C X C Nu C C C C
1
CHAPTER 11 - REACTIONS OF ALKYL HALIDES NUCLEOPHILIC SUBSTITUTIONS AND ELIMINATIONS: The Heart of Organic Reaction Mechanisms A general scheme of the two types of reactions is shown. In nucleophilic substitution, a nucleophile (anion or neutral species) replaces the halide. Nu
C
X
C
Nu
+ X-
In elimination, a base abstracts a proton from the carbon adjacent to the halide carbon and a pi bond forms when the halide leaves, resulting in an alkene product. H Base
C
C
C
C
X No mechanism for these reactions is implied in the above schematic representations. But these reactons have been extensively studied and two mechanisms will be examined for each of these reactions. Nucleophilic Substitution 1. SN2 - Bimolecular Nucleophilic Substitution Walden Inversion Late in the nineteenth century, German chemist Paul Walden discovered he could change one optical isomer into its enantiomer in a two step sequence and back again to the original isomer.
1
2 O
* CH C
C CH2 HO
OH (-)-malic acid
O
O
PCl5
C CH2
OH
HO
* CH C
OH Cl (+)-chlorosuccinic acid
Ag2O, H2 O O * CH C
C CH2 HO
O
Ag2O, H2 O O
PCl5
C CH2
OH
Cl
O
HO
* CH C OH
(-)-chlorocuccinic acid
O OH
(+)-malic acid
Since there is no relationship between (+) and (-) and absolute configuration, it was not known at the time in which step inversion of configuration took place. Further study was done in the 1920's - with compounds with only one functional group that could be reacting. *
TsCl
CH3COO -
*
O-H H (+)-1-phenyl-2-propanol [a]D = +33.0˚
O-Ts
H
* H
AcO
H2O, HOCH3COO -
* H
*
OAc
TsO
*
TsCl H
HO
H
(-)-1-phenyl-2-propanol [a]D = -33.0˚ O S O
TsCl = H3C toayl chloride = p-toluenesulfonyl chloride
Cl
AcO- = CH3 acetate =
O C
O-
In the first reaction (with TsCl) the O-H bond is broken and the C-O bond remains intact. The inversion of configuration takes place in the second step (with acetate). The hydrolysis of esters was studied, and it was known that the
2
3
-O---C=O is the bond that is broken. The only step that involves the asymmetric carbon is the nucleophilic substitution step. Conclusion was that nucleophilic substitution of 1˚ and 2˚ halides and tosylates takes place with inversion of configuration. What is the mechanism that is occurring here? Other studies of the reaction revealed several things, primarily about the kinetics and stereochemistry of the reaction. Kinetics Kinetics studies the rate of the reaction. Is it fast or slow. How is rate affected by the concentration of the reactants. HO
-
+ CH3 -Br
HO-CH3 + Br-
At a given temperature and concentration of reactants, this reaction occurs at a certain rate. In this reaction the rate changed with a change in concentration of either the hydroxide or the halide. Rate ~ [HO-] [CH3Br] Rate = k [HO-] [CH3Br] where k is "rate constant" for the reaction. This reaction is second order. If double the conc. of hydroxide, double the reaction rate. If halve the concentration of the alkyl halide, halve the reaction rate. It is given the name SN2 - substitution, nucleophilic, bimolecular Takes place in one step - CONCERTED - no intermediate, one transition state. So we know that the reaction takes place with inversion of configuration and is a one step concerted reaction that includes both the nucleophile and the alkyl halide substrate in the one step.
3
4
H HO
-
+ CH3 -Br
dHO
C
dBr
HO-CH3 + Br-
HH
transition state The potential energy diagram is as follows: Transition state
Energy Starting material Products Progress of reaction
What are other variables in the reaction scheme - other than concentration of reactants - that have an effect on the reaction? To change the rate of a reaction, must do something to lower the activation energy (DG+ +) Two possibilities are shown below.
4
5
Transition state
Energy
Starting material Products Progress of reaction Lower energy of transition state and the energy of activiation gets smaller.
Transition state
Energy
Starting material Products Progress of reaction Raise energy of starting material and the energy of activiation gets smaller.
What structural changes in the reactants or changes in reaction conditions can be made to affect the peed of reaction?. We will look at possible variables.
5
6
Substrate Change R group of alkyl halide. Look at the transition state. Very crowded with new group coming in before halide has left. Steric hindrance plays a role. CH3-X best > 1˚ > 2˚ > 3˚ Primary halides have lower transition state energy. Tertiary halides have high transition state energy. SN2 reactions occur at relatively unhindered sites. In addition the two examples below DO NOT HAPPEN BY THIS MECHANISM.
X
Nu
-
X
Nu
X
Backside attacks on sp2 centers such as vinyl and benzene do not happen.
