Tim Berners-Lee. Université McGill

Transcription

1 Tim Berners-Lee Université McGill FORMAT DES PRÉSENTATIONS VISUEL CLIPS SONORES DÉMONSTRATIONS EXEMPLES PERTINENTS SUPPORTS PÉDAGOGIQUES FORMAT DES PRÉSENTATIONS VISUEL CLIPS SONORES DÉMONSTRATIONS EXEMPLES PERTINENTS SUPPORTS PÉDAGOGIQUES FORMAT DES PRÉSENTATIONS VISUEL 200 DIAPOSITIVES COURS DE 75 MINUTES CLIPS SONORES DÉMONSTRATIONS EXEMPLES PERTINENTS SUPPORTS PÉDAGOGIQUES 1

2 FORMAT DES PRÉSENTATIONS VISUEL CLIPS SONORES DÉMONSTRATIONS ΔE = Δm x c EXEMPLES PERTINENTS SUPPORTS PÉDAGOGIQUES TRANSITIONS ÉLECTRONIQUES FORMAT DES PRÉSENTATIONS VISUEL CLIPS SONORES DÉMONSTRATIONS EXEMPLES PERTINENTS SUPPORTS PÉDAGOGIQUES LE PRINCIPE DE LE CATELIER 2

3 Ca 5 (PO 4 ) 3 O Apatite Ca 5 (PO 4 ) 3 O Apatite 5 Ca PO O - 5 Ca PO O Ca 5 (PO 4 ) 3 O Apatite Ca 5 (PO 4 ) 3 O Apatite 5 Ca PO O Ca PO O O 2 O CFBrCl CFBrCl REPRÉSENTATIONS DE CRAM 3

4 STÉRÉOISOMÈRES Molécules où les atomes ont la même connectivité mais différentes organisations spatiales CFBrCl REPRÉSENTATIONS DE CRAM ISOMÈRES DE CONSTITUTION (STRUCTURE) Molécules ou les atomes la même formule brute mais différentes connectivités ISOMÈRES DE CONSTITUTION (STRUCTURE) Molécules ou les atomes la même formule brute mais différentes connectivités 4

5 O C 3 C 3 Androstenol FORMAT DES PRÉSENTATIONS VISUEL CLIPS SONORES DÉMONSTRATIONS EXEMPLES PERTINENTS SUPPORTS PÉDAGOGIQUES General Chemistry Principles and Modern Applications Petrucci et al. 10 th Edition $

6 Exercices ebook Version électronique Téléchargées avant le cours 6

7 Ariel Fenster Cours LC 101 Jussieu http// COOL COursesOnLine ipod 40 Salles de classe http//www. // Conférences OSS UPMC Cours 50,000 Étudiants 7

8 Lequel de ces scientifiques n'a pas reçu le prix Nobel? STUDENT RESPONSE SYSTEM "Clickers" 1. Ernest Rutherford 2. Niels Bohr 3. Dimitri Mendeleïev 4. Marie Curie 5. Linus Pauling Ernest Rutherf. 0% 0% 0% 0% 0% Niels Bohr Dimitri Mende. Marie Curie Linus Pauling Quelle relation existe-t-il il entre ces deux structures? 1. Isomères de constitution 2. Stéréoisomères 3. Identiques 4. Aucune idée Isomères de constit. 0% 0% 0% 0% Stéréoisomères Identiques Aucune idée ariel.fenster@mcgill.ca ca oss.mcgill.ca cool.mcgill.ca GENERAL CEMISTRY CEM-110 Chemical Symbol Lewis Symbols nucleus Dots valence electrons Si LEWIS STRUCTURES 8

9 Octet Rule Atoms surround themselves with eight electrons to achieve noble gas stability Al 2 O 3 Al 2 O 3 WRITING LEWIS STRUCTURES 1. Count total # of valence e - For +ve charge, remove e - For ve charge, add e - 9

10 2. Draw a plausible skeleton Symmetrical structures are preferred Least electronegative element usually in centre 3. Achieve an octet around each atom Convert lone pairs into double or triple bonds, if necessary NO 3 - NO 3 - N 5 valence e - x 1 = 5 O 6 valence e - x 3 = (-) 24 e - NO 3 - NO 3-10

