Chapter 8 YEOH

7/21/2019 Chapter 8 YEOH

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Chapter 8

Alkyl Halides

and

Elimination Reactions


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Alkyl Halides and Elimination Reactions

Elimination reactions involve the loss of elements from the

starting material to form a new bond in the product.


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In both reactions below, a base removes the

elements of an acid, HX, from the organic starting

material.


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Dehydrohalogenation: removal of the elements HX.

An example of elimination.


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The most common bases used in elimination reactions

are negatively charged oxygen compounds,

E.g.: HO and its alkyl derivatives, RO .


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To draw any product of dehydrohalogenation :

(i) Find the carbon.

(ii) Identify all carbons with H atoms.

(iii) Remove the elements of H and X from the and

carbons and form a bond.


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The double bond of an alkene consists of a bond and a bond.

AlkenesThe Products of Elimination


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Alkenes are classified according to the number of carbon

atoms bonded to the carbons of the double bond.

Figure 8.1: Classifying alkenes by the number of R groups bonded to the double bond


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Because of restricted rotation, two stereoisomers of 2-

butene are possible.

Cis-2-butene and trans-2-butene are diastereomers They are stereoisomers that are not mirror images of

each other.


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When the two groups on each endof a carbon-carbon

double bond are different from each other, two

diastereomers are possible.


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In general, trans alkenes are more stable than cis

alkenes

The groups bonded to the double bond carbons in transalkenes are further apart, reducing steric interactions.


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The stability of an alkene increases as the number of R

groups bonded to the double bond carbons increases.

The higher the percent s-character, the more readily an atom

accepts electron density.

sp2carbons are more able to accept electron density and sp3

carbons are more able to donate electron density.

Increasing the number of electron donating groups on a

carbon atom able to accept electron density makes the

alkene more stable.


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There are two mechanisms of eliminationE2 and E1.

E2 mechanismbimolecular elimination

E1 mechanismunimolecular elimination

The E2 and E1 mechanisms differ in the timingof bondcleavage and bond formation

E2 and SN2 reactions have some features in common,

as do E1 and SN1 reactions.

Mechanisms of Elimination


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The most common mechanism for dehydrohalogenation

is the E2 mechanism.

Second-order kinetics

E2 mechanism

rate = k[(CH3)3CBr][ OH]

The reaction is concertedall bonds are broken and

formed in a singlestep.


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E2: A Concerted Mechanism


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Figure 8.3:An energy diagram for an E2 reaction:


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There are close parallels between E2 and SN2 mechanisms

in how the identity of the base: the leaving group and the

solvent affect the rate.

The rate of the E2 reaction increases as the strength of thebase increases.

E2 reactions are generally run with strong, negatively

chargedbaseslike OH and OR.

Two strong sterically hindered nitrogen bases called DBN

and DBUare also sometimes used.


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Figure 8.4:An E2 elimination with DBN used as the base


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Example of polar aprotic solvents: MeCN (acetonitrile), DMF

(Dimethylformamide) and DMSO (Dimethylsulfoxide)


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The SN2 and E2 mechanisms differ in how the R group

affects the reaction rate.


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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 (i.e.,lowers Ea, which increases the rate of the reaction.


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Increasing the number of R groups on the carbon with the

leaving group forms more highly substituted, more stable

alkenes in E2 reactions.

Hence, 3 alkyl halides reacts faster than the 1 alkyl halide.


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When alkyl halides have two or more different carbons,

more than one alkene product is formed. One of the products usually predominates.

The majorproduct is the more stableproductthe one with

the more substituted double bond.

This phenomenon is called the Zaitsev rule.

The Zaitsev Rule


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A reaction is regioselective when it yields predominantly or

exclusively one constitutional isomer when more than one is

possible.

E2 reaction is regioselective.


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A reaction is stereoselective when it forms predominantly or

exclusively one stereoisomer when two or more are possible.

E2 reaction is stereoselective because one stereoisomer is

formed preferentially.


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Exhibits first-order kinetics:

E1 mechanism

rate = k[RX]

The E1 reaction proceeds via a two-step mechanism:

(i) the bond to the leaving group breaks

(ii) bond is formed.

(i) is the slow step, involving only the alkyl halide(unimolecular).


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Figure 8.6: Energy diagram for an E1 reaction.


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The rate of an E1 reaction increases as the number of R

groups on the carbon with the leaving group increases.

The strength of the base usually determines whether a

reaction follows the E1 or E2 mechanism.

Strong bases favor E2 reactions,

Weaker bases favor E1 reactions.


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E1 reactions are regioselective, favoring formation of the

more substituted, more stable alkene.

