Chemistry 125: Lecture 44January 27, 2010

Nucleophilic Substitutionand Mechanistic Tools:

Rate Law & Rate Constant

This

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F CH2

CH2 H :OH

"E2 Elimination"

ABNABN

AON

Make TwoBreak TwoF

H OH

CH2

CH2

Nucleophilic Substitutionand -Elimination

Chapter 7

F H :OH

F H OH

"Acid-Base"

F CH3 :OH

CH3 OHF

"SN2 Substitution"

ABN

Make & BreakSame

(Cf. Lecture 16)

All are Nucleophilic SubstitutionWilliamson Ether Synthesis (1852)

O- Na+ EtBr+ OEt Na+ Br-+

Finkelstein Reaction (1910)

Na+ Cl-Na+ I- + RCl RI +also RBr Na+ Br-

Menschutkin Reaction (1890)

Et3N + RI Et3N-R + I-+

ExchangeIons

(Double Decomposition)

CreateIons

DestroyIons

Breaking apartby solvent

Solvolysis

(CH3)3C-BrEtOH

HBr +(CH3)3C-OEt

Generalization

Meerwein Reagent (1940s)

RO- Na+ (CH3)3O BF4+ ROCH3 Na+ BF4 ++ - -

()acetone

*

*

*

*

*

LUMOHOMO

+ -

Generality of Nucleophilic Substitution

CH3

+

Nucleophile Substrate

Solvent

Nu: R-L Nu-R L(+) (-)

S-Adenosylmethionine

ARGININE

:

H

+

Biological Methylation(Post-Transcriptional Protein Modification, etc.)

Product

ARGININE

LeavingGroup

METHIONINEOH

ADENOSINE

ADENINEH OH

SubstituteNR2

for OH

SubstituteSR2 for “OH”

RIBOSE::

But different mechanisms are involved!

SubstituteNHR2 for SR2

SubstituteBase for NR3

SN2 Nucleophilic Substitution

Nucleophile Substrate

Solvent

Nu: R-L Nu-R L(+) (-)

the Pragmatic Logicof Proving a Mechanism

with Experiment & Theory

(mostly by disproving all alternative mechanisms)

ProductLeavingGroup

"It is an old maxim of mine that when you have excluded the impossible,

whatever remains, however improbable,

must be the truth."

SN2 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 A A then D Simultaneous

Concerted A/D D/A

PentavalentIntermediate

Nu LC

TrivalentIntermediate

CNu LC

TransitionState

Nu

Concerted A/D D/A

PentavalentIntermediate

Nu LC

TrivalentIntermediate

CNu LC

PentavalentTransition 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)

StructureX-Ray and Quantum Mechanics

HCH3

O

OC

STEREOCHEMISTRYKenyon and Phillips (1923)

H

PhCH2

CH3

CH

O Cl SO2 CH3

PhCH2

CH3

CH

O SO2 CH3

+33°+31°

O

PhCH2

CH3

CH

-7°

CH3CO

O

PhCH2

CH3

CH

O CH3C

O

OHPhCH2

CH3

CH

O CH3C

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

CH3

CH

Why not avoid acetate steps by

using -OH? Becauseit 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.

PentavalentIntermediate

Nu LC

TrivalentIntermediate

CNu LC

PentavalentTransition State

Stereochemistry

Rate Law

Rate Constant

StructureX-Ray and Quantum Mechanics

Tools for Testing(i.e. Excluding) Mechanisms:

NaOEt + EtBr EtOEt + NaBr

[NaOEt] ( fixed [EtBr] )

rate

Second Order (SN2)

d[EtO-]dt = k2 [EtO-] [EtBr]

Nu enters

Concerted A/D D/A

Initial rate-limiting dissociation in D/A would give a rate independent of [Nu], not SN2.

PentavalentIntermediate

Nu LC

TrivalentIntermediate

CNu LC

PentavalentTransition State

Not D/A

Nu enters

Analogy

EtO- + H+ EtOH

EtO: + H+ EtOHH

+

H

EtO- + EtBr EtOEt

EtO: + EtBr EtOEtH

+

H

NaOEt + EtBr EtOEt + NaBrEtOH

+ k1 [EtBr]+ k [EtOH] [EtBr]

First Order (D/A?)Pseudo First Order

pKa15.7

-1.7

k2 = 20,000 k

[NaOEt]

d[EtO-]dtra

te = k2 [EtO-] [EtBr]

Second Order (SN2)

~ const

at equilibrium

Is it reasonable to be so different?

Ratio should be much less drastic at

SN2 transition state.1017.4

Stereochemistry

Rate Law

Rate Constant

StructureX-Ray and Quantum Mechanics

Tools for Testing(i.e. Excluding) Mechanisms:

Rate Constant Dependance on

NucleophileLeavingGroup

SolventNu: R-L Nu-R L

(+) (-)

Product

145

0.82

0.0078

0.000012

~ 0.0005 ?

Substrate

Something else happens

LUMO

Surface Potential+26 to -25 kcal/mole

[1]

krel

(CH3)2CH

CH3CH2

CH3

(CH3)3CCH2

CH3CH2CH2

R

(CH3)3C

Cf. Table 7.1 p. 275

RBr + I-

acetone / 25°C

(CH3)2CHCH2

0.036

145x

>15x

128x1.2x

3000x

23x

C-Lantibonding

node

~same

H

Methyl Ethyl iso-Propyl

t-Butyl

-Methylation

Total Density (vdW)

Steric Hindrance

Methyl Ethyl iso-Propyl

t-Butyl

-Methylation

LUMO at 0.04LUMO at 0.06Total Density (vdW)

Methyl Ethyl iso-Propyl

t-Butyl

-Methylation

Surface Potential+26 to -25 kcal/mole

-Methylation

Neopentyl

Ethyl [1] n-Propyl 0.82

iso-Butyl 0.0360.000012

No way to avoid the third -CH3

increased strain in transition state

Cycloalkyl Halides (Table 7.2)

krelative

[1]

1.6

0.008

<0.0001

0.01

C HCC

Br

I

120° sp2

60°

90°

109°

strain in starting material

~109°

End of Lecture 44Jan. 27, 2010

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