Coverage:1. Substitution Reactions, SN1 and SN22. Elimination Reactions, E1 and E2

Chapter 11: Reactions of Alkyl Halides

Problems:25-39, 43,47,54

Goals:1. Know the detailed mechanisms of SN1, SN2, E1 and E22. Know what is a good nucleophile and what is a poor nucleophile3. Understand the concept of inversion of configuration (SN2)4. Know the kinetics associated with each reaction5. Know Zaitzev’s Rule and how it applies to the elimination reactions.6. Know what is a good leaving group in a reaction.

There are two fundamental types of reaction mechanisms to be covered:

Substitution SN2 and SN2Elimination E1 and E2

Definitions

Nucleophile – electron-rich species that attacks a nucleus which is positively chargedElectrophile – electron-poor species that is attacked by a nucleophile.

SN2

S – substitutionN – nucleophilic2 - bimolecular reaction

CHH

HXNu- C

HH

HNu + X-

Backside attack

Kinetics

Rate = k [Nu-] [RX] 2nd order reaction

k – rate constant characteristic of the reaction. The larger the k, the faster the reaction.

Effect of Substrate Structure

Alkyl Halide Type Relative Rate

CH3X methyl 3 x 106

CH3CH2X 10 1 x 105

CH3CH2CH2X 10 4 x 104

(CH3)2CH-X 20 2.5 x 103

(CH3)3CCH2-X 10 1(CH3)3C-X 30 ~0

Inversion of Configuration: As the nucleophile attacks, the three groups attached to the carbon undergo inversion, that is, they flip to the opposite side of the carbon.

E

Reaction Coordinate

TS

R-X + Nu R-Nu + X

Ea

Reaction Energy Diagram for SN2 Mechanism

Why this order of reactivity? What controls the relative rate of reaction of the various substrates?

Answer: Steric Hindrance to nucleophilic attack.

CCH3

CH3

CH3

XNu- No Reaction

The bulky methyl groups prevent backside attack by sterically hindering thenucleophile from attacking the electrophilic carbons. Contrast this situation to that of the CH3X group.

Nucleophiles

Nu: + R-X Nu-R+ + X-neutral

Nu:- + R-X Nu-R + X-charged

Reaction:

Nu- + CH3Br Nu-CH3+

+ Br-

Nu- Relative Rate pKa (conjugate acid)

HS- 125,000 7.04

CN- 125,000 9.31

I- 100,000 0.77CH3CH2O

- 25,000 16

OH- 16,000 15.7

Ph-O- 8,000 10

CH3CO2- 500 4.8

H2O 1

Strongest base

Conclusion: The strongest base is not the best nucleophile. In other words, basicity does not control nucleophilicity.

What controls nucleophilicity?

1. Polarizability.

F-

Electrons tightly held by the nucleus and not easilydistorted

CHH

H X C

HH

H

F- X

Transition state – not good bonding between carbon and fluorine atom. High energy

I- CHH

H X C

HH

H

I- X

Electrons loosely held by the nucleus and easilydistorted

Transition state – very good bonding between carbon and iodine atom. Lower energy

2. Solvation

F- has a high charge density due to small size

F- H O

H

..

H O

H

..

H

O H

..

• Highly solvated by water• Very stabilized• Less reactive in Sn2 reaction

H-bond

Generally, the larger nucleophile is the better one in a given group.

Halogen Nucleophiles:

I- > Br- > Cl- > F- HS- > HO-

Best Worst NH3 > H2O

3. Charge on Nucleophile

Charged nucleophiles are better than neutral nucleophiles in the same group.

HS- > H2S OH- > H2O NH2- > NH3

Table of Nucleophilic Strengths

Strong Nucleophile Moderate Weak

(CH3CH2)2P Br- F-

HS- NH3 H2O

I- CH3SCH3CH3OH

(CH3CH2)2NH Cl-

CN- CH3CO2-

OH-

CH3O-

Strongest

Weakest

4. Bulky Nucleophiles

CH3CH2O- (CH3)3CO->

Alkoxides Ions

Not Bulky Bulky Unhindered Hindered

Effect of Leaving Group

Leaving Group

C XNCN X+

• The LG is usually displaced with a negative charge. LGs that best stabilize the negative charge are best.

• Electronegative LGs, which polarize the C atom are also good.

• LGs should be polarizable to stabilize the Transition State

• In general, the weaker the base, the better the LG.

LG Rel. Reactivity pKa (conjugate acid)

60,000 -6.5

I- 30,000 -9.5Br- 10,000 -9F- 1 3.2OH- ~0 15.7NH2- ~0 35

S O

O

O

CH3

Strongest base

Weakest base

S O

O

O

CH3 Symbolized as -OTs

tosylate group

SO

O

O

CH3CH3:N C

S-O

O

O

CH3:N C-CH3

Halogens as Leaving Groups

I- > Br- > Cl- > F-

Stereochemistry of SN2

Recall that the nucleophile attacks from the backside.

