
无机化学
H
θ
热力学 标准态
当反应物或生成物都是气体时各物
质分压为1×105 Pa；
当反应物及生成物都是溶液状态时，
各物质的浓度为1mol•L-1；
固体和液体的标准态则指处于标准
压力下的纯物质。
Inorganic Chemistry
Inorganic Chemistry
第三章 化学热力学基础
无机化学
过程与途径的关系
30℃，1atm 始态 途径Ⅰ 80℃, 2atm 终态
途径Ⅱ 恒温过程 30 ℃，2atm
恒压过程
Inorganic Chemistry
第三章 化学热力学基础
无机化学
实际上，热力学的实际过程都是十分复杂 的，因此，相应的计算也十分困难，但我们在 处理时，可以利用状态函数的性质，把复杂过 程分解成相应的简单过程去简化计算。如：
3basic Structural Chemistry-第三章
( H aa ESaa )ca ( H ab ESab )cb 0 ( H ab ESab )ca ( H bb ESbb )cb 0
H2+的久期方程 关于ca、cb的线性齐次方程组，
得到非零解的条件：系数行列式为0。 H aa ESaa H ab ESab 0 H ab ESab H bb ESbb 二阶久期行列式
= 1
;
R = ∞,
S
a b
0
S的大小与R有关： Sab =（1+R+R2/3）e-R
最 小 的 Sab 最大的Sab
一 般 的 Sab
把Haa ，Hab ， S
J+K E1=EH + 1+S
a b关系代入得
J -K E2 = EH + 1-S E1＜EH ＜ E2 E1=（α+β）/ 1+S E2=（α-β）/ 1-S
3.1.化学键概 述
3.1.1 化学键的定义和类型 广义化学键的定义:化学键是将原子结合成物质世界的作 用力。与泛分子相对应的是泛化学键。
化学键定义:在分子或晶体中两个或多个原子间的强烈相 互作用，导致形成相对稳定的分子和晶体。
泛化学键：共价键、离子键、金属键和次级键 。 基本理论：分子轨道理论、价键理论和密度泛函理论。
Y X c E c 0 a a X Y E 0 cb cb
求极值，即为体系的能量E
Y X c E c 0 a a X Y E 0 cb cb
2 2 X ca H aa 2ca cb H ab cb H bb
Fundamentals of Structural Chemistry
《结构化学》教学大纲(英文版)
‘Structural Chemistry ’Course SyllabusCourse Code：09040001Course Category：Major BasicMajors：ChemistrySemester：SpringTotal Hours：54 Hours Credit：3Lecture Hours：54 HoursTextbooks：《Structural Chemistry》孙墨珑编著，东北林业大学出版社。
I.Introduction to Structural ChemistryThe major targets this course includes the followings: (1) to introduce the material structure of the basic concepts, basic theory, and basic methods for learning “Structural Chemistry”; (2) to explore the relationship between the microstructures and properties of atoms, molecules, and crystals; (3) to systematically clarify the essence of the periodic law of elements; (4) to deeply and qualitatively clarify the essence of the chemical bonds. This course introduces the basic principles of quantum mechanics and their applications in simple systems, structure of atoms, molecules, and crystals, symmetry of molecular orbitals, molecular orbital theory, and ligand field theory, etc. After learning this course, the students should be able to analyze and solve the basic chemistry problems from the point of view of quantum mechanics.II.Table of contentsSection I (Chapter 1) Basic knowledge of quantum mechanics1.1 Failures of classical mechanics1）Black-body radiation & Planck’s solution;2）Ph otoelectric effect & Einstein’s theory;3）Hydrogen spectrum & Bohr’s model.1.2Characteristics of the motion of microscopic particles1)Wave-particle duality;2)Uncertainty principle.1.3The basic postulates of quantum mechanics1)Postulate 1: wavefunction;2)Postulate 2: Hermitian operators;3)Postulate 3: Schrödinger equation;4)Postulate 4: linearity and superposition;5)Postulate 5: Pauli exclusion principle.1.4Applications of quantum mechanics in simple cases1)Free particle in one-dimensional (1D) box;2)Applications of the 1D-box model in simple chemical systems;3)Free particle in two-dimensional (2D) & three-dimensional (3D) box;4)Tunneling & scanning tunneling microscopy (STM).Section II (Chapter 2) Structures and properties of atoms2.1 One-electron atom: H atom1)The Schrödinger equation of H atoms;2)Solution of the Schrödinger equation of H atom.2.2Quantum numbers1)Principle quantum number, n;2)Angular momentum quantum number, l;3)Magnetic quantum number, m;4)Zeeman effect.2.3Wavefunction and electron cloud1)Radial distribution;2)Angular distribution;3)Spatial distribution.2.4 Structure of multi-electron atoms1)The Schrödinger equation of multi-electron atoms•Self-consistent field method;•Central field approximation.2)The building-up principles and electron configuration of multi-electron atoms•Pauli exclusion principle;•Principle of minimum energy;Hund’s rule.2.5Electron spin and Pauli exclusion principle2.6Atomic spectroscopy1)Orbital-spin coupling;2)Spectroscopic terms & term symbol;3)Derivation of atomic term.4)Hund’s rule on the spectroscopic terms;2.7Atomic properties1)Energy of ionization;2)Electron affinity;3)Electronegativity.Section III (Chapters 3-6) Structures and properties of molecules Chapter 3 Geometric structure of molecules─Molecular symmetry & symmetry point group3.1Symmetry elements and symmetry operations1)Symmetry elements and symmetry operations;2)Combination