1. Chapter 11
Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations
Suggested Problems -
1-20, 26-30,31, 36-8

2. Alkyl Halides In general, alkyl halides have good leaving groups.
Alkyl halides have relatively good leaving groups – the halide ions are easily displaced.
Substitution reactions are important in organic chemistry because they make it possible to convert readily available alkyl halides into a wide variety of other compounds.

3. Alkyl Halides Undergo Substitution and Elimination Reactions
Substitution reaction—the electronegative group is replaced by another group.
In a substitution reaction, the electronegative atom or electron-withdrawing group is replaced by another atom or group.
In an elimination reaction, the electronegative atom or electron-withdrawing group is eliminated, along with a hydrogen from an adjacent carbon.
The atom or group that is substituted or eliminated is called a leaving group. Halides are good leaving groups.
Elimination reaction—the electronegative group is eliminated along with a hydrogen.

4. A Substitution Reaction
More precisely called a nucleophilic substitution reaction because the atom replacing the halogen is a nucleophile.
We will see that there are two different mechanisms by which a substitution reaction an take place.
In both mechanisms, the nucleophile replaces the leaving group, so the substitution reaction is more precisely called a nucleophilic substitution reaction.
Here we have a reaction in which the bromide in an alkyl bromide is being replaced by the hydroxide ion. An alcohol is the resulting product.

5. What is the Mechanism of the Reaction?
The kinetics of a reaction - the factors that affect the rate of the reaction - can help determine the mechanism.
Much can be learned about the mechanism of a reaction by studying its kinetics.
For example, the rate of this nucleophilic substitution reaction, in which an alkyl bromide reacts with a hydroxide ion, depends on the concentrations of both reactants.
If the concentration of either of the two reactants, methyl bromide or hydroxide ion, is doubled, the rate of the reaction doubles.
If the concentration of both of the reactants is doubled, the rate of the reaction quadruples.
Because we know the relationship between the rate of the reaction and the concentration of the reactants, we can write a rate law for the reaction (upper equation).
The proportionality sign can be replaced by an equals sign and a proportionality constant (k).
This reaction is a second-order reaction because its rate depends linearly on the concentration of each of the two reactants.
The rate law tells us which molecules are involved in the transition state of the rate-determining step of the reaction. Thus, the rate law for the reaction of bromomethane with hydroxide ion tells us that both bromomethane and hydroxide ion are involved in the rate-determining transition state.
The proportionality constant is called a rate constant. The magnitude of the rate constant for a particular reaction indicates how difficult it is for the reactants to overcome the energy barrier of the reaction – that is, how hard it is to reach the transition state.
The larger the rate constant, the lower is the energy barrier and, therefore, the easier it is for the reactants to reach the transition state.
The reaction of bromomethane with hydroxide ion is an example of an SN2 reaction. The “S” stands for substitution, the “N” stands for nucleophilic, and the “2” stands for bimolecular. Bimolecular means that two molecules are involved in the transition state of the rate-determining-step.
The mechanism for the SN2 reaction was based on three pieces of experimental evidence:
 1) The rate of the substitution reaction depends on the concentration of the alkyl halide and on the concentration of the nucleophile, indicating that both reactants are involved in the transition state of the rate-determining-step.
an SN2 reaction

6. Relative Rates of an SN2 Reaction
2) As the alkyl group becomes larger or as the hydrogens of bromomethane are successively replaced with methyl groups, the rate of the substitution reaction with a given nucleophile becomes slower.

7. Inverted Configuration
3) The substitution reaction of an alkyl halide in which the halogen is bonded to an asymmetric center leads to the formation of only one stereoisomer, and the configuration of the asymmetric center in the product is inverted relative to its configuration in the reacting alkyl halide.
If the halogen is bonded to an asymmetric center,
the product will have the inverted configuration.

8. Summary of the Experimental Evidence
for the Mechanism of an SN2 Reaction
The rate of the reaction is dependent on the concentration of both the alkyl halide and the nucleophile.
The relative rate of the reaction is
1o alkyl halide > 2o alkyl halide > 3o alkyl halide
The configuration of the product is inverted compared to the configuration of the reacting chiral alkyl halide.

9. The Mechanism back-side attack
Using the preceding evidence, Hughes and Ingold proposed that an SN2 Reaction is a concerted reaction – it takes place in a single step – so no intermediates are formed.
The nucleophile attacks the back side of the carbon that bears the leaving group and displaces it.
A productive collision is one that leads to the formation of the product. A productive collision in an SN2 reaction requires the nucleophile to hit the carbon on the side opposite the side that is bonded to the leaving group. The carbon is said to undergo back-side attack.

