Lecture 12

Lecture 12 ELIMINATION REACTIONS

1

ELIMINATION REACTIONS An elimination reaction is one where starting material loses the elements of a small molecule such as HCl or H2O or Cl2 during the course of the reaction to form the product.

C

C

H

Cl

- HCl

C

C

TWO EXAMPLES FOLLOW 2

ELIMINATION REACTIONS TWO EXAMPLES Alkyl halide + strong base & heat CH3CH2CH2CH2

NaOH

Cl

CH3CH2

∆ NaOH

CH3CH2

CH CH3

CH3

∆

Cl

Alcohol + strong acid and heat H2SO4

CH3CH2CH2CH2 OH

∆

LOSS OF HCl CH CH2

CH CH CH3

LOSS OF H2O

CH3CH2 CH CH2 3

ELIMINATION REACTIONS H H | | X – C α- Cβ – Y | | H H

→

H H | | C=C | | H H

+ X–Y

In this reaction, alkene formation requires that X and Y be substituents on adjacent carbon atoms. 4

ELIMINATION REACTIONS By assigning X as the reference atom and identifying the carbon attached to it as α carbon, the atom Y is a substituent on the β carbon. Carbon succeedingly more remote from the reference atom are designated as γ , δ and so on. Our discussion will confine to β elimination reactions (also called as 1,2 eliminations). 5

THE REACTION IS A β-ELIMINATION The β-hydrogen is attached to the β-carbon. β-carbon

H C C Cl

α-carbon The functional group is attached to the α-carbon.

Since the β-hydrogen is lost this reaction is called a β-elimination. Reagent = a strong base6

MECHANISM THE BASE TAKES THE β-HYDROGEN B:

B

H C C : Cl .. :

H C C ..

: Cl .. : 7

Alkenes—The Products of Elimination Whenever the two groups on each end of a carbon-carbon double bond are different from each other, two diastereomers are possible.

8

Alkenes—The Products of Elimination In general, trans alkenes are more stable than cis alkenes because the groups bonded to the double bond carbons are further apart, reducing steric interactions.

9

Alkenes—The Products of Elimination The stability of an alkene increases as the number of R groups bonded to the double bond carbons increases.

10

Alkenes—The Products of Elimination The higher the percent s-character, the more readily an atom accepts electron density. Thus, sp2 carbons are more able to accept electron density and sp3 carbons are more able to donate electron density. Consequently, increasing the number of electron donating groups on a carbon atom able to accept electron density makes the alkene more stable. 11

Mechanisms of Elimination • There are two mechanisms of elimination E2 and E1 • E2 mechanism — bimolecular elimination • E1 mechanism— unimolecular elimination • The E2 and E1 mechanisms differ in the timing of bond cleavage and bond formation

12

Mechanisms of Elimination—E2 E2 reactions are generally run with strong, negatively charged bases like¯OH and ¯OR. The reaction exhibits second-order kinetics, and both the alkyl halide and the base appear in the rate equation i.e. rate = k[(CH3)3CBr][¯OH] The reaction is concerted—all bonds are broken and formed in a single step. 13

Mechanisms of Elimination—E2

The base removes a proton from the β carbon The electron pair in the β C-H bond forms the new π bond. The leaving group comes off with the electron pair in the C-X bond 14

The E2 Mechanism

R

.. – : O ..

H C

C : X: ..

Reactants

The E2 Mechanism

R

.. – : O ..

H C

C : X: ..

Reactants

The E2 Mechanism

R

δ..– O ..

H

Transition state C

C δ– : X: ..

The E2 Mechanism

R

.. O ..

H C

C .. – : X: ..

Products

Energy diagram for E2 reaction

19

Mechanisms of Elimination—E2 Because the bond to the leaving group is partially broken in the transition state, the better the leaving group the faster the E2 reaction

20

Mechanisms of Elimination—E2 The SN2 and E2 mechanisms differ in how the R group affects the reaction rate. As the number of R groups on the carbon with the leaving group increases, the rate of the E2 reaction increases.

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Mechanisms of Elimination—E2 The increase in E2 reaction rate with increasing alkyl substitution can be rationalized in terms of transition state stability. In the transition state, the double bond is partially formed. Thus, increasing the stability of the double bond with alkyl substituents stabilizes the transition state. 22

Mechanisms of Elimination—E2 Increasing the number of R groups on the carbon with the leaving group forms more highly substituted, more stable alkenes in E2 reactions. In the reactions on the next slide, since the disubstituted alkene is more stable, the 30 alkyl halide reacts faster than the 10 alkyl halide. 23

Mechanisms of Elimination—E2

24

Mechanisms of Elimination—E2

25

REGIOSELECTIVITY

A reaction is regioselective when it yields mostly one constitutional isomer when more than one product is possible.

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WHAT HAPPENS IF THERE IS MORE THAN ONE β-HYDROGEN ?

β

β’ H

H

C C C Cl Which one do we lose 27?

Zaitsev’s (Saytzeff) rule If more than one elimination product is possible, the mostsubstituted alkene is the major product (most stable). R2C=CR2 > R2C=CHR > RHC=CHR > H2C=CHR tetra

>

tri

>

di

> mono

28

Major (90%)

Minor (10%)

29

Stereochemistry of the E2 Reaction The transition state of an E2 reaction consists of four atoms from an alkyl halide— one hydrogen atom, two carbon atoms, and the leaving group (X)—all aligned in a plane. There are two ways for the C—H and C—X bonds to be coplanar.