rules of symmetry elements;3.2Point groups & symmetry classification of molecules3.3Point groups & groups multiplication3.4Applications of molecular symmetry1)Chirality & optical activity;2)Polarity & dipole moment.Chapter 4 S tructure of biatomic molecules (X2 & XY)4.1 Linear variation method and structure of H2+ ion1) Shrödinger equation of H2+ ion;2) Linear variation method;3) Treatment of H2+ ion using linear variation method;4) Solutions of H2+ ion.4.2 Molecular orbital theory and diatomic molecules1) Molecular orbital theory;2) Structure of homonuclear diatomic molecules (X2);3) Structure of heteronuclear diatomic molecules (XY).4.3 Valence bond (VB) theory and H2 moleculeChapter 5 Structure of polyatomic molecules (A)5.1 Structure of Methane (CH4)1) Delocalized molecular orbitals of methane (CH4);2) Localized molecular orbitals of methane (CH4).5.2 Molecular orbital hybridization1) Theory of molecular orbital hybridization;2) Construction of hybrid orbitals;3) Structure of AB n molecules;4) Molecular stereochemistry: valence shell electron-pair repulsion (VSEPR)model.5.3 Delocalized molecular orbital theory─Hückel molecular orbital (HMO) theory1) HMO method & conjugated systems;2) HMO treatment for butadiene;3) HMO treatment for cyclic conjugated polyene (C n H n);4) Molecular diagrams;5) Delocalized π bonds.5.4 Structure of electron deficient molecules5.5 Symmetry of molecular orbitals and symmetry rules for molecular reactions5.6 Molecular spectroscopy1)Infrared absorption spectroscopy: molecular vibrations;2)Raman scattering spectroscopy: molecular vibrations;3)Fluorescence spectroscopy: electronic transitions;4)NMR spectroscopy: nuclear magnetic resonances.Chapter 6 Structure of polyatomic molecules (B), coordination compounds 6.1 Crystal field theory6.2 CO and N2 coordination complexes6.3 Organic metal complexes1) Zeise’s salts;2) Sandwich complexes.6.4 Clusters1) Transition-metal cluster compounds2) Carbon clusters and nanotubesSection IV (Chapters 7-9) Structure of crystalsChapter 7 Basics of crystallography7.1 Periodicity and lattices of crystal structure1) Characteristics of crystal structure;2) Lattices and unit cells;3) Bravais lattices and unit cells of crystals;4) Real crystals & crystal defects.7.2 Symmetry in crystal structure1) Symmetry elements and symmetry operations;2) Point groups (32) and space groups (230).7.3 X-Ray diffraction of crystals1) X-ray diffraction of crystals•Laue equation;•Bragg’s law;•Reciprocal lattice.2) Instrumentation of X-ray diffraction;3) Applications of X-Ray diffraction•Single crystal diffraction: crystal structure determination;•Powder diffraction: qualitative & quantitative analysis of crystalline materialsChapter 8 Crystalline solids, I: metals and alloys8.1 Close Packing of Spheres1) Close packing of identical spheres;2) Packing density;3) Interstices.8.2 Structures and Properties of Pure Metals8.3 Structures and Properties of AlloyChapter 9 Crystalline solids, II: ionic crystals9.1 Packing of Ions;9.2 Crystal Structure of Some Typical Ionic Compounds9.3 Trend of Variation of Ionic Radii9.4 Pauling Rule of Ionic Crystal Structure9.5 Crystals of Functional Materials1) Nonlinear optical materials;2) Magnetic materials;3) Conductive polymers;4) Semiconductors: band gap and photocatalysisIII.Table of ScheduleReferences[1] 王荣顺主编，东北师范大学等，《结构化学》，高等教育出版社，2003年。
Chapter-3-Structure-Stereochemistry-Alkanes
Solution
The staggered conformations of isopentane are A, C, and E; the eclipsed conformations are B, D, and F.
Conformations A and E have the lowest energy because they have one fewer gauche interaction than conformation C; hence, either of these conformations is present in greater concentration than conformation C.
SOLVED PROBLEM 3-11
Draw the most stable conformation of trans-1-ethyl-3-methylcyclohexane.
Solution
First, we draw the two conformations.
Both of these conformations require one group to be axial while the other is equatorial. The ethyl group is bulkier than the methyl group, so the conformation with the ethyl group equatorial is more stable. These chair conformations are in equilibrium at room temperature, and the one with the equatorial ethyl group predominates.
