Chemistry 125: Lecture 45January 29, 2010

Nucleophilic Substitutionand Mechanistic Tools:

Rate Law, Rate Constant,Structure This

For copyright notice see final page of this file

Stereochemistry

Rate Law

Rate Constant

StructureX-Ray and Quantum Mechanics

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

LeavingGroupSubstrate

Rate Constant Dependance onSolvent

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

Product

80

1,000

10,000

16,000

126,000

Nucleophile

[1]

krel

Br-

F-

H2O

HO-

Cl-

Nu

HS-Sec. 7.4d, Table 7.3

I- 80,000

-8

-9

7

-10

3.2

15.7

-1.7

pKa (NuH+)

For first-row elementsnucleophilicity (attack

C-L )

parallels basicity (attack H+).Both require high HOMO.

But as atoms get bigger, they get better at attacking

C-L (compared to attacking H+)

Solvent

LeavingGroupSubstrate

Rate Constant Dependance onNu: R-L Nu-R L

(+) (-)

Nucleophile

80

1,000

10,000

16,000

126,000

[1]

krel

Br-

F-

H2O

HO-

Cl-

Nu

HS-Sec. 7.4dg

I- 80,000

-8

-9

7

-10

3.2

15.7

-1.7

pKa (NuH+) krel

CH3I in H2O

[1]

14

160

krel

CH3Br in Acetone

11

5

[1]

harderto break H-bonds

to smaller ions

Polar solvents accelerate reactions that generate (or concentrate) charge,

and vice versa.

Sen

sibl

e

Backw

ards

Nucleophile

Solvent

Rate Constant Dependance onNu: R-L Nu-R L

(+) (-)

Sec. 7.4e

Substrate

goodRSO2O- -3

LeavingGroup

bad

good

good

v. bad

bad

good

Br-

F-

H2O

HO-

Cl-

HS-

I- v. good

-8

-9

7

-10

3.2

15.7

-1.7

L pKa (LH+)Weak bases are

good leaving groups(H like R,

as expected)

R-OH v. bad

R-OH2+ good

R-OSO2R’ good(Kenyon/Phillips)

(acid catalysis)

OH Leaving-Group-Trick Lore (sec 7.4f)pKa ~16 pKa -1.7

OH2+

Ether / HBr

Good Leaving Group

Good Nucleophile

OH Leaving-Group-Trick Lore (sec 7.4f)

Cf. Kenyon & Phillips (1923)

OSO2R

pKa -3

OH Leaving-Group-Trick Lore (sec 7.4f)

OSOCl

gases

OH Leaving-Group-Trick Lore (sec 7.4f)

OPXn

OH Leaving-Group-Trick Lore (sec 7.4f)

OP+

1

2

Maximizing Synthetic Speed to support PET scanning (from Loudon, Org. Chem.)

e- e+ 18O= + 7 MeV proton

- neutron 18F- t1/2

110 min

http://en.wikipedia.org/wiki/Positron_emission_tomography

Need to get 18F where cancer isor

11C t1/2 ~

20 min

13N t1/2 ~

10 min

15O t1/2 ~

2 min

positron

Connecting coincident scintillations tells where 18F’s were.

and you have to do it within a couple

hours.

Yale PET

What tosynthesize?

Problems :K+

trifluormethanesulfonate

(Triflate)

Maybe it would suck up 2-Fluoroglucose as well.

Rapid metabolism of tumor sucks up glucose.

AcO = CH3C-O-(acetate protecting group)

O

F tied up by H-bondingand by K+cation

inversion gives wrong configuration

HO a horrid leaving group

wrong C-OH could be attacked

Glucose

start withMannose

2-FluoroglucoseKF18

SN2?

Cl-SO2CF3

well known for sugarsK+ F18-

CH3CN(aprotic solvent)

H2O H+

Mannose to 2-F18Glucose - ASAP

F18

Office Hours 3-4:30

this afternoon181 SCL

Loudon, Organic Chemistry

Stereochemistry

Rate Law

Rate Constant

StructureX-Ray and Quantum Mechanics

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

So far we’ve just been beating up on the D/A mechanism

(trivalent C intermediate)

though there are cases (SN1) where it in fact applies.

The tougher problem is to distinguish between concerted and A/D with a very weakly stabilized intermediate.

(see supplementary reading on Course website)

Might there be Pentavalent A/D Intermediate instead of a Concerted SN2 Transition State?