11 NO 3 - C 3 NCO C N C O? C N C O C N C O Formal Charges Tells us where electrons are located. elps us predict the most plausible structure. Formal Charges Tells us where electrons are located F.C. = # of valence e - # of l.p. e - ½ # of bonding e - NO + N O + NO + N O + F.C. (N) = 5 2 ½ 6 = 0 F.C. (O) = 6 2 ½ 6 = +1 Note F.C. (N) + F.C. (O) = +1 The sum of the formal charges must be equal to overall charge of the species 11

12 RESONANCE FORMS RESONANCE FORMS Several Lewis structures can be written, but the true one is a hybrid Resonance forms have the same connectivity but different electron distribution O 3 Determining the "best" Resonance form Writing resonance structures 1. When writing resonance structures the connectivity cannot be altered (only lone pair electrons and electrons in double and triple multiple bond can be moved). N 2 O 2. A structure with the lowest magnitudes of formal charges is preferred (greater contribution to the hybrid). 3. A structure with a negative charge on the most electronegative atom is preferred. 12

13 Consider of the four forms that can be written for N 2 0 What is the relationship between structure 1 and 4? 1. Resonance forms 2. Identical 3. Structural isomers Resonance form. 0% 0% 0% Identical Structural iso. 05 STRUCTURAL ISOMERS What is the relationship between structure 1,2 and 3? Structural isomers have the same molecular formula but different connectivities 1. Resonance forms 2. Identical 3. Structural isomers Resonance form. 0% 0% 0% Identical Structural iso. 05 RESONANCE FORMS N 2 O Resonance forms have the same connectivity but different electron distribution

14 Which of the four forms that can be written for N 2 0 is (are) the most plausible BEST N 2 O 1. (1) 2. (2) 3. (3) 4. (4) % 0% 0% 0% Which of the structures best represent methyl isocyanate Which of the structures best represent methyl isocyanate 1. A C N C O 0 A C N C O B +1 C N 0 C 0 +1 O C B 3. C 4. All A 0% 0% 0% 0% B C 05 All C 3 NCO Exceptions to the Octet Rule 1. Incomplete Octet C N C O 0 BEST C N C O +1 C N 0 C 0 +1 O -1 14

15 BF 3 F B F F 6 e - 0 BF 3 0 F B F F 0 6 e - 0 BETTER! Exceptions to the Octet Rule SO Incomplete Octet 2. Expanded d Octet t SO 4 2- Exceptions to the Octet Rule 1. Incomplete Octet 2. Expanded d Octet t BETTER! 3. Paramagnetic Species 15

16 NO NO Odd # of electrons GENERAL CEMISTRY CEM-110 V.S.E.P.R. VSEPR Theory Molecules adopt the geometry which maximizes the distance between electron pairs around a central atom, and thus minimizes electrostatic repulsions. VSEPR Theory Molecular shape depends on The number of electron pairs The type of electron pairs 16

17 VSEPR Molecules containing only bonding pairs around the central atom 2 e - pairs Linear The shape depends only on the number of electron pairs 2 e - pairs Linear Formula AB 2 Bond Angle e - pairs Linear 3 e - pairs Trigonal Example BeI e - pairs Trigonal Formula AB 3 Bond Angle e - pairs Linear 3 e - pairs Trigonal 4 e - pairs Tetrahedral Example BF

18 4 e - pairs Tetrahedral 2 e - pairs Linear 3 e - pairs Trigonal 4 e - pairs Tetrahedral Formula AB 4 Bond Angle Example C e - pairs Trigonal bipyramidal 5 e - pairs Trigonal bipyramidal 2 e - pairs Linear 3 e - pairs Trigonal 4 e - pairs Tetrahedral 90 0 Formula AB 5 Bond Angle 90, 120 Example PF e - pairs Trigonal bipyramidal 6 e - pairs Octahedral 6 e - pairs VSEPR Examples Octahedral Formula AB 6 Bond Angle Predict the geometry of the following molecules Example SF