Cl
Leaving group Good leaving groups are those species that are able to accommodate negative charge. Weak bases are good leaving groups. (Conjugate bases of strong acids) TsO- ~ I- > Br- > Cl- >> F- >> HO-, RO- > H2N-
Best
Bad
If the negative charge on the leaving group is more stabilized as it is leaving, then the transition state is more stable. So... better leaving group, reaction goes faster. (See second energy diagram above).. Attacking Nucleophile Nucleophiles can be anionic or neutral, as long as they have a lone pair of electrons. Anionic nucleophile Nu
+
R-Y
Neutral nucleophile R-Nu +
Y
Nu
6
+
R-Y
R-Nu +
Y
7
Examples of nucleophiles: HS-, CN-, I-, CH3O-, HO-, Cl-, NH3, H2O, CH3OH Nucleophilicity if the affinity of a species for carbon atoms in an SN2 reaction, but it is hard to measure absolutely. When reaction conditions change, relative nucleophilicity may change. In general anions are better nucleophiles than neutral species. Within a group on the periodic table, nucleophilicity increases as you go down the group. HO- < HS- < HSeIf comparing a set of nucleophiles, all with the same attacking atom, nucleophilicity parallels basicity. (Must know pKa of conjugate acids). So CH3OHO-
O-
O R
C
O-
H2O
Stronger nucleophiles react faster in SN2 reactions. Solvent Types of solvents: polar protic - ROH, HOH, RNH2 polar aprotic - DMF*, DMSO*, HMPA,* acetonitrile non polar - Hexanes, CCl4 (* See text for structures of these solvents)
Polar protic solvents (with an H that may supply a proton) "solvate" both cation and anion species. That is, the cation and the anion are surrounded with solvent molecules so as to stabilize the + or - charge on these ions. The ions are not particularly free to attack in a reaction and the lowering of the energy of the species (that are part of the starting materials in the system) leads to a higher DG+ +.
7
8 H-OR RO-H
Nu
H-OR
Since we want the anion (nucleophile) to be free to attack in an SN2 reaction this situation is not good. Non-polar solvents will solvate neither the cation nor the anion because they have little affinity for each other. However, such solvents (e.g. hexane) are totally inappropriate for these reactions that involve salts (e.g. NaCN, KN3 , NaOCH3 , etc.) since salts are completely insoluble in non-polar solvents. But the best situation is the use of polar aprotic ("without protons") because these solvate the cations but leave the anionic nucleophile free - in effect raising the energy of the starting material.... therefore lowers DG+ +. And, many salts are quite soluble in these solvents. For the reaction below the relative rates in various solvents was determined. -
-
CH3CH2CH2CH2Br + N3 ---------> CH3CH2CH2CH2N3 + Br HMPA Relative rate 200,000 Solvent
CH3CN 5,000
DMF 2,800
DMSO 1,300
H2O 7
CH3OH 1
TO SUMMARIZE FOR SN2. Substrate: Steric hindrance raises energy of transition state/ Therefore CH3 > 1˚ > 2˚ > 3˚ Nucleophile: More reactive nucleophile, higher energy starting material therefore lower activation energy, reaction faster Leaving group: Best leaving groups form more stable anions
8
9
Solvent:
Polar aprotic solvents solvate cation but leave anion (nucleophile) free and of higher starting energy --> lower DG+ +.
Not all nucleophilic substitution reactions follow these parameters. Under certain conditions 3˚ halides react faster, and the stereochemistry is different. There is another mechanism that occurs under different reaction conditions than those favored for SN2. Unimolecular Nucleophilic Substitution Consider the reaction: CH3 H3C C Br
CH3 H3C C OH
HOH
CH3
CH3
Tertiary halides in neutral water as nucleophile (and solvent) react about one million times faster that CH3Br in HOH. In fact when various halides are studied under these reaction conditions, the following relative rates are observed. R groups Relative rate
3˚ > 6 1.2 x 10
2˚ 12
>
1˚ 1
> CH3Br 1
Examine same aspects of this reaction Kinetics Rate depends only on concentration of halides (or tosylate) Rate = k [RX] Nucleophile does not take part in the rate determining step. The substrate must lose the anion first in the slowest rate-determining step and then the nucleophile comes in. Since only the substrate is involved in the rate determining step, this reaction is first order. It is still a nucleophilic substitution. So we give it the name
SN1, substitution nucleophilic unimolecular
9
10
This is a two step reaction, two transition states with an intermediate species (carbocation)
Transition State
Transition State
Carbocation Intermefiate
Product
Starting Materials
The mechanism for the above reaction is as shown CH3 H3C C
CH3 H3C C
Br
CH3
CH3 CH3 + H3C C
+ Br
H O H
CH3
CH3 H H3C C O H CH3
The product of these two steps is a protonated species called an oxonium ion. There is a very fast third step in this reaction (not in all SN1 reactions) that is a deprotonation step. CH3 H H3C C O H CH3
CH3 H3C C OH
Br
CH3
The actual energy diagram for this reaction would look like this.
10
11
Transition State
Transition State
Carbocation Intermefiate
Transition State Protonated Product
Starting Materials
Not all SN1 reactions have the last deprotonation step
Product
Stereochemistry Look at the carbocation intermediate carefully
C
Br
C
Carbocation is planar with the two lobes of the empty p orbital perpendicular to the plane of the carbon and atoms attached to it. Would predict that nucleophile could come in from either side equally well and would get a racemic mixture. Most SN1 reactions do not give a complete racemate. There is approximately 020% inversion (excess of the inverted product). Example: A reaction that gives 20% inversion has 80% racemized product. retained stereochemistry 80% racemate Totals
inverted stereochemistry 20%
40% 40%
40% 60%
This is explained as per schematic below
11
12
Nu C
Br
Br
C
C
This face is still blocked to some extent
Eventually a free C+ and both faces are open
There is some time when the face from which the Br- is leaving is blocked by the large anionic leaving group and therefore extra time when the incoming nucleophile attacks from the back - leading to more inverted stereochemical product. Substrate Reaction rate depends on carbocation stability. More stable the C+, the faster the reaction 3˚ > 2˚ ~ benzylic ~ allylic > 1˚ > methyl Benzylic and allylic are equally reactive in SN1 and SN2 reactions.
