Chemistry 125: Lecture 40 January 15, 2010 Predicting Rate Constants, and Reactivity - Selectivity...

Chemistry 125: Lecture 40January 15, 2010

Predicting Rate Constants, and Reactivity - Selectivity Relation.

Rates of Chain Reactions. This

For copyright notice see final page of this file

-10

-8

-6

-4

-2

0

2

Ph Me Et iPr tBu Allyl

H

Me

Et

iPr

tBu

R-X BDE : Alkyl Variation in Detail

X

R

Me-R

Et-R

i Pr-R

t Bu-R

H-R

If this trend is due to radical stabilization by substitution, other X-R bond strengths should show the same trend.

BD

E r

elat

ive

to C

H3X

(kc

al/m

ole)

12.2

8.3

5.0

2.3

t Bu-R

MolecularMechanics

Strain Energies inStarting Material

t-Bu t-Bu

van der Waals Energy 26.9 kcal/mole5.2

“Idealized”Bond

Lengthsand

Angles

“Relaxed”Structure

Crunch!“steric hindrance”

van der Waals Energy drop by 16.8 to 5.2 kcal/mole

comes at the expense of bond stretching and bending.

van der Waals Energy drop by 16.8 to 5.2 kcal/mole

1.52Å

1.57Å

109.5°

112.3°

comes at the expense of bond stretching and bending.(which total 4.8 kcal/mole)

van der Waals Energy drop: 16.8 to 5.2 kcal/mole

ResultantTotal Strain (incl. 2.2 torsion) 12.2 kcal/mole

-10

-8

-6

-4

-2

0

2


H

Me

Et

iPr

tBu

R-X BDE : Alkyl Variation in Detail

X

R

For X = alkylalmost all of the Me to t-Bu change is

due to strain energy inthe starting material.

Me-R

Et-R

i Pr-R

t Bu-R

H-R

If this trend is due to radical stabilization by substitution, other X-R bond strengths should show the same trend.

BD

E r

elat

ive

to C

H3X

(kc

al/m

ole)

12.2

8.3

5.0

2.3

t Bu-R

0.8

1.51.9

2.3Me-R

molecularmechanics

strain energies

1.9

1.5

0.8

0

9.9

kcal

/mol

e di

ff. i

n in

itia

l Str

ain

w

hole

8.9

kca

l/m

ole

diff

. in

BD

E

Dit

to1.

5

2.6

But not for H-R

-10

-8

-6

-4

-2

0

2


Me

Et

iPr

tBu

R-X BDE : Corrected for R-X Strain

X

R

Me-REt-Ri Pr-Rt Bu-R

BD

Eco

rr r

elat

ive

to C

H3X

(kc

al/m

ole)

Alternative to hypothesis of radical stabilization by substitution

Intrinsic C-C bond strength (corrected for strain) is practically insensitive to substitution.

-10

-8

-6

-4

-2

0

2


H

Me

Et

iPr

tBu


X

R


H-R


But C-H bonds are weakened by alkylation of the carbon.


BD

Eco

rr r

elat

ive

to C

H3X

(kc

al/m

ole)

-7

-5

-3

-1

1

3

5


H

Me

Et

iPr

tBu

Cl

Br

I


X

R



H-RBut C-H bonds are weakened by alkylation of the carbon.


While C-Cl and C-Br are strengthened by alkylation of the carbon.

Cl-RBr-R

I-R

BD

Eco

rr r

elat

ive

to C

H3X

(kc

al/m

ole)

No one I know of understands this, but the textbooks are clearly wrong.

Can we use energies of stable structures that we “understand” to

infer the energies of other structures (e.g. transition states),so as to predict reactivity?

How can we predict activation energy?

Might exothermic reactions be faster than endothermic ones?

How can we predict activation energy?

This is no easy task a priori, especially when interaction with

solvent is important.But often one can say something

sensible about relative values of Ea (or G‡).

Compared to What?

The Hammond Postulate (1955)

George S. Hammond (1921-2005)

“If two states, as for example, a transition state and an unstable interme-diate, occur consecutively during a reaction process and have nearly the same energy content, their interconversion will involve only a small reorganization of the molecular structures.”

