Post on 27-Apr-2015
Coverage:1. Substitution Reactions, SN1 and SN22. Elimination Reactions, E1 and E2
Chapter 11: Reactions of Alkyl Halides
Problems:25-39, 43,47,54
Goals:1. Know the detailed mechanisms of SN1, SN2, E1 and E22. Know what is a good nucleophile and what is a poor nucleophile3. Understand the concept of inversion of configuration (SN2)4. Know the kinetics associated with each reaction5. Know Zaitzev’s Rule and how it applies to the elimination reactions.6. Know what is a good leaving group in a reaction.
There are two fundamental types of reaction mechanisms to be covered:
Substitution SN2 and SN2Elimination E1 and E2
Definitions
Nucleophile – electron-rich species that attacks a nucleus which is positively chargedElectrophile – electron-poor species that is attacked by a nucleophile.
SN2
S – substitutionN – nucleophilic2 - bimolecular reaction
CHH
HXNu- C
HH
HNu + X-
Backside attack
Kinetics
Rate = k [Nu-] [RX] 2nd order reaction
k – rate constant characteristic of the reaction. The larger the k, the faster the reaction.
Effect of Substrate Structure
Alkyl Halide Type Relative Rate
CH3X methyl 3 x 106
CH3CH2X 10 1 x 105
CH3CH2CH2X 10 4 x 104
(CH3)2CH-X 20 2.5 x 103
(CH3)3CCH2-X 10 1(CH3)3C-X 30 ~0
Inversion of Configuration: As the nucleophile attacks, the three groups attached to the carbon undergo inversion, that is, they flip to the opposite side of the carbon.
Why this order of reactivity? What controls the relative rate of reaction of the various substrates?
Answer: Steric Hindrance to nucleophilic attack.
CCH3
CH3
CH3
XNu- No Reaction
The bulky methyl groups prevent backside attack by sterically hindering thenucleophile from attacking the electrophilic carbons. Contrast this situation to that of the CH3X group.
Nucleophiles
Nu: + R-X Nu-R+ + X-neutral
Nu:- + R-X Nu-R + X-charged
Reaction:
Nu- + CH3Br Nu-CH3+
+ Br-
Nu- Relative Rate pKa (conjugate acid)
HS- 125,000 7.04
CN- 125,000 9.31
I- 100,000 0.77CH3CH2O
- 25,000 16
OH- 16,000 15.7
Ph-O- 8,000 10
CH3CO2- 500 4.8
H2O 1
Strongest base
Conclusion: The strongest base is not the best nucleophile. In other words, basicity does not control nucleophilicity.
What controls nucleophilicity?
1. Polarizability.
F-
Electrons tightly held by the nucleus and not easilydistorted
CHH
H X C
HH
H
F- X
Transition state – not good bonding between carbon and fluorine atom. High energy
I- CHH
H X C
HH
H
I- X
Electrons loosely held by the nucleus and easilydistorted
Transition state – very good bonding between carbon and iodine atom. Lower energy
2. Solvation
F- has a high charge density due to small size
F- H O
H
..
H O
H
..
H
O H
..
• Highly solvated by water• Very stabilized• Less reactive in Sn2 reaction
H-bond
Generally, the larger nucleophile is the better one in a given group.
Halogen Nucleophiles:
I- > Br- > Cl- > F- HS- > HO-
Best Worst NH3 > H2O
3. Charge on Nucleophile
Charged nucleophiles are better than neutral nucleophiles in the same group.
HS- > H2S OH- > H2O NH2- > NH3
Table of Nucleophilic Strengths
Strong Nucleophile Moderate Weak
(CH3CH2)2P Br- F-
HS- NH3 H2O
I- CH3SCH3CH3OH
(CH3CH2)2NH Cl-
CN- CH3CO2-
OH-
CH3O-
Strongest
Weakest
4. Bulky Nucleophiles
CH3CH2O- (CH3)3CO->
Alkoxides Ions
Not Bulky Bulky Unhindered Hindered
Effect of Leaving Group
Leaving Group
C XNCN X+
• The LG is usually displaced with a negative charge. LGs that best stabilize the negative charge are best.
• Electronegative LGs, which polarize the C atom are also good.
• LGs should be polarizable to stabilize the Transition State
• In general, the weaker the base, the better the LG.
LG Rel. Reactivity pKa (conjugate acid)
60,000 -6.5
I- 30,000 -9.5Br- 10,000 -9F- 1 3.2OH- ~0 15.7NH2- ~0 35
S O
O
O
CH3
Strongest base
Weakest base
S O
O
O
CH3 Symbolized as -OTs
tosylate group
SO
O
O
CH3CH3:N C
S-O
O
O
CH3:N C-CH3
Halogens as Leaving Groups
I- > Br- > Cl- > F-
Stereochemistry of SN2
Recall that the nucleophile attacks from the backside.