Leaving group Same as SN2. TsO- ~ I- > Br- > Cl- ~ H2O -OH is a very poor leaving group under SN1 or SN2 conditions - or anything else. But if OH is first protonated (acidic conditions, to give -OH2 +), the neutral HOH is the leaving group. This is a fairly good leaving group. Nucleophile Nucleophile has no kinetic effect on the SN1 reaction. Does not enter into reaction mechanism until after the rate-determining formation of C+.
12
13
Solvent Solvents have effects on SN1 as they do in SN2 reactions but the reasons are different. In an SN2 reaction the effect of the solvent in due to the stabilization or destabilization of the nucleophilic attacking agent. In SN1 reactions the effect in on the stabilization or destabilization of the transition state. Remember that the transition state in an SN1 reaction is a carbocation; so anything that can stabilize this ion will lower the transition state energy and thus the energy of activation. The reaction takes place more rapidly. Polar solvents stabilize the carbocation by solvation. Solvent molecules orient themselves around the carbocation such that the "negative" ends of the solvent point toward the cation. H
H O
H O C+
H O H
O
H O H H
H H It is absolutely essential that the solvents for SN1 reaction be protic –there must be effective hydrogen bonding to help pull off the leaving group. SN1 reactions are therefore invariably run in aqueous, alcoholic, or carboxylic acid solvents. We can see these effects in the relative rates of the reaction below going from ethanol to water CH3 H3C C Cl
+
CH3 H3C C OR
ROH
CH3 Ethanol Relative reactivity
1
+ HCl
CH3 40% Water/ 60% Ethanol 100
80%Water/ 20% Ethanol 14,000
13
Water 100,000
14
TO SUMMARIZE FOR SN1: Substrate: Substrates that yield stable carbocations are best 3˚ > 2˚ > 1˚> CH3 Leaving group: Best leaving groups form more stable anions. Order the same as for SN2. Nucleophile: Nucleophile should be non-basic to prevent elimination, Otherwise has no effect on reaction rate. Neutral nucleophiles are OK. Solvent:
Polar & protic solvents are best for SN1 reactions. They stabilize both the carbocation intermediate and the leaving group.
Elimination Reactions Elimination is a competitive reaction of substitution of alkyl halides, especially when the nucleophile is also a strong base. These reactions are more complex for several reasons. 1. Regiochemistry - What products result from elimination of HX from an unsymmetrical halide? 2. Product mixtures: The product is almost always a mixture. Can usually predict which will be the major product. Zaitsev's Rule When more than one product is possible from an elimination reaction, the more highly substituted alkene will be the major product. This is the more stable product. Alexander Zaitsev based this rule on numerous experimental data in 1875. CH3 CH3CH 2 C CH3 Br 2-bromo-2-methylbutane
CH3 CH3CH C 70%
+ CH3
H3C C CH3CH 2
CH2
30%
3. Mechanism: As with substitution, two reaction pathways are possible. 1. E2 - Bimolecular elimination This is the more common (and important) of the two mechanisms that occurs when an alkyl halide is treated with a strong base.
14
15
Base
H C
+ Base-H + X-
C X Alkene
The reaction takes place in one step (as in SN2). Kinetic studies show that both the base and the alkyl halides enter into the rate-determining step. Rate = k x [RX] x [Base] Secondly, experiments show that the four atoms involved the reaction, the hydrogen, the two carbons and the leaving halide must be in the same plane, called periplanar. Two geometries are possible: syn periplanar, where the hydrogen and the halide are on the same side of the molecule; and anti periplanar, where the hydrogen and the halide are on opposite sides of the molecule. Anti periplanar is the favored of the two geometries. H R
R
R
R
X
Newman projection
H
X
R
R
If we examine the Newman projection of the C-C bond we can see what is meant by anti periplanar. The H, the two carbons and the X are all in the same plane and the H and the X are anti to each other. This arrangement is necessary so that as th two carbons change from sp3 to sp2 hybridization, the forming p orbitals are aligned to overlap.
The syn periplanar arrangement also allows for orbital overlap as the reaction proceeds, but this is a higher energy conformation than the anti and thus is usually not favored. R
R
We can use the dehydrobromination of 2-bromobutane to illustrate what we have learned so far.
15
16
KOH
H3C CH CH2CH 3 H2C Ethanol Br
H3C CHCH2CH3+
CH3 C
C
1-Butene ~10%
H C
+
H
H
H3C H
cis 2-Butene ~30%
C
trans 2-Butene ~60%
Mechanism: HO-
Path 1:
H H2C
C CH CH2CH 3 H2 Br Path 2:
H2C
CHCH2CH3
HO-
H CH
H3C
C H
H C
CH3
CH3
H
Br
H3C
CH3 C
+
C
H
C H
What are some of the stereochemical consequences of the periplanar geometry necessary for an E2 elimination? If meso-1,2-dibromo-1,2-diphenylethane (common name stilbene dibromide) undergoes E2 elimination of one mole of HBr, the product is only one alkene, pure E-alkene. Base H
H
Ph Br
Br
H
Ph =
Ph
Ph PhenylPh Br Convince yourself that if the alternate HBr is eliminated, the product is the same. Elimination reactions of halogenated cyclohexane rings illustrates the stereochemical consequences of the anti periplanar requirement for E2 reactions.
