Stereochemistry, SN1 at a chiral center.
R
R
ionization
R
MeOH
X
OMe
- XR'
R"
chiral halide
optically active
R
MeO
-H +
R' R"
achiral
carbocation
optically inactive
R'
R"
R'R"
newly generated chiral centers
racemic mixture
optically inactive
racemization
Frequent complication: the Leaving Group will tend to block approach of the
nucleophile leading to more inversion than retention for the SN1
Stereochemistry SN2, Inversion at a Chiral Center
R3
R3
Y-
X
+ X-
Y
R2
R1
Examples
Inversion, frequently (but
not always) the R,S
designator changes
R2
R1
H
CH3O -
H
Br
+ Br
H3CO
H3C
C2H5
(S)-2-bromobutane
Here is the
inversion motion!
CH3
C2H5
(R)-2-methoxybutane
CH3
CH3
H
Br
C2H5
CH3O
H
C2H5
-
Another Example
CH3
CH3
Br
H
H3C
H
NaOCH3
SN2
H
The chiral center
will undergo
inversion.
OCH3
H3C
H
C2H5
C2H5
The non-reacting
chiral C will not
change.
Williamson ether synthesis
CH3
H3C
NaOCH3
H
H
Br
C2H5
SN2
CH3
H3C
H
CH3O
H
C2H5
How to understand the configurations: simply replace the Br with the OCH3
(retention). Now swap any two substituents (here done with H and OCH3) on the
reacting carbon to get the other configuration (inversion). Done.
Stereochemistry, SN2
Two things happening here:
1)Substitution of iodide, 127I, with labeled iodide,
131I.
2) Change in stereochemistry
Substitution
Recall iodide a good nucleophile, acetone an aprotic
solvent resulting in highly reactive iodide ion.
SN2 Stereochemistry: Inversion
Inverted configuration
Comparison of SN1 and SN2 mechanisms.
Substitution vs. Loss of Optical Activity
Stereochemistry: RI represents the R configuration of the alkyl iodide;
RI represents the S configuration.
Substitution: I is the normal 127I isotope; I is the tagged 131I iodine isotope.
If racemization: “SN1”
IRI RI RI RI RI
RI RI RI RI RI
RI RI RI RI RI
Only 20%
reacts
100% optically pure
If inversion: SN2
RI RI RI RI RI
RI RI RI RI RI
IOnly 20%
reacts
RI RI RI RI RI
20% substituted, 20% racemized,
20 % of optical purity lost (80%
optically pure). Rate of Loss of
optical activity = Rate of
substitution.
RI RI RI RI RI
RI RI RI RI RI
20% substituted, 40%
racemized, 40% optical purity
lost (60% optically pure). Rate
of loss of optical activity = 2 x
Rate of substitution.
Effect of Structure of the Haloalkane on Rates
Recall
SN1
Stability of resulting
carbocation, hyperconjugation
Ease of ionization
CH3X
CH3CH2X
Methyl
primary
(CH3)2CHX
secondary
Rate of SN1 Reactions
(CH3)3CX
tertiary
Now for SN2
SN2
Steric Hinderance, difficulty of
approach for nucleophile
CH3X
CH3CH2X
Methyl
primary
(CH3)2CHX
secondary
(CH3)3CX
tertiary
Rate of Reactions
Summary:
Methyl, primary use SN2 mechanism due to steric ease.
Tertiary uses SN1 mechanism due to stability of carbocations
Secondary utilizes SN1 and/or SN2 – depending on solvent and
nucleophile.
Recall: Resonance Stabilization of Carbocations
Allylic and benzylic carbocations are stabilized by resonance.
SN2
Both
SN1
Leaving Group
Recall that the leaving group becomes more negative.
X
R-X
Generally, the best leaving groups are groups that can stabilize
that negative charge: weak bases; conjugate bases of strong
acids.
X
H-X
Base Strength
Br
Example:
R-OH
NR
H
R-OH
Br
R-OH2
R-Br
Solvents
Polar solvents stabilize ions, better stabilization if the charge is compact.
