nucleophilic substitution and -elimination. substitution process nucleophiles have a pair of...
Post on 20-Dec-2015
221 views
TRANSCRIPT
![Page 1: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/1.jpg)
Nucleophilic Substitution and -elimination
![Page 2: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/2.jpg)
Substitution Process
Nucleophiles have a pair of electrons which are used to form a bond to the electrophile.
A Leaving Group departs making room for the incoming nucleophile.
Nucleophiles can also frequently function as Lewis bases.
The electrophile can function as Lewis acid.
Note that the nucleophile converts a lone pair into a bond and becomes more positive by +1
Note that the bond from C to the Leaving Group is collapsed into a lone pair on the Leaving Group which becomes more negative by -1.
![Page 3: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/3.jpg)
-Elimination
H
Lv
base
+ H-Lv
A pi bond is created.
Instead of substitution a base can remove both the leaving group and an adjacent hydrogen creating a pi bond. Recall dehydrohalogenation.
![Page 4: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/4.jpg)
Competition between Nucleophilic Substitution and -elimination.
Br
+ Na+ C2H5O-
H
nucleophilicsubstitution
OEt
+ Br-
Note the change in charges on the nucleophile and the Leaving group
First the nucleophilic substitution. The ethoxide attacks the carbon bearing the bromine.
![Page 5: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/5.jpg)
Now the -elimination.
Br
+ Na+ C2H5O-
H
Now the elimination. The ethoxide (base) attacks the hydrogen on a carbon adjacent to the carbon bearing the Br ().
elimination
+ C2H5OH + Br-
Since we are using Br as the leaving group this could also be called a dehydrohalogenation.
![Page 6: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/6.jpg)
Summary.
![Page 7: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/7.jpg)
Formal Charges and Nucleophilic Substitution
In the free nucleophile the pair of electrons is a lone pair belongs exclusively to the nucleophile.
In the product, it is a bond and shared.
The result is the nucleophile increases its charge by +1
Conversely the leaving group converts a shared pair of electrons (a bond) into unshared electrons (lone pair). The charge of the leaving group becomes more negative by -1.
![Page 8: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/8.jpg)
NH
H
C
H
Br
HH
C
H
Br
HH
H2N
NH
H
C
H
Br
HH
C
H
Br
HH
H3N
H
Negative Nucleophile
Neutral NucleophileOther things being equal, the more basic species will be a better nucleophile. NH2
- is a better nucleophile than NH3
N: from -1 to 0 ; Br: from 0 to -1
N: from 0 to +1; Br: from 0 to -1
Negative Nucleophile, positive leaving group
Br C
H
OH2
HH
C
H
HH
Br OH2
Br: from -1 to 0; O: from +1 to 0
![Page 9: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/9.jpg)
Two Nucleophilic Substitution Mechanisms: SN1 & SN2
SN2 mechanism: substitution, nucleophilic, 2nd order
Backside attack
Hydrogens flip to the other side. Inversion
of configutationExamine important points….
Look at energy profile next…
![Page 10: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/10.jpg)
Energy Profile, SN2
![Page 11: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/11.jpg)
SN1 reaction: substitution, nucleophilic, first order.
Step 1, Ionization,
Rate determining step.
Step 2, Nucleophile reacts with Electrophile.
Note stereochemistry: nucleophile can bond to either side of carbocation. Get both configurations.
Protonated ether.
Now the alternative mechanism: SN1
CH3OH + (CH3)3C-Br CH3OC(CH3)3 + H + + Br-
![Page 12: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/12.jpg)
Step 3, lesser importance, deprotonation of the ether.
Next, energy profile….
![Page 13: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/13.jpg)
Energy Profile of SN1, two steps.Slow step to form carbocation. Rate determining. Examine important points…..
Carbocation, sp2
Fast step to form product.
![Page 14: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/14.jpg)
Kinetics: SN1 vs. SN2SN1, two steps.
SN2, one step.
![Page 15: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/15.jpg)
Effect of Nucleophile on Rate:Structure of Nucleophile
SN1: Rate Determining Step does not involve nucleophile. Choice of Nucleophile: No Effect
SN2: Rate Determining Step involves nucleophile. Choice of nucleophile affects rate.
Note the solvent for this comparison: alcohol/water. Talk about it later…
Frequently, better nucleophiles are stronger bases.
Compare
Compare
But compare the halide ions!!
In aq. solution F – more basic than I -. (HI stronger acid.) But iodide is better nucleophile.
![Page 16: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/16.jpg)
We need to discuss Solvents
Classifications
Polar vs non-polar solvents, quantified by dielectric constant. Polar solvents reduce interaction of positive and negative ions.