10. Why Back-Side Attack?
The nucleophile attacks from the back side in simplest terms because the leaving group blocks approach of the nucleophile to the front side.
MO theory explains why back-side attack is required. In order to form a bond, the HOMO of one species must interact with the LUMO of the other.
Therefore, when the nucleophile approaches the alkyl halide to form a new bond, the non-bonding MO of the nucleophile (its HOMO) must interact with the empty σ* antibonding MO associated with the C-Br bond (its LUMO).
In a back-side attack, a bonding interaction occurs between the nucleophile and the larger lobe of the σ* antibonding MO.

11. If Front-Side Attack
But when the nucleophile approaches the front side of the carbon, both a bonding and an antibonding interaction occur, so the two cancel each other and no bond forms.
Therefore, an SN2 reaction can be successful only if the nucleophile approaches the sp3 carbon from its back side.

12. Why Bimolecular?
The mechanism shows that the alkyl halide and the nucleophile are both in the transition state of the one-step reaction.
Therefore, increasing the concentration of either of them makes their collision more probable, so the rate of the reaction will depend on the concentration of both.

13. Why Do Methyl Halides React the Fastest and Tertiary the Slowest?
Bulky substituents attached to the carbon that undergoes back-side attack will decrease the nucleophile’s access to the back side of the carbon and will therefore decrease the rate of the reaction.
Steric effects are effects caused by the fact that groups occupy a certain volume of space.
A steric effect that decreases reactivity is called steric hindrance.
The relative reactivities of alkyl halides in an SN2 reaction are sensitive to the steric congestion around the reacting center.
Thus the relative reactivities of alkyl halides in an SN2 reaction are: methyl halide > 1o alkyl halide > 2o alkyl halide > 3o alkyl halide (too unreactive to undergo SN2 reaction).
The three alkyl groups of a tertiary alkyl halide make it impossible for the nucleophile to come within bonding distance of the tertiary carbon, so tertiary alkyl halides are unable to undergo SN2 reactions.
steric hindrance

14. Both are Primary Alkyl Halides but They React at Different Rates
The rate of an SN2 reaction depends not only on the number of alkyl groups attached to the carbon that is undergoing nucleophilic attack, but also on their size.
For example, bromoethane is more than twice as reactive as bromopropane in an SN2 reaction, because propyl is bulkier than ethyl and provides more steric hindrance to back-side attack.

15. Why Steric Hindrance Decreases the Rate
Note the difference in the energy of the transition states for the reactions of an unhindered bromomethane and a more sterically hindered 2o alkyl bromide.

16. Although it is Primary, it Reacts Very Slowly
While 1-bromo-2,2-dimethylpropane is a primary alkyl halide, it undergoes SN2 reactions very slowly because its single alkyl group is unusually bulky.

17. Why the Configuration of the Product is Inverted
The inversion of configuration that is observed in an SN2 reaction is also called a Walden inversion.
Because an SN2 reaction occurs with inversion of configuration, only one substitution product is formed when an alkyl halide whose halogen atom is bonded to an asymmetric center undergoes an SN2 reaction.
The configuration of that product is inverted relative to the configuration of the alkyl halide.

18. The Weakest Base is the Best Leaving Group
There are a number of factors that affect SN2 reactions.
One is the nature of the leaving group.
Generally speaking, the weaker the base, the better the leaving group.
Thus, the halide in alkyl iodides is replaced most readily in SN2 reactions and the halide in alkyl fluorides is replaced least easily if it can be replaced at all.
The reason leaving propensity depends on basicity is that weak bases are stable bases; they readily bear the electrons they formerly shared with a proton.
Therefore, they do not share their electrons well.
Thus a weak base is not bonded as strongly to the carbon as a strong base would be, and a weaker bond is more readily broken.
The fluoride ion is such a strong base that alkyl fluorides essentially do not undergo SN2 reactions.
Weak bases are stable bases

19. The Rate of an SN2 Reaction is Affected by the Leaving Group
Note the different reaction rates observed in Sn2 reactions for the four halides.
Which halide is the weakest base?
Which halide is the best leaving group?

20. Base Strength and Nucleophile Strength
A second factor that affects SN2 reactions is the nature of the nucleophile.
Basicity is a measure of how well a compound (a base) shares its lone pair with a proton. The stronger the base, the better it shares its electrons.
Nucleophilicity is a measure of how readily a compound (a nucleophile) is able to attack an electron-deficient atom.
The better the nucleophile, the faster the rate of the SN2 reaction.
Because the nucleophile attacks an sp3 carbon in the rate-determining step of an SN2 reaction, the rate of the reaction will depend on the strength of the nucleophile: the better the nucleophile, the faster the rate of the SN2 reaction.
If the attacking atoms are of the same size, generally, the better the base, the better the nucleophile.
If atoms are in the same row, the strongest base is the best nucleophile.