Polar Protic solvents stabilize both anions (nucleophiles) and cations
(carbocations). Accelerate SN1 reactions where charge is generated in
the Rate Determining Step.
R-X  [R + --- X -]  R + + X -
Stabilized.
Stabilized.
Polar aprotic solvents usually stabilize cations more effectively than
anions (nucleophiles). Anions (nucleophiles) are left highly reactive.
Accelerates SN2 reactions where an anion (nucleophile) is a reactant.
Nuc - + R-X  [Nuc----R----X] -  Nuc-R + X Not Stabilized.
Note that it is the energy of the transition state relative to the reactant
which affects the rate of the forward reaction (but not the equilibrium).
Rearrangements for SN1: 1,2 Shift
Recall carbocations can rearrange (1,2 shift) to yield a more stable
carbocation. Occurs in SN1 – but not SN2 – reactions.
Initial Ionization in protic solvent.
1,2 shift converting 2o
carbocation to 3o benzylic
Nucleophile attacks.
Deprotonate to to yield ether
Next elimination…
Return to b elimination: competes
with nucleophilic substitution.
The competition:
SN1
and/or
SN2
Zaitsev Rule, prefer to form the more substituted alkene (more stable).
Mechanistic Possibilities to eliminate the H+ and X-
H
+
H
+
X
X
Possible Sequences for bond making/breaking…
•
Regard the alkyl halide as an acid. First remove H+ producing a
carbanion , then in a second step remove X- producing the alkene.
or
•
First remove X- producing a carbocation, then in a second step
remove H+ yielding the alkene. E1
or
•
Remove H and X in one step to yield the alkene.
E2
• There are two idealized mechanisms for belimination reactions
• E1 mechanism: at one extreme, breaking of the RLv bond to give a carbocation is complete before
reaction with base to break the C-H bond
– only R-Lv is involved in the rate-determining step (as in
SN1)
• E2 mechanism: at the other extreme, breaking of
the R-Lv and C-H bonds is concerted (same time)
– both R-Lv and base are involved in the rate-determining
step (as in SN2)
E1 Mechanism
– ionization of C-Br gives a carbocation intermediate
CH3
CH3 -C-CH3
s low, rate
determinin g
Br
CH3
CH3 -C-CH3 +
+
Br
(A carb ocation
in termediate)
– proton loss from the carbocation intermediate to a
base (for example, the solvent) gives the alkene
H
H3 C
CH3
O: + H-CH2 -C-CH3
+
fast
H
+
CH3
O H + CH2 =C-CH3
H3 C
Energy Profile for E1
mechanism, carbocations.
Rate Determining
Step; formation of
the carbocation.
Alkyl Halide
 (E1)
Alkyl Halide
 (Addition)
Reaction can
occur in either
direction…..
Alkene + HX
Alkene + HX
E2 Mechanism
breaking of the R-Lv and C-H bonds is concerted
CH3 O:
-
CH3 O
H
H
C
C
Lv
-H an d -Lv are anti and cop lanar
(dih edral an gle 180°)
Needs Strong Base
C
C
Lv
E2
Kinetics of E1 and E2
• E1 mechanism
– reaction occurs in two steps
– the rate-determining step is carbocation formation
involving only RLv
– the reaction is 1st order in RLv and zero order in base
Rate =
d[RLv]
= k[ RLv]
dt
• E2 mechanism
– reaction occurs in one step involving both RLv and
the base.
– reaction is 2nd order; first order in RLv and 1st order
in base
d[RLv]
Rate =
dt
= k[ RLv][ Base]
Regioselectivity of E1/E2
• E1: major product is the more stable alkene (more
substituted, more resonance)
• E2: with strong base, the major product is the more
stable (more substituted, more resonance) alkene
Special notes about sterically hindered bases such as tert-butoxide,
(CH3)3CO -.
E2 – anti-Zaitsev: with a strong, sterically hindered base the major product is
often the less stable (less substituted) alkene. Reason: hydrogens on less
substituted carbons are more accessible.