16
CH3
17
H
H
H H
H H
H
H H
H H
H
H Cl
H
H
H
H
H H
H
H In the Newman projections, we can see that the halide must be in an axial position in order to undergo E2 elimination. H
Cl
H
Cl
Newman projection with Cl in an equatorial position
Newman projection with Cl in an axial position
Examine the reactions of the isomers menthyl chloride and neomenthyl chloride under E2 conditions and find that neomenthyl chloride reacts 200x faster.
17
18
Cl H3C
H3C
CH3 CH CH3 Neomenthyl chloride
OEt EtOH fast
CH3 CH H CH3 Cl Menthyl chloride
H3C CH(CH3)2
H3C
CH(CH3)2
ring flip slow CH(CH3)2
CH(CH3)2
H
OEt CH3
Cl
EtOH fast
CH3
In order to have anti periplanar conformation in menthyl chloride the ring must flip from the energetically preferred conformer that has all substituents in equatorial positions. This requires significant energy and is therefore slow. The subsequent elimination occurs rapidly just as the elimination of the neomenthyl chloride isomer but the overall reaction of menthyl chloride is slower. The deuterium isotope effect also supports the one-step E2 mechanism. In the two reactions below, the compound from which a deuterium is eliminated reacts more slowly than that from which a hydrogen is eliminated. The C-D bond is stronger (thus requires more energy to break) than the C-H bond. This experimental result shows that breaking the C-H (or C-D) bond is part of the rate-determining step.
18
19 H C
Base CH2Br
C
H
Faster reaction
CH2
Slower reaction
H
D C
CH2
Base CH2Br
C
D
D
E1 - Unimolecular Elimination Like SN1 reactions, E1 reactions begin with the unimolecular dissociation resulting in a carbocation intermediate. In the second step of this two-step mechanism, an H+ is lost rather than substitution occurring. CH3 H3C C CH3
H3C Cl slow = rate H3C determining step
H C C H H
:Base
H3C H3C
C
CH2 + substitution product
SN1 and E1 processes usually take place in competition. The same substrates are favored in both of these reactions. E1 eliminations do not have the periplanar requirements of an SN2 reaction and if two elimination products are possible, we can expect Zaitsev's rule to hold and the more stable (more substituted) alkene will predominate. (See examples page 422 in text)
19
20
Comparison of Nucleophilic Substitution and Elimination Mechanisms Halide
SN1
SN2
1˚ RCH 2X
HIGHLY FAVORED
NO
2˚ R2CHX
Can occur with benzylic and allylic halides stable C+
3˚ R3CX
FAVORED IN HYDROXYLIC SOLVENTS ROH, HOH
Occurs in competition with E2 with nucleophiles that are not strong bases N3 -, CN-, SR NO
E1
E2
NO
Occurs with strong bases esp. bulky
Can occur with benzylic and allylic halides -
Favored when strong bases used OH,- OR, NH2
Occurs in competition with SN1
FAVORED WHEN BASES USED
Examined from a different point of view --Primary (1˚) RCH 2X
Primarily SN2 substitution
Secondary (2˚) R2 CHX
SN2 with non-basic nucleophiles E2 with strong bases
Tertiary (3˚) R3 CX
Mostly E2 (SN1 and E1 in non-basic solvents - under solvolysis conditions.
20
21 CHEM 233 HANDOUT ACID
pKa
CONJUGATE BASE
H-CH3
55
-
H-C=C-
36
-
H-NH2
35
H-C=CR
25
H-OC(CH3)3
18
H-OCH3
16
H-OH
15.72
H O
10
H-CN
9.2
very,very strong base
CH3
methyl anion
C=C-
vinylidene
-
NH2
amide
-
acetylide
C=CR
OC(CH3 )3 t-butoxide
-
OCH3
-
OH O
methoxide hydroxide
-
cyanide sulfhydryl
CN
H-SH
7.0
H-OC(O)R
5
-
H-F
3.2
-
fluoride
H-Cl
-7
-
chloride
HClO4
-10
SH
OC(O)R carboxylate F Cl
-
ClO4
strong bases
phenoxide
-
very strong acid
very strong
-
Increasing Basicity
Increasing Acidity
very,very weak acid
perchlorate very weak base
moderately strong bases
NOTES: Conjugate bases of weak acids are very strong bases. The bases listed on the left also have nucleophilic properties and can react as nucleophiles under certain reaction conditions. It is very difficult to make predictions on nucleophilicity based solely on basicity as outlined in this chart. Comparisons of nucleophilicity can be made for similar species in the same reaction under the same reaction conditions as outlined below. These are all for the reaction Nu:- +CH3 Br --> CH3 Nu + Br-
For example, we can compare species within a periodic group. I- > Br- > Cl- >> F - ; HSe- > HS - > HO- ; CH3S- > CH 3O- ; for similar species, nucleophilicity increases as you go down the periodic group. Also, if the attacking atom in the species is the same, nucleophilicity and basicity are related. So for the oxygen species, -Ot-Bu > -OCH3 > -OH > -OPhen >OC(O)R. The strongest base is the strongest nucleophile. Cross-comparison is difficult. KNOW THIS: For 1˚ halides and tosylates, the major reaction will be substitution (SN 2) except when bulky bases like -OC(CH 3)3 are used. Then the major product will be elimination (E2). For 3˚ halides and tosylates, the major reaction with any anionic base/nucleophile will be elimination. Substitution will be the major product ONLY with very weak, non-anionic bases which are also used as the solvent (HOH) or (ROH). For 2˚ halides and tosylates, mixtures occur but the reaction can be made to favor either substitution or elimination by the rules above. In all cases, some of the unwanted product is produced but can be minimized by setting reaction conditions. 21
CHAPTER 11 - REACTIONS OF ALKYL HALIDES NUCLEOPHILIC SUBSTITUTIONS AND ELIMINATIONS: The Heart of Organic Reaction Mechanisms A general scheme of the two types of reactions is shown. In nucleophilic substitution, a nucleophile (anion or neutral species) replaces the halide. Nu
C
X
C
Nu
+ X-
In elimination, a base abstracts a proton from the carbon adjacent to the halide carbon and a pi bond forms when the halide leaves, resulting in an alkene product. H Base
C
C
C
C
X No mechanism for these reactions is implied in the above schematic representations. But these reactons have been extensively studied and two mechanisms will be examined for each of these reactions. Nucleophilic Substitution 1. SN2 - Bimolecular Nucleophilic Substitution Walden Inversion Late in the nineteenth century, German chemist Paul Walden discovered he could change one optical isomer into its enantiomer in a two step sequence and back again to the original isomer.