21. Base Strength and Nucleophile Strength
A species with a negative charge is a stronger base and a better nucleophile than a species that has the same attacking atom but is neutral. Thus OH- is a stronger base and a better nucleophile than water.
A negatively charged atom is a stronger base and a better nucleophile than the same atom that is neutral.

22. Polarizability
As you go down a row in the periodic table the larger atoms are more polarizable, so they are better nucleophiles but they are weaker bases.
If the attacking atoms are very different in size, the polarizability of the atom comes into play.
Because the electrons are farther apart in the larger atom, they are not as tightly held and can, therefore, move more freely toward a positive charge.
As a result, the electrons are able to overlap the orbital of carbon from farther away.
This results in a greater degree of bonding in the transition state, which makes the transition state more stable.

23. Base Strength and Nucleophile Strength
F– is the best nucleophile in an aprotic solvent because it is the strongest base.
I– is the best nucleophile in a protic solvent because it more polarizable and it is poorly solvated.
The question is, does the greater polarizability that helps the larger atoms to be better nucleophiles make up for the decreased basicity that causes them to be poorer nucleophiles? The answer depends on the solvent.
If the reaction is carried out in an aprotic polar solvent – a polar solvent without a hydrogen bonded to oxygen or nitrogen – the direct relationship between basicity and nucleophilicity is maintained – the strongest bases are the best nucleophiles.
Therefore, iodide ion, the weakest base of the halides, is the poorest nucleophile.
If, however, the reaction is carried out in a protic solvent, the relationship between basicity and nucleophilicity is inverted.
The largest atom is the best nucleophile even though it is the weakest base.
Therefore, iodide ion, the weakest base, is the best nucleophile of the halide ions in a protic solvent.
How does one explain this discrepancy?
The strongest base is the best nucleophile unless
they differ in size, and
they are in a protic polar solvent.

24. Ion–Dipole Interactions
The reason for the inverted nucleophilicity cited above is that strong bases (eg. fluoride) are more highly solvated in polar solvents than weaker bases.
The solvent is thus better in shielding the nucleophilicity of fluoride.
Because the solvent shields the nucleophile, at least one of the ion-dipole interactions must be broken before the nucleophile can participate in an SN2 reaction.
Weak bases interact weakly with protic solvents, whereas strong bases interact strongly because they are better at sharing their electrons.
It is easier, therefore, to break the ion-dipole interactions between an iodide ion (a weak base) and the solvent than between a fluoride ion (a stronger base) and the solvent.
In a protic solvent, therefore, an iodide ion, even though it is a weaker base, is a better nucleophile than a fluoride ion.
Stronger bases form stronger ion–dipole interactions.
Ion–dipole interactions have to be broken before the nucleophile can react.

25. DMF and DMSO
An aprotic polar solvent does not have any hydrogens with partial positive charges to form ion-dipole interactions.
The molecules of an aprotic polar solvent (such as DMF or DMSO) have a partial negative charge on their surface that can solvate cations, but the partial positive charge is on the inside of the molecule, and therefore less accessible to solvate anions.
Fluoride ion, therefore, is a good nucleophile in DMSO and a poor nucleophile in water.
They can solvate a cation (+) better than they can solvate an anion (–).

26. Decreases Nucleophilicity
Steric Hindrance
Decreases Nucleophilicity
Even though the tert-butoxide ion is a stronger base, it is a poorer nucleophile because nucleophilic attack is more sterically hindered.
Nucleophilicity is affected by steric effects.
A bulky nucleophile has a harder time approaching the back side of a carbon.
Basicity, on the other hand, is relatively unaffected by steric effects because a base removes an unhindered proton.
Thus bulky nucleophiles, while more basic, are poorer nucleophiles.

27. SN2 Reactions Can Be Used to Make a Variety of Compounds
Many different kinds of nucleophiles can react with alkyl halides.
Therefore, a wide variety of organic compounds can be synthesized by means of SN2 reactions.
In looking at the above list, one might wonder why the reverse reactions don’t take place.
One need only consider the leaving propensities of the two groups to determine the answer.
In the case of Cl- and OH-, Cl- is the much weaker base (HCl is the stronger acid).
Because it is a weaker base, Cl- is a better leaving group.
Consequently, OH- can displace Cl- (a good leaving group) in the forward reaction ,but Cl- cannot displace OH- (a poor leaving group) in the reverse is reaction.
The reactions are irreversible because a strong base displaces a weak base.