Also
E2 vs SN2: In competition of SN2 vs E2: steric bulk in either the alkyl halide
or the base/nucleophile prevents the SN2 reaction and favors the E2.
Stereochemistry of E2
• E2 is most favorable (lowest activation energy)
when H and Lv are anti and coplanar
CH3 O:
-
D
H
C
E
C
A
Lv
B
-H an d -Lv are anti and cop lanar
(dih edral an gle 180°)
CH3 O
H
A
B
C
D
E
C
Lv
Examples of E2 Stereochemistry
Explain both regioselectivity and relative rates of reaction.
CH3O -
Cl
cis
Faster
reaction
+
Major product.
Zaitsev product
But
CH3O -
Cl
trans
Slower
reaction
+
Only product
Anti-Zaitsev
Principles to be used in analysis
Stereochemical requirement: anti conformation for departing groups. This
means that both must be axial.
Dominant conformation: ring flipping between two chair conformations,
dominant conformation will be with iso propyl equatorial.
First the cis isomer.
Reactive Conformation; H
and Cl are anti to each other
-
CH3 O:
H
2
H
E2
6
H
+ CH3 OH + :Cl
H
1
Cl
In order for the H and the
Cl to be anti, both must be
in axial positions
1-Isopropylcycloh exene
Iso-propyl groups is in more stable
equatorial position. Dominant
conformation is reactive conformation.
Now the trans
In the more stable chair of the trans isomer, there is no H anti and
coplanar with Lv, but there is one in the less stable chair
HH
HH
66
HH
Cl
Cl
11
HH
Cl
Cl
22
HH Unreactive
conformation
More
s
table
chair
More s table chair
(no
(noHHisisanti
antian
andd
coplanar
coplanartotoCl)
Cl)
66
11
22
HH
Reactive but only with
the H on C 6
HH
HH
Less
Lessstable
stablech
chair
air
(H
(Hon
oncarb
carbon
on66isis
an
antitiand
andcop
coplan
lanar
artotoCl)
Cl)
Most of the compound exists in the
unreactive conformation. Slow reaction.
Cl
H
-
CH3 O:
6
H
2
1
H
E2
+ CH3 OH + Cl
H
(R)-3-Is op ropylcyclohexene
Anti Zaitsev
Example, Predict Product
Ph
base
H3C
Ph
H3C
Ph
CH3
H
H3C
Br
Ph
Problem!: Fischer projection diagram represents an eclipsed structure.
Task: convert to a staggered structure wherein H and Br are anti and predict
product. We will convert to a Newman and see what we get…
Ph
H
rotate upper
chiral C by 180 Ph
Ph
CH3
H
CH3
H3C
H
Br
H3C
Br
=
H3C
Br
Ph
H3C
rotate 120
further
Ph
CH3
H
Ph
H3C
Br
Ph
H & Br not anti
yet!
CH3
Br
H3C
H
Ph
H3C
Ph
H3C
Ph
Ph
Ph
H3C
base
H3C
Br
H
Ph
Ph
Now anti and
we can see
where the pi
bond will be.
Alternative Approach: CAR
Ph
base
H3C
Ph
H3C
Ph
CH3
H
H3C
Br
Ph
The H and Br
will be
leaving: just
indicate by
disks.
Anti
Geometry
A
CR
Meso or
Racemic??
This may be recognized
as one of the
enantiomers of the
racemic mixture.
A
C < -- > R
Relationship
works in both
directions.
Should get cis
isomer.
Note: As we have said
before it may take some
work to characterize a
compound as “racemic” or
“meso”.
E1 or E2
Alkyl halide
Primary
E1 (Carbocation)
E2
E2 is favored.
RCH2 X
E1 does not occur.
Primary carbocations are
so unstable, they are never
observed in solution.
Secondary
R2 CHX
Main reaction with weak
bases such as H 2O, ROH.
Main reaction with strong
bases such as OH - and OR -.
Tertiary
R3 CX
Main reaction with weak
bases such as H 2O, ROH.
Main reaction with strong
bases such as OH - and OR -.
