• 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

Alcohols

Hydrogen Bonding

Three ethanol molecules.

Hydrogen Bonding & boiling pointIncreases boiling point, higher temperature needed to separate the molecules.

Hexane 69 deg.

1-pentanol 138

1,4-butanediol 230

Ethanol 78 deg

Dimethyl ether 24

Earlier Discussion of Acidity

Methanol Ethanol 2-Propanol 2-Methyl-2-propanol

Increasing Hinderance of Solvation

RO-H RO – (solvated) + H + (solvated)

Increasing Acidity of the alcohol

Recall: H2O + Na Na+ + OH- + ½ H2(g)

Alcohols behave similarly

ROH + Na Na+ + OR- + ½ H2(g)

Alkoxide, strong base, strong nucleophile (unless sterically

hindered)

Also: ROH + NaH Na+ + OR- + ½ H2(g)

Increasing Basicity of Alkoxide Anion, the conjugate base

Alkoxide ion, base

Alkoxides can be produced in several ways…

-OH as a Leaving Group

R-OH + H + R-OH2+

Protonation of the alcohol sets-up a good leaving group, water.

Poor leaving group, hydroxide ion.

Another way to turn the –OH into a leaving group…

Conversion to Alkyl Halide,HX + ROH RX + H2O

When a carbocation can be formed (Tertiary, Secondary alcohols) beware of rearangement. SN1

Expect both configurations.

When a carbocation cannot be formed. Methanol, primary. SN2

RCH2OH RCH2OH2

X -H +

RCH2X

R3COH R3COH2 R3C + + H2O

X -H +

R3CX

But sometimes experiment does not agree with our ideas…

Observed reaction CH2OH

HX

CH2X

X

The problem:

•Rearrangement of carbon skeleton which usually indicates carbocations.

•Reacting alcohol is primary; do not expect carbocation.

•Time to adjust our thinking a bit….

H3C

CH2OH

H + H3C

CH2OH2+

CH2.......OH2

H3C CH3

H2O

X-X

Not a primary carbocation

Other ways to convert: ROH RXWe have used acid to convert OH into a good leaving group

There are other ways to accomplish the conversion to the halide.

RCH2-OHPBr3

RCH2-O(H)PBr2

Br -

RCH2Br + HOPBr2

primary, secondary

RCH2-OHSOCl2

RCH2-OS(O)ClCl -

RCH2Cl + SO2

primary, secondary, tertiary

amine

R3COH R3COH2 R3C + + H2O

X -H +

R3CX

Leaving group.

Leaving group.Next, a very useful alternative to halide…

An alternative to making the halide: ROH ROTs

p-toluenesulfonyl chloride

Tosyl chloride

TsCl

ROH +S OO

Cl

S OO

O

CH3 CH3

R Tosylate group, -OTs, good leaving

group, including the

oxygen.

The configuration of the R group is unchanged.

Preparation from alcohols.

Example

CH3

H OH

C2H5

C3H7 CH3

TsCl

CH3

H OTs

C2H5

C3H7 CH3

Preparation of tosylate.

Retention of configuration

Substitution on a tosylate

The –OTs group is an excellent leaving group

Acid Catalyzed Dehydration of an Alcohol, discussed earlier as reverse of hydration

Protonation, establishing of good leaving group.

Elimination of water to yield carbocation in rate determining step.

Expect tertiary faster than secondary.

Rearrangements can occur.

Elimination of H+ from carbocation to yield alkene.

Zaitsev Rule followed.

Secondary and tertiary alcohols, carbocations

Primary alcohols

Problem: primary carbocations are not observed. Need a modified, non-carbocation mechanism.

Recall these concepts:

1. Nucleophilic substitution on tertiary halides invokes the carbocation but nucleophilic substitution on primary RX avoids the carbocation by requiring the nucleophile to become involved immediately.

2. The E2 reaction requires the strong base to become involved immediately.

Note that secondary and tertiary protonated alcohols eliminate the water to yield a carbocation because the carbocation is relatively stable. The carbocation then undergoes a second step: removal of the H+.

The primary carbocation is too unstable for our liking so we combine the departure of the water with the removal of the H+.

What would the mechanism be???

Here is the mechanism for acid catalyzed dehydration of Primary alcohols

1. protonation

2. The carbocation is avoided by removing the H at the same time as H2O departs (like E2).

As before, rearrangements can be done while avoiding the primary carbocation.

Principle of Microscopic Reversibility

Same mechanism in either direction.

Pinacol Rearrangement: an example of stabilization of a carbocation by an adjacent lone

pair.

Overall:

MechanismReversible protonation.

Elimination of water to yield tertiary carbocation.

1,2 rearrangement to yield resonance stabilized cation.

Deprotonation.

This is a protonated

ketone!

Oxidation

Primary alcohol

RCH2OH RCH=O RCO2H

Na2Cr2O7

Na2Cr2O7

Na2Cr2O7 (orange) Cr3+ (green) Actual reagent is H2CrO4, chromic acid.

Secondary

R2CHOH R2C=O

Tertiary

R3COH NR

KMnO4 (basic) can also be used. MnO2 is produced.

The failure of an attempted oxidation (no color change) is evidence for a tertiary alcohol.

Na2Cr2O7

OH

CH2OH

HO

Na2Cr2O7

acid

OH

CO2H

O

Example…

Oxidation using PCC

Primary alcohol

RCH2OH RCH=OPCC

PCCSecondary

R2CHOH R2C=O

Stops here, is not oxidized to carboxylic acid

Periodic Acid Oxidation

OH

OH

glycol

HIO4

O

O

two aldehydes

+ HIO3

OH

O HIO4

O

O

aldehydes

HO

carboxylic acid

+ HIO3

O

O HIO4

O

OHO

carboxylic acidcarboxylic acid

OH + HIO3

OH

O 2 HIO4

OH

O

OH

OHO

O

+ 2 HIO3

Mechanistic Notes

Cyclic structure is formed during the reaction.

Evidence of cyclic intermediate.

Sulfur Analogs, Thiols

Preparation

RI + HS- RSH

SN2 reaction. Best for primary, ok secondary, not tertiary (E2 instead)

Acidity

H2S pKa = 7.0

RSH pKa = 8.5

Oxidation