This stimulated organic chemists to think about transition states and try to generalize plausibly

about reaction coordinates.

by p

erm

issi

on, E

. Men

ger

The more exothermic a reaction - the more similar the transition state to starting material

(in both energy and structure)

StartingMaterial

Product

endoproduct

At least among one-step reactions

that are closely analogous, such as

X• + H-R.

.X-H + •R…..

The more exothermic a reaction - the more similar the transition state to starting material

(in both energy and structure)

There is “likely” a continuum

between starting material and product with respect to the factors that

influence stability.

endoproduct

An effect mostly influencing the

energy of the product of an endothermic

reaction should have a similar (slightly

smaller) influence on its (late) transition

state.

Rates of slower reactions should be more sensitive to overall G!

An effect mostly influencing the

energy of the product of an exothermic

reaction should have a small influence on its (early) transition

state.

e.g. resonance stabilization

HPhCH2

PhCH2

H

PhCH2

Reactivity/Selectivity “Principle”

More Reactive

Less Reactive

More Selective

Less Selective

‡

G 1

G Reactivity‡

k e -G /RT 10

13 - 3/4 G /sec‡ ‡

G Relative Reactivity‡

k1 e -G /RT‡

1

k2 e -G /RT‡

2

k1/k2 = e -G /RT 10

-3/4 G‡ ‡

‡

G 2

‡

Gor G -4/3 (log(k) -13)

‡

or G -4/3 log(k1/k2)‡

Rates converge with increasing T Increase discrimination by lowering T !

Reactivity & Selectivity

+ X•H3C-CH2-CH3

+ HXH3C-CH2CH2

•

+ HX

•H3C-CH-CH3

43% X = Cl• 57%

8% X = Br• 92%

‡G 4/3 log(4) = 0.8 kcal/mole

‡G 9/3 log(35) = 4.5 kcal/mole@ 330°C( N.B. : Br• shows greater selectivity despite increased T )

k1° 6 k2° 2Note correction for

number of primary & secondary H atoms.

k2°/k1° = 4

k2°/k1° = 35(3 14)

n-Pr-X i-Pr-X

136.3103.287.571.3

1°

35.22.1

13.629.8

FClBrI

101.1”””

Reactivity & Selectivity

2°

37.74.6

11.127.3

98.6”””

136.3103.287.571.3

+ X•

+ HX + HXH2C-CH2CH3

•

H3C-CH2-CH3

•H3C-CH-CH3


More Reactive

Less Reactive

More Selective

Less Selective

22.14.6 Cl

713.611.1Br

1° abstraction

2° abstraction

11.1

13.6

-4.6

-2.1

k2/k1 ~ 4 k2/k1 ~ 35

G‡ 4.5G‡ 0.8


Less Reactive

More Selective

713.611.1Br

1°

2°

11.1

13.6

k2/k1 ~ 35

G‡ 4.5

How can transition states be more different than

products are?

Br H •R

Br• H R

Br– H• R+

Maybe polar character helps transition states.

(2° cations are much more stable than 1°)

G 2.5!

Sometimes factors involved in stabilizing Transition States can

be different from those involved in stabilizing either starting materials or products.

Chain H-X Addition to Alkene

C=C

H-X

•X •C-CX

C-CX H

cyclicmachinery

X• + C=C(+ X-H)

X-C-C-H+ X•

0

10

-10

-20

-30

-40

-50 RX HX116 13581 10368 8752 71

FClBrI

X-C-C•(+ X-H)

83146 99

Chain H-X Addition to Alkene

83

Only HBr works fast enough in

both steps.

Average Bond Energies

146135

8399

116

281 298 = 17

But only HBr works.Why?

End of Lecture 40Jan. 15, 2010

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Chemistry 125: Lecture 40 January 15, 2010 Predicting Rate Constants, and Reactivity - Selectivity...

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Transcript of Chemistry 125: Lecture 40 January 15, 2010 Predicting Rate Constants, and Reactivity - Selectivity...