SN1 Reactions
t-Butyl bromide undergoes solvolysis when boiled in methanol:
Solvolysis: “cleavage by solvent” nucleophilic substitution reaction in
which the solvent serves as the nucleophile
CH3CCH3
CH3
Br
+ CH3OH
CH3CCH3
CH3
OCH3
+ HBr
SN1 Reactions
The reaction between t-BuBr and methanol does NOT occur via an SN2 mechanism because:
t-BuBr: tertiary alkyl halide
too hindered to be SN2 substrate
CH3OH: weak nucleophile
Solvolysis reactions occur via an SN1 mechanism:
SN1 Reactions
SN1 Reactions substitution nucleophilic unimolecular
Rate = k[R-X]1st order overall
1st order in [R-X]zero order in [Nuc]
Only R-X is present in the transition state for the rate determining stepNucleophile is NOT present in RDS
SN1 Reactions
General Mechanism:
Step 1: Formation of carbocation is the RDS.
Step 2: Attack of the nucleophile
R C
R"
R'
X R C
R"
R'
X
+ NucR C
R"
R'
R C
R"
R'
Nuc
SN1 Reactions
Reaction Energy Diagram for SN1 Reactions:
Formation of a carbocation is highly endothermic.According to Hammond’s Postulate, the transition state most closely resembles the carbocation.
SN1 Reactions
The reactivity of a substrate in an SN1 reaction depends on the stability of the carbocation formed:
3o > 2o > 1o > methyl
Allylic and benzylic halides often undergo SN1 reactions because the resulting carbocations are resonance stabilized.
CH
CH
H CH2 Br
HC
HC
H
HC
HC
H
CH2 Br
CH
HH
CH
CH
C H
H
CH
CH
H CH2 Br
HC
HC
H
HC
HC
H
CH2 Br
CH
HH
CH
CH
C H
H
+
+
SN1 Reactions
SN1 reactions involve: weak nucleophile
H2O not OH-
CH3OH not CH3O-
Substrates that form stable carbocation intermediates:3o, benzylic, or allylic halide are most favored
2o (sometimes)
SN1 Reactions
Example: Draw the mechanism for the reaction of t-butyl bromide with methanol.
SN1 Reactions
SN1 Reactions-Stereochemistry
The carbocation ion intermediate formed during an SN1 reaction is sp2 hybridized and planar. The nucleophile can attack from either side of
the carbocation.A mixture of both possible enantiomers forms.
SN1 reactions occur with racemization: a process that gives both enantiomers (not
necessarily in equal amounts) of the productRacemization occurs because both retention and inversion of configuration take place.
SN1 Reactions-Stereochemistry
CH3
C
OCH2CH
3
CH(CH3)
2
CH2CH
3
CH3
C
CH(CH3)
2
CH2CH
3
OCH2CH
3
CCH
2CH
3
CH(CH3)
2CH3 C
CH2CH
3
CH(CH3)
2CH3
+
C
CH3CH
2OH
CH2CH
3
CH(CH3)
2CH3
C
CH3CH
2OH
CH2CH
3
CH(CH3)
2CH3
Attack from top
Attack from bottom
- H+
- H+
SN1 Reactions-Stereochemistry
When the nucleophile attacks from the side where the leaving group was originally, retention of configuration occurs.
CH3
CH3
CH3
C
CH(CH3)
2
CH2CH
3
C
Br
CH(CH3)
2
CH2CH
3
C
CH2CH
3
CH(CH3)
2
OCH2CH
3
OCH2CH
3
CH3
CH3
CH3
C
CH(CH3)
2
CH2CH
3
C
Br
CH(CH3)
2
CH2CH
3
C
CH2CH
3
CH(CH3)
2
OCH2CH
3
OCH2CH
3
NaOCH2CH3
Attack from top
(R)(R)
SN1 Reactions-Stereochemistry
When the nucleophile attacks from the back side (opposite to the original leaving group), inversion of configuration occurs.
CH3
CH3
CH3
C
CH(CH3)
2
CH2CH
3
C
Br
CH(CH3)
2
CH2CH
3
C
CH2CH
3
CH(CH3)
2
OCH2CH
3
OCH2CH
3
CH3
CH3
CH3
C
CH(CH3)
2
CH2CH
3
C
Br
CH(CH3)
2
CH2CH
3
C
CH2CH
3
CH(CH3)
2
OCH2CH
3
OCH2CH
3
(R) (S)
NaOCH2CH3
Attack from bottom
SN1 Reactions-Stereochemistry
For most SN1 reactions, the leaving group partially blocks the front side of the carbonium ion more inversion of configuration less retention of configuration
SN1 Reactions-Rearrangements
Carbocations often undergo rearrangements, forming more stable cations. Structural changes resulting in a new
bonding sequence within the molecule
The driving force for a rearrangement is the formation of a more stable intermediate. 1o or 2o carbocation rearranges to a
more stable 3o carbocation or resonance-stabilized carbocation
SN1 Reactions-Rearrangements
A mixture of products often forms as a result of rearrangements during SN1 reactions. NOTE: Rearrangements cannot occur
during SN2 reactions since an intermediate is not formed.