19 BeCl 2 What is the molecular geometry of BeCl 2? Lewis Structure Be Cl Cl 1. Linear 2. Trigonal planar 3. Tetrahedral 4. Trigonal bipyramidal 5. Octahedral 0% 0% 0% 0% 0% Linear Trigonal planar Tetrahedral Trigonal bipyramidal Octahedral 05 BeCl 2 N 4 + Lewis Structure Be Cl Cl Lewis Structure N + Two electron pairs around Be Linear molecule What is the molecular geometry of N 4+? 1. Linear 2. Trigonal planar 3. Tetrahedral 4. Trigonal bipyramidal 5. Octahedral 0% 0% 0% 0% 0% N N Lewis Structure Four electron pairs around N Linear Trigonal planar Tetrahedral Trigonal bipyramidal Octahedral 05 Tetrahedral molecule 19

20 N 4 + SF 6 48 e - Lewis Structure F F F S F F F Tetrahedral molecule What is the molecular geometry of SF 6? SF 6 48 e - 1. Linear 2. Trigonal planar 3. Tetrahedral 4. Trigonal bipyramidal 5. Octahedral 0% 0% 0% 0% 0% F F Lewis Structure F S F F F Six electron pairs around S Linear Trigonal planar Tetrahedral Trigonal bipyramidal Octahedral 05 Octahedral molecule SF 6 VSEPR molecules containing both, bonding and lone pairs, around the central atom The shape depends on the respective number of pairs and their position Octahedral molecule 20

21 Lone pairs (l.p.) are not as localized as bonding pairs (b.p.), and thus cause stronger repulsions Trigonal bipyramidal VSEPR Lone Pair Examples Predict the geometry of the following molecules 21

22 2 O 2 O Lewis Structure Ọ. electron pairs geometry tetrahedral Bonding Pairs 2 Lone Pairs What is the molecular geometry of 2 O? 1. Linear 2. Bent 3. Pyramidal 4. T-Shaped 5. See-saw 6. Square planar 0% 0% 0% 0% 0% 0% Linear Bent Pyramidal T-Shaped See-saw Square planar 05 Lewis Structure BrF e F.... F.... F Br F BrF e - Lewis Structure F + F.Br F Bonding Pairs 4 Lone Pairs F 22

23 BrF e - electron pairs geometry trigonal bipyramidal molecular geometry BrF e - electron pairs geometry trigonal bipyramidal 90 o lp-bp 90 o lp-bp Br Br + see-saw trigonal pyramidal BrF e - XeF 2 22 e - electron pairs geometry trigonal bipyramidal molecular geometry see-saw Lewis Structure. F Xe..... F... Br + What is the molecular geometry of XeF 2? XeF 2 22 e - 1. Linear 2. Trigonal planar 3. Tetrahedral 4. Trigonal bipyramidal 5. Octahedral 0% 0% 0% 0% 0% Linear Trigonal planar Tetrahedral Trigonal bipyramidal Octahedral 05 Lewis Structure Bonding Pairs 2 Lone Pairs F Xe..... F... 23

24 XeF 22 e - 2 XeF 22 e - 2 electron pairs geometry trigonal bipyramidal electron pairs geometry trigonal bipyramidal molecular geometry linear XeF 2 22 e - electron pairs geometry trigonal bipyramidal molecular geometry linear Molecules and Ions containing Multiple Bonds C 3 NCO For geometrical considerations, a multiple bond can be treated as if it were a single bond Recall 24

25 C 3 NCO C 3 NCO Recall e - groups 4 e - geometry tetrahedral molecular geometry tetrahedral Angle ~109.5 Recall C 3 NCO C 3 NCO Recall e - groups 3 Recall e - groups 3 Molecular geometry bent Angle ~120 C 3 NCO C 3 NCO Recall Recall e - groups 2 25

26 C 3 NCO C 3 NCO Recall e - groups 2 e - geometry linear molecular geometry linear Angle ~180 GENERAL CEMISTRY CEM-110 VALENCE BOND TEORY Valence Bond Theory ydrogen 2 The covalent bond results from the overlap of atomic orbitals containing i one unpaired electron each 1s 1s 26

