Lecture 21 Organic Chemistry 1
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1 CEM 232 Organic Chemistry I at Chicago Lecture 21 Organic Chemistry 1 Professor Duncan Wardrop March 30,
2 Self Test Question Ethanolysis of alkyl halide 1 gives ether 2 as one product; however, several other constitutional isomers of 2 are also isolated. Which of the following structures can not be one of those products? int: The carbocation intermediate is stabilized by resonance. Draw all possible resonance structures to determine which carbons are electrophilic. C 3 C 2 O C 3 C 2 O ethanolysis OC 2 C A B C D OC 2 C 3 OC 2 C 3 OC 2 C University of Illinois at Chicago CEM 232, Spring 2010 Slide 2 2
3 CEM 232 Organic Chemistry I at Chicago Chapter 10 Conjugation in Alkadienes & Allylic Systems Sections Sections
4 Conjugation conjugare = latin verb: to link or yoke together conjugated not conjugated conjugated conjugated conjugated π-system = contains p-orbitals conjugation: overlap between adjacent π-systems π-systems must be adjacent in order to overlap at Chicago CEM 232, Spring 2010 Slide 4 4
5 Conjugated π-systems allylic carbocation: carbocation adjacent to a vinyl group C C C Review: Allyl Group C C C at Chicago CEM 232, Spring 2010 Slide 5 5
6 Conjugated π-systems allylic carbocation: carbocation adjacent to a vinyl group C C C Review: Allyl Group C C C vinylic carbons at Chicago CEM 232, Spring 2010 Slide 6 6
7 Conjugated π-systems allylic carbocation: carbocation adjacent to a vinyl group C C C Review: Allyl Group C C C vinylic carbons vinylic hydrogens at Chicago CEM 232, Spring 2010 Slide 7 7
8 Conjugated π-systems allylic carbocation: carbocation adjacent to a vinyl group C C C Review: Allyl Group C C C vinylic carbons vinylic hydrogens allylic carbon at Chicago CEM 232, Spring 2010 Slide 8 8
9 Conjugated π-systems allylic carbocation: carbocation adjacent to a vinyl group C C C Review: Allyl Group C C C vinylic carbons vinylic hydrogens allylic carbon allylic hydrogens at Chicago CEM 232, Spring 2010 Slide 9 9
10 Conjugated π-systems allylic radical: radical adjacent to a vinyl group C C C at Chicago CEM 232, Spring 2010 Slide 10 10
11 Conjugated π-systems conjugated alkadienes: two adjacent vinyl groups (alkenes) C C C C at Chicago CEM 232, Spring 2010 Slide 11 11
12 Allylic SN1 Allylic alkyl chlorides undergo SN1 substitution faster than saturated alkyl chlorides C 3 C 2 O ethanolysis OC 2 C 3 k rel = 123 C 3 C 2 O ethanolysis OC 2 C 3 k rel = 1 at Chicago CEM 232, Spring 2010 Slide 12 12
13 CEM 232 Organic Chemistry I at Chicago Chapter 10 Allylic Effects on Reactivity Sections: You are responsible for section
14 Allylic SN1 allylic carbocations are more stable (lower energy) than non-conjugated carbocations lower energy intermediate = lower energy transition state (TS) OC 2 C 3 OC 2 C 3 lower energy TS = lower Ea = faster reaction differences in ground state energies have no effect since they are present in both carbocations and products at Chicago CEM 232, Spring 2010 Slide 14 14
15 Stabilization of Allylic Carbocations Allylic carbocations are stabilized by delocalization of the positive charge. Two models can explain delocalization: 1. Resonance model 2. Orbital overlap model C 3 C 3 C 3 C 3 at Chicago CEM 232, Spring 2010 C 3 δ + δ + C 3 Slide 15 15
16 Stabilization of Allylic Carbocations Allylic carbocations are stabilized by delocalization of the positive charge. Two models can explain delocalization: 1. Resonance model 2. Orbital overlap model geometry = planar bond angles = 120º all C atoms sp2 hybridized 2 conj. p-systems: p-bond & 2p orbital overlap = new MO two electrons delocalized over 3 carbon atoms positive charge shared equally between terminal C atoms at Chicago CEM 232, Spring 2010 Slide 16 16
