Nucleophilic Substitution and -elimination

Substitution Process

Nucleophiles have a pair of electrons which are used to form a bond to the electrophile.

A Leaving Group departs making room for the incoming nucleophile.

Nucleophiles can also frequently function as Lewis bases.

The electrophile can function as Lewis acid.

Note that the nucleophile converts a lone pair into a bond and becomes more positive by +1

Note that the bond from C to the Leaving Group is collapsed into a lone pair on the Leaving Group which becomes more negative by -1.

-Elimination

H

Lv

base

+ H-Lv

A pi bond is created.

Instead of substitution a base can remove both the leaving group and an adjacent hydrogen creating a pi bond. Recall dehydrohalogenation.

Competition between Nucleophilic Substitution and -elimination.

Br

+ Na+ C2H5O-

H

nucleophilicsubstitution

OEt

+ Br-

Note the change in charges on the nucleophile and the Leaving group

First the nucleophilic substitution. The ethoxide attacks the carbon bearing the bromine.

Now the -elimination.

Br

+ Na+ C2H5O-

H

Now the elimination. The ethoxide (base) attacks the hydrogen on a carbon adjacent to the carbon bearing the Br ().

elimination

+ C2H5OH + Br-

Since we are using Br as the leaving group this could also be called a dehydrohalogenation.

Summary.

Formal Charges and Nucleophilic Substitution

In the free nucleophile the pair of electrons is a lone pair belongs exclusively to the nucleophile.

In the product, it is a bond and shared.

The result is the nucleophile increases its charge by +1

Conversely the leaving group converts a shared pair of electrons (a bond) into unshared electrons (lone pair). The charge of the leaving group becomes more negative by -1.

NH

H

C

H

Br

HH

C

H

Br

HH

H2N

NH

H

C

H

Br

HH

C

H

Br

HH

H3N

H

Negative Nucleophile

Neutral NucleophileOther things being equal, the more basic species will be a better nucleophile. NH2

- is a better nucleophile than NH3

N: from -1 to 0 ; Br: from 0 to -1

N: from 0 to +1; Br: from 0 to -1

Negative Nucleophile, positive leaving group

Br C

H

OH2

HH

C

H

HH

Br OH2

Br: from -1 to 0; O: from +1 to 0

Two Nucleophilic Substitution Mechanisms: SN1 & SN2

SN2 mechanism: substitution, nucleophilic, 2nd order

Backside attack

Hydrogens flip to the other side. Inversion

of configutationExamine important points….

Look at energy profile next…

Energy Profile, SN2

SN1 reaction: substitution, nucleophilic, first order.

Step 1, Ionization,

Rate determining step.

Step 2, Nucleophile reacts with Electrophile.

Note stereochemistry: nucleophile can bond to either side of carbocation. Get both configurations.

Protonated ether.

Now the alternative mechanism: SN1

CH3OH + (CH3)3C-Br CH3OC(CH3)3 + H + + Br-

Step 3, lesser importance, deprotonation of the ether.

Next, energy profile….

Energy Profile of SN1, two steps.Slow step to form carbocation. Rate determining. Examine important points…..

Carbocation, sp2

Fast step to form product.

Kinetics: SN1 vs. SN2SN1, two steps.

SN2, one step.

Effect of Nucleophile on Rate:Structure of Nucleophile

SN1: Rate Determining Step does not involve nucleophile. Choice of Nucleophile: No Effect

SN2: Rate Determining Step involves nucleophile. Choice of nucleophile affects rate.

Note the solvent for this comparison: alcohol/water. Talk about it later…

Frequently, better nucleophiles are stronger bases.

Compare

Compare

But compare the halide ions!!

In aq. solution F – more basic than I -. (HI stronger acid.) But iodide is better nucleophile.

We need to discuss Solvents

Classifications

Polar vs non-polar solvents, quantified by dielectric constant. Polar solvents reduce interaction of positive and negative ions.

Water > EtOH > Acetic acid > hexane

Solvents. Another Classification

Protic vs aprotic solvents. Protic solvents have a (weakly) acidic hydrogen having a positive charge which stabilize anions.