固态化学讲义(麻省理工)3
and two others with nodes (antibonding states), i.e., three molecular orbitals in all. (See fig. 3.)As the length of the chain is increased, the number of electronic states into which the atomic 2s state splits also increases, the number of states always equaling the number of atoms. The same occurs when lithium chains are placed side-by-side or stacked on top of each other, so that finally the space lattice of the lithium crystal is obtained. It is of great significance that these electronic states have energies which are bounded by an upper and lower limiting value (see fig. 4). Within these limits the states form an energy band of closely spaced values (one gram of lithium contains nearly 1023 atoms). Similarly, energy bands can also result from overlapping p and d orbitals. The electronic states (orbitals) within an energy band are filled progressively by pairs of electrons in the same way that the orbitals of an atom were filled in accordance with the Pauli principle. This means that for lithium the electronic states of the 2s band will be exactly half-filled.It is of interest to consider why lithium atoms or Li2 molecules combine to form a metal lattice. In the lithium lattice the smallest distance between neighboring atoms is3.03 x 10–10 m, which is larger than in the Li2 molecule. This reflects the fact that bonds between pairs of atoms in the metal are weaker than they are in the molecule. Nevertheless, the metallic form of lithium is more stable than the molecular form because in the metal one atom has many more neighbors than in the Li2 molecule. As a result, the binding energy per gram atom of lithium (i.e., per 6.92 g of lithium) is163 kJ for the metal lattice, but only 56 kJ for one mole of molecule.[The possibility of hybridization (first advanced by L. Pauling to explain metallic bonding) is also a likely factor for the formation of metallic bonds. Thus strong bonds can be formed when the valence electron clouds become concentrated along the direction in which the bonding partners are situated. According to Pauling the situationtable, can often replace each other in arbitrary proportions without altering either the lattice type or the structure of the energy bands. This explains why such metals tend to form a complete series of solid solutions. Metallic alloys consist of such solid solutions or of heterogeneous mixtures of such solutions. Within certain limits, even metal atoms of different valence can be interchanged in a lattice.Band Structure of MetalsAccording to the above considerations the band structure of Li metal can be represented as shown in fig. 5.According to previous reasoning, the 2s band has N states (N = number of atoms) and accommodates n 2s electrons (where n is the number of electrons per atom in the 2s state times N). Thus, this band has only half of the states filled since each state can accommodate two electrons of opposite spin (Pauli exclusion principle). In accordance with the Aufbau principle, the lowest energy states of the band are filled first and the upper states remain empty – but can readily be occupied by electrons upon thermal excitation or the application of an electric field. Since the width of the energy band is of the order of a few volts, spacings of states within the band are of the order of~10–20 eV (1 eV = 1.6 x 10–19 J), electrons can readily acquire the energy necessary to move into excited states, be accelerated, and move through the metal as conducting electrons. Partly filled bands thus constitute conduction bands.The conduction mechanism in Mg, for example, appears complicated by the fact that each 3s state of the valence shell in the atoms is doubly occupied (3s2). Thus the 3s band must be filled completely and no electronic conduction would in principle be expected. Electronic conduction, however, is observed because of a partial overlap of the 3s and the empty 3p bands. With this overlap, electrons can be activated into empty 3p states and exhibit conduction, as in the partly filled s band in Li.filledsolidfilled levelsFig. 6 Band structure of insulators and semiconductors (molecular crystals); the conditions depicted reflect a molar crystal of carbon (diamond).Both insulators and semiconductors have the same basic band structure – the primary) between the valence and the, generically, are materials with very high resistivity (see T able I), comprising glasses, polymers, refractories, composites, liquids and gases. In the present context, an insulator is a molecular crystal, such as diamond (C) or sapphire, with a band gap ) in excess of 4 eV (arbitrary value). Generally such materials will not conduct electricity since their valence band is filled and the energy required to transfer electronsfrom the valence band to the empty conduction band is far in excess of both the thermal energies at room temperature and the energy provided by radiation of the visible spectrum (~2 eV). Therefore, insulators (in single crystal form) are normally transparent (colorless); however, if light is excessively or totally scattered at internal heterogeneities (such as grain boundaries), they may be translucent and even opaque. It should also be recognized that impurities (Cr3+ in Al2O3) or particular point defects (color centers) may impart a color to