Chapter 11: Reactions of Alkyl Halides
Chapter 11: Reactions of Alkyl Halides Coverage: 1. Substitution Reactions, S N1 and S N2 2. Elimination Reactions, E1 and E2 Problems:25-39, 43,47,54
Goals: 1. Know the detailed mechanisms of SN1, SN2, E1 and E2 2. Know what is a good nucleophile and what is a poor nucleophile 3. Understand the concept of inversion of configuration (SN2) 4. Know the kinetics associated with each reaction 5. Know Zaitzev’s Rule and how it applies to the elimination reactions. 6. Know what is a good leaving group in a reaction.
11-1
There are two fundamental types of reaction mechanisms to be covered: Substitution Elimination
SN2 and SN2 E1 and E2
Definitions Nucleophile – electron-rich species that attacks a nucleus which is positively charged Electrophile – electron-poor species that is attacked by a nucleophile. SN2 S – substitution N – nucleophilic 2 - bimolecular reaction
Nu-
HH C X H Backside attack
Nu
C
H H
+ X-
H 11-2
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. Kinetics Rate = k [Nu-] [RX]
2 nd 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 CH3CH2X CH3CH2CH2X (CH3)2CH-X (CH3)3CCH2-X (CH3)3C-X
methyl 10 10 20 10 30
3 x 106 1 x 105 4 x 104 2.5 x 103 1 ~0 11-3
Reaction Energy Diagram for SN2 Mechanism TS
Ea E
R-X + Nu
R-Nu + X
Reaction Coordinate
11-4
Why this order of reactivity? What controls the relative rate of reaction of the various substrates? Answer: Steric Hindrance to nucleophilic attack.
H3C H3C
Nu-
No Reaction
C X
H3C The bulky methyl groups prevent backside attack by sterically hindering the nucleophile from attacking the electrophilic carbons. Contrast this situation to that of the CH3X group. Nucleophiles +
R-X
à
Nu-R+ + X-
Nu:- + charged
R-X
à
Nu-R
Nu: neutral
+ X11-5
Reaction: Nu-
+
NuHSCNICH3CH2OOHPh-OCH3CO2H2O
Nu-CH3+
CH3Br Relative Rate 125,000 125,000 100,000 25,000 16,000 8,000 500 1
+
Br-
pKa (conjugate acid) 7.04 9.31 0.77 16 15.7 10 4.8
Strongest base
Conclusion: The strongest base is not the best nucleophile. In other words, basicity does not control nucleophilicity. What controls nucleophilicity? 11-6
1. Polarizability.
F-
HH
HH C H
F-
X
C
X
H Electrons tightly held by the nucleus and not easily distorted
Transition state – not good bonding between carbon and fluorine atom. High energy
HH I-
HH C H
X
I-
C
X
H Electrons loosely held by the nucleus and easily distorted
Transition state – very good bonding between carbon and iodine atom. Lower energy 11-7
2. Solvation F - has a high charge density due to small size
.. H
O
• Highly solvated by water • Very stabilized • Less reactive in Sn2 reaction
H
..
F-
H
O H
H-bond H H
..O
Generally, the larger nucleophile is the better one in a given group. Halogen Nucleophiles: I- > Br- > Cl- > F-
HS- > HO -
Best
NH3 > H2O
Worst
11-8
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 Strongest
Moderate
(CH3CH2)2P HSI-
BrNH3 CH3SCH3
(CH3CH2)2NH CN-
ClCH3CO2-
Weak FH2O CH3OH
OHWeakest
CH3O-
11-9
4. Bulky Nucleophiles CH3CH2O-
Not Bulky Unhindered
>
(CH3)3CO-
Bulky Hindered
Alkoxides Ions
Effect of Leaving Group
N
C
X
N
C
+
X
Leaving Group 11-10
• 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)
O S O
H3C
60,000
-6.5
30,000 10,000 1 ~0 ~0
-9.5 -9 3.2 15.7 35
O
IBrFOHNH2-
Weakest base
Strongest base
11-11
O H3C
S O
Symbolized as -OTs
O
tosylate group O :N C
H3C
O
CH3
S O
O :N C-CH3
-O
S
CH3
O
Halogens as Leaving Groups I- > Br- > Cl- > F11-12
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 H3C H
HOC CH3CH2
CH3 H Br
(S)-2-bromobutane
HO
C
CH 3 H X
CH2CH3
HO
C
+
X
CH2CH3
(R)-2-butanol
SN2 reaction proceeds with 100% inversion of configuration – termed Walden inversion.