CH3CHCHCH
3
CH3CH
2CCH
3
CH3CHCHCH
3
Br
CH3
CH3
CH3
OCH2CH
3
OCH2CH
3CH3CHCHCH
3
CH3CH
2CCH
3
CH3CHCHCH
3
Br
CH3
CH3
CH3
OCH2CH
3
OCH2CH
3
CH3CHCHCH
3
CH3CH
2CCH
3
CH3CHCHCH
3
Br
CH3
CH3
CH3
OCH2CH
3
OCH2CH
3
+CH3CH2OH
Rearranged productRearrangement occurs via
hydride shift.
SN1 Reactions-Rearrangements
Common rearrangements: Hydride shift (~H)
the movement of a hydrogen atom and its bonding pair of electrons
Methyl shift (~CH3)the movement of a methyl group and its bonding pair of electrons
Alkyl shift (~R)the movement of any alkyl group and its bonding pair of electrons
SN1 Reactions-Rearrangements
Hydride Shift Mechanism: Step 1: Formation of carbocation and
rearrangement:
CH3CHCCH
3 CH3CH
2CCH
3
CH3CHCHCH
3
CH3 CH
3
CH3
Br
CH3 CH
3
CH3
OCH2CH
3OCH
2CH
3H
C C CH3
Br H
H CH3
C
H
C CH3
H
CH3
C
H
C CH3
CH3
H
CH3CHCCH
3 CH3CH
2CCH
3
CH3CHCHCH
3
CH3 CH
3
CH3
Br
CH3 CH
3
CH3
OCH2CH
3OCH
2CH
3H
C C CH3
Br H
H CH3
C
H
C CH3
H
CH3
C
H
C CH3
CH3
H
+
CH3CHCCH
3 CH3CH
2CCH
3
CH3CHCHCH
3
CH3 CH
3
CH3
Br
CH3 CH
3
CH3
OCH2CH
3OCH
2CH
3H
C C CH3
Br H
H CH3
C
H
C CH3
H
CH3
C
H
C CH3
CH3
H+
+ Br -
~H2o
3o
SN1 Reactions-Rearrangements
Hydride Shift Mechanism: Step 2: Nucleophile attack and loss of
proton (if needed)
CH3CHCCH
3 CH3CH
2CCH
3
CH3CHCHCH
3
CH3 CH
3
CH3
Br
CH3 CH
3
CH3
OCH2CH
3OCH
2CH
3H
C C CH3
Br H
H CH3
C
H
C CH3
H
CH3
C
H
C CH3
CH3
H OCH2CH
3
H
CH3CHCCH
3 CH3CH
2CCH
3
CH3CHCHCH
3
CH3 CH
3
CH3
CH3
CH3
Br
CH3 CH
3
CH3
OCH2CH
3OCH
2CH
3H
C C CH3
Br H
H CH3
C
H
C CH3
H
CH3
C
H
C CH3
CH3
H OCH2CH
3
H
C
H
H
C CH3
CH3
OCH2CH
3
H
C
H
H
C CH3
CH3
OCH2CH
3
CH3CHCCH
3 CH3CH
2CCH
3
CH3CHCHCH
3
CH3 CH
3
CH3
CH3
CH3
Br
CH3 CH
3
CH3
OCH2CH
3OCH
2CH
3H
C C CH3
Br H
H CH3
C
H
C CH3
H
CH3
C
H
C CH3
CH3
H
C
H
H
C CH3
CH3
OCH2CH
3
H
C
H
H
C CH3
CH3
OCH2CH
3
+ CH3CH2OH+
CH3CH2OHCH3CH2OH2
+ +
SN1 Reactions-Rearrangements
Example of a Methyl Shift (~CH3):
CH3CCH2Br
CH3
CH3
EtOH
CH3CCH2OCH2CH3
CH3
CH3
CH3CCH2CH3
OCH2CH3
CH3EtOH
CH3CCH2Br
CH3
CH3
SN1 Reactions-Rearrangements
Mechanism of ~CH3: Step 1: Simultaneous (because primary
carbocation is unstable) shift of methyl group and loss of leaving group:
3o carbocation formed preferentially
CH3C
CH3
CH3
C
H
H
Br~CH3 CH3CCH2CH3
CH3
+ Br-
SN1 Reactions-Rearrangements
Mechanism of ~CH3: Step 2: Attack of nucleophile and loss
of proton (if needed)
CH3CCH2CH3
CH3 CH3CH2OHCH3CCH2CH3
CH3
CH3CH2OH
CH3CH2OHCH3CCH2CH3
CH3
OCH2CH3
CH3CH2OH2+ +
Important! Important! Important!