ROH --- Br - --- HOR

Alcohols are protic solvents Aprotic solvents

CH3CN acetone tolueneIncreasing polarity

Role of Solvents

Some solvents can stabilize ions, reducing their reactivity.

Many nucleophiles are ions, anions.

Protic solvents can stabilize anions. Protic solvents have (weakly) acidic hydrogens bearing a positive charge. Anions may be stabilized

Methanol, protic solvent, stabilizing the fluoride ion, reducing its nucleophilicity.

Small, compact anions (like fluoride ion) are especially well stabilized and have reduced nucleophilicity. Iodide ion is large diffuse charge and less stabilization occurs.

Halide ion problem

Iodide ion Bromide ion Chloride ion Fluoride ion

basicity

Protic solvent solvation

nucleophilicity

The problem: basicity and nucleophilicity of the halide ions do not parallel each other in protic solvents.

nucleophilicity

The explanation. Fluoride most stabilized in protic solvents reducing its nucleophilicity.

Summary for Halide Ions

Iodide ion Bromide ion Chloride ion Fluoride ion

Nucleophilicity in aprotic solvents

Protic solvent solvation

Nucleophilicity in protic solvents

basicityBut in aprotic solvents.

Protic solvents.

basicity

Protic solvent solvation

Stereochemistry, SN1 at a chiral center.

R

X

R"R'

ionization

- X -

chiral halideoptically active

R

R"R'achiralcarbocationoptically inactive

MeOH

-H +

R R

R" R"R' R'

OMe MeO

newly generated chiral centersracemic mixtureoptically inactive

Frequent complication: the Leaving Group will tend to block approach of the nucleophile leading to more inversion than retention for the SN1

racemization

Stereochemistry SN2, Inversion at a Chiral Center

R1

R3

R2

XY -

R1

R3

R2

Y + X -

Inversion, frequently (but not always) the R,S designator changes

Examples

C2H5

H

H3C

BrCH3O -

C2H5

H

CH3

H3CO + Br -

(R)-2-methoxybutane(S)-2-bromobutane

CH3

C2H5

H Br

CH3

C2H5

CH3O H

Here is the inversion motion!

Another Example

CH3

Br H

C2H5

H3C H

NaOCH3

CH3

H OCH3

C2H5

H3C HSN2

Williamson ether synthesis

H

CH3

Br

H

C2H5

H3CNaOCH3

SN2 CH3O

CH3

H

H

C2H5

H3C

The chiral center will undergo inversion.

The non-reacting chiral C will not change.

How to understand the configurations: simply replace the Br with the OCH3 (retention). Now swap any two substituents (here done with H and OCH3) on the reacting carbon to get the other configuration (inversion). Done.

Stereochemistry, SN2

Substitution

Recall iodide a good nucleophile, acetone an aprotic solvent resulting in highly reactive iodide ion.

Inverted configuration

SN2 Stereochemistry: Inversion

Two things happening here:

1)Substitution of iodide, 127I, with labeled iodide, 131I.

2) Change in stereochemistry

Comparison of SN1 and SN2 mechanisms. Substitution vs. Loss of Optical Activity

RI RI RI RI RIRI RI RI RI RI

Stereochemistry: RI represents the R configuration of the alkyl iodide;

RI represents the S configuration.

Substitution: I is the normal 127I isotope; I is the tagged 131I iodine isotope.

I -

If racemization: “SN1”

RI RI RI RI RIRI RI RI RI RI

Only 20% reacts 20% substituted, 20% racemized,

20 % of optical purity lost (80% optically pure). Rate of Loss of optical activity = Rate of substitution.

RI RI RI RI RIRI RI RI RI RI

I -

If inversion: SN2

RI RI RI RI RIRI RI RI RI RI

Only 20% reacts 20% substituted, 40%

racemized, 40% optical purity lost (60% optically pure). Rate of loss of optical activity = 2 x Rate of substitution.

