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http://196.3.3.103/courses/index.html

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ALKYL HALIDES – ALKYL HALIDES – ELIMINATION REACTIONSELIMINATION REACTIONS

ALKYL HALIDES UNDERGO ELIMINATION OF HX WHEN TREATED WITH BASE. THE PRODUCTS ARE ALKENES.

ELIMINATION REACTIONS USUALLY REQUIRE FORCING CONDITIONS, I.E. HEAT AND STRONG BASE.

THE ELIMINATION REACTIONS WHICH ALKYL HALIDES UNDERGO ARE KNOWN AS I,2-ELIMINATIONS OR ELIMINATIONS.

CH3

CCH3

CH3

Br KOH, EtOHheat

CH3

CCH3 CH2

+ HBr

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ALKYL HALIDES: ELIMINATION REACTIONSALKYL HALIDES: ELIMINATION REACTIONS

The elements of H-X are lost from neighboring carbon atoms and a C=C is formed. The head carbon of the alkyl halide is termed (“alpha”) and the carbon atom or atoms next to it are designated (“beta”).The halogen atom is lost from the carbon, and the hydrogen from one of the carbons.

CH3

CCH3

CH3

Br KOH, EtOHheat

CH3

CCH3 CH2

+ HBr

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ALKYL HALIDES: ELIMINATION REACTIONSALKYL HALIDES: ELIMINATION REACTIONS

THE TWO MOST IMPORTANT MECHANISMS BY WHICH ALKYL HALIDES UNDERGO ELIMINATION REACTIONS ARE:

1. THE E1 MECHANISM (UNIMOLECULAR);

2. THE E2 MECHANISM (BIMOLECULAR).

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ELIMINATION REACTIONS OF ALKYL HALIDES: ELIMINATION REACTIONS OF ALKYL HALIDES: THE UNIMOLECULAR MECHANISM (E1)THE UNIMOLECULAR MECHANISM (E1)

The slow, rate determining step entails one species – the alkyl halide.The rate of the reaction = k[alkyl halide]Note the carbocation intermediate

H

CR C

R

X

R

R

H

CR C

R

R

R

slow, r.d.s.(a)

(b)

H

CR C

R

R

R

B

fast

R

CR C

R

R

+ X

+ B-H

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ELIMINATION REACTIONS OF ALKYL HALIDES: THE UNIMOLECULAR MECHANISM (E1)

A carbocation intermediate is formed when alkyl halides undergo elimination via the E1 (unimolecular) mechanism.3o alkyl halides are likely to lose HX via this mechanism.For t-butyl bromide in aqueous alcoholic KOH:

H

CH C

H

Br

CH3

CH3

H

CH C

H

CH3

CH3

slow, r.d.s.(a)

(b)

H

CH C

H

CH3

CH3

HO

fast

H

CH C

CH3

CH3

+ Br

+ H-O-H

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ELIMINATION REACTIONS OF ALKYL HALIDES: THE BIMOLECULAR MECHANISM (E2)

THIS IS A CONCERTED REACTION.BOND FORMATION AND BOND BREAKING TAKE

PLACE SIMULTANEOUSLY.THE RATE DETERMINING STEP ENTAILS THE BASE AND

THE ALKYL HALIDE.RATE = k[alkyl halide][base]

R

CR C

H

R

X

R

B R

CR C

R

R+ B-H X+

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THE BIMOLECULAR MECHANISM (E2)A VERY IMPORTANT FEATURE

For an alkyl halide to undergo elimination via the E2 mechanism, the H and X groups must be anti to each other and be in the same plane with each other and the carbon atoms to which they are attached.

THE ELEMENTS OF H-X MUST BEANTIPERIPLANAR.

R

CR C

R

R

+ B-H X+X

R RH

R R

B

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OTHER ASPECTS OF E1 AND E2 REACTIONSOTHER ASPECTS OF E1 AND E2 REACTIONS

1. THE DISTINCTION BETWEEN THE E1 AND E2 MECHANISMS IS NOT AS CLEAR AS THE DISTINCTION BETWEEN THE SN1 AND SN2 MECHANISMS.