How do you know if the carbocation forms first and then rearrangement occurs or if formation of carbocation and rearrangement occur simulataneously???
In general: Secondary (2o) halides form the carbocation first and then rearrangement occurs (i.e. 2 steps)
In general: Primary halides undergo simultaneous formation of carbocation and rearrangement. (Primary carbocation is quite unstable!)
SN1 Reactions-Rearrangements
Example: Propose a mechanism for the following reaction.
CH3
Cl
CH3
OCH2CH
3
CH3
OCH2CH
3
CH3
Cl
CH3
OCH2CH
3
CH3
OCH2CH
3
CH3
Cl
CH3
OCH2CH
3
CH3
OCH2CH
3
CH2CH3OH
+
SN1 vs. SN2
SN2
Strong nucleophile
Primary or methyl halide
Polar aprotic solvents (acetone, CH3CN, DMF)
Inversion at chiral carbon
No rearrangements
Weak nucleophile (may also be solvent)
Tertiary,allylic, benzylic halides
Polar protic solvent (alcohols, water)
Racemization of optically active compound
Rearranged products
SN1
E1 Reactions
An elimination reaction involves the loss of two atoms or groups of atoms from a substrate, usually forming a new bond.
Elimination reactions can occur via a first order (E1) or a second order (E2) process.
H
C
C C
H
CH3 Br
CH2CH
3
CH2CH
3
CH
3C
H CH2CH
3
CH2CH
3
H
C
C C
H
CH3 Br
CH2CH
3
CH2CH
3
CH
3C
H CH2CH
3
CH2CH
3
Na+ -OCH3
CH3OH
+ Br -
E1 Reactions
E1 reactions: Elimination, unimolecular
1st order kineticsRate = k[R-X]RDS transition state involves a single molecule
General conditions:3o and 2o halidesweak bases
E1 ReactionsE1 Mechanism:Step 1: Formation of carbocation (RDS)
Step 2: Base abstracts proton
C C
R""
X
R'
R
R"
H
C C
R""
R'
R
R"
H
X
C C
R""
R'
R
R"
H
B C C
R""
R
R"
R'H B
cis and trans
E1 Reactions
E1 reactions almost always occur together with SN1 reactions.
CH3CCH3
CH3
Br
EtOH H2C C
CH3
CH3
(E1)
+
CH3CCH3
CH3
OCH2CH3
(SN1)
E1 Reactions
CH3CH2-O-H
H
+
E1 Reactions
Once formed, a carbonium ion can: recombine with the leaving group react with a nucleophile forming a
substitution product (SN1) lose a proton to form an alkene (E1) rearrange to form a more stable
carbocation and then:react with nucleophilelose a proton to form an alkene
E2 Reactions
E2 reactions: Elimination, bimolecular
2nd order kineticsRate = k[R-X][B-]RDS transition state involves two molecules
General conditions:3o and 2o halidesstrong bases
E2 Reactions In the presence of a strong base,
elimination generally occurs in a concerted reaction via an E2 mechanism
B -
E2 Reactions
SN2 reactions require an unhindered methyl or 1o halide steric hinderance prevents nucleophile
from attacking 3o halides and forming the substitution product
E2 reactions generally involve the reaction between a 3o and 2o alkyl halides and a strong base.
E2 Reactions The reaction of t-butyl bromide with methoxide
ion gives only the elimination product.
The base attacks the alkyl bromide much faster than the bromide can ionize.
E2 Reactions
Many alkyl halides can eliminate in more than one way. Mixture of alkenes produced
E2 Reactions
Saytzeff Rule: When two or more elimination products
can be formed, the product with the most highly substituted double bond will usually predominate.
R2C=CR2 > R2C=CHR > RHC=CHR and R2C=CH2 > RHC=CH2
E2 Reactions
Example: Draw the structures for all possible products of the following reaction. Which one will predominate?
Br CH3
NaOCH2CH3
EtOH
E2 Reactions
E2 reactions follow a concerted mechanism: bonds breaking and forming
simultaneously
specific geometry required to allow overlap of orbitals of bonds being broken and bonds being formed
E2 reactions commonly involve an anti-coplanar conformation.
E2 Reactions
E2 Reactions
Example: Predict the structure of the elimination product formed by the following reaction.
C C
HBr
H
CH3
PhPh
NaOCH3
CH3OH
C
C
Ph =
C
H
CH3
PhBr
HPh
CPh
H
CH3
Ph
E1 vs E2
E1
Weak base 30 > 2o Good ionizing
solvent polar, protic
(water, alcohols)
Saytzeff product No required
geometry
Rearranged products possible
Strong base required
3o > 2o
Solvent polarity not important
Saytzeff product Coplanar leaving
groups (usually anti) No rearrangements
E2