100% optically pure

Effect of Structure of the Haloalkane on Rates

CH3X CH3CH2X (CH3)2CHX (CH3)3CX

Methyl primary secondary tertiary

SN1 RecallStability of resulting carbocation, hyperconjugation

Ease of ionization

Rate of SN1 Reactions

Now for SN2

CH3X CH3CH2X (CH3)2CHX (CH3)3CX

Methyl primary secondary tertiary

SN2

Steric Hinderance, difficulty of approach for nucleophile

Rate of Reactions

Summary:

Methyl, primary use SN2 mechanism due to steric ease.

Tertiary uses SN1 mechanism due to stability of carbocations

Secondary utilizes SN1 and/or SN2 – depending on solvent and nucleophile.

Recall: Resonance Stabilization of Carbocations

Allylic and benzylic carbocations are stabilized by resonance.

SN2 SN1Both

Leaving GroupRecall that the leaving group becomes more negative.

R - X X

Generally, the best leaving groups are groups that can stabilize that negative charge: weak bases; conjugate bases of strong acids.

Base Strength

Example:

R-OH NRBr

BrR-OH R-OH2 R-Br

H

H - X X

Solvents

Polar solvents stabilize ions, better stabilization if the charge is compact.

Polar Protic solvents stabilize both anions (nucleophiles) and cations (carbocations). Accelerate SN1 reactions where charge is generated in the Rate Determining Step.

R-X [R + --- X -] R + + X -

Polar aprotic solvents usually stabilize cations more effectively than anions (nucleophiles). Anions (nucleophiles) are left highly reactive. Accelerates SN2 reactions where an anion (nucleophile) is a reactant.

Nuc - + R-X [Nuc----R----X] - Nuc-R + X -

Stabilized. Stabilized.

Note that it is the energy of the transition state relative to the reactant which affects the rate of the forward reaction (but not the equilibrium).

Not Stabilized.

Rearrangements for SN1: 1,2 Shift

Recall carbocations can rearrange (1,2 shift) to yield a more stable carbocation. Occurs in SN1 – but not SN2 – reactions.

Initial Ionization in protic solvent.

1,2 shift converting 2o carbocation to 3o benzylic

Nucleophile attacks.Deprotonate to to yield ether

Next elimination…

Return to elimination: competes with nucleophilic

substitution.

Zaitsev Rule, prefer to form the more substituted alkene (more stable).

The competition: SN1 and/orSN2

Mechanistic Possibilities to eliminate the H+ and X-

HX

+ H + X

Possible Sequences for bond making/breaking…

• Regard the alkyl halide as an acid. First remove H+ producing a carbanion , then in a second step remove X- producing the alkene.

or

• First remove X- producing a carbocation, then in a second step remove H+ yielding the alkene. E1

or

• Remove H and X in one step to yield the alkene. E2

• There are two idealized mechanisms for -elimination reactions

• E1 mechanism:E1 mechanism: at one extreme, breaking of the R-Lv bond to give a carbocation is complete before reaction with base to break the C-H bond– only R-Lv is involved in the rate-determining step (as in

SN1)

• E2 mechanism:E2 mechanism: at the other extreme, breaking of the R-Lv and C-H bonds is concerted (same time)– both R-Lv and base are involved in the rate-determining

step (as in SN2)

E1 Mechanism

– ionization of C-Br gives a carbocation intermediate

– proton loss from the carbocation intermediate to a base (for example, the solvent) gives the alkene

CH3-C-CH3

Br

CH3

CH3-C-CH3

CH3

Br

slow, ratedetermining

+(A carbocation intermediate)

+

O:

H

H3CH-CH2-C-CH3

CH3

O

H

H3CH CH2=C-CH3

CH3fast

+

++ +

Rate Determining Step; formation of the carbocation.

Energy Profile for E1 mechanism, carbocations.

Alkyl Halide (E1) Alkene + HX

Alkyl Halide (Addition) Alkene + HX

Reaction can occur in either direction…..