11-13
SN1 Mechanism
1- Unimolecular
CH3
CH3
+
H3C C Br
H2O
H3C C OH
CH3
+
H Br
CH3
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
slow
H3 C C Br CH3
+
H2O
Br
+
H
H3 C C OH2+ CH3
CH3 CH3
+
CH3
CH3 H3 C C OH2+
H3C C+ CH3
CH3 H3 C C+
CH3
Br
CH3 H3C C OH CH3
Br
11-14
Reaction Energy Diagram for SN1 TS#1
TS#2
R+ + XE
NuH
R-X
TS#3
R-NuH+ + XR-Nu + H-X Reaction Coordinate
First step is rate-determining (highest activation energy). Rate = k [R-X] First order reaction Rate-determining Step is Unimolecular
11-15
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
E
10 carbocation à less stable à higher E a à forms slower 30 carbocation à more stable à lower E a à forms faster Reaction Coordinate 11-16
Allylic and Benzylic Substrates are very reactive in SN1 reactions. Why so?? They form resonance-stabilized carbocations. Br
+
+
Allyl bromide H2O
OH2 +
OH
Br
H Br
What about benzyl bromide? Write a mechanism showing its reaction with water. CH2Br
+
H2O
? 11-17
Leaving Groups: The same factors that favor SN2 leaving groups also Favor SN1 leaving groups, i.e. if a LG is good for SN2, it is good for SN1. Stereochemistry of SN1
OCH3 CH3OH
H3C B
C
CH2 CH3
(CH3)2CH
Br
+ H 3C
C CH2 CH3
(CH3 )2CH
H 3C (CH3 )2 CH
C
S isomer ~50%
CH2CH3
A
S isomer CH3OH
Planar, symmetric carbocation
H3C (CH3 )2 CH
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
CH2CH3 C OCH3
R isomer ~50% 11-18
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. Br
OCH2 CH3
CH3 CHCHCH3 CH3
CH3CH2OH
OCH2CH3
+
CH3CHCHCH3 CH3
CH3CHCHCH3 CH3
+
H Br
Practice: Write a mechanism for this reaction to account for both products.
11-19
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. O CH3CCH 3
Acetone
O CH3C N
Acetonitrile
CH3SCH 3
Dimethylsulfoxide (DMSO)
SN1 Polar, protic solvents are best. Protic – possess OH or NH group These solvents promote formation of ions through H-bonding. H2 O
CH 3OH
Water
Methanol
CH3CH2OH
Ethanol 11-20
Summary of SN1 and SN2 Topic
SN 2
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
11-21
E2 Elimination Requirements: Alkyl substrate with a good leaving group possessing ß-hydrogen
Hß Cß Cα X The ß-hydrogen is bonded to the ß-carbon, which is bonded to the α-carbon which is bonded to the leaving group X! A strong base is also required. Any of these will do: OH- < CH3O- < CH3CH2O- < (CH3)3CO- < NH2Weakest
Strongest
11-22
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. In Addition, the Hß and the leaving group X must be anti-coplanar for rapid reaction.
OHH X
+
H2O
+
X
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.
11-23
CH 3
: O ..
Overlap develops in the T.S. if the anti-coplanar relationship is maintained.
H .. H R
CH 3
C
C
R H
.. :Br .. :
Kinetics
.. O .. H H R
The overlap stabilizes the T.S. and the reaction takes place faster.
. C
. C
Rate = k [B-][R-X]
2 nd Order Reaction
R H .. :Br .. :
E2 Ball and Stick Movie
11-24
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
R
R
H
R
H
R
H
R
H
H
R
R
R
R
R
H
H
H
H
H
tetrasubstituted > trisubstituted > disubstituted > monosubstituted > unsubstituted
11-25
OHH CH3 H
CH3 H X
30 substrate
H
CH3
H
CH3
+
H2O
+
X
Disubstituted Alkene - more stable
OHH H H3C
H H X
10 substrate
H3 C
H
H
H
+
H2O
+
X
Monosubstituted alkene – less stable
11-26
What about substrates that can yield more than one possible product such as 2-bromobutane. a
HO-
b
a
H3C
H H H H3C C C C H
H
H
+
C C
H
CH3
H Br H
H C C
H3C
CH3
81%
b
+ CH3 CH2
H2O
+
Br
H C C
H
H
19% The more stable alkene predominates in the product mixture. Zaitzev’s Rule: In elimination reactions, the more highly substituted, more stable alkene is usually the major product. This product is referred to as the Zaitzev Product 11-27
E2 reactions with Diastereomers Ph Br
s H
H3C
s H
OH-
Ph
Ph
H
H
or
Ph
H3C
(1S,2S)-1-bromo-1,2-diphenylpropane
Ph
CH3
Ph
trans
cis
In the above reaction, only 1 product forms. Which one?
