Transcript Physical Organic Chemistry

Islamic University in Madinah
Department of Chemistry
Physical Organic Chemistry
CH-4
Nucleophilic aromatic substitution &
Elimination reactions
Prepared By
Dr. Khalid Ahmad Shadid
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 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.
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
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- > (CH3)2CH- > CH3CH2- > CH3-‫فعالية الكاربوكاتيونات‬
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.
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
 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
UNIMOLECULAR ELIMINATION REACTIONS E1
 Carbocation intermediate rearranged by migration of Hydrogen
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
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.
BIMOLECULAR ELIMINATION REACTIONS E2
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
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 - TS
S N2
weaker bases
less steric hindrance
lower temperature
E2
stronger bases
more steric hindrance
higher temperature
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
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 conditions
some 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 conditions
some E1 occurs
The effect of directing in Elimination reactions
Hofmann’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

Hofmann’s rule:
The major alkene product has
fewer alkyl groups bonded to the carbons of the
double bond (the less highly substituted 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
The effect of directing in Elimination reactions
Hofmann’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
base
+
(CH3)2CBrCH2CH3 ..........> (CH3)2C=CHCH3
I
+
H2C=C(CH3)CH2CH
II
EtO- 70% and 30%; Me3CO- 28% and 72% , Et3CO- 12% and 88%
3
 When substrate is a an ammonium salts, sulfur or quaternary phosphonium, will
produce less substituted alkene (Hofmann)
CH3CH2CH(S+Me2)CH3 ..........>
CH3CH=CHCH3 (26%) + CH3CH2CH=CH2
CH3CH(N+Me3)CH2CH2CH3 ......> CH2=CHCH2CH2CH3 (Major) + CH3CH=CHCH2CH3 minor)
 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 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.
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