the transparent insulator crystals. The color arises because of partial absorption of white light and selective transmission of the other portions of the visible spectrum.Semiconductors: The conventional semiconductors, silicon (Si) and germanium (Ge), have a band gap (E g) of 1.1 and 0.7 eV respectively and therefore absorb visible radiation; they are opaque (fig. 7). Considering the statistical nature of the thermal energy distribution in the solid matrix (Maxwell-Boltzmann), a significant number of electrons in the valence band will, at room temperature, acquire sufficient energy to cross the existing energy gap and thus provide for semiconductivity. The conductivity will therefore increase with temperature, contrary to metallic systems, until electron scattering effects, due to increased lattice vibrations (which decrease the mobility of electrons), begin to dominate.The value of semiconductors for solid state device fabrication lies in the fact that the number and type of conducting electric charge carriers [electrons are n-type (negative), holes are p-type (positive)] can be controlled through incorporation of appropriate dopant elements. Thus the substitutional incorporation of Group V elements (Sb, As, P) provides for shallow donor levels in the band gap at about 0.01 eV from the conduction band. The substitutional incorporation of Group III elements (B, Al) generates acceptor levels in the band gap at about 0.01 eV from the valence band. The two types of impurities are almost completely ionized at room temperature and give rise to extrinsicE h (transparent to light)Insulators (carbon)Impurity and defect levels in the energygap may give rise to selective absorptionof light (colour the object).E h (visible light is absorbed)(-)(+)excited electron holeSemiconductor (Si)Fig. 7 Optical bebavior of insulators and semiconductorsn-type and p-type conductivity – the basis for the formation of diodes and transistors (fig. 8).Of increasing importance are compound GaAs, InSb, InP and GaP (compounds of Group III and Group V elements). Together these compounds provide eight valence electrons and, by sp form a diamond-like, covalent crystal structure with semiconductor properties. These* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *Table I. ELECTRICAL RESISTIVITIES OFMETALS AND NONMETALS AT 20_C*______________________________________________________________________________Resistivity,Resistivity, Metals10–8 ohm–m**Nonmetals ohm–m**______________________________________________________________________________Silver 1.6SemiconductorsCopper 1.67Silicon1000.0Gold 2.3Germanium0.09Aluminum 2.69InsulatorsMagnesium 4.4Diamond1010–1011Sodium 4.61Quartz 1.2 x 1012Tungsten 5.5Ebonite 2 x 1013Zinc 5.92Sulfur 4 x 1013Cobalt 6.24Mica9 x 1013Nickel 6.84Selenium 2 x 1014Cadmium7.4Paraffin wax 3 x 1016Iron9.71Tin12.8Lead20.6Uranium29Zirconium41Manganin44Titanium55Lanthanum5996%Iron–4%Si62Cerium78Nichrome100*From American Institute of Physics Handbook, Dwight E. Gray, ed. McGraw-Hill, New York(1963), pp. 4–90; 9–38.**Note the different units in the two columns.* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *EXERCISE FOR THE IDLE MIND1. A solid is found to have an energy band gap (E g) of 3 eV. What is the likely colorof this solid in transmitted sunlight?2.An optically transparent solid appears green in transmitted sunlight. What do youexpect the band gap (E g) of this solid to be (in eV)?3.AlN and GaSb are compounds, solid at room temperature. On the basis ofbonding considerations and data provided in the P/T, attempt to predictdifferences in the properties of these solids.4.Account for the conductivity of (a) Na (metallic) and (b) Mg (metallic) on the basisof appropriate energy diagrams.5.What is the radiation of longest wavelength which is still capable of beingtransmitted through:(a)Si with E g = 1.1 eV(b)Ge with E g = 0.7 eV, and(c)the compound GaAs with E g = 1.43 eV?6.Explain the difference between extrinsic and intrinsic semiconductors.7.Semiconductors in single crystal form are usually produced by solidification ofmelts. To achieve extrinsic semiconductivity it is customary to add to the melt“doping elements” which are substitutionally incorporated [replace, for example, a silicon atom in the crystal (ordered structure)]. If the doping element is P, which has 5 valence electrons, and it replaces a silicon atom (whose 4 valence electrons are normally immobilized because of bond formation), each P atom will be able to contribute one electron to conduction; its other four valence electrons take part in bond formation. Assume you add 3 mg P to 50 g silicon and form a crystal from it in which the P atoms are uniformly distributed; what is the number of conduction electrons/cm3 in the doped crystal? (You may neglect the volume of thesubstitutionally replaced Si atoms and assume that only electrons from P atoms contribute to conduction.)8.Potassium (K) and beryllium (Be) are metals which exhibit good electricalconductivity. Explain for both elements the reasons for the observed conductivity on the basis of the band structure.9.The energy gap (E g) in zinc oxide (ZnO) is 3.2 eV.(a)Is this material transparent to visible radiation?