Ph Br
H
H3C
H Ph
≡
H3C H Ph
Br H Ph
rotate
Ph H3C
Br Ph
H
H
Note: H and Br are anti-coplanar in a staggered conformation. 11-28
Ph H3C
Br Ph H
H
Ph
Ph
H3C
H
100% cis isomer OHThe 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 à 100% cis product (1S,2R)-1-bromo-1,2-diphenylpropane à 100% trans product Homework Problem: Show that the (1R,2R) yields the cis and the (1R,2S) yields only the trans. 11-29
E2 in Cyclohexane Systems Br OH-
Recall that there are two possible conformations of bromocyclohexane H
Anti, coplanar H and Br
Br H H
H Br
The equatorial conformation if favored, but it does not provide the necessary anti, coplanar relationship of the H and Br. But the axial conformation does! So it reacts.
11-30
OHH + H2O +
H
Br-
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
base, E2
CH3
Br
trans CH3
CH3 base, E2
cis
Br
11-31
E1 Mechanism CH3
CH3
CH3OH
H3C C Br
H2C C
+
HBr
CH3
CH3
This reaction cannot be E2 because there is no strong base present
CH3 H3C C
slow
Br
CH3 H3C C
CH3
+
Br
CH3
CH3OH H
CH3
H2 C C
+
CH3 H2 C C CH3
CH3
Kinetics
Rate = k [R-X]
+
CH3OH2+
1st order
Substrate Reactivity 30 > 20 > 10 (parallels carbocation stability!) 11-32
E1 always competes with SN1 CH3
CH3
H3 C C Br
H3C C
CH3
+
Br
CH3
CH3OH as nucleophile
CH3OH as base
CH3 H3 C C OCH3
CH3 H2 C C CH3
CH3
SN1
E1 Mixture of Products
11-33
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 Requirements
Anti-coplanar H and X
None required
Rearrangements
No
Yes 11-34
Goals: 1. Know the detailed mechanisms of SN1, SN2, E1 and E2 2. Know what is a good nucleophile and what is a poor nucleophile 3. Understand the concept of inversion of configuration (SN2) 4. Know the kinetics associated with each reaction 5. Know Zaitzev’s Rule and how it applies to the elimination reactions. 6. Know what is a good leaving group in a reaction.
11-1
There are two fundamental types of reaction mechanisms to be covered: Substitution Elimination
SN2 and SN2 E1 and E2
Definitions Nucleophile – electron-rich species that attacks a nucleus which is positively charged Electrophile – electron-poor species that is attacked by a nucleophile. SN2 S – substitution N – nucleophilic 2 - bimolecular reaction
Nu-
HH C X H Backside attack
Nu
C
H H
+ X-
H 11-2
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. Kinetics Rate = k [Nu-] [RX]
2 nd 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 CH3CH2X CH3CH2CH2X (CH3)2CH-X (CH3)3CCH2-X (CH3)3C-X
methyl 10 10 20 10 30
3 x 106 1 x 105 4 x 104 2.5 x 103 1 ~0 11-3
Reaction Energy Diagram for SN2 Mechanism TS
Ea E
R-X + Nu
R-Nu + X
Reaction Coordinate
11-4
Why this order of reactivity? What controls the relative rate of reaction of the various substrates? Answer: Steric Hindrance to nucleophilic attack.
H3C H3C
Nu-
No Reaction
C X
H3C The bulky methyl groups prevent backside attack by sterically hindering the nucleophile from attacking the electrophilic carbons. Contrast this situation to that of the CH3X group. Nucleophiles +
R-X
à
Nu-R+ + X-
Nu:- + charged
R-X
à
Nu-R
Nu: neutral
+ X11-5
Reaction: Nu-
+
NuHSCNICH3CH2OOHPh-OCH3CO2H2O
Nu-CH3+
CH3Br Relative Rate 125,000 125,000 100,000 25,000 16,000 8,000 500 1
+
Br-
pKa (conjugate acid) 7.04 9.31 0.77 16 15.7 10 4.8
Strongest base
Conclusion: The strongest base is not the best nucleophile. In other words, basicity does not control nucleophilicity. What controls nucleophilicity? 11-6