1o good nucleophiles, aprotic solvents
1o
2o good nucleophiles but also poor bases, 2o lower hinderance, better nucleophile than
aprotic solvents
base
30
3o lower hinderance, better nucleophile than
R-Nuc
base
SN1
R-Nuc SN2
good
nucleophile
R-X
weak nucleophile
ionization
Rearrange ?
R+ + X-
1o,
strong
base
2o, 3o polar solvents,
weak nucleophiles,
weak bases
alkene
E2
weak base
alkene
1o strong, bulky bases
E1
2o strong bases
1o
3o strong bases
2o heat, more hindered
3o heat, more hindered
Recall Halohydrins and Epoxides
Cl2, H2O
Cl
base
OH
Cl
H2O
Cl
Creation of Nucleophile
Internal SN2 reaction
with inversion
O
RO
ROH
H
O
Creation of
good
leaving
group.
O
H
Attack by poor
nucleophile
OH
Neighboring Group Effect
• Mustard gases
– contain either S-C-C-X or N-C-C-X
N
S
Cl
Cl
Bis(2-chloroethyl)sulfide
(a sulfur mustard gas)
Cl
Cl
Bis(2-chloroethyl)methylamine
(a nitrogen mustard gas)
– what is unusual about the mustard gases is that they undergo
hydrolysis rapidly in water, a very poor nucleophile
Cl
S
Cl
+
2 H2 O
HO
S
OH + 2 HCl
– the reason is neighboring group participation
by the adjacent heteroatom
:
slow , rate
d etermining
S
Cl
Cl
an internal
SN 2 reaction
Good nucleophile.
S
Cl
Cl
S
+
Cl
A cyclic
sulfonium ion
:
+
+
+ : O-H
H
fas t
a second
S N 2 reaction
S
Cl
H
+O
H
– proton transfer to “solvent” completes the reaction
From an old quiz
5. Provide a clear, unambiguous mechanism to explain
the following stereochemical results.
Complete structures of intermediates, if any, should be shown.
Use curved arrow notation consistently.
SCH3
H
SCH3
H2O
CH3
H
SCH3
CH3
H3C
H
+
H3C
H3C
H
H
Cl
H
OH
CH3
OH
Here is the crux of the matter: how can the non-reacting carbon change its
configuration??? Further it does not always change but only if configuration of the
reacting carbon changes!! We got a mixture of enantiomers, a racemic mixture.
Something strange is happening!!
Expect sulfur to attack the C-Cl, displacing the Cl and forming a three membered ring.
Like this…
S
H
CH3
CH3
SCH3
S
CH3
H3C
H
H2O
Cl
But we have to be careful with stereochemistry
OH
We have to put the molecule in the correct conformation.
SCH3
SCH3
H
H
H
CH3
=
H3C
CH3
H
=
H3C
H
H3C
CH3
H
Cl
Cl
Cl
SCH3
S and Cl are
eclipsed, not
anti.
Reactive conformation reached by 180 rotation around C-C bond
H
Cl
CH3
H
H
H3C
SCH3
H
H3C
CH3
S
CH3
And then the ring is opened by
attack of water
But let’s pause for a moment. Our reactant was optically active with two chiral carbons.
Recall the problem: If reaction occurs only at the C bearing the Cl the other should
remain chiral! Hmmmm?
But now notice that the intermediate sulfonium ion is achiral. It has a mirror plane of
symmetry. Only optically inactive products will result.
Two modes of attack by water.
OH2
SCH3
H
HO
H
H3C
H
H3C
H
CH3
CH3
H3C
CH3
=
H
S
CH3
H
H
CH3
=
H3C
SCH3
HO
H
SCH3
180 rotation
OH
Enantiomers,
racemic
mixture
And…
OH2
SCH3
H
H
H
H3C
CH3
OH
H
H3C
H
H3CS
S
CH3
=
H3C
CH3
H
H3C
H
=
CH3
H
H3CS
CH3
OH
180 rotation
Again note the ring structure is achiral and that we must,
of course, produce optically inactive product.
OH