Water > EtOH > Acetic acid > hexane
![Page 17: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/17.jpg)
Solvents. Another Classification
Protic vs aprotic solvents. Protic solvents have a (weakly) acidic hydrogen having a positive charge which stabilize anions.
ROH --- Br - --- HOR
Alcohols are protic solvents Aprotic solvents
CH3CN acetone tolueneIncreasing polarity
![Page 18: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/18.jpg)
Role of Solvents
Some solvents can stabilize ions, reducing their reactivity.
Many nucleophiles are ions, anions.
Protic solvents can stabilize anions. Protic solvents have (weakly) acidic hydrogens bearing a positive charge. Anions may be stabilized
Methanol, protic solvent, stabilizing the fluoride ion, reducing its nucleophilicity.
Small, compact anions (like fluoride ion) are especially well stabilized and have reduced nucleophilicity. Iodide ion is large diffuse charge and less stabilization occurs.
![Page 19: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/19.jpg)
Halide ion problem
Iodide ion Bromide ion Chloride ion Fluoride ion
basicity
Protic solvent solvation
nucleophilicity
The problem: basicity and nucleophilicity of the halide ions do not parallel each other in protic solvents.
nucleophilicity
The explanation. Fluoride most stabilized in protic solvents reducing its nucleophilicity.
![Page 20: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/20.jpg)
Summary for Halide Ions
Iodide ion Bromide ion Chloride ion Fluoride ion
Nucleophilicity in aprotic solvents
Protic solvent solvation
Nucleophilicity in protic solvents
basicityBut in aprotic solvents.
Protic solvents.
basicity
Protic solvent solvation
![Page 21: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/21.jpg)
Stereochemistry, SN1 at a chiral center.
R
X
R"R'
ionization
- X -
chiral halideoptically active
R
R"R'achiralcarbocationoptically inactive
MeOH
-H +
R R
R" R"R' R'
OMe MeO
newly generated chiral centersracemic mixtureoptically inactive
Frequent complication: the Leaving Group will tend to block approach of the nucleophile leading to more inversion than retention for the SN1
racemization
![Page 22: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/22.jpg)
Stereochemistry SN2, Inversion at a Chiral Center
R1
R3
R2
XY -
R1
R3
R2
Y + X -
Inversion, frequently (but not always) the R,S designator changes
Examples
C2H5
H
H3C
BrCH3O -
C2H5
H
CH3
H3CO + Br -
(R)-2-methoxybutane(S)-2-bromobutane
CH3
C2H5
H Br
CH3
C2H5
CH3O H
Here is the inversion motion!
![Page 23: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/23.jpg)
Another Example
CH3
Br H
C2H5
H3C H
NaOCH3
CH3
H OCH3
C2H5
H3C HSN2
Williamson ether synthesis
H
CH3
Br
H
C2H5
H3CNaOCH3
SN2 CH3O
CH3
H
H
C2H5
H3C
The chiral center will undergo inversion.
The non-reacting chiral C will not change.
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.
![Page 24: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/24.jpg)
Stereochemistry, SN2
Substitution
Recall iodide a good nucleophile, acetone an aprotic solvent resulting in highly reactive iodide ion.
Inverted configuration
SN2 Stereochemistry: Inversion
Two things happening here:
1)Substitution of iodide, 127I, with labeled iodide, 131I.
2) Change in stereochemistry
![Page 25: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/25.jpg)
Comparison of SN1 and SN2 mechanisms. Substitution vs. Loss of Optical Activity
RI RI RI RI RIRI RI RI RI RI
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.
I -
If racemization: “SN1”
RI RI RI RI RIRI RI RI RI RI
Only 20% reacts 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 RIRI RI RI RI RI
I -
If inversion: SN2
RI RI RI RI RIRI RI RI RI RI
Only 20% reacts 20% substituted, 40%
racemized, 40% optical purity lost (60% optically pure). Rate of loss of optical activity = 2 x Rate of substitution.
100% optically pure
![Page 26: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/26.jpg)
Effect of Structure of the Haloalkane on Rates
CH3X CH3CH2X (CH3)2CHX (CH3)3CX
Methyl primary secondary tertiary
SN1 RecallStability of resulting carbocation, hyperconjugation
Ease of ionization
Rate of SN1 Reactions
![Page 27: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/27.jpg)
Now for SN2
CH3X CH3CH2X (CH3)2CHX (CH3)3CX
Methyl primary secondary tertiary
SN2
Steric Hinderance, difficulty of approach for nucleophile
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.
![Page 28: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/28.jpg)
Recall: Resonance Stabilization of Carbocations
Allylic and benzylic carbocations are stabilized by resonance.
