CHEM Lecture 6

Transcription

1 EM 494 Special Topics in hemistry Illinois at hicago EM Lecture 6 Prof. Duncan Wardrop October 15, 2012

2 Midterm Papers Factors that ontrol ydrocarbon Acidity Factors that ontrol ydrocarbon onformation Platonic ydrocarbons (ubane and Beyond) ydrocarbon hemistry of Gasoline ydrocarbon hemistry of Anesthesia hemistry of the Perfect Pasta Sauce EM 494, Spring 2010 Slide 2

3 EM 494 Special Topics in hemistry Illinois at hicago Nucleophilic Substitution at Saturated arbon hapter 19

4 Structure of arbocations 3 2p Z carbocations can be stabilized by inductive effects and hyperconjugation carbocations are high energy intermediates; hard, but not impossible to isolate carbon is sp 2 -hybridized with a single, unoccupied 2pZ orbital; 6 valence electrons planar structure : three bonds to carbon are at 120º angles from each other and 90º to empty p-orbital; VSEPR nucleophiles add to either lobe of the empty p-orbital; since it is flat, there is no preference to which side nucleophile adds EM 494, Spring 2010 Slide 4

5 1. Inductive Effects Stability of arbocations electron withdrawal or electron donation that is transmitted through σ- bonds; polarization of σ-bonds 1º cation δ+ 3 δ+ δ+ electron donation through σ-bonds toward carbocation delocalizes charge (spreads out) - σ-bonds are more polarizable, therefore donate more electron density through σ-bonds more - σ-bonds = more stable carbocation EM 494, Spring 2010 Slide 5

6 1. Inductive Effects Stability of arbocations Since - σ-bonds are more polarizable than - bonds, the additional of more alkyl groups leads to stabilization of the cation 2º cation 3º cation δ+ 3 δ+ 3 δ+ δ+ 3 δ+ 3 3 δ+ EM 494, Spring 2010 Slide 6

7 2. yperconjugation Stability of arbocations stabilizing interaction that results from the interaction of the electrons in a σ-bond ( or bond ) with an adjacent empty (or partially filled) orbital. Leads to the formation of an extended molecular orbital that increases the stability of the system filled σ orbital 1º cation empty p orbital EM 494, Spring 2010 stabilization results from σ-donation to empty p orbital of planar carbocation electron donation through σ-bonds toward carbocation delocalizes charge (spreads out) methyl cations cannot be stabilized by hyperconjugation since σ-bonds are perpendicular to the empty p orbital Slide 7

8 2. yperconjugation Stability of arbocations filled σ orbital empty 2p Z orbital empty p orbital Energy - bonding (filled) σ 2p Z Stabilization resulting from hyperconjugation EM 494, Spring 2010 Slide 8

9 Stability of arbocations º carbocation 2 - bond hyperconjugative donors 3º carbocation 3 - bond hyperconjugative donors 3º carbocation 3 - or - bond hyperconjugative donors EM 494, Spring 2010 Slide 9

10 i>icker Question Rank the following carbocations in order of increasing stability? a. b. c. d º 3º 1º methyl A. a,b,c,d B. c,d,b,a. d,c,a,b D. b,c,a,d E. d,a,c,b EM 494, Spring 2010 Slide 10

11 Stabilizing Effects on arbocations smallest inductive effect no hyperconjugation EM 494, Spring 2010 largest inductive effect most hyperconjugation Slide 11

12 ow arbocation Stability Effects Rate of Reaction more stable (lower energy) carbocation = more stable (lower energy) transition state (ammond Post.) = lower activation energy (Ea) = faster reaction EM 494, Spring 2010 Slide 12

13 Why are 1 & 2 Alcohols Less Reactive? O -X O E A too high O O + E a E a simple 1 and 2 alcohols do not undergo substitution by the SN1 mechanism since methyl and primary carbocations are too high in energy to be intermediates in nucleophilic substitution reactions = transition state R + 2 O an alternative mechanism is required... EM 494, Spring 2010 Slide 13

14 Bimolecular Substitution - S N 2 Mechanism 3 O Br fast 3 O Br 3 slow (rate-determining) Step 1 Protonation Step 2 Nucleophilic Attack δ- Br 3 O δ+ 3 + O Br -O bond breaks at the same time the nucleophile (Br) forms the -X bond RDS is nucleophilic attack; bimolecular, therefore Ingold notation = S N 2 fewer steps does not mean faster reaction EM 494, Spring 2010 Slide 14

15 Self Test Question Which rate equation below best describes the rate determining step (RDS) in an S N 2 mechanism? 3 O Br A. rate = k[oxonium ion] δ- Br 3 O δ+ B. rate = k[carbocation]. rate = k[oxonium ion][halide] D. rate = k[carbocation][halide] 3 Br E. rate = k[alcohol][x] EM 494, Spring 2010 Slide 15