17 Allylic Carbocation: Unequal Charge Distribution 3 C C 3 2 C 2 O Na 2 CO 3 hydrolysis 3 C C 3 C 3 2 C + O O C 3 85% 15% C 3 C 3 2 O Na 2 CO 3 hydrolysis 3 C C 3 C 3 2 C + O O C 3 85% 15% The same products are isolated from each reaction since both proceed through the same carbocation intermediate. at Chicago CEM 232, Spring 2010 Slide 17 17
18 Allylic Carbocation: Unequal Charge Distribution 3 C C 3 2 C 2 O Na 2 CO 3 hydrolysis 3 C C 3 C 3 2 C + O O C 3 85% 15% C 3 C 3 2 O Na 2 CO 3 hydrolysis 3 C C 3 C 3 2 C + O O C 3 85% 15% δ + δ + δ + most substituted carbon = most positive charge = most electrophilic = lowest Ea upon nucleophile capture at Chicago CEM 232, Spring 2010 Slide 18 18
19 Allylic Carbocation: Unequal Charge Distribution 3 C C 3 2 C 2 O Na 2 CO 3 hydrolysis 3 C C 3 C 3 2 C + O O C 3 85% 15% C 3 C 3 2 O Na 2 CO 3 hydrolysis 3 C C 3 C 3 2 C + O O C 3 85% 15% δ + δ + δ + these are only resonance structures = neither actually exists; this does not represent two carbocation intermediates in equilibrium with each other at Chicago CEM 232, Spring 2010 Slide 19 19
20 Unequal Charge Distribution = Unequal Product Distribution TS 2 O δ+ δ + TS δ + δ + O 2 one carbocation intermediate O 85% two transition states (TS) lower energy TS = + on most substituted C lower energy TS = lower Ea = faster rxn O 15% at Chicago CEM 232, Spring 2010 Slide 20 20
21 Self Test Question Predict the major product for the following reaction. A B C 3 O C 3 O C 3 O δ + δ + δ + C 3 O heat C D OC 3 E C 3 O University of Illinois at Chicago CEM 232, Spring 2010 Slide 21 21
22 SN2 Reactions of Allylic alides Allylic halides undergo SN2 faster than nonallylic halides 2 C NaI acetone 2 C I k rel = 80 3 C NaI acetone 3 C at Chicago CEM 232, Spring 2010 I k rel = 1 Since their are no carbocation intermediates in either reaction, how can we explain this observation? Slide 22 22
23 Allylic SN2 is Faster: Two Arguments 1. steric hinderance (VWF) 2. molecular orbital interactions I I since backside attach is required, the nucleophile is less sterically hindered when adjacent carbon is sp 2 -hybridized at Chicago CEM 232, Spring 2010 Slide 23 23
24 Allylic SN2 is Faster: Two Arguments 1. steric hinderance (VWF) 2. molecular orbital interactions 1s AO nodal plane e - probability = 0 = no bond antibonding (σ*) MO 1s AO bonding (σ) MO molecular orbitals (MOs) are formed by mixing atomic orbitals (AOs) at Chicago CEM 232, Spring 2010 Slide 24 24
25 Allylic SN2 is Faster: Two Arguments 1. steric hinderance (VWF) 2. molecular orbital interactions Generalization: New bonds are formed by overlap between LUMO of electrophile and OMO of nucleophile C LUMO (lowest unoccupied molecular orbital) C C OMO (highest occupied molecular orbital) at Chicago CEM 232, Spring 2010 Slide 25 25
26 Allylic SN2 is Faster: Two Arguments 1. steric hinderance (VWF) 2. molecular orbital interactions Electrons in OMO of nucleophile flow into the empty LUMO of the electrophile O C Nucleophile (OMO) electrophile (LUMO) since the LUMO is an antibonding orbital (σ*), adding electrons to this orbital weakens the C bond until it breaks bonding overlap (constructive interference) of OMO must be from side opposite to C bond to form C O bond O C at Chicago CEM 232, Spring 2010 Slide 26 26