(b)Do you expect this material to be a conductor at room temperature? (Givethe reasons for your answer.)18. A chemical analysis indicates that a silicon crystal weighing 100 g contains 33 mgof aluminum (Al) which is substitutionally incorporated (the Al atoms replace some Si atoms in the crystal).(a)Is this crystal n-type or p-type? (Explain in one sentence.)(b)What is the number of extrinsic charge carriers (per cm3) in this crystal?19.An unknown material is transparent to light of frequencies ($) up to 1.3 x 1014 s–1.Draw a meaningful schematic band structure for this material.20.We know that, in semiconductors, charge carriers can be thermally activated fromthe valence band into the conduction band. The number of thermally activatedelectrons (n e) per cm3 is given by:n e+A T3ń2e*E gńkT(where A = 5 x 1015 cm–3 for silicon). Determine for pure silicon (Si) the number of electrons/cm3 in the conduction band at 500°C.21. A crystal of germanium (E g = 0.7 eV) is found to be n-type with 5 x 1018 mobilecharge carriers/cm3 (at room temperature).(a)Draw a schematic energy band diagram that reflects the indicatedproperties. (Label pertinent features in the diagram.)(b)Do you expect this crystal to be transparent or opaque to radiation of $=1 x 1015 s–1?22.Draw three energy band structures representing respectively (and identifiably) an:(1)extrinsic n-type semiconductor(2)an insulator(3)and a metal.23. A 50 kWatt radio transmitter emits radio waves with a wavelength of 300m. Howmany photons does it emit per minute (1 Joule = 1 Watt per sec)?24. A material exhibits an “optical band edge” (transition from absorption of light totransmission) at $ = 5 x 1014 Hz (s–).(a)Draw a diagram which reflects the indicated optical behavior.(b)What do you expect the color of this material to be when viewed in daylight?(c)What is the band gap (E g) of this material?25. A sample of germanium (Ge), weighing 30 g, is found to contain 54 mg of arsenic(As). Determine for this sample the mobile charge carrier density (carriers/cm3) at room temperature. (Assume As to be substitutionally incorporated in Ge and that all As atoms are ionized at room temperature; you may neglect any intrinsiccharge contributions.)。
[化学课件]有机化学英文课件chapter3
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WilliamH.Brown ChristopherS.Foote
M acintosh PICT M acintosh PICT im age form at im age form at
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Stereoisomerism and Chirality
is not supported
M acin to sh P IC T im ag e fo rm at
is n o t su p p o rted
M a c in to s h P IC T im a g e fo rm a t
is n o t s u p p o rte d
Chiral Center
Chirality
Chiral: from the Greek, cheir, hand
• an object that is not superposable on its mirror image
Achiral: an object that lacks chirality; one that lacks handedness
Constitutional isomers: isomers with a different connectivity
Stereoisomers: isomers with the same connectivity but a different orientation of their atoms in space
Elements of Symmetry
Symmetry in objects
Elements of SFra bibliotekmmetry Plane of symmetry (cont’d)
College Chemistry Chapter 03
Tro's "Introductory Chemistry", Chapter 3 11
Gases, Continued
• Because there is a lot of empty space, the particles can be squeezed closer together. Therefore, gases are compressible. • Because the particles are not held in close contact and are moving freely, gases expand to fill and take the shape of their container, and will flow.
• A mixture of table sugar and black pepper = heterogeneous • A mixture of sugar dissolved in water = homogeneous • Oil and vinegar salad dressing = heterogeneous
Practice—Classify the Following as Homogeneous or Heterogeneous
• Table sugar. • A mixture of table sugar and black pepper. • A mixture of sugar dissolved in water.
Tro's "Introductory Chemistry", Chapter 3 10
Gases
• In the gas state, the particles have complete freedom from each other. • The particles are constantly flying around, bumping into each other and the container. • In the gas state, there is a lot of empty space between the particles.
有机化学 英文课件 chapter(3)
Energy Diagram
A two-step reaction with one intermediate
6-12
Developing a Reaction Mechanism
How it is done
• design experiments to reveal details of a particular chemical reaction
6-14
Electrophilic Additions
• hydrohalogenation using HCl, HBr, HI • hydration using H2O in the presence of H2SO4 • halogenation using Cl2, Br2 • halohydrination using HOCl, HOBr • oxymercuration using Hg(OAc)2, H2O followed by
Addition is regioselective
• regioselective reaction: an addition or substitution reaction in which one of two or more possible products is formed in preference to all others that might be formed
• summary of the relationships between DG0, DH0, DS0, and the position of chemical equilibrium
M acin to sh P IC T im ag e fo rm at
Chapter 3 Stereochemistry
H
Cl
Cl
H
Cl
Cl
Cl
Cl
Cl
H
H
Cl
cis-1,2-dichlorocyclopentane achiral Same compounds
trans-1,2-dichlorocyclopentane chiral Different compounds, Enantiomers (对映体)
H H Cl F F Cl H H
180°
mirror
F Cl H H H H Cl F
S2 = i
H3C H H H3C
H CH3
H CH3
CH3 H H
I 90°
H CH3
CH3 H
CH3 H
IV
CH3
S4 I = III
CH3 H H
H3C H H H3C
H CH3
II = IV
CH3 H
II
CH2CH3 H3C Br C H
CH2CH3 H3C H C H
OC
H
H
H
H
COOH
Cl
Cl
σ:
0
1
1
1
All molecules with a plane of symmetry are achiral． 具有对称面的分子没有手性
(2) symmetry axis (简单对称轴)：Ｃn
H C H H H
Examples:
How to determine whether an object is chiral or achiral?