Zaitsevs rule applies to E1 reactions also.


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SN1 and E1 reactions have exactly the same first step

formation of a carbocation.

SN1 vs E1 Reactions

E1 reactions often occur with a competing SN1 reaction, E1

reactions of alkyl halides are much less useful than E2

reactions.


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The transition state of an E2 reaction consists of four atoms from an

alkyl halideall aligned in a plane.

There are two ways for the CH and CX bonds to be coplanar.

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

Allows the incoming base and leaving group to be further away from

each other.


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Figure 8.7: Two possible geometries for the E2 reaction.


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The stereochemical requirement of an anti periplanar geometry in an

E2 reaction has important consequences for compounds containing

six-membered rings.

For E2 elimination, the C-Cl bond must be anti periplanar to the CH

bond on a carbon, and this occurs only when the H and Cl atoms

are both in the axialposition.

The requirement for trans diaxial geometry means that elimination

must occur from the less stable conformer, B.


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Figure 8.8: The trans diaxial geometry for the E2 elimination in chlorocyclohexane

E2 d h d h l ti f i d t 1 hl 2 th l l h


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E2 dehydrohalogenation of cis- and trans-1-chloro-2-methylcyclohexane.

This cis isomer exists as two conformations, A and B, each of whichas one group axial and one group equatorial.

E2 reaction must occur from conformation B, which contains an axial

Cl atom.


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B has two different axial hydrogens, labeled Haand Hb, E2

reaction occurs in two different directions to afford two

alkenes.

The major product contains the more stable trisubstituteddouble bond, as predicted by the Zaitsev rule.


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The transisomer of 1-chloro-2-methylcyclohexane exists as two

conformers:

C, having two equatorial substituents, and

D, having two axial substituents.

E2 reaction must occur from D, since it contains an axial Cl

atom.


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D has only one axial H, E2 reaction occurs only in one

direction to afford a single product.

This is not predicted by the Zaitsev rule.


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Cl

CH3

H

H H

CH3CH3

-H

+

-H+

Cl

CH3

H

H H

-H+

CH3

H

cis-1-chloro-2-

methylcyclohexane

trans-1-chloro-2-

methylcyclohexane

majorminor

Only product, NOT

predicted by

Zaitsev rule

- HCl - HCl

- HCl

Stereochemistry of the E2 Reaction


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A single elimination reaction produces a bond of an alkene.

Two consecutive elimination reactions produce two bondsof an alkyne.

E2 Reactions and Alkyne Synthesis


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Two elimination reactions are needed to remove two

moles of HX from a dihalide substrate.

Two different starting materials can be useda vicinaldihalideor a geminal dihalide.


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Stronger bases are needed to synthesize alkynes by

dehydrohalogenation.

Typical base used is NH2(amide) ORH- (hydride)


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Reason : The transition state for the second elimination

reaction includes partial cleavage of the CH

bond.

However, the carbon atom is sp2hybridized

Sp2hybridized CH bonds are stronger than sp3hybridized

CH bonds.

A stronger base is needed to cleave this bond.

?


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Figure 8.9: Example of dehydrohalogenation of dihalides to afford alkynes.


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E1 or E2?

Tertiary > Secondary

Weak base

Good ionizing solvent

Rate = k[halide]

Zaitsev product

No required geometry

Rearranged products

Tertiary > Secondary

Strong base required

Solvent polarity not

important

Rate =k[halide][base]

Zaitsev product

Coplanar leaving groups

(usually anti-periplanar) No rearrangements

Substitution or Elimination?


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Substitution or Elimination?

Strengthof base / Nu: determines order:

Strong Nu:, bimolecular, SN2 or E2.

1halide usually SN2.

3halides: E2 (strong base)

SN1 & E1 (weak base) 2halides: SN2 & E2 (strong base)

SN1 & E1 (weak base)

High temperature favors elimination. Bulkybases favor elimination.

Good nucleophiles, but weak bases (Br-, I-), favorsubstitution


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For each of the following reactions:

(i) Give the expected product(s) including the correct stereochemistry.

(ii) Indicate the type of mechanism (SN1, SN2, E1 and/or E2).

Cl

NaOH

CH3

I CH3

H

H2O


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Ph

Br

CH3

t-BuOH

(CH3)3CO-K+

H

Br

H

CH3 -CN


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CH2CH3

CH3OH


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Q: When 1-iodo-1-methylcyclohexane is treated withNaOCH2CH3, the more highly substituted alkene

product predominates. When KOC(CH3

)3

is usedinstead, the less highly substituted alkene productpredominates. Offer an explanation.

A:.

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Chapter 8 YEOH