27 Allylic SN2 is Faster: Two Arguments 1. steric hinderance (VWF) 2. molecular orbital interactions C LUMO of C-X bond can adopt a coplanar arrangement with p- orbitals of p-bond = electron delocalization over three orbitals = lower energy LUMO = lower activation energy = faster reaction Why does lower energy LUMO result in lower activation energy? at Chicago CEM 232, Spring 2010 Slide 27 27
28 Allylic SN2 is Faster: Two Arguments 1. steric hinderance (VWF) 2. molecular orbital interactions LUMO of C-X bond can adopt a coplanar arrangement with p- orbitals of p-bond = electron delocalization over three orbitals = lower energy LUMO = lower activation energy = faster reaction Why does lower energy LUMO result in lower activation energy? at Chicago CEM 232, Spring 2010 Slide 28 28
29 SN2 Usually of 1º Allylic alides SN1 competes with SN2 when electrophile is a 2º or 3º alkyl halides. O 3 C O C 3 phenoxide 1º allylic chloride only product O 3 C O + O C 3 phenoxide C 3 2º allylic chloride C 3 C 3 C 3 mixture at Chicago CEM 232, Spring 2010 Slide 29 29
30 Self Test Question Unlike phenoxide (pka of phenol = 10), tert-butoxide (pka of tertbutanol = 18) does not give SN2 as major products. What is the major product? A B O O O + C O O + D elimination predominates when nucleophiles are stronger bases than hydroxide (pka of 2O = 15.7) E University of Illinois at Chicago CEM 232, Spring 2010 Slide 30 30
31 Allylic Free Radicals Allylic radicals are also stabilized by electron delocalization through resonance at Chicago CEM 232, Spring 2010 Slide 31 31
32 Allylic Free Radicals Free radical stabilities are related to bond dissociation energies (BDE) or bond strength at Chicago CEM 232, Spring 2010 Slide 32 32
33 Allylic Free Radicals allylic C- bonds are weaker than non-conjugated C- = less endothermic homolysis of allylic C- bonds because allylic free radical intermediate is more stable reactions that for allylic free radical happen faster than those that generate nonconjugated free radicals at Chicago CEM 232, Spring 2010 Slide 33 33
34 Self Test Question Based on what you ve learned about conjugated carbocation, predict the major product of the following free-radical chlorination. A D 2, light B E C 4 C University of Illinois at Chicago CEM 232, Spring 2010 Slide 34 34
35 alogenation Initiation homolysis + light (hν) Propagation hydrogen abstraction C C 3 C C 3 C C 3 halogen abstraction C C 3 + at Chicago CEM 232, Spring 2010 Slide 35 35
36 Allylic alogenation Initiation homolysis + light (hν) Propagation C C C 3 C 2 weakest C- C C hydrogen abstraction fastest -abs halogen abstraction C C C 3 C 2 most stable radical C C + C 3 C 2 C 3 C 2 at Chicago CEM 232, Spring 2010 Slide 36 36
37 Allylic alogenation most stable at Chicago CEM 232, Spring 2010 Slide 37 37
38 Allylic alogenation in Synthesis Although halogenation of allylic carbons can be regioselective, large mixtures are still obtained when there are more than one set of allylic hydrogens. Therefore, this method should only be in synthetic problems when the following conditions are met: 1. All the allylic hydrogens in the starting alkene must be equivalent. C 3 C C 2, hv 2. Both resonance forms of the allylic radical must be equivalent. C 2 C C C 2 C C 2 C at Chicago CEM 232, Spring 2010 Slide 38 38
39 Allylic alogenation in Synthesis Although halogenation of allylic carbons can be regioselective, large mixtures are still obtained when there are more than one set of allylic hydrogens. Therefore, this method should only be in synthetic problems when the following conditions are met: 1. All the allylic hydrogens in the starting alkene must be equivalent. 2. Both resonance forms of the allylic radical must be equivalent. 2, hv at Chicago CEM 232, Spring 2010 Slide 39 39