1. Polarizability.
F-
HH
HH C H
F-
X
C
X
H Electrons tightly held by the nucleus and not easily distorted
Transition state – not good bonding between carbon and fluorine atom. High energy
HH I-
HH C H
X
I-
C
X
H Electrons loosely held by the nucleus and easily distorted
Transition state – very good bonding between carbon and iodine atom. Lower energy 11-7
2. Solvation F - has a high charge density due to small size
.. H
O
• Highly solvated by water • Very stabilized • Less reactive in Sn2 reaction
H
..
F-
H
O H
H-bond H H
..O
Generally, the larger nucleophile is the better one in a given group. Halogen Nucleophiles: I- > Br- > Cl- > F-
HS- > HO -
Best
NH3 > H2O
Worst
11-8
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 Strongest
Moderate
(CH3CH2)2P HSI-
BrNH3 CH3SCH3
(CH3CH2)2NH CN-
ClCH3CO2-
Weak FH2O CH3OH
OHWeakest
CH3O-
11-9
4. Bulky Nucleophiles CH3CH2O-
Not Bulky Unhindered
>
(CH3)3CO-
Bulky Hindered
Alkoxides Ions
Effect of Leaving Group
N
C
X
N
C
+
X
Leaving Group 11-10
• 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)
O S O
H3C
60,000
-6.5
30,000 10,000 1 ~0 ~0
-9.5 -9 3.2 15.7 35
O
IBrFOHNH2-
Weakest base
Strongest base
11-11
O H3C
S O
Symbolized as -OTs
O
tosylate group O :N C
H3C
O
CH3
S O
O :N C-CH3
-O
S
CH3
O
Halogens as Leaving Groups I- > Br- > Cl- > F11-12
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 H3C H
HOC CH3CH2
CH3 H Br
(S)-2-bromobutane
HO
C
CH 3 H X
CH2CH3
HO
C
+
X
CH2CH3
(R)-2-butanol
SN2 reaction proceeds with 100% inversion of configuration – termed Walden inversion.
11-13
SN1 Mechanism
1- Unimolecular
CH3
CH3
+
H3C C Br
H2O
H3C C OH
CH3
+
H Br
CH3
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
slow
H3 C C Br CH3
+
H2O
Br
+
H
H3 C C OH2+ CH3
CH3 CH3
+
CH3
CH3 H3 C C OH2+
H3C C+ CH3
CH3 H3 C C+
CH3
Br
CH3 H3C C OH CH3
Br
11-14
Reaction Energy Diagram for SN1 TS#1
TS#2
R+ + XE
NuH
R-X
TS#3
R-NuH+ + XR-Nu + H-X Reaction Coordinate
First step is rate-determining (highest activation energy). Rate = k [R-X] First order reaction Rate-determining Step is Unimolecular
11-15
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
E
10 carbocation à less stable à higher E a à forms slower 30 carbocation à more stable à lower E a à forms faster Reaction Coordinate 11-16
Allylic and Benzylic Substrates are very reactive in SN1 reactions. Why so?? They form resonance-stabilized carbocations. Br
+
+
Allyl bromide H2O
OH2 +
OH
Br
H Br
What about benzyl bromide? Write a mechanism showing its reaction with water. CH2Br
+
H2O
? 11-17
Leaving Groups: The same factors that favor SN2 leaving groups also Favor SN1 leaving groups, i.e. if a LG is good for SN2, it is good for SN1. Stereochemistry of SN1
OCH3 CH3OH
H3C B
C
CH2 CH3
(CH3)2CH
Br
+ H 3C
C CH2 CH3
(CH3 )2CH
H 3C (CH3 )2 CH
C
S isomer ~50%
CH2CH3
A
S isomer CH3OH
Planar, symmetric carbocation
H3C (CH3 )2 CH
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
CH2CH3 C OCH3
R isomer ~50% 11-18
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. Br
OCH2 CH3
CH3 CHCHCH3 CH3
CH3CH2OH
OCH2CH3
+
CH3CHCHCH3 CH3
CH3CHCHCH3 CH3
+
H Br
Practice: Write a mechanism for this reaction to account for both products.