1
2 O
* CH C
C CH2 HO
OH (-)-malic acid
O
O
PCl5
C CH2
OH
HO
* CH C
OH Cl (+)-chlorosuccinic acid
Ag2O, H2 O O * CH C
C CH2 HO
O
Ag2O, H2 O O
PCl5
C CH2
OH
Cl
O
HO
* CH C OH
(-)-chlorocuccinic acid
O OH
(+)-malic acid
Since there is no relationship between (+) and (-) and absolute configuration, it was not known at the time in which step inversion of configuration took place. Further study was done in the 1920's - with compounds with only one functional group that could be reacting. *
TsCl
CH3COO -
*
O-H H (+)-1-phenyl-2-propanol [a]D = +33.0˚
O-Ts
H
* H
AcO
H2O, HOCH3COO -
* H
*
OAc
TsO
*
TsCl H
HO
H
(-)-1-phenyl-2-propanol [a]D = -33.0˚ O S O
TsCl = H3C toayl chloride = p-toluenesulfonyl chloride
Cl
AcO- = CH3 acetate =
O C
O-
In the first reaction (with TsCl) the O-H bond is broken and the C-O bond remains intact. The inversion of configuration takes place in the second step (with acetate). The hydrolysis of esters was studied, and it was known that the
2
3
-O---C=O is the bond that is broken. The only step that involves the asymmetric carbon is the nucleophilic substitution step. Conclusion was that nucleophilic substitution of 1˚ and 2˚ halides and tosylates takes place with inversion of configuration. What is the mechanism that is occurring here? Other studies of the reaction revealed several things, primarily about the kinetics and stereochemistry of the reaction. Kinetics Kinetics studies the rate of the reaction. Is it fast or slow. How is rate affected by the concentration of the reactants. HO
-
+ CH3 -Br
HO-CH3 + Br-
At a given temperature and concentration of reactants, this reaction occurs at a certain rate. In this reaction the rate changed with a change in concentration of either the hydroxide or the halide. Rate ~ [HO-] [CH3Br] Rate = k [HO-] [CH3Br] where k is "rate constant" for the reaction. This reaction is second order. If double the conc. of hydroxide, double the reaction rate. If halve the concentration of the alkyl halide, halve the reaction rate. It is given the name SN2 - substitution, nucleophilic, bimolecular Takes place in one step - CONCERTED - no intermediate, one transition state. So we know that the reaction takes place with inversion of configuration and is a one step concerted reaction that includes both the nucleophile and the alkyl halide substrate in the one step.
3
4
H HO
-
+ CH3 -Br
dHO
C
dBr
HO-CH3 + Br-
HH
transition state The potential energy diagram is as follows: Transition state
Energy Starting material Products Progress of reaction
What are other variables in the reaction scheme - other than concentration of reactants - that have an effect on the reaction? To change the rate of a reaction, must do something to lower the activation energy (DG+ +) Two possibilities are shown below.
4
5
Transition state
Energy
Starting material Products Progress of reaction Lower energy of transition state and the energy of activiation gets smaller.
Transition state
Energy
Starting material Products Progress of reaction Raise energy of starting material and the energy of activiation gets smaller.
What structural changes in the reactants or changes in reaction conditions can be made to affect the peed of reaction?. We will look at possible variables.
5
6
Substrate Change R group of alkyl halide. Look at the transition state. Very crowded with new group coming in before halide has left. Steric hindrance plays a role. CH3-X best > 1˚ > 2˚ > 3˚ Primary halides have lower transition state energy. Tertiary halides have high transition state energy. SN2 reactions occur at relatively unhindered sites. In addition the two examples below DO NOT HAPPEN BY THIS MECHANISM.
X
Nu
-
X
Nu
X
Backside attacks on sp2 centers such as vinyl and benzene do not happen.