E2 Mechanism

CH3O:-

C C

H

Lv

CH3OH

C C

Lv-H and -Lv are anti and coplanar

(dihedral angle 180°)

breaking of the R-Lv and C-H bonds is concerted

Needs Strong Base

E2

Kinetics of E1 and E2• E1 mechanism

– reaction occurs in two steps– the rate-determining step is carbocation formation

involving only RLv– the reaction is 1st order in RLv and zero order in base

• E2 mechanism– reaction occurs in one step involving both RLv and

the base.– reaction is 2nd order; first order in RLv and 1st order

in base d[RLv]

dtRate = = k[RLv][Base]

d[RLv]

dtRate = = k[RLv]

Regioselectivity of E1/E2

• E1: major product is the more stable alkene (more substituted, more resonance)

• E2: with strong base, the major product is the more stable (more substituted, more resonance) alkene

Special notes about sterically hindered bases such as tert-butoxide, (CH3)3CO -.E2 – anti-Zaitsev: with a strong, sterically hindered base the major product is often the less stable (less substituted) alkene. Reason: hydrogens on less substituted carbons are more accessible.

AlsoE2 vs SN2: In competition of SN2 vs E2: steric bulk in either the alkyl halide or the base/nucleophile prevents the SN2 reaction and favors the E2.

• E2 is most favorable (lowest activation energy) when H and Lv are anti and coplanar

CH3O:-

C C

H

Lv

CH3OH

C C

Lv-H and -Lv are anti and coplanar

(dihedral angle 180°)

Stereochemistry of E2

AB

D E

E

DA

B

Examples of E2 Stereochemistry

Cl

CH3O -

+

cis Major product.

Zaitsev product

Cl

CH3O -

+

trans

Only product

Anti-Zaitsev

Explain both regioselectivity and relative rates of reaction.

But

Faster reaction

Slower reaction

In order for the H and the Cl to be anti, both must be in axial positions

First the cis isomer. Reactive Conformation; H and Cl are anti to each other

Iso-propyl groups is in more stable equatorial position. Dominant

conformation is reactive conformation.

CH3O:-

H

H

H

H

Cl

CH3OH :Cl

1-Isopropyl-cyclohexene

2

1

6 + +E2

Principles to be used in analysis

Stereochemical requirement: anti conformation for departing groups. This means that both must be axial.

Dominant conformation: ring flipping between two chair conformations, dominant conformation will be with iso propyl equatorial.

In the more stable chair of the trans isomer, there is no H anti and coplanar with Lv, but there is one in the less stable chair

More stable chair (no H is anti and coplanar to Cl)

Less stable chair(H on carbon 6 is

anti and coplanar to Cl)

2

2

11

6 6Cl

H

H

H

H

Cl

HH

HH

Now the trans

Unreactive conformation

Reactive but only with the H on C 6

Most of the compound exists in the unreactive conformation. Slow reaction.

H

Cl

HH

HCH3O:

-

CH3OH Cl

(R)-3-Isopropyl-cyclohexene

21

6E2

+ +

Anti Zaitsev

More stable chair (no H is anti and coplanar to Cl)

Less stable chair(H on carbon 6 is

anti and coplanar to Cl)

2

2

11

6 6Cl

H

H

H

H

Cl

HH

HH

Example, Predict Product

Problem!: Fischer projection diagram represents an eclipsed structure.

Task: convert to a staggered structure wherein H and Br are anti and predict product. We will convert to a Newman and see what we get…

Ph

H CH3

Ph

H3C Br

base

Ph

H CH3

Ph

H3C Br

Ph

H CH3

Ph

H3C Br

=

Ph

H3C H

Ph

H3C Br

rotate upper chiral C by 180

H3C

Ph

H

Br

Ph

H3C

CH3

H Ph

Ph

H3C Br

H

CH3

Ph

Br

Ph

H3C

rotate 120further

H3C Ph

H3C Phbase

H3C Ph

H3C Ph

H & Br not anti yet!

Now anti and we can see where the pi bond will be.

Ph

H CH3

Ph

H3C Br

base

Alternative Approach: CAR

The H and Br will be leaving: just indicate by disks.

Meso or Racemic??

Anti Geometry

CRA

Relationship works in both directions. Should get cis isomer.

H3C Ph

H3C Ph

Note: As we have said before it may take some work to characterize a compound as “racemic” or “meso”.

This may be recognized as one of the enantiomers of the racemic mixture.

AC < -- > R

RCH2X

R2CHX

R3CX

Alkyl halide E1 E2

Primary

Secondary

Tertiary

E1 does not occur.Primary carbocations areso unstable, they are neverobserved in solution.