PentavalentIntermediate

Nu LCNu LC

TransitionState

Quantum Mechanics says Transition State

for OH- attackingless crowded CH3OH.

1.88 Å 1.88 Å

Quantum Mechanics says Transition State

for H2O attacking protonated t-BuOH.

2.64 Å 2.64 Å

But neither reaction is practical in the laboratory!What does experiment say? X-ray?

Might there be Pentavalent A/D Intermediate instead of a Concerted SN2 Transition State?

Problem: Neither Transition State nor Intermediate would hold together long enough to study.

PentavalentIntermediate

Nu LCNu LC

TransitionState

HOYWAT

JACS 4354-4371 (2005)

Nu LC Held in place by molecular framework

+

HOYWAT+

JACS 4354-4371 (2005)

CH3

O OC

+CH3-O(CH3)2 BF4

-

: :ARE THERE BONDS HERE?

OCH3+NOT elongated to reflect superposed

average of two “bell-clapper” structures.

F3BF- BF3

2.64 Å 2.64 Å

Powerful alkylating agent like

“Meerwein’s . Reagent”

Et3O+ BF4-

5.02 Å

4.86 Å shortened by 0.16 Å

BUT without central C+ etc. shortened by 0.21 Å!

4.96 Å

4.75 Å

125° 114°125° 113°Pentavalent C

attraction?

Eclipsed repulsion

bent inbent in

etc.

Pentavalence seemed to be a safe inference

5.02 Å

4.86 Å

5.08 Å

5.00 ÅO

SiO2

CF3

124° 116°125° 113°

Central O only slightly repulsive compared to C+.

Eclipsed repulsion

126° 112°4.92 Å

4.56 ÅBF2

BF2 does seem to suck in CH3O groups.

Eclipsed repulsion

125° 113°5.02 Å

4.86 Å

Double minimum with stronger nucleophile O-

(higher HOMO& lower LUMO)

127° 109°

4.84 Å

1.47 Å 2.99 Å

CF3CF3

-O

K+and . nearby

1.88 Å 1.88 Å

Range

Bonded O (or S)

seems to “use up” the vacant AO.

CCH3O OCH3+

CCH3O OCH3

+

Higher neighbor HOMOs favor tetravalence.

For F withdrawal dominates donation.

Compared to what?

AX

A-X

dis

tanc

es (

Å)

Compound

Short & Long X A X Distances

Hnonbondreference

B“tight”Bonded?

Btetracoordinate

C+

C+

~ equalsymmetrical

very differentunsymmetrical

B“loose”

like H

No sign of stabilityfor pentavalent

SN2 “intermediate”

transistion state transistion state (as calculated by q. mech.)

Pressed in

by HCH3 repulsion

F CH2

CH2 H :OH

"E2 Elimination"

ABNABN

AON

F

H OH

CH2

CH2

E2 -Elimination Text sec. 7.9

Rate influenced by [base], nature of leaving group,

H isotope (kinetic isotope effect)

C HH OC H O

DD D

kH > kD but only if bond is weakened in rate-determining transition state

Heavier atom, lower ZPE

see Lecture 8: frames 21-22

ZPE(kinetic)

E2 -Elimination Text sec. 9.5

Rate influenced by [base], nature of leaving group,

H isotope (kinetic isotope effect)

Stereochemistry (Anti) sec 9.5G

Regiochemistry (Zaitsev/Hammond) sec 9.5F

E2 vs. SN2 (Sterics & Base Strength) sec 9.5G

N C and RC C as Nucleophiles

Cyanide Table 9.6 p. 396

Acetylide Sec. 14.7B pp. 665-666

Ethylene Oxide as C2 Source

Synthesis GamesStudy Problem 14.2 p. 666-667

Problems 14.18-14.23 p. 666-668

0

-1

1

-2

-3

-4

log

(fra

ctio

n of

R-B

r co

nver

ted

to H

OR

/min

)

(CH3)3C(CH3)2CH

SN1

Hughes Ingold

(1933-1940)

CH3 CH3CH2

EtOH/H2O (4:1)55°C

NaOH + R-Br HO-R + NaBr

k2 (M-1min-1)

concerted displacement

slowed by crowding

k1 (min-1)

D/Aaccelerated

by crowding,

(CH3)3C+cation stabilization,

polar solvent

(0.01 M)

plus ~19%

E2

Rate extrapolated from lower temperature.