物体与其镜象是否重合 重合 (superimposable)：非手性 (achiral) 不重合(nonsuperimposable) ：手性 (chiral) 2) Chirality and enantiomerism in organic molecules
Structural_Chemistry_Chapter_4-2
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3
3
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（1）B2 （5×2=10电子）
KK（2σg）2（2σu）2 (1πu) 1 (1πu) 1 特点：无σ键，只有π1+π1 键，BO=1，顺磁性。
_
乙烯的LUMO: πg
+
乙烯的HOMO: πu
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δ
AO以“面对面”方式形成成键轨
道δg和反键轨道δu ，它们都有两个包含 键轴的节面, 区分在于有无垂直于键轴 的节面：
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分 子 轨 道 式 与 键 级
将电子填入MO，可写出分子的电子组态， 也称为“分子轨道式”. 分子中有不成对电子时 表现为顺磁性. 由此还可定义键级BO： BO =（成键电子数-反键电子数）/2 = 净成键电子对数目
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3σu (强反键) σ*2pz π*2pxπ*2py 1πg 3σg(弱成键或非键) π2px π2py σ2pz σ*2s 2σu (弱反键或非键) σ2s 2σg(强成键) O2 F2 B 2 C2 N2 1πu
Chapter3 立体化学
Organic Chemistry
相对构型（D-L）
Shandong University at Weihai
以甘油醛为参照标准而确定的构型称为相对构型（D－L）
Organic Chemistry
Shandong University at Weihai
Organic Chemistry Shandong University at Weihai
Organic Chemistry
第三章 立体化学
Shandong University at Weihai
McMurry’s Organic Chemistry Chapter 9
刑其毅等 有机化学 Chapter 3
Organic Chemistry Shandong University at Weihai
Fischer投影式
1. 碳链尽量放在垂直的方向上，氧化态高的 在上面，氧化态低的在下面，其他的放在 水平方向上； 2. 横向的基团位于平面的前方，竖向的基团 位于平面的后方；
3. 交叉点表示碳原子
Organic Chemistry Shandong University at Weihai
Fischer投影式
Shandong University at Weihai
Organic Chemistry
3.2.1 手性中心 手性分子碳原子
Shandong University at Weihai
Organic Chemistry Shandong University at Weihai
Asymmetric carbon Chiral carbon 如果分子的手性是由于原子和基团围绕某一点的非对称 排列而产生的，这个点就是手性中心（Chiral Center）
化学原理Chemistry课件post3atomicstructure
( Ephoton = DE = RH
1 n2i
1 n2f
)
Ephoton = 2.18 x 10-18 J x (1/25 - 1/9)
Ephoton = DE = 1.55 x 10-19 J
Ephoton = h c/ l
l = h c / Ephoton
l = 6.63 x 10-34 (J•s) x 3.00 x 108 (m/s)/1.55 x 10-19J
Nobel Prize in 1929
What is the de Broglie wavelength (in nm) associated with a 2.5 g Ping-Pong ball traveling at 15.6 m/s?
l = 6.63 x 10-34 / (2.5 x 10-3 x 15.6)
1. Application of Schrödinger to Hydrogen Atom
z
(x,y,z)
Ĥ = E Ĥ 2 2 e2
2m
r
r
y
2 2m
2
e2 r
(x,y,z)
= E (x,y,z)
After transferred into sphere coordinate
x x r sin cos y r sin sin z r cos
Ĥ2 2m2V2: Laplacian Operator
2 2 2
2 x2 y2 z2
Uncertainty principle (1927, W K Heisenberg): DX Dp h/4
It is impossible to determine precisely both the position and momentum of a particle at the same time.
《化学chapter》PPT课件
3.1 Relationship of stereoisomers and chiral centers
Stereoisomer discrimination: two types 1) Enantiomer discrimination
The difference between the homochiral interaction
(R…R) and heterochiral interaction (R…S).
Distinguished by difference of the retention times.
Chiral GC column
Chiral HPLC columns
3.2 Determination of enantiomer & diastereomer composition
Cyclodextrin:
3.1 Relationship of stereoisomers and chiral centers e.g n = 2 (identical), N = 22 = 4, are there actually 4 stereoisomers?
If the two stereogenic centers in a molecule are identical, there are only 3 stereoisomers (a pair of enantiomers plus a meso compound).