SN2 SN1Both
![Page 29: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/29.jpg)
Leaving GroupRecall that the leaving group becomes more negative.
R - X X
Generally, the best leaving groups are groups that can stabilize that negative charge: weak bases; conjugate bases of strong acids.
Base Strength
Example:
R-OH NRBr
BrR-OH R-OH2 R-Br
H
H - X X
![Page 30: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/30.jpg)
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 -
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 -
Stabilized. 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).
Not Stabilized.
![Page 31: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/31.jpg)
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…
![Page 32: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/32.jpg)
Return to elimination: competes with nucleophilic
substitution.
Zaitsev Rule, prefer to form the more substituted alkene (more stable).
The competition: SN1 and/orSN2
![Page 33: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/33.jpg)
Mechanistic Possibilities to eliminate the H+ and X-
HX
+ H + 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
![Page 34: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/34.jpg)
• There are two idealized mechanisms for -elimination reactions
• E1 mechanism:E1 mechanism: at one extreme, breaking of the R-Lv 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: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)
![Page 35: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/35.jpg)
E1 Mechanism
– ionization of C-Br gives a carbocation intermediate
– proton loss from the carbocation intermediate to a base (for example, the solvent) gives the alkene
CH3-C-CH3
Br
CH3
CH3-C-CH3
CH3
Br
slow, ratedetermining
+(A carbocation intermediate)
+
O:
H
H3CH-CH2-C-CH3
CH3
O
H
H3CH CH2=C-CH3
CH3fast
+
++ +
![Page 36: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/36.jpg)
Rate Determining Step; formation of the carbocation.
Energy Profile for E1 mechanism, carbocations.
Alkyl Halide (E1) Alkene + HX
Alkyl Halide (Addition) Alkene + HX
Reaction can occur in either direction…..
![Page 37: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/37.jpg)
E2 Mechanism
CH3O:-
C C
H
Lv
CH3OH
C C
Lv-H and -Lv are anti and coplanar
(dihedral angle 180°)
breaking of the R-Lv and C-H bonds is concerted
Needs Strong Base
![Page 38: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/38.jpg)
E2
![Page 39: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/39.jpg)
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
• 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]
dtRate = = k[RLv][Base]
d[RLv]
dtRate = = k[RLv]
![Page 40: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/40.jpg)
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.
AlsoE2 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.
![Page 41: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/41.jpg)
• E2 is most favorable (lowest activation energy) when H and Lv are anti and coplanar
CH3O:-
C C
H
Lv
CH3OH
C C
Lv-H and -Lv are anti and coplanar
(dihedral angle 180°)
Stereochemistry of E2
AB
D E
E
DA
B
![Page 42: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/42.jpg)
Examples of E2 Stereochemistry
Cl
CH3O -
+
cis Major product.
Zaitsev product
Cl
CH3O -
+
trans
Only product
Anti-Zaitsev
Explain both regioselectivity and relative rates of reaction.
But
Faster reaction
Slower reaction
![Page 43: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/43.jpg)
In order for the H and the Cl to be anti, both must be in axial positions
First the cis isomer. Reactive Conformation; H and Cl are anti to each other
Iso-propyl groups is in more stable equatorial position. Dominant
conformation is reactive conformation.
CH3O:-
H
H
H
H
Cl
CH3OH :Cl
1-Isopropyl-cyclohexene
2
1
6 + +E2
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.
![Page 44: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/44.jpg)
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
More stable chair (no H is anti and coplanar to Cl)
Less stable chair(H on carbon 6 is
anti and coplanar to Cl)
2
2
11
6 6Cl
H
H
H
H
Cl
HH
HH
Now the trans
Unreactive conformation
Reactive but only with the H on C 6
Most of the compound exists in the unreactive conformation. Slow reaction.
H
Cl
HH
HCH3O:
-
CH3OH Cl
(R)-3-Isopropyl-cyclohexene
21
6E2
+ +
Anti Zaitsev
More stable chair (no H is anti and coplanar to Cl)
Less stable chair(H on carbon 6 is
anti and coplanar to Cl)
2
2
11
6 6Cl
H
H
H
H
Cl
HH
HH
![Page 45: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/45.jpg)
Example, Predict Product
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 CH3
Ph
H3C Br
base
Ph
H CH3
Ph
H3C Br
Ph
H CH3
Ph
H3C Br
=
Ph
H3C H
Ph
H3C Br
rotate upper chiral C by 180
H3C
Ph
H
Br
Ph
H3C
CH3
H Ph
Ph
H3C Br
H
CH3
Ph
Br
Ph
H3C
rotate 120further
H3C Ph
H3C Phbase
H3C Ph
H3C Ph
H & Br not anti yet!