11-19
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. O CH3CCH 3
Acetone
O CH3C N
Acetonitrile
CH3SCH 3
Dimethylsulfoxide (DMSO)
SN1 Polar, protic solvents are best. Protic – possess OH or NH group These solvents promote formation of ions through H-bonding. H2 O
CH 3OH
Water
Methanol
CH3CH2OH
Ethanol 11-20
Summary of SN1 and SN2 Topic
SN 2
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
11-21
E2 Elimination Requirements: Alkyl substrate with a good leaving group possessing ß-hydrogen
Hß Cß Cα X The ß-hydrogen is bonded to the ß-carbon, which is bonded to the α-carbon which is bonded to the leaving group X! A strong base is also required. Any of these will do: OH- < CH3O- < CH3CH2O- < (CH3)3CO- < NH2Weakest
Strongest
11-22
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. In Addition, the Hß and the leaving group X must be anti-coplanar for rapid reaction.
OHH X
+
H2O
+
X
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.
11-23
CH 3
: O ..
Overlap develops in the T.S. if the anti-coplanar relationship is maintained.
H .. H R
CH 3
C
C
R H
.. :Br .. :
Kinetics
.. O .. H H R
The overlap stabilizes the T.S. and the reaction takes place faster.
. C
. C
Rate = k [B-][R-X]
2 nd Order Reaction
R H .. :Br .. :
E2 Ball and Stick Movie
11-24
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
R
R
H
R
H
R
H
R
H
H
R
R
R
R
R
H
H
H
H
H
tetrasubstituted > trisubstituted > disubstituted > monosubstituted > unsubstituted
11-25
OHH CH3 H
CH3 H X
30 substrate
H
CH3
H
CH3
+
H2O
+
X
Disubstituted Alkene - more stable
OHH H H3C
H H X
10 substrate
H3 C
H
H
H
+
H2O
+
X
Monosubstituted alkene – less stable
11-26
What about substrates that can yield more than one possible product such as 2-bromobutane. a
HO-
b
a
H3C
H H H H3C C C C H
H
H
+
C C
H
CH3
H Br H
H C C
H3C
CH3
81%
b
+ CH3 CH2
H2O
+
Br
H C C
H
H
19% The more stable alkene predominates in the product mixture. Zaitzev’s Rule: In elimination reactions, the more highly substituted, more stable alkene is usually the major product. This product is referred to as the Zaitzev Product 11-27
E2 reactions with Diastereomers Ph Br
s H
H3C
s H
OH-
Ph
Ph
H
H
or
Ph
H3C
(1S,2S)-1-bromo-1,2-diphenylpropane
Ph
CH3
Ph
trans
cis
In the above reaction, only 1 product forms. Which one?
Ph Br
H
H3C
H Ph
≡
H3C H Ph
Br H Ph
rotate
Ph H3C
Br Ph
H
H
Note: H and Br are anti-coplanar in a staggered conformation. 11-28
Ph H3C
Br Ph H
H
Ph
Ph
H3C
H
100% cis isomer OHThe 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 à 100% cis product (1S,2R)-1-bromo-1,2-diphenylpropane à 100% trans product Homework Problem: Show that the (1R,2R) yields the cis and the (1R,2S) yields only the trans. 11-29
E2 in Cyclohexane Systems Br OH-
Recall that there are two possible conformations of bromocyclohexane H
Anti, coplanar H and Br
Br H H
H Br
The equatorial conformation if favored, but it does not provide the necessary anti, coplanar relationship of the H and Br. But the axial conformation does! So it reacts.
11-30
OHH + H2O +
H
Br-
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
base, E2
CH3
Br
trans CH3
CH3 base, E2
cis
Br
11-31
E1 Mechanism CH3
CH3
CH3OH
H3C C Br
H2C C
+
HBr
CH3
CH3
This reaction cannot be E2 because there is no strong base present
CH3 H3C C
slow
Br
CH3 H3C C
CH3
+
Br
CH3
CH3OH H
CH3
H2 C C
+
CH3 H2 C C CH3
CH3
Kinetics
Rate = k [R-X]
+
CH3OH2+
1st order
Substrate Reactivity 30 > 20 > 10 (parallels carbocation stability!) 11-32
E1 always competes with SN1 CH3
CH3
H3 C C Br
H3C C
CH3
+
Br
CH3
CH3OH as nucleophile
CH3OH as base
CH3 H3 C C OCH3
CH3 H2 C C CH3
CH3
SN1
E1 Mixture of Products
11-33
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 Requirements
Anti-coplanar H and X
None required
Rearrangements
No
Yes 11-34