28. Synthesizing an Amine
Ethylamine and methyl iodide undergo an SN2 reaction.
The product of the reaction is a secondary amine that is predominantly in its basic (neutral) form because the pH of the basic solution is greater than the pKa of the protonated amine. The secondary amine can undergo an SN2 reaction with another equivalent of methyl iodide, forming a tertiary amine.
The tertiary amine can react with methyl iodide in yet another SN2 reaction. The final product of the reaction is a quaternary ammonium iodide.

29. Worked Example
What is the product of a nucleophilic substitution reaction of (S)-2-bromohexane with acetate ion, CH3CO2- ?
Assume that inversion of configuration occurs, and show the chemistry of both the reactant and product

30. Worked Example Solution:
Identify the leaving group and the chirality center
Draw the product carbon skeleton
Invert the configuration at the chirality center
Replace the leaving group (bromide) with the nucleophilic reactant (acetate)

31. Worked Example
What is the product obtained from the SN2 reaction of OH– with (R)-2-bromobutane?
Show the chemistry of the reactant and the product
Solution:

32. Worked Example
Draw the product formed in an SN2 reaction between 1-bromobutane and NaI
Solution:

33. A Substitution Reaction
A tertiary alkyl halide and a poor nucleophile
Given the fact that water is a poor nucleophile and this alkyl halide is sterically hindered to back-side attack, one might expect the rate of this reaction to be very slow.
In fact, this reaction is surprisingly fast – one million times faster than the reaction of water with bromomethane.
This suggests that a different mechanism is involved.
The reaction is surprisingly fast, so it must be taking place by a different mechanism.

34. The Rate Depends Only on the Concentration of the Alkyl Halide
For the reaction of 2-bromo-2-methylpropane with water, doubling the concentration of the alkyl halide doubles the rate of the reaction, but changing the concentration of the nucleophile (water) has no effect on its rate.
This allows one to write the following rate law: rate = k [alkyl halide].
The rate of the reaction depends linearly on the concentration of only one reactant, so the reaction is a first-order reaction.
Because the rate law for the reaction of 2-bromo-2-methylpropane with water differs from the rate law for the reaction of bromomethane with hydroxide ion, the two reactions must have different mechanisms. The reaction between 2-bromo-2-methylpropane and water is an SN1 reaction.
The “1” in SN1 corresponds to a unimolecular reaction. Unimolecular means that only one molecule is involved in the transition state of the rate-determining step.
The mechanism for an SN1 reaction is based on the following experimental evidence:
1) The rate law shows that the rate of the reaction depends only on the concentration of the alkyl halide, so only the alkyl halide is involved in the transition state of the rate-determining step.
an SN1 reaction

35. Relative Rates of an SN1 Reaction
Generally only tertiary alkyl halides
undergo SN1 reactions.
Tertiary alkyl halides undergo SN1 reactions but methyl halides and primary alkyl halides do not. Similarly secondary alkyl halides do not undergo SN1 reactions.
Thus, methyl halides, primary and secondary alkyl halides undergo only SN2 reactions.

36. The Product is a Pair of Enantiomers
The substitution reaction of an alkyl halide in which the halogen is bonded to an asymmetric center forms two stereoisomers: one with the same relative configuration as that of the reacting alkyl halide, and the other with the inverted configuration.
If the halogen is bonded to an asymmetric center, the product will be a pair of enantiomers.

37. Summary of the Experimental Evidence
for the Mechanism of an SN1 Reaction
The rate of the reaction depends only on the
concentration of the alkyl halide.
2. Tertiary alkyl halides react the fastest.
This slide summarizes the experimental evidence for the mechanism of an SN1 reaction.
3. If the halogen is attached to an asymmetric center the product will be a pair of enantiomers.

38. The Mechanism
The leaving group departs before the nucleophile approaches.
The mechanism for an SN1 reaction is delineated here. The leaving group departs before the nucleophile approaches.
In the first step, the carbon-halogen bond beaks and the previously shared pair of electrons stays with the halogen. A carbocation intermediate is formed.
In the second step, the nucleophile (in this case water) reacts rapidly with the carbocation to form a protonated alcohol.
The protonated alcohol then loses a hydrogen to a water molecule.
Most SN1 reactions are solvolysis reactions; the nucleophile is the solvent.

