Transcript ppt
Chemistry 125: Lecture 44 January 26, 2011 This Nucleophilic Substitution and Mechanistic Tools: Stereochemistry, Rate Law, Substrate, Nucleophile For copyright notice see final page of this file SN2 Nucleophilic Substitution Generality of Nucleophilic Substitution Solvent Nu: R-L Nucleophile Substrate (+) (-) Nu-R L Product Leaving Group But there are different mechanisms! the Pragmatic Logic of Proving a Mechanism with Experiment & Theory (mostly by disproving all alternative mechanisms) "It is an old maxim of mine that when you have excluded the impossible, whatever remains, however improbable, must be the truth." The Adventure of the Beryl Coronet SN2 Nucleophilic Substitution Nu: R-L (+) Nu-R (-) L Break bond Make bond the Pragmatic Logic (Dissociation) (Association) Simultaneous of Proving a Mechanism D then A A then D “Concerted” with Experiment &(make-as-you-break) Theory (mostly by disproving all alternative mechanisms) Concerted Nu C L Transition State A/D Nu C L Pentavalent Intermediate D/A C Trivalent Intermediate Concerted a Nu A/D b C a L c Transition State Nu enantiomers a b b C D/A L c Pentavalent Intermediate Nu C Nu c Trivalent Intermediate chiralfor chiral achiral Unlikely Which is it very exothermic normally? processStereochemical Implications! ( (Hammond implausibility) Tools for Testing (i.e. Excluding) Mechanisms: Stereochemistry (J&F sec 7.4b) Rate Law (J&F sec 7.4a) Rate Constant (J&F sec 7.4cdefg) Structure X-Ray and Quantum Mechanics Nucleophilic Substitution Walden N Inversion + RL (1898) - “the most L +astounding RN discovery in stereochemistry since the groundbreaking work of van’t Hoff.” E. Fischer N + C L L N +C C+ L N Displacement Replacement STEREOCHEMISTRY Kenyon and Phillips (1923) nucleophilic substitution at S O H PhCH2 CH +33° CH3 (A/D, A favored by vacant d orbital of S) PhCH CH CH3 -OH CH3 O SO2 PhCH2 CH +31° Why notBackside avoid AttackHin CH 3 nucleophilic acetate steps by substitution nucleophilic O substitution at C=O using OH? Because at saturated O C CH3 (A/D, A favored by C. *) C it attacks H (the only step involving chiral C) Same as O O O starting O C CH3 material? O H O C CH3 HO OH PhCH2 CH PhCH2 CH OH PhCH2 CH CH3 CH3 CH3 Inversion! -7° -32° (R) (S) Proves nothing Cl SO2 CH3 Concerted Nu C L Pentavalent Transition State A/D Nu C L Pentavalent Intermediate D/A C Trivalent Intermediate Trivalent intermediate could be attacked from either face racemization, not inversion. Tools for Testing (i.e. Excluding) Mechanisms: Stereochemistry Rate Law Rate Constant Structure X-Ray and Quantum Mechanics NaOEt + EtBr EtOEt + NaBr rate d[EtO-] dt Second Order (SN2) = k2 [EtO-] [EtBr] 0 [NaOEt] ( fixed [EtBr] ) Concerted A/D D/A Nu enters Nu C L Pentavalent Transition State Nu C Nu enters L Pentavalent Intermediate C Trivalent Intermediate Initial rate-limiting dissociation in D/A would give a rate independent of [Nu], not SN2. Not D/A NaOEt + EtBr Analogy EtO- + EtBr H+ EtOH EtOEt H H pKa 15.7 -1.7 EtOEt + NaBr 1017.4 at equilibrium d[EtO-] rate EtO: + H ++ + EtBr H EtOH EtOEt EtOH dt Ratio should be much less drastic at early SN2 transition state. Second Order (SN2) = k2 [EtO-] [EtBr] + k1 [EtBr] [EtOH] [EtBr] ~ const Pseudo First Order First(D/A?) Order k2 = 20,000 k 0 [NaOEt] Is it reasonable to be so different? Tools for Testing (i.e. Excluding) Mechanisms: Stereochemistry Rate Law Rate Constant Structure X-Ray and Quantum Mechanics Rate Constant Dependance on Nu: R-L Nucleophile Substrate 23x Leaving Group R krel CH3 145 CH3CH2 [1] CH3CH2CH2 0.82 1.2x (CH3)2CH (CH3)2CHCH2 3000x Solvent Nu-R L Product RBr + Iacetone / 25°C C-L antibonding node 128x ~same H 0.036 >15x 0.000012 LUMO 145x 0.0078 (CH3)3C ~ 0.0005 ? (CH3)3CCH2 (-) (+) Surface Potential +26 to -25 kcal/mole Something else happens e.g. J&F Table 7.1 p. 275 Methyl Ethyl iso-Propyl Steric Hindrance Total Density (vdW) -Methylation t-Butyl Methyl Ethyl iso-Propyl LUMO at 0.06 LUMO at 0.04 Total Density (vdW) -Methylation t-Butyl Methyl Ethyl iso-Propyl Surface Potential +26 to -25 kcal/mole -Methylation t-Butyl Ethyl [1] n-Propyl 0.82 iso-Butyl 0.036 -Methylation Neopentyl 0.000012 No way to avoid the third -CH3 Nu C L C Transition State Backside Attack Planar Trivalent Intermediate Might it be possible to have frontside attack? or formation of a non-planar cation? (remember planar BH3) C Nu L Transition State Frontside Attack C+ Nonplanar Trivalent Intermediate “In 1939 Bartlett and Knox published the account of their work on the bridgehead chloride, apocamphyl chloride. I believed then, and I believe now, that this was a fantastically influential paper. For thirty years afterwards, no one really accepted any mechanism unless it had been tested out on a bridgehead case. Indeed, the Bartlett-Knox paper shaped the interests and viewpoint of many chemists about the kind of physical organic they wanted to do.” John D. Roberts Caltech 1975 Molecule specifically designed and prepared to test these mechanistic questions Bartlett and Knox * (J.Am.Chem.Soc. - 1939) “bridgehead” chloride Cl boat c-hexane with a bridge bicyclo[2.2.1]heptane Flattening would generate highly strained angles (estimated >23 kcal/mole). Cation would not be planar. Backside of s*C-Cl is inaccessible, and inversion would be impossible. Attack would have to be frontside. Bartlett and Knox * (J.Am.Chem.Soc. - 1939) “C=C bonds cannot originate from such a bridgehead.” (Bredt’s Rule) Although there are -H atoms, they are not in the anti position necessary to allow sCH - s*C-X overlap during elimination of H-X to form C=C. Horrid Overlap! gauche H H H Would competition from loss of HCl make it impossible to measure the expected really slow rate of substitution? Bartlett and Knox * (J.Am.Chem.Soc. - 1939) C Nu L C+ >>106 slower than typical backside attack >109 slower than from Et(CH3)2C-Cl 60°cooler and without Ag+ pull on Cl instead of pushing at C R-Cl: + Ag+ R+ + AgCl ( ) Bartlett and Knox * (J.Am.Chem.Soc. - 1939) Cycloalkyl Halides (e.g. J&F Table 7.2) krelative ~109° 60 strain° in starting material 90 ° 109 ° [1] <0.0001 ??? Br C 120 C ° C H sp2 I increased strain in transition state 0.008 1.6 0.01 OK bent Rate Constant Dependance on Nu: Nucleophile Nu Solvent R-L (+) Nu-R Leaving Substrate Group krel (-) L Product pKa (NuH+) H2O [1] -1.7 F- 80 3.2 Cl- 1,000 -8 Br- 10,000 -9 HO- 16,000 15.7 I- 80,000 -10 HS- 126,000 7 For first-row elements nucleophilicity (attack sC-L ) parallels basicity (attack H+). Both require high HOMO. But as atoms get bigger, they get better at attacking sC-L (compared to attacking H+) e.g. J&F Sec. 7.4d, Table 7.3 Rate Constant Dependance on Nucleophile Nu R-L Leaving Substrate Group krel pKa [1] -1.7 F- 80 3.2 Cl- 1,000 -8 Br- 10,000 -9 HO- 16,000 15.7 I- 80,000 -10 HS- 126,000 7 Nu-R L Polar solvents accelerate reactions that generate (or concentrate) charge, and vice versa. (NuH+) krel CH3I in H2O harder [1] to break 14 H-bonds to smaller ions 160 krel CH3Br in Acetone Backwards H2O (-) (+) 11 5 Sensible Nu: Solvent [1] e.g. J&F Sec. 7.4dg End of Lecture 44 Jan. 26, 2011 Copyright © J. M. McBride 2011. Some rights reserved. Except for cited third-party materials, and those used by visiting speakers, all content is licensed under a Creative Commons License (Attribution-NonCommercial-ShareAlike 3.0). Use of this content constitutes your acceptance of the noted license and the terms and conditions of use. Materials from Wikimedia Commons are denoted by the symbol . Third party materials may be subject to additional intellectual property notices, information, or restrictions. The following attribution may be used when reusing material that is not identified as third-party content: J. M. McBride, Chem 125. License: Creative Commons BY-NC-SA 3.0