无机化学英文课件：Chapter 3 Chemical kinetics
Example：The reaction of O3 with NO
Activation energy of a reverse reaction Ea ( reverse) = Eac - E(Ⅱ)
ΔrHm= E(Ⅱ) - E(Ⅰ)= [Eac - Ea(reverse)] -[Eac -Ea(forward)]
Example： NO(g) + O3(g)→ NO2(g)+ O2(g)
NO Its activated complex O O O，has high potential energy Eac , is very unstable, and decomposes quickly into product molecules NO2 and O2。
Subtracting lnk1 from lnk2 and rearranging it
gives
ln k2 k1
=
Ea R
⎜⎜⎝⎛
1 T1
−1 T2
⎟⎟⎠⎞
The orders of magnitude of Ea are between 40 ~400 kJ·mol-1 . For most reactions , values
4200
3.61×10−4
1200 3.61 × 10 −4
5400
3.68 ×10 −4
The ratio is constant. Therefore, the reaction
有机化学英文课件chapter3
Stereoisomerism and Chirality
Macintosh PICT Macintosh PICT im age form at im age form at is not supported is not supported
Chapter 3
3-2
Isomers
Isomers: different compounds with the same molecular formula
M a c in to s h P IC T im a g e fo r m a t
is n o t s u p p o r te d
3-13
R,S Convention
3. Atoms participating in a double or triple bond are considered to be bonded to an equivalent number of similar atoms by single bonds
• an achiral object has at least one element of symmetry • plane of symmetry: an imaginary plane passing
through an object dividing it so that one half is the mirror image of the other half • center of symmetry: a point so situated that identical components are located on opposite sides and equidistant from that point along the axis passing through it
Lecture3-06
The model of animal cells1. Models of membrane structureQuickTime?and aPhoto - JPEG decompressor are needed to see this picture.Boundary between two glial cells: the plasma membraneCell 1Cell 2!J.D. Robertson (1959):The TEM showing: thetrilaminar appearance ofPM;Unit membrane model!S.J. Singer and G.Nicolson (1972):Fluid-mosaic model!K. Simons et al(1997):Lipid rafts model ;Functional rafts on cellmembranes.Nature 387:569-572Models of membrane structureCell membraneMembrane proteins and lipids can be confinedto s specific domainLipid rafts are rich in choleterol and sphingolipids2. The chemical composition of membranesComposition of biomembranePhospholipid moleculeFour major phospholipids in mammalian plasma membranesGlycolipids are found on the surface of all plasma membranesSome functions of phospholipids in cell signalingPhospholipasecleavage sitesLiposomes: phospholipid vesiclesApplication: gene transfer; as a carrier.Membrane protein attachment via lipidsPalmitic acid (18 saturated fatty acid)geranylgeranyl groupMany membrane proteins are glycosylatedon the non-cytosolic side of the membraneN-linked glycosylationO-linked glycosylationDetergents disrupt the lipid bilayer and solubilize membrane proteinsdetergentssystemsMembrane proteins and lipids can be confinedto s specific domainVarious ways to restrict the movement of specific membrane proteinsPlasma membrane proteins in red blood cellsPlasma membrane proteins in red blood cellsThe cortical region of the cytosol consists of a complicated cytoskeletal network rich in actin filaments, as illustrated in RBCThe surface of lymphocyteThe glycocalyx(cell coat)is formed by carbohydrates projecting from membrane lipids and proteins.3. Characteristics of biomembraneFluorescence Recovery After Photobleaching(FRAP) to demonstrate the lateral diffusion of membrane lipidsFLIP: Fluorescence Loss In PhotobleachingThe mobility of membrane proteinsexperimentally by the mixing of membrane proteins that occurs when two cells are tagged with different fluorescent labels and then induced to fuseintegral proteins can bePM contains lipid rafts that are enriched in sphingolipids, Cl and someMembrane proteins and lipids can be confinedto a specific domainLipid rafts are signaling centers?4. An overview of membrane functionsA. PM define the boundaries of the cell and organelles.B. Compartmentalization:membranes form continuous sheets thatenclose intracellular compartments.C. Transporting solutes:membrane proteins facilitate the movementof substances between compartments.D. Responding to external signals:membrane receptors transducesignals from outside the cell in response to specific ligands.E. Intercellular interaction:membrane mediate recognition andinteraction between adjacent cells by cell-to-cell communication and junction.F. Locus for biochemical activities:membrane provide a scaffold thatorganizes enzymes for effective interaction.G. Energy transduction:membranes transduce photosyntheticenergy, convert chemical energy to ATP, and store energy in ion and solute gradients.Integrating cells to tissues。
大学化学03-lecture-preview
© 2019 Pearson Education, Inc.
Stoichiometry
Chemical Equations
Chemical equations are concise representations of chemical reactions.
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g)
Products appear on the right side of the equation.
© 2019 Pearson Education, Inc.
Stoichiometry
Anatomy of a Chemical Equation
Ca: 1(40.08 amu) + Cl: 2(35.453 amu)
110.99 amu
• Formula weights are generally reported for ionic compounds.
© 2019 Pearson Education, Inc.
Stoichiometry
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g)
The states of the reactants and products are written in parentheses to the right of each compound.
© 2019 Pearson Education, Inc.
© 2019 Pearson Education, Inc.