What happens when a single enantiomer with a reactive chiral carbon undergoes SN2 reaction???

Answer: It undergoes inversion of configuration

C

CH3H

CH3CH2

BrHO-C

CH3H

CH2CH3

OH XXC

CH3H

CH2CH3

OH +

(S)-2-bromobutane (R)-2-butanol

SN2 reaction proceeds with 100% inversion of configuration – termedWalden inversion.

SN1 Mechanism 1- Unimolecular

CH3

CCH3

CH3

Br OH2

CH3

CCH3

CH3

OH H Br+ +

30 Halide Solvent 30 alcohol

Solvolysis: Solvent acts as nucleophile and reacts with substrate.What is the mechanism? Not SN2! Remember, a 30 substrate is unreactive in SN2 Answer: SN1

CH3

CCH3

CH3

Br slowCH3

CCH3

CH3

Br

CH3

CCH3

CH3

OH2

CH3

CCH3

CH3

OH2+

CH3

CCH3

CH3

OH2+Br CH3

CCH3

CH3

OH H Br

+ +

+ +

+

Reaction Energy Diagram for SN1

R-X

R+ + X-

NuH

R-NuH+ + X-

R-Nu + H-X

E

Reaction Coordinate

TS#1 TS#2

TS#3

First step is rate-determining (highest activation energy).

Rate = k [R-X] First order reactionRate-determining Step is Unimolecular

Effect of Substrate Structure on SN1 Reactivity

30 > 20 > 10 > CH3X

Most Least Reactive

Why this order

• 30 substrates form stable 30 carbocations in rate-determining step. They form faster with a lower Energy of Activation

30 carbocation more stable lower Ea forms faster

10 carbocation less stable higher Ea forms slower

Reaction Coordinate

E

Allylic and Benzylic Substrates are very reactive in SN1 reactions.

Why so??

They form resonance-stabilized carbocations.

Br

OH2

OH2+ BrOH

H Br

+ +

Allyl bromide

What about benzyl bromide? Write a mechanism showing its reactionwith water. CH2Br

OH2+?

Leaving Groups: The same factors that favor SN2 leaving groups alsoFavor SN1 leaving groups, i.e. if a LG is good for SN2, it is good for SN1.

Stereochemistry of SN1

C

Br

(CH3)2CH

CH3CH2CH3

C(CH3)2CH

CH3CH2CH3

C

(CH3)2CH

CH3 CH2CH3

OCH3

C

OCH3

(CH3)2CHCH3 CH2CH3

CH3OH

CH3OH

B

A

+

Planar, symmetric carbocation

S isomer

S isomer~50%

R isomer~50%

The carbocation is attacked both from the top and the bottom by the nucleophilic methanol, resulting in a near racemic mixture of enantiomers as products

Reactions that proceed by the SN1 reaction often undergo rearrangements.Why?

Carbocations are intermediates and may undergo 1,2 ~H or 1,2 ~CH3 shifts.

CH3CHCHCH3

CH3

Br

CH3CH2OHCH3CHCHCH3

CH3

OCH2CH3

CH3CHCHCH3

CH3

OCH2CH3

H Br

+

+

Practice: Write a mechanism for this reaction to account for both products.

Solvents in SN1 and SN2 Reactions

SN2 Polar, aprotic solvents are best.

Aprotic - no OH or NH group present These solvents cannot H-bond to nucleophile and therefore the nucleophile is more reactive.

CH3CCH3

O

CH3C N

O

CH3SCH3

Acetone Acetonitrile Dimethylsulfoxide (DMSO)

SN1 Polar, protic solvents are best.

Protic – possess OH or NH group These solvents promote formation of ions through H-bonding.

OH2 CH3OH CH3CH2OH

Water Methanol Ethanol

Summary of SN1 and SN2

Topic SN2 SN1

Kinetics Rate=k[R-X][Nu] Rate=k[RX]

Nucleophile Strong Nu required Weak Nu required, usually solvent

Substrate Polar, aprotic Polar, protic

Stereochemistry 100% inversion Racemization

Rearrangements No Yes

E2 Elimination

Requirements: Alkyl substrate with a good leaving group possessing ß-hydrogen

C

H

C

X

ß

ß

The ß-hydrogen is bonded to the ß-carbon, which is bonded to the -carbonwhich is bonded to the leaving group X!

A strong base is also required. Any of these will do:

OH- < CH3O- < CH3CH2O- < (CH3)3CO- < NH2-

Weakest Strongest

Mechanism: The E2 mechanism is a one-step mechanism with bond-breaking and Bond-making taking place at the same time; termed a concerted mechansim. InAddition, the Hß and the leaving group X must be anti-coplanar for rapid reaction.