Depends on [OH-]

SN1 and E1 sec. 9.6

Product Determined After Rate 9.6B

By Competition for Short-Lived Cation

Rearrangement of Short-Lived Cation 9.6C

Net Inversion from Short-Lived Ion Pair 9.6D

EtOH/H2O (4:1)55°C

NaOH + tBu-Br HO-t-Bu + NaBr(0.01 M)

+ CH2=C(CH3)2

E2 or E1? How do you tell?

Overall rate (and % alkene) depends on [OH-]

Kinetic Isotope Effect shows whether H is being transferred in rate-determining step.

CH3-Br + OH-

5. (5 min) Give a real example of the influence of a change of reactant structure on the ratio of SN2 to E2 products. Be as specific and quantitative as you can. (You will need to show the ratios for two different reactants.)

(CH3)3C-Br + OH-

Perspectives on Drastic Product Ratios

Synthetic Organic Chemist : Reliable High-Yield Tool

Physical-Organic Chemist : Definitive Ea DifferenceUnambiguous interpretation of cause

e.g. syn- vs. anti-hydrogenation of acetylene

e.g. Steric retardation of SN2 / 105 acceleration for t-Butyl via SN1

Perspectives on 50:50 Product Ratios

Physical-Organic Chemist : Valuable “Borderline” Reference

Synthetic Organic Chemist : Deadly Influence on 12-Step Synthesis

(1/2)12 = 0.02% Yield

(Might provide optimizable lead)

Allows Sensitive Tests of Subtle Influences.e.g. isotope effect by competition

A lesson from E2 Elimination

If Step 1 (motion) is rate-limiting, H- and D-transfer products should

form in equal amounts. (because their motions should be equally fast)

If Step 2 (atom shift) is rate-limiting, more H-transfer product should form.

kH/kD > 1 (kinetic “isotope effect”)

In a Very Viscous Solvent Can Short-Range Motion Constitute a Rate- (and Product-) Determining Step?

Generates steric hindrance & requires moving radicals past N2

N

N

CH3

CH3

H3C CD3

CD3

CD3

•

•

UV LightCH3

CH3

H3C

CD3

CD3

CD3

Radical-PairCombination

CH3

CH3

H3C CD2 CD3

CD3

•

•

DD

(1) Rotate N2 + C4D9

(2) Shift D atomexothermic/easy/fast

N

N

Radical-Pair“Disproportionation”

(1) Rotate N2 + C4H9

(2) Shift H atomexothermic/easy/faster

CD3

CD3

CD3

CH3

H3CCH3 HCH2

•

•

Jo David’s Question:

NN

NN

t-Butylhydrazine

CH3

CH3

H3C NH

NH2

CH3

CH3

H3C Cl NH2

NH2

(prepare from)

CH3

CH3

H2C

?To do his project, Jo David needed to prepare this compound.

E2 >> SN2

CH3

CH3

H3C N

N

CD3

CH3

CD3

CD3

Smith-Lakritz

It is very common to change a C=X double bond into

C=O and H2X (we’ll discuss this soon)

C=N-R C=O + H2N-R

NN- +

t-Butylhydrazine

CH3

CH3

H3C NH

NH2

CH3

CH3

H3C MgCl

??? Jo David FineApril-October 1971

O

CH3

CH3

H3C N

N

CD3

CH3

CD3

CD3

Jo David Fine

Jo David Fine Notebook p. 91 (October 1971)

Jo David is now a respectedprofessor of dermatology at Vanderbilt University, whose son has graduated from Yale.

Happy Ending:

Crucial Lesson (from S. Nelsen, U. Wisc.)

CH3

CH3

H3C NH

NH2

CH3

CH3

H3C Cl NH2

NH2

CH3

CH3

H2C95%

5% SN1When you need a compound, % yield isn’t everything!

HCl salt easilypurified by

crystallization

E1 / E2

Major product a gas,just “goes away”

CH3

CH3

H3C N

N

CD3

CH3

CD3

CD3

Happy Ending:

Jo David Fine’s successor found that in fluid solvents, there was more H- than D-transfer (atom transfer is rate-limiting), but that in very viscous solvents at low temperature this “kinetic isotope effect” disappeared (there were equal amounts of H- and D-transfer), because motion

had indeed become rate-limiting.

End of Lecture 45Jan. 29, 2010

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