2. 3O AND 2O ALKYL HALIDES WILL ELIMINATE H-X VIA BOTH THE E1 AND E2 MECHANISMS.THE ELIMINATION OF H-X FROM 1O ALKYL HALIDES TAKES PLACE VIA THE E2 MECHANISM ONLY.FOR BOTH E1 AND E2 MECHANISMS, THE RATES FOLLOW THE TREND:3O R-X > 2O R-X > 1O R-X (do not react via E1)

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OTHER ASPECTS OF E1 AND E2 REACTIONS3. FOR MANY ALKYL HALIDES, THERE ARE TWO

POSSIBLE ELIMINATION PRODUCTS.

THE 3O ALKYL HALIDE BELOW HAS THREE CARBONS; TWO ARE IDENTICAL METHYL (CH3) GROUPS, AND THE THIRD IS A METHYLENE (CH2) GROUP.

LET US EXAMINE THE ELIMINATION OF H-Br FROM THIS COMPOUND VIA THE E1 MECHANISM.

CC

Br

CH3

C CH3

HH

H

H

H

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ELIMINATION PRODUCTS: E1 MECHANISMELIMINATION PRODUCTS: E1 MECHANISM

CC

Br

CH3

C CH3

HH

H

H

H

slow, r.d.s

CC

CH3

C CH3

HH

H

H

H

STEP I

STEP 2

CC

CH3

C CH3

HH

H

H

H

+ Br

OH

CC

CH3

CCH3

HH

HH

more substituted alkenemore stableSAYTZEFF PRODUCT

CC

CH3

C CH3

H

HH

H

H

HO

CC

CH3C

CH3

H

H H

H

less substituted alkeneless stableHOFMANN PRODUCT

carbocation

and/or

+ H2O

+ H2O

Two products can result from the loss of H-Br

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ELIMINATION PRODUCTS: E2 MECHANISMELIMINATION PRODUCTS: E2 MECHANISM

THE 2O ALKYL HALIDE SHOWN BELOW HAS TWO CARBONS WHICH ARE NOT IDENTICAL.

ONE IS A METHYL (CH3) GROUP AND THE OTHER IS A

METHYLENE (CH2) GROUP.

LET US EXAMINE THE ELIMINATION OF H-Br FROM THIS COMPOUND VIA THE E2 MECHANISM.

CC

Br

H

C HH

H3CH

H

H

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ELIMINATION PRODUCTS: E2 MECHANISMELIMINATION PRODUCTS: E2 MECHANISMTWO PRODUCTS CAN FORM VIA THE E2 MECHANISM

CC

Br

H

C H

H

H3CH

H

H

HO

CC H

CH3

H

CH3

more substituted alkenemore stableSAYTZEFF PRODUCT

and/or

+ Br

+ H2O

CC

Br

H

C H

H

H3CH

H

H

OH

CC H

H

CH3CH2

H

+ Br

+ H2O

less substituted alkeneless stableHOFMANN PRODUCT

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ELIMINATION PRODUCTS:HOFMANN VS. SAYTZEFFELIMINATION PRODUCTS:HOFMANN VS. SAYTZEFFTHE PROPORTION OF THE LESS SUBSTITUTED ALKENE (HOFMANN PRODUCT) CAN BE INCREASED BY USING A VERY BULKY BASE. TWO EXAMPLES OF BULKY BASES ARE SHOWN

CH3

CH3C

CH3

O K potassium t-butoxide

t-BuOK

CH2CH3

CCH3CH2

CH2CH3

O K potassium 3-ethyl-3-pentoxideEtCEt

Et

O K

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ELIMINATION PRODUCTS: HOFMANN VS. SAYTZEFFELIMINATION PRODUCTS: HOFMANN VS. SAYTZEFF

BULKY BASES INCREASE THE PROPORTION OF THE LESS SUBSTITUTED ALKENE (HOFMANN PRODUCT) FORMED IN ELIMINATION REACTIONS.

The H’s on the less substituted carbon are more sterically accessible to the base than are the H’s on the more substituted carbon. When the base is very bulky, then the H’s on the less substituted carbon are almost exclusively removed, and the less substituted (Hofmann) alkene product predominates.

CC

Cl

CH3

CH

H3CCH3

less substituted alkeneHOFMANN PRODUCT

H

HH

CC

CH3

CH

H3CCH3

HH

97%

EtCEtEt

O K

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ELIMINATION PRODUCTS: HOFMANN VS. SAYTZEFFELIMINATION PRODUCTS: HOFMANN VS. SAYTZEFF

STERIC ACCESSIBILITY OF THE H AFFECTS THE OUTCOME OF ELIMINATION REACTIONS.If the H on the carbon whose elimination leads to the more substituted alkene is very crowded, then the proportion of the less substituted alkene product will be high.