H

XOH2 X

OH-

+ +

Notice that H and X are anti to each other and the four atoms, H-C-C-X, are coplanar in the reactant. This situation stabilizes the transition state leading to the alkene.

C C

H

Br

R

R

H H

OCH3 :..

..

..

..: :

Overlap develops in the T.S. if theanti-coplanar relationship is maintained.

The overlap stabilizes the T.S. and the reaction takes place faster.

C C

H

Br

R

RH

H

OCH3 ..

..

..: :

..

. .

Kinetics

Rate = k [B-][R-X] 2nd Order Reaction

E2 Ball and Stick Movie

Substrate Reactivity

30 > 20 > 10

Why this order of reactivity?

30 substrates yield more stable alkenes and therefore react faster. 10 substrates yield unstable alkenes and react more slowly.

Recall: Alkene Stability

tetrasubstituted > trisubstituted > disubstituted > monosubstituted > unsubstituted

R

R

R

R

R

R

H

R

R

H

H

R

R

H

H

H

H

H

H

H

H

H

CH3

H

HX HH

HCH3OH2 X

OH-

+ +

10 substrate Monosubstituted alkene – less stable

H

H

H

CH3

CH3

X CH3H

CH3HOH2 X

OH-

+ +

30 substrate Disubstituted Alkene - more stable

What about substrates that can yield more than one possible product such as2-bromobutane.

ba

a

b

C

H

C

H

C

BrH

CH3

H

H

H

HO-

C CCH3

H

H

CH3

C CCH3

H H

CH3

C CCH3CH2

H

H

H

OH2 Br

+

+ +

81%

19%

The more stable alkene predominates in the product mixture.

Zaitzev’s Rule: In elimination reactions, the more highly substituted, morestable alkene is usually the major product. This product is referred to as the Zaitzev Product

E2 reactions with Diastereomers

Br H

HCH3

Ph

Ph

H

CH3

Ph

Ph

H

CH3

Ph

Ph

(1S,2S)-1-bromo-1,2-diphenylpropane trans cis

OH-

In the above reaction, only 1 product forms. Which one?

Br H

HCH3

Ph

Ph

BrH

CH3H

Ph Ph

Br

H

CH3

H

Ph

Ph

rotate

Note: H and Br are anti-coplanar in a staggeredconformation.

or

s

s

OH-

Br

H

CH3

H

Ph

Ph CH3 H

PhPh

100% cis isomer

The E2 reaction is an example of a stereospecific reaction.

Stereospecific Reaction: a reaction in which different stereoisomers of a given reactant yield different stereoisomeric products.

(1S,2S)-1-bromo-1,2-diphenylpropane 100% cis product(1S,2R)-1-bromo-1,2-diphenylpropane 100% trans product

Homework Problem: Show that the (1R,2R) yields the cis and the (1R,2S)yields only the trans.

E2 in Cyclohexane Systems

Br

OH-

Recall that there are two possible conformations of bromocyclohexane

The equatorial conformation if favored, but it does not provide the necessaryanti, coplanar relationship of the H and Br.

But the axial conformation does! So it reacts.

BrH

H Br

H

H

Anti, coplanar H and Br

Br

H

H

OH-

+ H2O + Br-

Conclusion: In order for the H and Br to be anti-coplanar, both must be in axial positions. This geometry is referred to as trans-diaxial.

Homework: The following two geometric isomers yield different alkenes as products. Why? Write mechanisms to account for each product.

CH3

Br

base, E2CH3

CH3

Br

base, E2

CH3

trans

cis

E1 Mechanism

C

CH3

CH3

CH3

BrCH3OH

CH2 C

CH3

CH3

BrH+

This reaction cannot be E2 because there is no strong base present

C

CH3

CH3

CH3

Br C

CH3

CH3

CH3

C

CH3

H2C

CH3

H

CH3OH

CH2 C

CH3

CH3

Br

CH3OH2+

+

+ +

slow

Kinetics Rate = k [R-X] 1st order

Substrate Reactivity 30 > 20 > 10 (parallels carbocation stability!)

E1 always competes with SN1

C

CH3

CH3

CH3

Br C

CH3

CH3

CH3

Br+

C

CH3

CH3

CH3

OCH3CH2 C

CH3

CH3

CH3OH as baseCH3OH as nucleophile

SN1 E1

Mixture of Products

Summary of E1 and E2

Topic E2 E1

Kinetics 2nd order 1st order

Base Strong required Weak required

Substrate 30 > 20 > 10 30 > 20 > 10

Solvent Type not critical Polar, aprotic for ionization

Orientation Zaitzev rule Zaitzev Rule

Conformation Anti-coplanar None requiredRequirements H and X

Rearrangements No Yes