Physical Organic ChemistryCH-4

Nucleophilic aromatic substitution & Elimination reactions

Prepared By

Dr. Khalid Ahmad Shadid

Islamic University in MadinahDepartment of Chemistry

Nucleophilic aromatic substitution reactions

Electrophilic substitution reaction generally occur in an aromatic compounds. Aryl halides are less reactive in Nucleophilic substitution reaction due to:

high electron density in benzene ring.

bond in C-X stronger and shorter

Aryl cation unstable therefore no SN1

There is no transition state with same plane of the ring C-Br hence no SN2

Nucleophilic aromatic substitutions reaction occur in Addition-Elimination reaction.

The electron withdrawing group EWG in ortho and para position to hydrogen will stabilize carbanion ion.

No reaction without EWGs.

Chlorobenzen will never react with sodiumethoxide, but it will react with EWG like notro.

Nucleophilic aromatic substitution reactions

Nucleophilic aromatic substitution reactions Another example; the substitution reaction of chlorine by hydroxyl. The

reaction temperature decrease when number of EWG increase

If EWG in meta position, the reaction will not give a product

Benzyne mechanism

The aromatic halides like chlorobenzene and bromobenzene are not react with nucleophiles in normal condition, but will react while benzyne intermediate form.

Benzyne intermediate occur in:

1. Dieles-Alder reaction

2. When there is no alpha hydrogen in reactant.

3. In isotops labeling

Elimination Reactions

Elimination reaction: to eliminate two atoms, two groups, or one atom and one group without substituted with another atom or group.

The elimination of HX molecule from alkyl derivatives. While X is a halogen or ester… etc. the hydrogen atom on adjacent carbon with X

Elimination reactions and nucleophilic substitution are similar in cases of affecting factors.

Hence it’s a competitive reaction which produce alkenes (β-elimination)

Elimination Reactions

α- elimination: elimination of groups from one carbon and produce carbene

First step: formation of carbocation

الكاربوكاتيونات  <   -3C-  >  )CH3(2CH(CH3 )فعالية

CH3CH2-  >  CH3-

UNIMOLECULAR ELIMINATION REACTIONS E1

In this reaction the substrate will determine the rate of reaction

Substrate K = rate of reaction

Mechanism:

Second step : Lose of proton to form double bond

Double bond form When a proton near a positive charge (by elimination of proton)

Due of carbocation formation in this type of reaction SN1 reaction will form also.

UNIMOLECULAR ELIMINATION REACTIONS E1

Reaction of tertiary butyl bromide with alkoxide ion to form prppene:

1st step: (rate determining step)

C-X cleavage due to a good leaving halide group to form carbocation

2nd step: (fast step)

A proton elimination with a strong base to firm alkene

UNIMOLECULAR ELIMINATION REACTIONS E1

UNIMOLECULAR ELIMINATION REACTIONS E1 E1 and SN1 are similar in reaction: happen in an ionized solvent and good leaving group.

2-chloro-2-metheylbutane to give different alkenes

UNIMOLECULAR ELIMINATION REACTIONS E1

Formation more stable carbocation Intermediate carbocation of E1 and SN1 can rearrange to more stable intermediate

Example: solvolysis of neopentyl iodide to form 2-methyl-2-butane. Happen when methyl group migrate, hence carbocation intermediate converted from primary to tertiary more stable

Carbocation intermediate rearranged by migration of Hydrogen

UNIMOLECULAR ELIMINATION REACTIONS E1

BIMOLECULAR ELIMINATION REACTIONS E2

In this reaction the substrate and nucleophile will determine the rate of reaction

Elimination of bimolecule in the same time in one step

Happen when adjacent proton to leaving group. Base will eliminate proton and C-X cleavage by leaving group then formation of a double bond

E2 and SN2 are competitive. When base is Nucleophile.

To reduce competitive and to increase E2 we use non nucleophilic base

The condition of SN2 always will form elimination.

The reaction of 2-bromopropane with sodium ethoxide in ethanol. The elimination rate depend on both substrate and nucleophile. Then its second order reaction.

BIMOLECULAR ELIMINATION REACTIONS E2 Kinetic and mechanism

Structural effects

E2 depend on a good leaving group like halides, ammonium ions, sulphoniume. Like SN2 .

E2 prefers Strong base .

SN2 prefers weak base I - , (except for nonpolar and aprotic)

BIMOLECULAR ELIMINATION REACTIONS E2

For Alkyl groups:

1. C-H single bond in beta position of Leaving group.

2. E2 easily happen primary R< secondary R< tertiary R

3. E2 can react fast with tertiary alkyl not like SN2 due to steric hindrance.

4. E2 reaction is fast because there is no steric hindrance unless base molecule is big.

BIMOLECULAR ELIMINATION REACTIONS E2

BIMOLECULAR ELIMINATION REACTIONS E2

Structural effects H-X elimination from alkene halides or Arene halides (both strong

bonds) are less reactive than alkyl halides. This can happen in a few conditions like alkene preparation.