Stoichiometry
Using Moles
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[
h2 d 2 1 2 − 2 + kx ψ = E v ψ 2 2 8 π μ dx
]
有解的条件为：Ev = (v+1/2)h
v = 0,1,2,· · · 振动量子数
e：谐振子的经典振动频率
1 k υ e= 2 π μ 19
√
例：
1 k υ e= 2π μ
C C 2260–2100 cm-1 C = C 1680–1600 cm-1 C - C 1680–1600 cm-1 C = O 1780–1650 cm-1 C = N 1650–1550 cm-1 C = C 1680–1600 cm-1
William E. Moerner
/nobel_prizes/chemistry/laureates/2014/
8
9
分子中原子运动与运动状态
z
(x1,y1,z1) (xn,yn,zn)
(x1+x,y1,z1)
o z
(xn+x,yn,zn)
y
结构化学
结构化学基础
1
第三章
共价键与双原子分子
2
3.5 3.5.0 3.5.1 3.5.2
双原子分子的光谱 光谱与微观粒子内部运动状态 双原子分子的转动光谱 双原 （振——转光谱的精细结构）
3
3.5.0 光谱与微观粒子运动状态 E 光谱 h = = i - j 微观粒子能级变化
/nobel_prizes/chemistry/laureates/2008/
7
2014 Nobel prize award (super-resolved fluorescence microscopy)
Eric Betzig
Stefan W. Hell
d z1 d zn (x1,y1,z1) dy1 dyn dx1 dxn (xn,yn,zn)
z
x
(x1,y1,z1) (xn’,yn’,zn’) (xn,yn,zn)
o
y
(x1’,y1’,z1’)
o
y z
(x1,y1,z1) (xn,yn,zn) (xn,yn+dy,zn)
x
x
(x1,y1-dy,z1)
故刚性转子Hamilton算符为
^2 M ^= T ^ +V ^= H 2I
2 ^ M ψ = Eψ 2I
(刚性转子只有动能)
其Schrodinger方程为：
M2 = J(J+1)ħ2 J = 0，1，2· · ·
转动量子数
Er = J(J+1)ħ2/2I = J(J+1)h2/82I = J(J+1)Bhc
√
kC-C, kC=C, kCC 大小关系？ C=C, C=N, C=O 大小关系？
20
红外吸收光谱：
选律：① v = ± 1 ② 极性分子(对双原子分子适用) ③ = abd 0 能级变化(v v+1)： E = Ev+1 - Ev = (v +1+1/2)he - (v + 1/2)he = he
o x
10
y
分子运动自由度
d z1 dzn(x ,y ,z ) n n n (x1,y1,z1) dy1 dyn dx1 dxn
z
f总 单原子 双原子 3 6
ft 3 3 3
fr 0 2 2
fv 0 1 3n-5
o
y
x
n原子分子 (x1,y1,z1,· · · , xn,yn,zn) 3n个独立坐标 3n个自由度
6 分子振动 分子转动 ,ESR NMR
2008 Nobel prize award (Green Fluorescent Protein )
Osamu Shimomura
Martin Chalfie
Roger Y. Tsien
/programs/view/RlWSOCTiFks/
m1r1 = m2r2 r = r1 + r2 r1 = m2r/(m1+m2) r2 = m1r/(m1+m2)
故 I = m1r12 + m2r22 = m1m2r2/(m1+m2) = r2 平动：T=mv2/2=p2/2m
2 ^ p ^= T 2m
2 ^ M ^= 转动：T=I2/2=M2/2I T 2I M= I,角动量；,角速度 13
总 = t + r + v + e + n
~10-4~10-2eV < 400 cm-1 ~100~102eV 8000~160000 cm-1
5
光谱区间、可检测手段与微观粒子运动状态对应关系 A. 光谱区间与可检测方法
B. 光谱区间及其对应的微观粒子运动状态
x-射线 远紫外 紫外 可见 近红外 红外 远红外 内层电子跃迁 外层电子跃迁 微波 无线电波
21
振转耦合
EV R 1 h2 (n )hv 2 J ( J 1) 2 8 I
E
1 J’ = 0 2 3
v=1
4
J 0, J 1,
v ve ,
J 0,1, 2,...
v ve 2B( J 1), J 0,1, 2,... J 1, 2,3,...
1 ’ = 0 2
1s1 2p1
4
E
T
3
T
1s2
4
1 =0
3 2
(a)
R
(b)
24
E
H2分子由1s2 → 1s1 2p1 E
1 ’ = 0 2
1s1 2p1
4
E
T
3
T
1s2
4
1 =0
3 2
(a)
R
(b)
25
E
作业： 3.15、3.17、3.19、3.22
26
i
4
j → i
微观粒子运动状态
1 ’ = 0
2
3
j
4
3 2
原子核 状态n
电子 状态e 平动t
原子状 态a 转动r 振动v
4
1 =0
R
总 = t + r + v + e + n
10-2 ~100eV 160~8000 cm-1 ~10-18eV
核裂变、聚变 或衰变
J 1, v ve 2BJ ,
v=0
4 3 2
1 J=0
22 (a)
R
3.5.3 分子的电子光谱 紫外可见光谱仪示意图:
Franck-Condon原理：振动能级间跃迁强度最 高的谱线是与相同核间距对应有最高几率密 度的振动态间的跃迁（垂直跃迁）。
23
H2分子由1s2 → 1s1 2p1 E
多原子 3n 直线型
多原子 3n 非直线型
3
3
3n-6
11
例：H2O与CO2的振动模式与振动自由度
3657 cm-1
1595 cm-1
3756 cm-1
668 cm-1
668 cm-1
1373 cm-1
12
2420 cm-1
3.5.1 双原子分子的转动光谱 (刚性转子模型) m2 m1 O B A r1 r2
B= h 8 π 2 Ic
转动常数
14
跃迁选律：J＝±1
光谱的波数
15
16
应用：测定异核双原分子的键长；同位素效应等性质
例：H35Cl的远红外光谱线的波数分别为21.18, 42.38, 63.54, 84.72, 105.91cm-1，试求其转动惯量及 核间距
17
18
3.5.2 双原子分子的振动光谱 模型：简谐振子 = (r) 势能：V = ½ k(r - re)2 = ½ kx2 re：平衡距离 x：分子核间距与平衡核间距之差 k：力常数，表示化学键的强弱 Schrö dinger方程