E2 is favored.

Main reaction with strong bases such as OH- and OR-.

Main reaction with weak bases such as H2O, ROH.

Main reaction with strong bases such as OH- and OR-.

Main reaction with weak bases such as H2O, ROH.

E1 or E2

(Carbocation)

ionization

1o,

2o, 3o polar solvents, weak nucleophiles, weak bases

1o strong, bulky bases

2o strong bases

3o strong bases

1o good nucleophiles, aprotic solvents

2o good nucleophiles but also poor bases, aprotic solvents

30

SN2SN1

E2 E1

Rearrange ?

1o

2o heat, more hindered

3o heat, more hindered

1o

2o lower hinderance, better nucleophile than base

3o lower hinderance, better nucleophile than base

R - X R + + X -

good nucleophile

strong base

R-Nuc

alkene

weak nucleophile

weak base

alkene

R-Nuc

Recall Halohydrins and EpoxidesCl2, H2O Cl H2O

Cl

OH

base Cl

O

O

H

OH

ROH

OH

RO

Creation of Nucleophile Internal SN2 reaction with inversion

Creation of good leaving group.

Attack by poor nucleophile

Neighboring Group Effect

• Mustard gases – contain either S-C-C-X or N-C-C-X

– what is unusual about the mustard gases is that they undergo hydrolysis rapidly in water, a very poor nucleophile

ClS

Cl 2H2O HOS

OH 2HCl+ +

Bis(2-chloroethyl)sulfide(a sulfur mustard gas)

Bis(2-chloroethyl)methylamine(a nitrogen mustard gas)

ClS

Cl ClN

Cl

– the reason is neighboring group participation by the adjacent heteroatom

– proton transfer to “solvent” completes the reaction

ClS

Cl

ClS O-H

H

ClS

ClS

O

H

H

Cl+

+

A cyclic sulfonium ion

an internal SN2 reaction

slow, ratedetermining

++

a secondSN2 reaction

fast+

:

:

:

Good nucleophile.

5. Provide a clear, unambiguous mechanism to explain the following stereochemical results. Complete structures of intermediates, if any, should be shown. Use curved arrow notation consistently.

H

SCH3

CH3

Cl

H3C H

H2O H

SCH3

CH3

OH

H3C H

H3C

SCH3

H

OH

H CH3

+

Expect sulfur to attack the C-Cl, displacing the Cl and forming a three membered ring. Like this…

But we have to be careful with stereochemistry

H

SCH3

CH3

Cl

H3C H

S

CH3

H2O

S

CH3

OH

Here is the crux of the matter: how can the non-reacting carbon change its configuration??? Further it does not always change but only if configuration of the reacting carbon changes!! We got a mixture of enantiomers, a racemic mixture. Something strange is happening!!

From an old quiz

H

SCH3

CH3

Cl

H3C H

We have to put the molecule in the correct conformation.

=H

SCH3

CH3

Cl

H3C HCl SCH3

H CH3H3C H

=

Reactive conformation reached by 180 rotation around C-C bond

Cl

SCH3H3C

CH3

H

H

SCH3

H3C CH3

HH

And then the ring is opened by attack of water

S and Cl are eclipsed, not anti.

But let’s pause for a moment. Our reactant was optically active with two chiral carbons.

Recall the problem: If reaction occurs only at the C bearing the Cl the other should remain chiral! Hmmmm?But now notice that the intermediate sulfonium ion is achiral. It has a mirror plane of symmetry. Only optically inactive products will result.

H

SCH3

CH3

OH

H3C HSCH3

H3C CH3

HH

OH2

SCH3H3C

CH3

H

H =

SCH3

H CH3HH3C

=

180 rotation

HO

HO

Two modes of attack by water.

And…

SCH3

H3C CH3

HH

OH2

H3CS

H3C

CH3

H

H OH

H3C

SCH3

H

OH

H CH3

=

H3CS

H3CH

CH3H

OH

=

180 rotation

Again note the ring structure is achiral and that we must, of course, produce optically inactive product.

Enantiomers, racemic mixture