E2 can be favored over SN2 by:

1. strong base nucleophile

2. Big nucleophile

3. Increase alkyl substitution on alpha carbon

4. Increase temperature High temperature without solvent

Stereochemistry of E2 Reactions

E2 is stereoselective

The Competition between Elimination and Substitution

SN2 and E2

favored over SN1 and E1 by a strong base/Nu

SN2 is slowed by steric hindrance, but E2 is not

strong base, strong Nu

strong base means E2, not SN1

SN2 and E2

Stronger bases favor E2 over SN2

stronger base

weaker base

SN2 and E2

higher temperatures favor elimination

G = H - TS

SN2

weaker basesless steric hindrancelower temperature

E2

stronger basesmore steric hindrancehigher temperature

SN1 and E1

favored over SN2/E2 by absence of strong base/Nuoften neutral or acidic conditions

tertiary or secondary substrates in polar solvents

SN1 is usually major, but some E1 always occurs also

Methyl Substrates: CH3L

SN2 only

Primary Substrates: RCH2L

good for SN2 with almost any nucleophile

no SN1/E1

can cause E2 with a sterically hindered strong base

potassium tert-butoxide (KOt-Bu)

Secondary Substrates: R2CHL

SN2 favored with good Nu that is not too basic(especially in aprotic solvents)

CH3CO2–, RCO2

–, CN –, RS –

E2 favored with strong bases

HO –, RO – (NaOH, NaOEt)

SN1 favored by absence of good Nu in polar solvent

often neutral or acidic conditionssome E1 product is usually formed

a solvolysis reaction

Tertiary Substrates

no SN2 (too hindered)

E2 favored with strong bases

HO –, RO – (NaOH, NaOEt)

SN1 favored by absence of good Nu in polar solvent

often neutral or acidic conditionssome E1 occurs

Hofmann’s rule: The major alkene product has fewer alkyl groups bonded to the carbons of the double bond (the less highly substituted product).

The effect of directing in Elimination reactionsHofmann’s and Zaitsev’s Rule

unsimilar alkyls on alkyl halides like 2-chloro-2-methylbutane, can form one alkene or more. Depending on the relativity rate of beta elimination

The use of HO - or NH2 - will form more stable alkene which contain less number of Hydrogen and more number of alkyl groups bonded to double bond carbon alkene is Zaitsev’s product.

The other product which contain more number of hydrogen is Hofmann’s product

Zaitsev’s Rule: The major alkene product is the one with more alkyl groups on the carbons of the double bond (the more highly substituted product).

Change of Base in this reaction will change yields

Base + (CH3)2CBrCH2CH3 ..........>  (CH3)2C=CHCH3    +    H2C=C(CH3)CH2CH3

I II

EtO- 70% and 30%; Me3CO- 28% and 72% , Et3CO- 12% and 88%

The effect of directing in Elimination reactionsHofmann’s and Zaitsev’s Rule

Hofmann’s and Zaitsev’s products will vary and depends on:

1. How easily of proton elimination from two adjacent beta carbons near leaving group

2. the stability of olefins produces

3. Effect of strain on replacing Leaving group

4. How base is big in elimination rxn

base

CH3CH2CH(S+Me2)CH3 ..........> CH3CH=CHCH3 (26%) + CH3CH2CH=CH2

CH3CH(N+Me3)CH2CH2CH3 ......> CH2=CHCH2CH2CH3 (Major) + CH3CH=CHCH2CH3 minor)

When substrate is a an ammonium salts, sulfur or quaternary phosphonium, will produce less substituted alkene (Hofmann)

Steric hindrance on alkyl halihes will prevent proton elimination hence the products is Zaitsev’s products which contain many substituted groups. The product with less substituted is more preferred (Hofmann)

Elimination and substitution reaction with Alcohols and ethers occur only in a strong acids.

Alkenes preparation from alcohols by E1and E2 reactions will depend on alcohol, acid, solvent and temperature.

Elimination Reactions with acidic catalyst

Elimination Reactions with acidic catalyst

Reaction Tertiary butyl alcohol in E1:

1st step: reversible and fast addition of proton to hydroxyl to make it a good leaving group

2nd step: C-O cleavage and H2O as a good leaving group to form carbocation. Rate determining step.

3rd convert carbocation to alkene by eliminate proton using water

GOOD LUCK