Now anti and we can see where the pi bond will be.
![Page 46: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/46.jpg)
Ph
H CH3
Ph
H3C Br
base
Alternative Approach: CAR
The H and Br will be leaving: just indicate by disks.
Meso or Racemic??
Anti Geometry
CRA
Relationship works in both directions. Should get cis isomer.
H3C Ph
H3C Ph
Note: As we have said before it may take some work to characterize a compound as “racemic” or “meso”.
This may be recognized as one of the enantiomers of the racemic mixture.
AC < -- > R
![Page 47: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/47.jpg)
RCH2X
R2CHX
R3CX
Alkyl halide E1 E2
Primary
Secondary
Tertiary
E1 does not occur.Primary carbocations areso unstable, they are neverobserved in solution.
E2 is favored.
Main reaction with strong bases such as OH- and OR-.
Main reaction with weak bases such as H2O, ROH.
Main reaction with strong bases such as OH- and OR-.
Main reaction with weak bases such as H2O, ROH.
E1 or E2
(Carbocation)
![Page 48: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/48.jpg)
ionization
1o,
2o, 3o polar solvents, weak nucleophiles, weak bases
1o strong, bulky bases
2o strong bases
3o strong bases
1o good nucleophiles, aprotic solvents
2o good nucleophiles but also poor bases, aprotic solvents
30
SN2SN1
E2 E1
Rearrange ?
1o
2o heat, more hindered
3o heat, more hindered
1o
2o lower hinderance, better nucleophile than base
3o lower hinderance, better nucleophile than base
R - X R + + X -
good nucleophile
strong base
R-Nuc
alkene
weak nucleophile
weak base
alkene
R-Nuc
![Page 49: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/49.jpg)
Recall Halohydrins and EpoxidesCl2, H2O Cl H2O
Cl
OH
base Cl
O
O
H
OH
ROH
OH
RO
Creation of Nucleophile Internal SN2 reaction with inversion
Creation of good leaving group.
Attack by poor nucleophile
![Page 50: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/50.jpg)
Neighboring Group Effect
• Mustard gases – contain either S-C-C-X or N-C-C-X
– what is unusual about the mustard gases is that they undergo hydrolysis rapidly in water, a very poor nucleophile
ClS
Cl 2H2O HOS
OH 2HCl+ +
Bis(2-chloroethyl)sulfide(a sulfur mustard gas)
Bis(2-chloroethyl)methylamine(a nitrogen mustard gas)
ClS
Cl ClN
Cl
![Page 51: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/51.jpg)
– the reason is neighboring group participation by the adjacent heteroatom
– proton transfer to “solvent” completes the reaction
ClS
Cl
ClS O-H
H
ClS
ClS
O
H
H
Cl+
+
A cyclic sulfonium ion
an internal SN2 reaction
slow, ratedetermining
++
a secondSN2 reaction
fast+
:
:
:
Good nucleophile.
![Page 52: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/52.jpg)
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.
H
SCH3
CH3
Cl
H3C H
H2O H
SCH3
CH3
OH
H3C H
H3C
SCH3
H
OH
H CH3
+
Expect sulfur to attack the C-Cl, displacing the Cl and forming a three membered ring. Like this…
But we have to be careful with stereochemistry
H
SCH3
CH3
Cl
H3C H
S
CH3
H2O
S
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!!
From an old quiz
![Page 53: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/53.jpg)
H
SCH3
CH3
Cl
H3C H
We have to put the molecule in the correct conformation.
=H
SCH3
CH3
Cl
H3C HCl SCH3
H CH3H3C H
=
Reactive conformation reached by 180 rotation around C-C bond
Cl
SCH3H3C
CH3
H
H
SCH3
H3C CH3
HH
And then the ring is opened by attack of water
S and Cl are eclipsed, not anti.
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.
![Page 54: Nucleophilic Substitution and -elimination. Substitution Process Nucleophiles have a pair of electrons which are used to form a bond to the electrophile](https://reader038.vdocument.in/reader038/viewer/2022102808/56649d4b5503460f94a287d7/html5/thumbnails/54.jpg)
H
SCH3
CH3
OH
H3C HSCH3
H3C CH3
HH
OH2
SCH3H3C
CH3
H
H =
SCH3
H CH3HH3C
=
180 rotation
HO
HO
Two modes of attack by water.
And…
SCH3
H3C CH3
HH
OH2
H3CS
H3C
CH3
H
H OH
H3C
SCH3
H
OH
H CH3
=
H3CS
H3CH
CH3H
OH
=
180 rotation
Again note the ring structure is achiral and that we must, of course, produce optically inactive product.
Enantiomers, racemic mixture