What happens when a single enantiomer with a reactive chiral carbon undergoes SN2 reaction???
Answer: It undergoes inversion of configuration
C
CH3H
CH3CH2
BrHO-C
CH3H
CH2CH3
OH XXC
CH3H
CH2CH3
OH +
(S)-2-bromobutane (R)-2-butanol
SN2 reaction proceeds with 100% inversion of configuration – termedWalden inversion.
SN1 Mechanism 1- Unimolecular
CH3
CCH3
CH3
Br OH2
CH3
CCH3
CH3
OH H Br+ +
30 Halide Solvent 30 alcohol
Solvolysis: Solvent acts as nucleophile and reacts with substrate.What is the mechanism? Not SN2! Remember, a 30 substrate is unreactive in SN2 Answer: SN1
CH3
CCH3
CH3
Br slowCH3
CCH3
CH3
Br
CH3
CCH3
CH3
OH2
CH3
CCH3
CH3
OH2+
CH3
CCH3
CH3
OH2+Br CH3
CCH3
CH3
OH H Br
+ +
+ +
+
Reaction Energy Diagram for SN1
R-X
R+ + X-
NuH
R-NuH+ + X-
R-Nu + H-X
E
Reaction Coordinate
TS#1 TS#2
TS#3
First step is rate-determining (highest activation energy).
Rate = k [R-X] First order reactionRate-determining Step is Unimolecular
Effect of Substrate Structure on SN1 Reactivity
30 > 20 > 10 > CH3X
Most Least Reactive
Why this order
• 30 substrates form stable 30 carbocations in rate-determining step. They form faster with a lower Energy of Activation
30 carbocation more stable lower Ea forms faster
10 carbocation less stable higher Ea forms slower
Reaction Coordinate
E
Allylic and Benzylic Substrates are very reactive in SN1 reactions.
Why so??
They form resonance-stabilized carbocations.
Br
OH2
OH2+ BrOH
H Br
+ +
Allyl bromide
What about benzyl bromide? Write a mechanism showing its reactionwith water. CH2Br
OH2+?
Leaving Groups: The same factors that favor SN2 leaving groups alsoFavor SN1 leaving groups, i.e. if a LG is good for SN2, it is good for SN1.
Stereochemistry of SN1
C
Br
(CH3)2CH
CH3CH2CH3
C(CH3)2CH
CH3CH2CH3
C
(CH3)2CH
CH3 CH2CH3
OCH3
C
OCH3
(CH3)2CHCH3 CH2CH3
CH3OH
CH3OH
B
A
+
Planar, symmetric carbocation
S isomer
S isomer~50%
R isomer~50%
The carbocation is attacked both from the top and the bottom by the nucleophilic methanol, resulting in a near racemic mixture of enantiomers as products
Reactions that proceed by the SN1 reaction often undergo rearrangements.Why?
Carbocations are intermediates and may undergo 1,2 ~H or 1,2 ~CH3 shifts.
CH3CHCHCH3
CH3
Br
CH3CH2OHCH3CHCHCH3
CH3
OCH2CH3
CH3CHCHCH3
CH3
OCH2CH3
H Br
+
+
Practice: Write a mechanism for this reaction to account for both products.
Solvents in SN1 and SN2 Reactions
SN2 Polar, aprotic solvents are best.
Aprotic - no OH or NH group present These solvents cannot H-bond to nucleophile and therefore the nucleophile is more reactive.
CH3CCH3
O
CH3C N
O
CH3SCH3
Acetone Acetonitrile Dimethylsulfoxide (DMSO)
SN1 Polar, protic solvents are best.
Protic – possess OH or NH group These solvents promote formation of ions through H-bonding.
OH2 CH3OH CH3CH2OH
Water Methanol Ethanol
Summary of SN1 and SN2
Topic SN2 SN1
Kinetics Rate=k[R-X][Nu] Rate=k[RX]
Nucleophile Strong Nu required Weak Nu required, usually solvent
Substrate Polar, aprotic Polar, protic
Stereochemistry 100% inversion Racemization
Rearrangements No Yes
E2 Elimination
Requirements: Alkyl substrate with a good leaving group possessing ß-hydrogen
C
H
C
X
ß
ß
The ß-hydrogen is bonded to the ß-carbon, which is bonded to the -carbonwhich is bonded to the leaving group X!
A strong base is also required. Any of these will do:
OH- < CH3O- < CH3CH2O- < (CH3)3CO- < NH2-
Weakest Strongest
Mechanism: The E2 mechanism is a one-step mechanism with bond-breaking and Bond-making taking place at the same time; termed a concerted mechansim. InAddition, the Hß and the leaving group X must be anti-coplanar for rapid reaction.