Cl
Leaving group Good leaving groups are those species that are able to accommodate negative charge. Weak bases are good leaving groups. (Conjugate bases of strong acids) TsO- ~ I- > Br- > Cl- >> F- >> HO-, RO- > H2N-
Best
Bad
If the negative charge on the leaving group is more stabilized as it is leaving, then the transition state is more stable. So... better leaving group, reaction goes faster. (See second energy diagram above).. Attacking Nucleophile Nucleophiles can be anionic or neutral, as long as they have a lone pair of electrons. Anionic nucleophile Nu
+
R-Y
Neutral nucleophile R-Nu +
Y
Nu
6
+
R-Y
R-Nu +
Y
7
Examples of nucleophiles: HS-, CN-, I-, CH3O-, HO-, Cl-, NH3, H2O, CH3OH Nucleophilicity if the affinity of a species for carbon atoms in an SN2 reaction, but it is hard to measure absolutely. When reaction conditions change, relative nucleophilicity may change. In general anions are better nucleophiles than neutral species. Within a group on the periodic table, nucleophilicity increases as you go down the group. HO- < HS- < HSeIf comparing a set of nucleophiles, all with the same attacking atom, nucleophilicity parallels basicity. (Must know pKa of conjugate acids). So CH3OHO-
O-
O R
C
O-
H2O
Stronger nucleophiles react faster in SN2 reactions. Solvent Types of solvents: polar protic - ROH, HOH, RNH2 polar aprotic - DMF*, DMSO*, HMPA,* acetonitrile non polar - Hexanes, CCl4 (* See text for structures of these solvents)
Polar protic solvents (with an H that may supply a proton) "solvate" both cation and anion species. That is, the cation and the anion are surrounded with solvent molecules so as to stabilize the + or - charge on these ions. The ions are not particularly free to attack in a reaction and the lowering of the energy of the species (that are part of the starting materials in the system) leads to a higher DG+ +.
7
8 H-OR RO-H
Nu
H-OR
Since we want the anion (nucleophile) to be free to attack in an SN2 reaction this situation is not good. Non-polar solvents will solvate neither the cation nor the anion because they have little affinity for each other. However, such solvents (e.g. hexane) are totally inappropriate for these reactions that involve salts (e.g. NaCN, KN3 , NaOCH3 , etc.) since salts are completely insoluble in non-polar solvents. But the best situation is the use of polar aprotic ("without protons") because these solvate the cations but leave the anionic nucleophile free - in effect raising the energy of the starting material.... therefore lowers DG+ +. And, many salts are quite soluble in these solvents. For the reaction below the relative rates in various solvents was determined. -
-
CH3CH2CH2CH2Br + N3 ---------> CH3CH2CH2CH2N3 + Br HMPA Relative rate 200,000 Solvent
CH3CN 5,000
DMF 2,800
DMSO 1,300
H2O 7
CH3OH 1
TO SUMMARIZE FOR SN2. Substrate: Steric hindrance raises energy of transition state/ Therefore CH3 > 1˚ > 2˚ > 3˚ Nucleophile: More reactive nucleophile, higher energy starting material therefore lower activation energy, reaction faster Leaving group: Best leaving groups form more stable anions
8
9
Solvent:
Polar aprotic solvents solvate cation but leave anion (nucleophile) free and of higher starting energy --> lower DG+ +.
Not all nucleophilic substitution reactions follow these parameters. Under certain conditions 3˚ halides react faster, and the stereochemistry is different. There is another mechanism that occurs under different reaction conditions than those favored for SN2. Unimolecular Nucleophilic Substitution Consider the reaction: CH3 H3C C Br
CH3 H3C C OH
HOH
CH3
CH3
Tertiary halides in neutral water as nucleophile (and solvent) react about one million times faster that CH3Br in HOH. In fact when various halides are studied under these reaction conditions, the following relative rates are observed. R groups Relative rate
3˚ > 6 1.2 x 10
2˚ 12
>
1˚ 1
> CH3Br 1
Examine same aspects of this reaction Kinetics Rate depends only on concentration of halides (or tosylate) Rate = k [RX] Nucleophile does not take part in the rate determining step. The substrate must lose the anion first in the slowest rate-determining step and then the nucleophile comes in. Since only the substrate is involved in the rate determining step, this reaction is first order. It is still a nucleophilic substitution. So we give it the name
SN1, substitution nucleophilic unimolecular
9
10
This is a two step reaction, two transition states with an intermediate species (carbocation)
Transition State
Transition State
Carbocation Intermefiate
Product
Starting Materials
The mechanism for the above reaction is as shown CH3 H3C C
CH3 H3C C
Br
CH3
CH3 CH3 + H3C C
+ Br
H O H
CH3
CH3 H H3C C O H CH3
The product of these two steps is a protonated species called an oxonium ion. There is a very fast third step in this reaction (not in all SN1 reactions) that is a deprotonation step. CH3 H H3C C O H CH3
CH3 H3C C OH
Br
CH3
The actual energy diagram for this reaction would look like this.
10
11
Transition State
Transition State
Carbocation Intermefiate
Transition State Protonated Product
Starting Materials
Not all SN1 reactions have the last deprotonation step
Product
Stereochemistry Look at the carbocation intermediate carefully
C
Br
C
Carbocation is planar with the two lobes of the empty p orbital perpendicular to the plane of the carbon and atoms attached to it. Would predict that nucleophile could come in from either side equally well and would get a racemic mixture. Most SN1 reactions do not give a complete racemate. There is approximately 020% inversion (excess of the inverted product). Example: A reaction that gives 20% inversion has 80% racemized product. retained stereochemistry 80% racemate Totals
inverted stereochemistry 20%
40% 40%
40% 60%
This is explained as per schematic below
11
12
Nu C
Br
Br
C
C
This face is still blocked to some extent
Eventually a free C+ and both faces are open
There is some time when the face from which the Br- is leaving is blocked by the large anionic leaving group and therefore extra time when the incoming nucleophile attacks from the back - leading to more inverted stereochemical product. Substrate Reaction rate depends on carbocation stability. More stable the C+, the faster the reaction 3˚ > 2˚ ~ benzylic ~ allylic > 1˚ > methyl Benzylic and allylic are equally reactive in SN1 and SN2 reactions.