40 Allylic alogenation in Synthesis Although halogenation of allylic carbons can be regioselective, large mixtures are still obtained when there are more than one set of allylic hydrogens. Therefore, this method should only be in synthetic problems when the following conditions are met: 1. All the allylic hydrogens in the starting alkene must be equivalent. 2. Both resonance forms of the allylic radical must be equivalent. O 2, hv O O O at Chicago CEM 232, Spring 2010 Slide 40 40
41 NBS: Another Radical omination Reagent O O N hv N + O N-bromosuccinimide (NBS) O + 2 N-bromosuccinimide (NBS) also causes allylic radical bromination only used for radical bromination advantages: NBS generates the radical initiator ( ) and maintains low concentration of 2 low [2] = suppresses competing addition across alkene at Chicago CEM 232, Spring 2010 Slide 41 41
42 NBS: Another Radical omination Reagent NBS is a substitute for 2/hv NBS generates radical initiator ( ) and maintain a low concentration of 2 prevents addition of across alkene to give vicinal dihalide at Chicago CEM 232, Spring 2010 Slide 42 42
43 Self Test Question Which allylic bromide could be prepared as the sole product by radical bromination of an alkene? A B C C 3 C 3 D 3 C University of only 1 set allylic s X more than one resonance form Illinois at Chicago CEM 232, Spring 2010 E Slide 43 43
44 Self Test Question Which allylic bromide could be prepared as the sole product by radical bromination of an alkene? 3 C 3 C A B C C 3 C 3 University of X more than 1 set allylic s X more than one resonance form for each Illinois at Chicago CEM 232, Spring 2010 D E 3 C Slide 44 44
45 Self Test Question Which allylic bromide could be prepared as the sole product by radical bromination of an alkene? A B C 3 C 3 C D 3 C University of only 1 set of allylic s X more than one resonance form Illinois at Chicago CEM 232, Spring 2010 E Slide 45 45
46 Self Test Question C 3 University of C 3 X more than 1 set allylic s X more than one resonance form for each Which allylic bromide could be prepared as the sole product by radical bromination of an alkene? C 3 C 3 Illinois at Chicago CEM 232, Spring 2010 A B C D E C 3 C 3 3 C Slide 46 46
47 Self Test Question Which allylic bromide could be prepared as the sole product by radical bromination of an alkene? A B C 3 C 3 C D 3 C University of only 1 set of allylic s only one resonance form Illinois at Chicago CEM 232, Spring 2010 E Slide 47 47
48 CEM 232 Organic Chemistry I at Chicago Next Lecture... Chapter 10: Sections Quiz This Week... Chapter 9 & Synthesis Problems 48
49 CEM 232 Organic Chemistry I at Chicago Exam Two Monday, April 5 6:00-7:15 p.m. 250 SES Chapters 6-10 (everything!) Makeup Exam: Monday, April 12th, time t.b.a. Makeup policy: There are no makeup exams without prior approval. Only students showing proof of a class conflict will have the option to take a makeup exam. To be added to the makeup list, you must me no later than Friday, Feb
50 CEM 232 Organic Chemistry I at Chicago Exam One Grade Distribution Q1. Ranking (50 points) Q2. Predict the Products (50 points) Q3. Arrow-Pushing Mechanism (50 points) Q4. Nomenclature (20 points) Q5. Drawing & Conformational Analysis (50 points) Q6. Functional Groups (30 points) 50
51 CEM 232 Organic Chemistry I at Chicago Exam One Policies Non-scientific calculators allowed only No cell phones, ipods or others electronic devices No molecular models Periodic table will be provided Seating will be assigned ing Your I.D. 51
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