Chemistry 125: Lecture 44 January 27, 2010 Nucleophilic Substitution and Mechanistic Tools: Rate Law & Rate Constant This For copyright notice see final page of this file

F CH 2 CH 2 H:OH "E2 Elimination" ABN AON Make Two Break Two F H OH CH 2 Nucleophilic Substitution and  -Elimination Chapter 7 F H:OH FH OH "Acid-Base" F CH 3 :OH CH 3 OHF  "S N 2 Substitution" ABN Make & Break Same (Cf. Lecture 16)

All are Nucleophilic Substitution Williamson Ether Synthesis (1852) O-O- Na + EtBr + OEt Na + Br - + Finkelstein Reaction (1910) Na + Cl - Na + I-I- + RCl RI + also RBr Na + Br - Menschutkin Reaction (1890) Et 3 N + RI Et 3 N-R + I - + Exchange Ions (Double Decomposition) Create Ions Destroy Ions Breaking apart by solvent Solvolysis (CH 3 ) 3 C-Br EtOH HBr + (CH 3 ) 3 C-OEt Generalization Meerwein Reagent (1940s) RO - Na + (CH 3 ) 3 O BF 4 + ROCH 3 Na + BF ()() acetone ** ** ** ** ** LUMO HOMO

+ - Generality of Nucleophilic Substitution CH 3 + Nucleophile Substrate Solvent Nu: R-L Nu-R L (+) (-)(-) S-Adenosylmethionine ARGININE : H + Biological Methylation (Post-Transcriptional Protein Modification, etc.) Product ARGININE Leaving Group METHIONINE OH ADENOSINE ADENINE H OH Substitute NR 2 for OH Substitute SR 2 for “OH” RIBOSE : : But different mechanisms are involved! Substitute NHR 2 for SR 2 Substitute Base for NR 3

S N 2 Nucleophilic Substitution Nucleophile Substrate Solvent Nu: R-L Nu-R L (+) (-)(-) the Pragmatic Logic of Proving a Mechanism with Experiment & Theory (mostly by disproving all alternative mechanisms) Product Leaving Group

"It is an old maxim of mine that when you have excluded the impossible, whatever remains, however improbable, must be the truth."

S N 2 Nucleophilic Substitution Nu: R-L Nu-R L (+) (-)(-) Break bond (Dissociation) (mostly by disproving all alternative mechanisms) Make bond (Association) the Pragmatic Logic of Proving a Mechanism with Experiment & Theory D then AA then DSimultaneous

Concerted A/D D/A Pentavalent Intermediate Nu L C Trivalent Intermediate C Nu L C Transition State

Nu Concerted A/D D/A Pentavalent Intermediate Nu L C Trivalent Intermediate C Nu L C Pentavalent Transition State Which is it normally? Unlikely for exothermic process (Hammond implausibility) Nu a b c a b c a b c enantiomers stereochemical proof !

Tools for Testing (i.e. Excluding) Mechanisms: Stereochemistry (sec 7.4b) Rate Law (sec 7.4a) Rate Constant (sec 7.4cdefg) Structure X-Ray and Quantum Mechanics

H CH 3 O O C STEREOCHEMISTRY Kenyon and Phillips (1923) H PhCH 2 CH 3 CH O ClSO 2 CH 3 PhCH 2 CH 3 CH O SO 2 CH 3 +33° +31° O PhCH 2 CH 3 CH -7° CH 3 CO O PhCH 2 CH 3 CH OCH 3 C O OH PhCH 2 CH 3 CH OCH 3 C O OH -32° Inversion! (R)  (S)  Backside Attack in nucleophilic substitution at S (A/D, A favored by vacant d orbital of S) nucleophilic substitution at C=O (A/D, A favored by  *) nucleophilic substitution at saturated C. Same as starting material? PhCH CH 3 CH Why not avoid acetate steps by using - OH? Because it attacks H. - OH (only step involving chiral C) H H Proves nothing

Concerted A/D D/A Trivalent intermediate could be attacked from either face  racemization, not inversion. Pentavalent Intermediate Nu L C Trivalent Intermediate C Nu L C Pentavalent Transition State

Stereochemistry Rate Law Rate Constant Structure X-Ray and Quantum Mechanics Tools for Testing (i.e. Excluding) Mechanisms:

NaOEt + EtBr EtOEt + NaBr [NaOEt] ( fixed [EtBr] ) rate Second Order (S N 2) d[EtO - ] dt = k 2 [EtO - ] [EtBr]

Nu enters Concerted A/D D/A Initial rate-limiting dissociation in D/A would give a rate independent of [Nu], not S N 2. Pentavalent Intermediate Nu L C Trivalent Intermediate C Nu L C Pentavalent Transition State  Not D/A Nu enters

Analogy EtO - + H +  EtOH EtO: + H +  EtOH H + H EtO - + EtBr  EtOEt EtO: + EtBr  EtOEt H + H NaOEt + EtBr EtOEt + NaBr EtOH + k 1 [EtBr]+ k  [EtOH] [EtBr] First Order (D/A?)Pseudo First Order pK a k 2 = 20,000  k  [NaOEt] d[EtO - ] dt rate = k 2 [EtO - ] [EtBr] Second Order (S N 2) ~ const at equilibrium Is it reasonable to be so different? Ratio should be much less drastic at S N 2 transition state

Stereochemistry Rate Law Rate Constant Structure X-Ray and Quantum Mechanics Tools for Testing (i.e. Excluding) Mechanisms:

Rate Constant Dependance on Nucleophile Leaving Group Solvent Nu: R-L Nu-R L (+) (-)(-) Product ~ ? Substrate Something else happens LUMO Surface Potential +26 to -25 kcal/mole [1] k rel (CH 3 ) 2 CH CH 3 CH 2 CH 3 (CH 3 ) 3 CCH 2 CH 3 CH 2 CH 2 R (CH 3 ) 3 C Cf. Table 7.1 p. 275 RBr + I - acetone / 25°C (CH 3 ) 2 CHCH 2  x >15x 128x 1.2x 3000x 23x C-L antibonding node ~same  H

MethylEthyliso-Propyl t-Butyl  -Methylation Total Density (vdW) Steric Hindrance

MethylEthyliso-Propyl t-Butyl  -Methylation LUMO at 0.04 LUMO at 0.06 Total Density (vdW)

MethylEthyliso-Propyl t-Butyl  -Methylation Surface Potential +26 to -25 kcal/mole

 -Methylation Neopentyl Ethyl [1]n-Propyl 0.82 iso-Butyl No way to avoid the third  -CH 3

increased strain in transition state Cycloalkyl Halides (Table 7.2) k relative [1] < C H C C Br I 120°sp 2 60° 90° 109° strain in starting material ~109°

End of Lecture 44 Jan. 27, 2010 Copyright © J. M. McBride 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).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