Leaving group Same as SN2. TsO- ~ I- > Br- > Cl- ~ H2O -OH is a very poor leaving group under SN1 or SN2 conditions - or anything else. But if OH is first protonated (acidic conditions, to give -OH2 +), the neutral HOH is the leaving group. This is a fairly good leaving group. Nucleophile Nucleophile has no kinetic effect on the SN1 reaction. Does not enter into reaction mechanism until after the rate-determining formation of C+.
12
13
Solvent Solvents have effects on SN1 as they do in SN2 reactions but the reasons are different. In an SN2 reaction the effect of the solvent in due to the stabilization or destabilization of the nucleophilic attacking agent. In SN1 reactions the effect in on the stabilization or destabilization of the transition state. Remember that the transition state in an SN1 reaction is a carbocation; so anything that can stabilize this ion will lower the transition state energy and thus the energy of activation. The reaction takes place more rapidly. Polar solvents stabilize the carbocation by solvation. Solvent molecules orient themselves around the carbocation such that the "negative" ends of the solvent point toward the cation. H
H O
H O C+
H O H
O
H O H H
H H It is absolutely essential that the solvents for SN1 reaction be protic –there must be effective hydrogen bonding to help pull off the leaving group. SN1 reactions are therefore invariably run in aqueous, alcoholic, or carboxylic acid solvents. We can see these effects in the relative rates of the reaction below going from ethanol to water CH3 H3C C Cl
+
CH3 H3C C OR
ROH
CH3 Ethanol Relative reactivity
1
+ HCl
CH3 40% Water/ 60% Ethanol 100
80%Water/ 20% Ethanol 14,000
13
Water 100,000
14
TO SUMMARIZE FOR SN1: Substrate: Substrates that yield stable carbocations are best 3˚ > 2˚ > 1˚> CH3 Leaving group: Best leaving groups form more stable anions. Order the same as for SN2. Nucleophile: Nucleophile should be non-basic to prevent elimination, Otherwise has no effect on reaction rate. Neutral nucleophiles are OK. Solvent:
Polar & protic solvents are best for SN1 reactions. They stabilize both the carbocation intermediate and the leaving group.
Elimination Reactions Elimination is a competitive reaction of substitution of alkyl halides, especially when the nucleophile is also a strong base. These reactions are more complex for several reasons. 1. Regiochemistry - What products result from elimination of HX from an unsymmetrical halide? 2. Product mixtures: The product is almost always a mixture. Can usually predict which will be the major product. Zaitsev's Rule When more than one product is possible from an elimination reaction, the more highly substituted alkene will be the major product. This is the more stable product. Alexander Zaitsev based this rule on numerous experimental data in 1875. CH3 CH3CH 2 C CH3 Br 2-bromo-2-methylbutane
CH3 CH3CH C 70%
+ CH3
H3C C CH3CH 2
CH2
30%
3. Mechanism: As with substitution, two reaction pathways are possible. 1. E2 - Bimolecular elimination This is the more common (and important) of the two mechanisms that occurs when an alkyl halide is treated with a strong base.
14
15
Base
H C
+ Base-H + X-
C X Alkene
The reaction takes place in one step (as in SN2). Kinetic studies show that both the base and the alkyl halides enter into the rate-determining step. Rate = k x [RX] x [Base] Secondly, experiments show that the four atoms involved the reaction, the hydrogen, the two carbons and the leaving halide must be in the same plane, called periplanar. Two geometries are possible: syn periplanar, where the hydrogen and the halide are on the same side of the molecule; and anti periplanar, where the hydrogen and the halide are on opposite sides of the molecule. Anti periplanar is the favored of the two geometries. H R
R
R
R
X
Newman projection
H
X
R
R
If we examine the Newman projection of the C-C bond we can see what is meant by anti periplanar. The H, the two carbons and the X are all in the same plane and the H and the X are anti to each other. This arrangement is necessary so that as th two carbons change from sp3 to sp2 hybridization, the forming p orbitals are aligned to overlap.
The syn periplanar arrangement also allows for orbital overlap as the reaction proceeds, but this is a higher energy conformation than the anti and thus is usually not favored. R
R
We can use the dehydrobromination of 2-bromobutane to illustrate what we have learned so far.
15
16
KOH
H3C CH CH2CH 3 H2C Ethanol Br
H3C CHCH2CH3+
CH3 C
C
1-Butene ~10%
H C
+
H
H
H3C H
cis 2-Butene ~30%
C
trans 2-Butene ~60%
Mechanism: HO-
Path 1:
H H2C
C CH CH2CH 3 H2 Br Path 2:
H2C
CHCH2CH3
HO-
H CH
H3C
C H
H C
CH3
CH3
H
Br
H3C
CH3 C
+
C
H
C H
What are some of the stereochemical consequences of the periplanar geometry necessary for an E2 elimination? If meso-1,2-dibromo-1,2-diphenylethane (common name stilbene dibromide) undergoes E2 elimination of one mole of HBr, the product is only one alkene, pure E-alkene. Base H
H
Ph Br
Br
H
Ph =
Ph
Ph PhenylPh Br Convince yourself that if the alternate HBr is eliminated, the product is the same. Elimination reactions of halogenated cyclohexane rings illustrates the stereochemical consequences of the anti periplanar requirement for E2 reactions.