CC

Br

CH3

C

H

CH

HH

H3C

CH3

CH3

more accessiblevery

crowded

HEt O Na

CC

CH3

C

H

C

H

H

H3C

CH3

CH3

H

less substituted alkeneHOFMANN PRODUCT

MAJOR

CC

CH3

C

H

CH

HH

H3C

CH3

CH3

more substituted alkeneSAYTZEFF PRODUCT

MINOR

+

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SUBSTITUTION VERSUS ELIMINATION: SSUBSTITUTION VERSUS ELIMINATION: SNN1 VS E11 VS E1

When substitution reactions are carried out on 3o alkyl halides (SN1 reactions), products

of elimination (alkenes) are almost inevitably formed. Let us consider the the following reaction.

CH2CH3C CH2CH3Br

CH3CH2 + H2OCH2CH3C CH2CH3OH

CH3CH2 + HBr

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SUBSTITUTION VERSUS ELIMINATION: SN1 VS E1

In this reaction the carbocation intermediate, once it is formed, can lose a proton by reaction with as weak a base as H2O to give appreciable quantities of the alkene (elimination)

product.

CH2CH3C CH2CH3Br

CH3CH2 + H2OCH2CH3C CH2CH3OH

CH3CH2 + HBr

C

CCH2CH3CH3CH2

CH3

H HH2O

C

CCH2CH3CH3CH2

H3C H

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SUBSTITUTION VERSUS ELIMINATION: E2 VS SSUBSTITUTION VERSUS ELIMINATION: E2 VS SNN22

IT IS EASIER TO CREATE CONDITIONS WHICH FAVOR THE E2 MECHANISM OVER THE SN2 MECHANISM, OR VICE VERSA.

VERY STRONG BASE (ETHOXIDE AS OPPOSED TO HYDROXIDE) RELATIVELY NON-POLAR SOLVENTS (E.G. ETHANOL IN PREFERENCE TO WATER) HIGHER TEMPERATURES, WILL FAVOR THE E2 MECHANISM OVER THE SN2 MECHANISM.

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ORGANOMETALLIC COMPOUNDS

COMPOUNDS IN WHICH A METAL IS DIRECTLY BONDED TO CARBON ARE KNOWN AS ORGANOMETALLIC COMPOUNDS. THE METAL-CARBON BOND IS POLARIZED AS SHOWN.

METALS ARE LESS ELECTRONEGATIVE THAN CARBON; LARGER DIFFERENCES IN ELECTRONEGATIVITY BETWEEN THE METAL AND CARBON INCREASE THE IONIC CHARACTER OF THE METAL-CARBON BOND. IONIC CHARACTER OF METAL CARBON BONDS FOLLOWS THE TREND

Na > Li > Mg > Al > Zn > Cd > Hg

C M

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ORGANOMETALLIC COMPOUNDSORGANOMETALLIC COMPOUNDSALKYL DERIVATIVES OF ALMOST ALL METALS HAVE BEEN PREPARED.

THESE ARE NAMED AS “ALKYLMETALS”

(CH3)2Hg DIMETHYLMERCURY(liquid; bp 92 oC; neurotoxin; environmental contaminant)

(CH3CH2)4Pb TETRAETHYLLEAD(liquid; bp ~ 220 oC; toxic; formerly used as a gasoline

additive)

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GRIGNARD REAGENTSGRIGNARD REAGENTSALKYLMAGNESIUM HALIDES, R-Mg-X, ARE KNOWN AS GRIGNARD REAGENTS.

GRIGNARD REAGENTS ARE PREPARED BY REACTING ALKYL HALIDES WITH EXCESS MAGNESIUM METAL IN DRY ALCOHOL-FREE DIETHYL ETHER OR TETRAHYDROFURAN (THF). DIETHYL ETHER AND THF ARE SOLVENTS. OCH3CH2

O CH2CH3

diethyl ether tetrahydrofuran (THF), a cyclic ether

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GRIGNARD REAGENTSGRIGNARD REAGENTSPREPARATIONR-X + Mg R-Mg-X (radical mechanism)Ease of formation follows the trends shown belowR-I > R-Br > R-Cl.CH3X > C2H5X >C3H7X

Grignard reagents are usually closely associated with two molecules of the ethereal solvent in whichthey have been prepared. R

Mg

X O

O Et

Et

Et

Et