Physical Organic Chemistry CH-4 Nucleophilic aromatic substitution

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Physical Organic ChemistryCH-4 Nucleophilic
aromatic substitution Elimination reactions
Islamic University in Madinah Department of
Chemistry

• Prepared By
• Dr. Khalid Ahmad Shadid

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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

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Nucleophilic aromatic substitution reactions

• 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.

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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

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Benzyne mechanism

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

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• Benzyne intermediate occur in
• 1. Dieles-Alder reaction
• 2. When there is no alpha hydrogen in reactant.
• 3. In isotops labeling

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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 its a competitive reaction which produce
  alkenes (ß-elimination)

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Elimination Reactions

• a- elimination elimination of groups from one
  carbon and produce carbene

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UNIMOLECULAR ELIMINATION REACTIONS E1

• In this reaction the substrate will determine the
  rate of reaction
• Substrate K rate of reaction
• Mechanism

First step formation of carbocation
Second step Lose of proton to form double bond
?????? ??????????????? (CH3)3C-  gt  (CH3)2CH-  
gt  CH3CH2-  gt  CH3-
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UNIMOLECULAR ELIMINATION REACTIONS E1

• 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.

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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

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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

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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

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UNIMOLECULAR ELIMINATION REACTIONS E1

• Carbocation intermediate rearranged by migration
  of Hydrogen

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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

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BIMOLECULAR ELIMINATION REACTIONS E2 Kinetic
and mechanism

• 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.

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Structural effects
BIMOLECULAR ELIMINATION REACTIONS E2

• 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)

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BIMOLECULAR ELIMINATION REACTIONS E2

• For Alkyl groups
• C-H single bond in beta position of Leaving
  group.
• E2 easily happen primary Rlt secondary Rlt tertiary
  R
• E2 can react fast with tertiary alkyl not like
  SN2 due to steric hindrance.
• E2 reaction is fast because there is no steric
  hindrance unless base molecule is big.

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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
• strong base nucleophile
• Big nucleophile
• Increase alkyl substitution on alpha carbon
• Increase temperature

High temperature without solvent
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Stereochemistry of E2 Reactions

• E2 is stereoselective

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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
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SN2 and E2
Stronger bases favor E2 over SN2
stronger base
weaker base
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SN2 and E2
higher temperatures favor elimination
?G ?H - T?S
SN2
E2
weaker bases less steric hindrance lower
temperature
stronger bases more steric hindrance higher
temperature
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SN1 and E1
favored over SN2/E2 by absence of strong
base/Nu often neutral or acidic conditions
tertiary or secondary substrates in polar solvents
SN1 is usually major, but some E1 always occurs
also
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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)
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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 conditions some E1
product is usually formed
a solvolysis reaction
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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 conditions some E1 occurs
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The effect of directing in Elimination
reactions Hofmanns and Zaitsevs 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 Zaitsevs
  product.
• The other product which contain more number of
  hydrogen is Hofmanns product

Hofmanns rule The major alkene product has
fewer alkyl groups bonded to the carbons of the
double bond (the less highly substituted product).
Zaitsevs Rule The major alkene product is the
one with more alkyl groups on the carbons of the
double bond (the more highly substituted
product).
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• Change of Base in this reaction will change yields

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The effect of directing in Elimination
reactions Hofmanns and Zaitsevs Rule

• Hofmanns and Zaitsevs products will vary and
  depends on
• How easily of proton elimination from two
  adjacent beta carbons near leaving group
• the stability of olefins produces
• Effect of strain on replacing Leaving group
• How base is big in elimination rxn

Base (CH3)2CBrCH2CH3 ..........gt 
(CH3)2CCHCH3        H2CC(CH3)CH2CH3
I                                        II
EtO-   70 and 30 Me3CO-    28 and 72   ,
Et3CO-   12 and 88
base
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• When substrate is a an ammonium salts, sulfur or
  quaternary phosphonium, will produce less
  substituted alkene (Hofmann)

CH3CH2CH(SMe2)CH3 ..........gt
CH3CHCHCH3 (26) CH3CH2CHCH2
CH3CH(NMe3)CH2CH2CH3 ......gt CH2CHCH2CH2CH3 
(Major) CH3CHCHCH2CH3 minor)

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

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Elimination Reactions with acidic catalyst

• 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.

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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

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GOOD LUCK