16
CH3
17
H
H
H H
H H
H
H H
H H
H
H Cl
H
H
H
H
H H
H
H In the Newman projections, we can see that the halide must be in an axial position in order to undergo E2 elimination. H
Cl
H
Cl
Newman projection with Cl in an equatorial position
Newman projection with Cl in an axial position
Examine the reactions of the isomers menthyl chloride and neomenthyl chloride under E2 conditions and find that neomenthyl chloride reacts 200x faster.
17
18
Cl H3C
H3C
CH3 CH CH3 Neomenthyl chloride
OEt EtOH fast
CH3 CH H CH3 Cl Menthyl chloride
H3C CH(CH3)2
H3C
CH(CH3)2
ring flip slow CH(CH3)2
CH(CH3)2
H
OEt CH3
Cl
EtOH fast
CH3
In order to have anti periplanar conformation in menthyl chloride the ring must flip from the energetically preferred conformer that has all substituents in equatorial positions. This requires significant energy and is therefore slow. The subsequent elimination occurs rapidly just as the elimination of the neomenthyl chloride isomer but the overall reaction of menthyl chloride is slower. The deuterium isotope effect also supports the one-step E2 mechanism. In the two reactions below, the compound from which a deuterium is eliminated reacts more slowly than that from which a hydrogen is eliminated. The C-D bond is stronger (thus requires more energy to break) than the C-H bond. This experimental result shows that breaking the C-H (or C-D) bond is part of the rate-determining step.
18
19 H C
Base CH2Br
C
H
Faster reaction
CH2
Slower reaction
H
D C
CH2
Base CH2Br
C
D
D
E1 - Unimolecular Elimination Like SN1 reactions, E1 reactions begin with the unimolecular dissociation resulting in a carbocation intermediate. In the second step of this two-step mechanism, an H+ is lost rather than substitution occurring. CH3 H3C C CH3
H3C Cl slow = rate H3C determining step
H C C H H
:Base
H3C H3C
C
CH2 + substitution product
SN1 and E1 processes usually take place in competition. The same substrates are favored in both of these reactions. E1 eliminations do not have the periplanar requirements of an SN2 reaction and if two elimination products are possible, we can expect Zaitsev's rule to hold and the more stable (more substituted) alkene will predominate. (See examples page 422 in text)
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20
Comparison of Nucleophilic Substitution and Elimination Mechanisms Halide
SN1
SN2
1˚ RCH 2X
HIGHLY FAVORED
NO
2˚ R2CHX
Can occur with benzylic and allylic halides stable C+
3˚ R3CX
FAVORED IN HYDROXYLIC SOLVENTS ROH, HOH
Occurs in competition with E2 with nucleophiles that are not strong bases N3 -, CN-, SR NO
E1
E2
NO
Occurs with strong bases esp. bulky
Can occur with benzylic and allylic halides -
Favored when strong bases used OH,- OR, NH2
Occurs in competition with SN1
FAVORED WHEN BASES USED
Examined from a different point of view --Primary (1˚) RCH 2X
Primarily SN2 substitution
Secondary (2˚) R2 CHX
SN2 with non-basic nucleophiles E2 with strong bases
Tertiary (3˚) R3 CX
Mostly E2 (SN1 and E1 in non-basic solvents - under solvolysis conditions.
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21 CHEM 233 HANDOUT ACID
pKa
CONJUGATE BASE
H-CH3
55
-
H-C=C-
36
-
H-NH2
35
H-C=CR
25
H-OC(CH3)3
18
H-OCH3
16
H-OH
15.72
H O
10
H-CN
9.2
very,very strong base
CH3
methyl anion
C=C-
vinylidene
-
NH2
amide
-
acetylide
C=CR
OC(CH3 )3 t-butoxide
-
OCH3
-
OH O
methoxide hydroxide
-
cyanide sulfhydryl
CN
H-SH
7.0
H-OC(O)R
5
-
H-F
3.2
-
fluoride
H-Cl
-7
-
chloride
HClO4
-10
SH
OC(O)R carboxylate F Cl
-
ClO4
strong bases
phenoxide
-
very strong acid
very strong
-
Increasing Basicity
Increasing Acidity
very,very weak acid
perchlorate very weak base
moderately strong bases
NOTES: Conjugate bases of weak acids are very strong bases. The bases listed on the left also have nucleophilic properties and can react as nucleophiles under certain reaction conditions. It is very difficult to make predictions on nucleophilicity based solely on basicity as outlined in this chart. Comparisons of nucleophilicity can be made for similar species in the same reaction under the same reaction conditions as outlined below. These are all for the reaction Nu:- +CH3 Br --> CH3 Nu + Br-
For example, we can compare species within a periodic group. I- > Br- > Cl- >> F - ; HSe- > HS - > HO- ; CH3S- > CH 3O- ; for similar species, nucleophilicity increases as you go down the periodic group. Also, if the attacking atom in the species is the same, nucleophilicity and basicity are related. So for the oxygen species, -Ot-Bu > -OCH3 > -OH > -OPhen >OC(O)R. The strongest base is the strongest nucleophile. Cross-comparison is difficult. KNOW THIS: For 1˚ halides and tosylates, the major reaction will be substitution (SN 2) except when bulky bases like -OC(CH 3)3 are used. Then the major product will be elimination (E2). For 3˚ halides and tosylates, the major reaction with any anionic base/nucleophile will be elimination. Substitution will be the major product ONLY with very weak, non-anionic bases which are also used as the solvent (HOH) or (ROH). For 2˚ halides and tosylates, mixtures occur but the reaction can be made to favor either substitution or elimination by the rules above. In all cases, some of the unwanted product is produced but can be minimized by setting reaction conditions. 21
