Synthesis of Alkenes via Elimination Reactions S:7- 6DehydrohalogenationReactions by an E2 mechanism are most usefulE1 reactions can be problematicE2 reaction are favored by:Secondary or tertiary alkyl halidesAlkoxide bases such as sodium ethoxide or potassium tert-butoxideBulky bases such as potassium tert-butoxide should be used for E2 reactions of primary alkyl halides

Chapter 7

SN2 or SN1?Primary or methylStrong nucleophile

Polar aprotic solvent

Rate = k[halide][Nuc]Inversion at chiral carbonNo rearrangements TertiaryWeak nucleophile (may also be solvent) Polar protic solvent, silver saltsRate = k[halide]Racemization of optically active compoundRearranged products =>

Chapter 7

Substitution or Elimination?Strength of the nucleophile determines order: Strong nuc. will go SN2 or E2.Primary halide usually SN2.Tertiary halide mixture of SN1, E1 or E2High temperature favors elimination.Bulky bases favor elimination.Good nucleophiles, but weak bases, favor substitution. =>

Chapter 7

Elimination ReactionsThe alkyl halide loses halogen as a halide ion, and also loses H+ on the adjacent carbon to a base.A pi bond is formed. Product is alkene.Also called dehydrohalogenation (-HX).

Chapter 7

E1 ReactionUnimolecular eliminationTwo groups lost (usually X- and H+)Nucleophile acts as baseAlso have SN1 products (mixture) =>

Chapter 7

E1 MechanismHalide ion leaves, forming carbocation.Base removes H+ from adjacent carbon.Pi bond forms. =>

Chapter 7

A Closer Look

Chapter 7

E1 Energy DiagramNote: first step is same as SN1=>

Chapter 7

E1 or E2?Tertiary > SecondaryWeak base Good ionizing solvent

Rate = k[halide]Saytzeff productNo required geometry

Rearranged productsTertiary > SecondaryStrong base requiredSolvent polarity not importantRate = k[halide][base]Saytzeff productCoplanar leaving groups (usually anti)No rearrangements =>

Chapter 7

E2 ReactionBimolecular eliminationRequires a strong baseHalide leaving and proton abstraction happens simultaneously - no intermediate. =>

Chapter 7

E2 Mechanism

Chapter 7

Relative Stabilities of Alkenes S:7- 3Generally cis alkenes are less stable than trans alkenes because of steric hinderance

Heat of HydrogenationS:7- 3 AThe relative stabilities of alkenes can be measured using the exothermic heats of hydrogenationThe same alkane product must be obtained to get accurate results

Chapter 7

Heats of hydrogenation of three butene isomers:

Fig7.2Overall Relative Stabilities of Alkenes S:7- 3F BThe greater the number of attached alkyl groups (i.e. the more highly substituted the carbon atoms of the double bond), the greater the alkenes stability

Chapter 7

Saytzeffs RuleIf more than one elimination product is possible, the most-substituted alkene is the major product (most stable).R2C=CR2 > R2C=CHR > RHC=CHR > H2C=CHR tetra > tri > di > mono

Chapter 7

Zaitsevs Rule: Formation of the Most Substituted Alkene is Favored with a Small Base S:7- 6 ASome hydrogen halides can eliminate to give two different alkene products

Chapter 7

Zaitzevs Rule: when two different alkene products are possible in an elimination, the most highly substituted (most stable) alkene will be the major productThis is true only if a small base such as ethoxide is used

The transition state in this E2 reaction has double bond characterThe trisubstituted alkene-like transition state will be most stable and have the lowest DG

Chapter 7

Kinetic control of product formation: When one of two products is formed because its free energy of activation is lower and therefore the rate of its formation is higherThis reaction is said to be under kinetic control

Fig. 7.6

Chapter 7

Formation of the Least Substituted Alkene Using a Bulky Base S:7- 6BBulky bases such as potassium tert-butoxide have difficulty removing sterically hindered hydrogens and generally only react with more accessible hydrogens (e.g. primary hydrogens)

Chapter 7

E2 Stereochemistry

Chapter 7

The Stereochemistry of E2 Reactions: The Orientation of Groups in the Transition State S:7- 6 CAll four atoms involved must be in the same planeAnti coplanar orientation is preferred because all atoms are staggered

In a cyclohexane ring the eliminating substituents must be diaxial to be anti coplanar

Chapter 7

Neomenthyl chloride and menthyl chloride give different elimination products because of this requirement

In neomenthyl chloride, the chloride is in the axial position in the most stable conformationTwo axial hydrogens anti to chlorine can eliminate; the Zaitzev product is major

Chapter 7

In menthyl chloride the molecule must first change to a less stable conformer to produce an axial chlorideElimination is slow and can yield only the least substituted (Hoffman) product from anti elimination

Chapter 7

Carbocation Stability and the Transition State S:7- 7BRecall the stability of carbocations is:

The second step of the E1 mechanism in which the carbocation forms is rate determiningThe transition state for this reaction has carbocation character Tertiary alcohols react the fastest because they have the most stable tertiary carbocation-like transition state in the second step

Chapter 7

The relative heights of DG for the second step of E1 dehydration indicate that primary alcohols have a prohibitively large energy barrier

Fig 7.7

Chapter 7

A Mechanism for Dehydration of Primary Alcohols: An E2 Reaction S:7- 7 CPrimary alcohols cannot undergo E1 dehydration because of the instability of the carbocation-like transition state in the 2nd stepIn the E2 dehydration the first step is again protonation of the hydroxyl to yield the good leaving group water

Chapter 7

Carbocation Stability and the Occurrence of Molecular Rearrangements S:7- 8Rearrangements During Dehydration of Secondary AlcoholsRearrangements of carbocations occur if a more stable carbocation can be obtainedExample

The first two steps are to same as for any E1 dehydration

Chapter 7

In the third step the less stable 2o carbocation rearranges by shift of a methyl group with its electrons (a methanide)This is called a 1,2 shift

Chapter 7

The removal of a proton to form the alkene occurs to give the Zaitzev (most substituted) product as the major one

A hydride shift (migration of a hydrogen with its electrons) can also occur to yield the most stable carbocation

Chapter 7

Carbocation rearrangements can lead to formation of different ring sizes

Chapter 7

Synthesis of Alkynes by Elimination Reactions S:7- 9Alkynes can be obtained by two consecutive dehydrohalogenation reactions of a vicinal dihalide

Chapter 7

In menthyl chloride the molecule must first change to a less stable conformer to produce an axial chlorideElimination is slow and can yield only the least substituted (Hoffman) product from anti elimination

Chapter 7

Carbocation Stability and the Transition State Recall the stability of carbocations is:

The second step of the E1 mechanism in which the carbocation forms is rate determiningThe transition state for this reaction has carbocation character Tertiary alcohols react the fastest because they have the most stable tertiary carbocation-like transition state in the second step

Chapter 7

Synthesis of Alkynes by Elimination Reactions Alkynes can be obtained by two consecutive dehydrohalogenation reactions of a vicinal dihalide

Chapter 7

Alkenes can be converted to alkynes by bromination and two consecutive dehydrohalogenation reactions

Geminal dihalides can also undergo consecutive dehydrohalogenation reactions to yield the alkyne

Chapter 7

Hydride Shift

Chapter 7

Methyl Shift

Chapter 7

Secondary Halides?Mixtures of products are common.=>

Chapter 7

Chapter 7