H
XOH2 X
OH-
+ +
Notice that H and X are anti to each other and the four atoms, H-C-C-X, are coplanar in the reactant. This situation stabilizes the transition state leading to the alkene.
C C
H
Br
R
R
H H
OCH3 :..
..
..
..: :
Overlap develops in the T.S. if theanti-coplanar relationship is maintained.
The overlap stabilizes the T.S. and the reaction takes place faster.
C C
H
Br
R
RH
H
OCH3 ..
..
..: :
..
. .
Kinetics
Rate = k [B-][R-X] 2nd Order Reaction
E2 Ball and Stick Movie
Substrate Reactivity
30 > 20 > 10
Why this order of reactivity?
30 substrates yield more stable alkenes and therefore react faster. 10 substrates yield unstable alkenes and react more slowly.
Recall: Alkene Stability
tetrasubstituted > trisubstituted > disubstituted > monosubstituted > unsubstituted
R
R
R
R
R
R
H
R
R
H
H
R
R
H
H
H
H
H
H
H
H
H
CH3
H
HX HH
HCH3OH2 X
OH-
+ +
10 substrate Monosubstituted alkene – less stable
H
H
H
CH3
CH3
X CH3H
CH3HOH2 X
OH-
+ +
30 substrate Disubstituted Alkene - more stable
What about substrates that can yield more than one possible product such as2-bromobutane.
ba
a
b
C
H
C
H
C
BrH
CH3
H
H
H
HO-
C CCH3
H
H
CH3
C CCH3
H H
CH3
C CCH3CH2
H
H
H
OH2 Br
+
+ +
81%
19%
The more stable alkene predominates in the product mixture.
Zaitzev’s Rule: In elimination reactions, the more highly substituted, morestable alkene is usually the major product. This product is referred to as the Zaitzev Product
E2 reactions with Diastereomers
Br H
HCH3
Ph
Ph
H
CH3
Ph
Ph
H
CH3
Ph
Ph
(1S,2S)-1-bromo-1,2-diphenylpropane trans cis
OH-
In the above reaction, only 1 product forms. Which one?
Br H
HCH3
Ph
Ph
BrH
CH3H
Ph Ph
Br
H
CH3
H
Ph
Ph
rotate
Note: H and Br are anti-coplanar in a staggeredconformation.
or
s
s
OH-
Br
H
CH3
H
Ph
Ph CH3 H
PhPh
100% cis isomer
The E2 reaction is an example of a stereospecific reaction.
Stereospecific Reaction: a reaction in which different stereoisomers of a given reactant yield different stereoisomeric products.
(1S,2S)-1-bromo-1,2-diphenylpropane 100% cis product(1S,2R)-1-bromo-1,2-diphenylpropane 100% trans product
Homework Problem: Show that the (1R,2R) yields the cis and the (1R,2S)yields only the trans.
E2 in Cyclohexane Systems
Br
OH-
Recall that there are two possible conformations of bromocyclohexane
The equatorial conformation if favored, but it does not provide the necessaryanti, coplanar relationship of the H and Br.
But the axial conformation does! So it reacts.
BrH
H Br
H
H
Anti, coplanar H and Br
Br
H
H
OH-
+ H2O + Br-
Conclusion: In order for the H and Br to be anti-coplanar, both must be in axial positions. This geometry is referred to as trans-diaxial.
Homework: The following two geometric isomers yield different alkenes as products. Why? Write mechanisms to account for each product.
CH3
Br
base, E2CH3
CH3
Br
base, E2
CH3
trans
cis
E1 Mechanism
C
CH3
CH3
CH3
BrCH3OH
CH2 C
CH3
CH3
BrH+
This reaction cannot be E2 because there is no strong base present
C
CH3
CH3
CH3
Br C
CH3
CH3
CH3
C
CH3
H2C
CH3
H
CH3OH
CH2 C
CH3
CH3
Br
CH3OH2+
+
+ +
slow
Kinetics Rate = k [R-X] 1st order
Substrate Reactivity 30 > 20 > 10 (parallels carbocation stability!)
E1 always competes with SN1
C
CH3
CH3
CH3
Br C
CH3
CH3
CH3
Br+
C
CH3
CH3
CH3
OCH3CH2 C
CH3
CH3
CH3OH as baseCH3OH as nucleophile
SN1 E1
Mixture of Products
Summary of E1 and E2
Topic E2 E1
Kinetics 2nd order 1st order
Base Strong required Weak required
Substrate 30 > 20 > 10 30 > 20 > 10
Solvent Type not critical Polar, aprotic for ionization
Orientation Zaitzev rule Zaitzev Rule
Conformation Anti-coplanar None requiredRequirements H and X
Rearrangements No Yes