27 Covalent Bond The electrons are no longer confined to a single s orbital around one atom. Instead, they can move over both orbitals for the entire molecule. Covalent Bond The electrons are no longer confined to a single s orbital around one atom. Instead, they can move over both orbitals for the entire molecule. GREATER STABILITY! ydrogen 2 Valence Bond Theory Examples s-s overlap 1s 1s ydrogen 2 ydrogen 2 1s 1s s-s overlap 1s s-s overlap 1s σ bond 27

28 Fluorine F 2 F F 1s 2s 2p 1s 2s 2p p p p-p overlap p-p head to head overlap Does diatomic helium e 2 exist? x 1. Yes 2. No 3. No Idea σ bond 0% 0% 0% p-p head to head overlap Yes No No Idea 05 elium elium e 1s e 1s e 1s e 1s Filled orbital no possible overlap Xe2 28

29 In O 2 the atoms are linked a? Multiple Bonds 1. Single bond 2. Double bond 3. Trriple bond 4. No idea 0% 0% 0% 0% Single bond Double bond Trriple bond No idea 05 Oxygen Oxygen. O O. 1 double covalent bond O O 1s 2s 2p 1s 2s 2p x x O O σ bond p x p x p-p head to head overlap p-p head to head overlap 29

30 Oxygen 1 σ bond O 1s 2s 2p y y O 1s 2s 2p p y p y p-p side to side overlap p-p side to side overlap p-p side to side overlap Overlap above and below the bond axis Oxygen 1 σ bond x π bond O O 1s 2s 2p 1s 2s 2p p-p side to side overlap 1 π bond 30

31 Double Bond In N 2 the atoms are linked a? 1 σ bond 1. Single bond 2. Double bond 3. Triple bond 4. No idea 0% 0% 0% 0% Maximum electron density along the the bond axis Single bond Double bond Triple bond No idea 05 Nitrogen Nitrogen. N 1 triple covalent bond N. N N 1s 2s 2p 1s 2s 2p Nitrogen 1 σ bond x N 1s 2s 2p N N σ bond N 1s 2s 2p p-p head to head overlap 31

32 Nitrogen 1 σ bond N N 1s 2s 2p 1s 2s 2p 2 π bonds p-p side to side overlap Vertical plane Overlap above and below the bond axis z Overlap above and below the bond axis x π bond y orizontal plane p-p side to side overlap π bond Nitrogen 1 σ bond Summary N N 1s 2s 2p 1s 2s 2p 2 π bonds Single bonds Always σ s-s overlap s-p overlap head to head p-p overlap head to head 32

33 Summary Single bonds Always σ Double bonds 1 σ + 1 π Triple bonds 1 σ + 2 π π bonds Summary Only after σ bonds Summary Which one of the following best describes the bonding in hydrogen cyanide, CN π bonds Only after σ bonds Only from p-p side to side overlap 1. 1σ and 1 π bond 2. 2σ and 1 π bond 3. 2 σ, 1 π bond and 1 lone pair 4. 2 σ, 2 π bond and 1 lone pair 5. 2 σ, 1 π bond and 2 lone i 0% 0% 0% 0% 0% 05 Describe the bonding in CN.. C N. 2 σ bonds 2 π bonds 1 lone pair STEREOCEMISTRY

34 STRUCTURAL ISOMERS Same molecular formulas but different connectivity. STRUCTURAL ISOMERS Same molecular formulas but different connectivity. CCl 2 C 2 CCl CCl 1,1-dichloroethene 1,2-dichloroethene C 2 2 Cl 2 STRUCTURAL ISOMERS Same molecular formulas but different connectivity. What is the relation between these two compounds? Cl C C C C Cl Cl Cl STEREOISOMERS Same molecular formulas and same connectivity but different orientation in space. 1. Identical 2. Constitutional isomers 3. Stereoisomers 4. Structural isomers 0% 0% 0% 0% Identical Constitutional isomers Stereoisomers Structural isomers 0 5 STEREOISOMERS CCl CCl 1,2-dichloroethene STEREOISOMERS CONFORMERS stereoisomers resulting from free rotation (or a flip) around carbon-carbon bonds. C C Cl Cl cis-1,2-dichloroethene Cl C C Cl trans-1,2-dichloroethene ENANTIOMERS stereoisomers that are mirror image of one another. DIASTEREOMERS stereoisomers that do not result from free rotation and are not mirror image of another. 34