16 EM 494 Special Topics in hemistry Illinois at hicago alogenation of Alkanes Methods and Mechanism hapter 39

17 alogenation of Alkanes Increasing Reactivity Fluorination (F 2 ): ighly Exothermic (Explosive!) hlorination ( 2 ): Exothermic Bromination (Br2): Slightly Exothermic Iodination (I2): Endothermic EM 494, Spring 2010 Slide 17

18 Step One Initiation via omolysis Radical hain Mechanism homolysis light (hν) + half-headed arrow fishhook movement of a single electron, not a pair homolysis (homolytic cleavage) cleavage of a covalent bond so that each atom in the bond retains one electron heterolysis (heterolytic cleavage) cleavage of a covalent bond so that one atom in the bond retains both electron free radical unpaired electron; stabilized by same factors that stabilized carbocations EM 494, Spring 2010 Slide 18

19 Radical hain Mechanism Step Two Propagation vis -Atom Abstraction chlorine radical hydrogen abstraction alkyl radical valence electrons + radical abstracts atom from most substituted atom alkyl radical is an intermediate in the mechanism alkyl radicals are stabilized by same factors that stabilize carbocations note that radical is generated - propagation! EM 494, Spring 2010 Slide 19

20 Radical hain Mechanism Step Three Propagation vis -Atom Abstraction chlorine molecule 3 alkyl radical halogen abstraction alkyl chloride chlorine radical alkyl radical abstract a halogen from a 2nd X2 molecule chlorine radical product continues on in chain; starts the cycle over again by abstracting hydrogen from another alkane radical chain mechanisms are faster than a stepwise mechanism which would require initiation in each step 3 + EM 494, Spring 2010 Slide 20

21 omplete Mechanism Initiation homolysis light (hν) + 7 valence electrons Propagation hydrogen abstraction valence electrons 3 halogen abstraction EM 494, Spring 2010 Slide 21

22 Structure of Alkyl Radical Intermediates 3 alkyl radicals can be stabilized by inductive effects and hyperconjugation; similar to carbocations radicals are high energy intermediates; 7 valence electrons; cannot be isolated sp 2 -hybridized; contain one empty p- orbital; unpaired electron in the p- orbital; approximately planar: three bonds to carbon are at ~120º angles from each other and ~90º to half-filled p-orbital stabilized by inductive effects and hyperconjugation Stability: 3º > 2º >> 1º > 3 EM 494, Spring 2010 Slide 22

23 Stabilizing Effects on Alkyl Radicals smallest inductive effect no hyperconjugation EM 494, Spring 2010 largest inductive effect most hyperconjugation Slide 23

24 Bromination is More Selective Than hlorination Br chlorination bromination = early transition state structures = late transition state structures Relative Rates (k rel ) of alogenation R 3 (tertiary, 3º) R 2 2 (secondary, 2º) Ea (chlorination) Ea (bromination) Ea (bromination) > Ea (chlorination) Bromination is more selective. R 3 (primary, 1º) ammond Postulate chlorine radicals are higher in energy than bromine radicals = transition states in chlorination are earlier= look more like reactants = less difference in TS energy = less selective = greater mixture bromine radicals are lower in energy than chlorine radicals = transition states in bromination are later= look more like products (radical interm.) = greater difference in TS energy = more selective = less of a mixture EM 494, Spring 2010 Slide 24

25 Quantifying Selectivity Relative Rates (k rel ) of alogenation R 3 R 2 2 (tertiary, 3º) (secondary, 2º) chlorination bromination Predicted Product Ratios R 3 (primary, 1º) Product Relative Yield Absolute Yield % = (krel) x (statistical factor) total chlorination A (2 2º s) 2 x 3.9 = /13.8 = 57% B (6 1º s) 6 x 1 = /13.8 = 43% Sum % A: 57% B: 43% bromination A (2 2º s) 2 x 82 = /170 = 96% B (6 1º s) 6 x 1 = /170 = 4% Sum % Br Br Br A: 96% B: 4% EM 494, Spring 2010 Slide 25

26 Self Test Question Determine the predicted product distribution for A in the following chlorination. Br 2 EM 494, Spring 2010 Br A + Br Relative Rates (k rel ) of alogenation R 3 R 2 2 R 3 (tertiary, 3º) (secondary, 2º) (primary, 1º) chlorination bromination B A. 99% B. 97%. 95% D. 93% E. 91% Slide 26

27 Defining Regioselectivity Regioselectivity (regioselective) A regioselective reaction is one in which one direction of bond making or breaking occurs preferentially over all other possible directions. Reactions are termed completely (100%) regioselective if the discrimination is complete, or partially (<100%), if the product of reaction at one site predominates over the product of reaction at other sites. The discrimination may also semi-quantitatively be referred to as high or low regioselectivity. IUPA ompendium of hemical Terminology 2nd Edition (1997) O Br N O 4, hν Regioselective hlorination......not stereoselective! EM 494, Spring 2010 Slide 27

28 Mechanism of Alkane hlorination Initiation homolysis light (hν) + 7 valence electrons Propagation hydrogen abstraction valence electrons 3 ~1,000,000 cycles per initiation step halogen abstraction EM 494, Spring 2010 Slide 28