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Stereochemistry of the E2 Reaction ORBITAL ALIGNMENT IN MOST CONCERTED REACTIONS THE ORBITALS BECOME PREALIGNED FOR A SMOOTH PROGRESSION OF EVENTS A coplanar arrangement allows a continuous movement of electrons from one end of the system to the other. 31

Stereochemistry of the E2 Reaction E2 elimination occurs most often in the anti periplanar geometry. This arrangement allows the molecule to react in the lower energy staggered conformation, and allows the incoming base and leaving group to be further away from each other.

32

Stereochemistry of the E2 Reaction

C

C

HCl

H

Cl

not common

H

H C

anti elimination

C Cl

Cl anti-coplanar

syn elimination

observed most often 33

Syn Elimination

34

Anti Elimination

35

STEREOCHEMISTRY

ACYCLIC HALIDES

36

ACYCLIC MOLECULES MAY HAVE TO ROTATE IN ORDER TO REACT

H

Cl H

Cl anti-coplanar 37

2-Bromo-3-phenylbutane RS SS enantiomers SR RR diastereomers

two stereocenters

* *CH CH

CH 3

CH3 Br

C CH 3

NaOEt EtOH

CH CH 3

+

*CH

CH CH 2

CH 3

Major Product Minor Product Zaitsev 38 Z or E ?

2S,3R-DIASTEREOMER CH 3 Ph

R

S

CH 3

CH 3 C C H H Br

Ph

Ph CH3

CH3 C C H Br

anti-coplanar

H

not observed

rotate H

C C

CH 3

NaOMe

Ph

MeOH

CH 3

C C

CH3 H

observed major product (Z)-2-phenyl-2-butene

(plus some 1-butene due to β-H on the CH3)39

MAKING THE 2S,3S-DIASTEREOMER CH 3 Ph

R

S

CH 3 C C H H Br H

Ph CH 3

CH 3 C C H Br

Ph C C CH 3

CH 3 H

(Z)

Ph

R

S

CH3

make diastereomer ( change one stereocenter )

S

S

CH3 C C H H Br

?

Z or E

Will the 2S,3S-diastereomer give the same product as its 2S,3R diastereomer? 40

2S,3S-DIASTEREOMER Ph CH3

S

S

Ph

CH3 C C H H Br

C C CH3

CH 3 CH3 C C H Ph Br

anti-coplanar

H

not observed

rotate H

CH3

NaOMe

CH3

MeOH

Ph

C C

CH3 H

observed major product

(E) -2-phenyl-2-butene (plus some 1-butene) 41

CONVERTING THE ALKYL HALIDE TO AN ALKENE

VISUALIZING THE PRODUCT THAT FORMS CH3 Ph

H (top)

H CH3 H Cl

alkyl halide

CH3

CH3

Ph

H

alkene

Cl (bottom)

The methyl groups (blue) are in back in both structures. The phenyl and the hydrogen (black) are in front in both. 42

ANOTHER VISUALIZATION OF THE REACTION 2S,3R

H

CH3

CH3

Ph Cl 2S,3S

back carbon

H

same side

H

H

CH3

front Ph carbon

C C CH3

same side H

CH3

CH3

H

C

CH3

Ph

C

Cl

same side

CH3

same side

Ph 43

CYCLIC HALIDES The stereochemical requirement of an anti periplanar geometry in an E2 reaction has important consequences for compounds containing six-membered rings. 44

Stereochemistry of the E2 Reaction Chlorocyclohexane exists as two chair conformers. Conformer A is preferred since the bulkier Cl group is in the equatorial position.

45

Stereochemistry of the E2 Reaction For E2 elimination, the C-Cl bond must be anti periplanar to the C—H bond on a β carbon This occurs only when the H and Cl atoms are both in the axial position. The requirement for axial geometry means that elimination must occur from the less stable conformer. 46

Stereochemistry of the E2 Reaction

47

Stereochemistry of the E2 Reaction

48

Stereochemistry of the E2 Reaction 1-Bromo-2-methylcyclohexane The expected result (naïve) : CH 3

CH 3 +

CH 3

Br

drawn flat without stereochemistry

major product Zaitsev

minor

The result actually depends on the stereochemistry of the starting material.

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cis stereoisomer The elimination needs to have H and Br anticoplanar H

H

H

CH3

CH3

Br

cis The other chair won’t work. Why?

major product +

Minor

(Zaitsev)

CH3

50

trans stereoisomer CH3

H H

CH3

H

H

Br

trans

The other chair won’t work. Why?

only product CH3

not formed 51

Reaction does not occur easily if anti-coplanar geometry cannot be achieved trans (e,e)

H

CH 3 CH 3

Br is not axial, no anti-coplanar H Br

C

NaOEt

no reaction

Et OH

CH 3

H3C

Br NaOEt

CH 3 CH 3

C CH 3

C

H H

cis (e,a)

EtOH

C

CH3 CH3

H

THESE RINGS WILL NOT INVERT 52

Both green hydrogens are anti to chlorine. Elimination by path (a) and (b) give different products

53

54

syn Elimination More difficult than anti-coplanar elimination, but do occur in some circumstances. Br H D

Na+OCH3

H

H D

+ CH3OH + NaBr

55

DO NOT DRAW THE ALKENES IN CHAIR CONFORMATION

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