35 STEREOISOMERS CONFORMERS stereoisomers resulting from free rotation (or a flip) around carbon-carbon bonds. ENANTIOMERS stereoisomers that are mirror image of one another. DIASTEREOMERS stereoisomers that do not result from free rotation and are not mirror image of another. ETANE Newman Projection STAGGERED 35

36 STAGGERED STEREOISOMERS CONFORMERS stereoisomers resulting from free rotation (or a flip) around carbon-carbon bonds. ENANTIOMERS stereoisomers that are mirror image of one another. ECLIPSED DIASTEREOMERS stereoisomers that do not result from free rotation and are not mirror image of another. 36

37 STEROISOMERS STEROISOMERS ENANTIOMERS NON-SUPERIMPOSABLE MIRROR IMAGES 37

38 STEREOISOMERS CONFORMERS stereoisomers resulting from free rotation (or a flip) around carbon-carbon bonds. STEREOISOMERS DIASTEREOMERS ENANTIOMERS stereoisomers that are mirror image of one another. DIASTEREOMERS stereoisomers that do not result from free rotation and are not mirror image of another. C C Cl Cl cis-1,2-dichloroethene Cl C C Cl trans-1,2-dichloroethene What is the relation between these two compounds? 1. Identical C 3 C Cl Cl 2. Constitutional isomers Cl C 3 C Cl What is the relation between these two compounds? 1. Identical Br C 3 C Cl 2. Constitutional isomers Cl C 3 C Br 3. Conformers 4. Diastereomers 5. Enantiomers Identical 0% 0% 0% 0% 0% 5 3. Conformers 4. Diastereomers 5. Enantiomers Identical Constitutional isomers 0% 0% 0% 0% 0% Conformers Diastereomers Enantiomers 5 What is the relation between these two compounds? C 3 C 3 C 3 C 3 O 1. Identical 2. Constitutional isomers 3. Conformers 4. Diastereomers 5. Enantiomers Identical Constitutional isomers 0% 0% 0% 0% 0% Conformers Diastereomers Enantiomers 0 5 C C 3 C 2 Carvone 38

39 O O C C C 3 C 2 2 C C 3 d- carvone l- carvone d-carvone d-carvone l-carvone R/S Convention Cahn - Ingold - Prelog RULES 1. Assign priority to the 4 groups attached. This is based on atomic NUMBER of the atom. 39

40 RULES 1. Assign priority to the 4 groups attached. This is based on atomic NUMBER of the atom. 2. Visualize the molecule with the LOWEST priority group AWAY from you. RULES 1. Assign priority to the 4 groups attached. This is based on atomic NUMBER of the atom. 2. Visualize the molecule with the LOWEST priority group AWAY from you. 3. Trace Clockwise or Counterclockwise. RULES 1. Assign priority to the 4 groups attached. This is based on atomic NUMBER of the atom. 2. Visualize the molecule with the LOWEST priority group AWAY from you. 3. Trace Clockwise or Counterclockwise. 40

41 IGEST ATOMIC # NEXT next IGEST ATOMIC # highest atomic # 41

42 R R RECTUS = RIGT S S SINISTER = LEFT 42

43 R/S Rules 1. Assign priority to the 4 groups attached. This is based on atomic NUMBER of the atom. 2. Visualize the molecule with the LOWEST priority group AWAY from you. The absolute configuration for the structure below is 1. R 2. S 3. Neither 0% 0% 0% R S Neither 5 Rules 1. Assign priority to the 4 groups attached. This is based on atomic NUMBER of the atom. 2. Visualize the molecule with the LOWEST priority group AWAY from you. Double or triple bonds are evaluated as C holding 2, 3 carbons (Solomons p 192) C C C C C Double or triple bonds are evaluated as C holding 2, 3 carbons (Solomons p 205) C 43

44 The absolute configuration for the structure below is C 3 C O C O O C Cl C C C 2 1. R 2. S Same for other species 0% 0% R S 5 44