39. Formation of the Carbocation
The Slow Step is
Formation of the Carbocation
Because the rate of an SN1 reaction depends only on the concentration of the alkyl halide, the first step must be the slow (rate determining) step.
The nucleophile is not involved in the rate-determining step, so its concentration has no effect on the rate of the reaction.  
Tertiary alkyl halides react the fastest; they form the most stable carbocations.
Secondary and primary alkyl halides do not undergo SN1 reactions.

40. Why a Pair of Enantiomers?
How does the mechanism for the SN1 reaction account for the three pieces of experimental evidence?
First because the alkyl halide is the only species that participates in the rate-determining step, the mechanism agrees with the observation that the rate of the reaction depends only on the concentration of the alkyl halide; it does not depend on the concentration of the nucleophile.
Second, the mechanism shows that a carbocation is formed in the rate-determining step. This explains why tertiary alkyl halides undergo SN1 reactions but primary and secondary alkyl halides do not. Tertiary carbocations are more stable than primary and secondary carbocations and, therefore, are the most easily formed.
Third, the positively charged carbon of the carbocation is sp2 hybridized which means the three bonds connected to it are in the same plane. In the second step of the SN1 reaction, the nucleophile can approach from either side of the plane, so some of the product will have the same configuration as the reacting alkyl halide and some will have an inverted configuration.

42. Why More Inverted Product
Typically in an SN1 reaction 50-70% of the product has the inverted configuration.
If the reaction does lead to equal amounts of the two stereoisomers, the reaction is said to take place with complete racemization.
When more of the inverted product is formed, the reaction is said to take place with partial racemization.
The higher propensity for inversion is due to the carbocation’s propensity to form an intimate ion pair with the leaving group. This eventually becomes a solvent-separated ion pair.

43. The Front Side is Partially Blocked
The nucleophile can attack any of the four species above, but it is hindered in its front-side approach to the intimate and solvent-separated ion pairs. Thus more of the back-side attack is observed.

44. Worked Example Configure the following substrate
Show the stereochemistry and identify the product that can be obtained by SN1 reaction with water (reddish brown = Br)

45. Worked Example Solution:
The S substrate reacts with water to form a mixture of R and S alcohols
Ratio of enantiomers is close to 50:50

46. The Weakest Base is the Best Leaving Group
Because the rate-determining step of an SN1 reaction is the formation of a carbocation, two factors affect the rate of reaction:
1) the ease with which the leaving group dissociates.
2) the stability of the carbocation formed.
There is a direct relationship between basicity and leaving propensity in an SN1 reaction - the weaker the base, the less tightly it is bonded to the carbon and he more easily the carbon-halogen bond can be broken. As a result, comparing alkyl halides with the same alkyl group, an alkyl iodide is the most reactive and an alkyl fluoride is the least reactive in both SN1 and SN2 reactions.

47. The Nucleophile has no Effect on the Rate
Because the nucleophile does not participate in an SN1 reaction until after the rate-determining step, the reactivity of the nucleophile has no effect on the rate of an SN1 reaction.
In most SN1 reactions, the solvent is the nucleophile. Methanol is both the nucleophile and the solvent in this reaction.
Reaction with a solvent is called solvolysis.

48. Benzylic and Allylic Halides
Undergo SN2 Reactions
Unless they are tertiary, allylic and benzylic halides readily undergo SN2 reactions.
Tertiary allylic and benzylic halides do not undergo SN2 reactions due to steric hindrance.

49. Benzyl and Phenyl Groups
A benzene ring attached to a methylene group is called a benzyl group.
A substituent that consists of just a benzene ring is called a phenyl group.

50. Benzylic and Allylic Halides
Undergo SN1 Reactions
Benzylic and allylic halides readily undergo SN1 reactions as well, because they form carbocations that are stabilized by electron delocalization.

51. Resonance Forms of Allylic and Benzylic Carbocations

52. The SN1 Reaction of Allylic Halides
Can Form Two Products
If the two resonance contributors of the allylic carbocation formed in an SN1 reaction are not mirror images (as in the preceding example), two substitution products will be formed.
This is another example of how electron delocalization can affect the nature of the products formed in a reaction.

53. Worked Example
Explain why 3-Bromo-1-butene and 1-bromo-2-butene undergo SN1 reaction at a similar rate, despite one being a secondary halide and the other being a primary halide
Solution:
The two bromobutenes form the same allylic carbocation in the rate limiting step

54. Vinylic and Aryl Halides Cannot Undergo SN2 Reactions
Vinylic and aryl halides do not undergo SN2 reactions. The nucleophile is repelled by the π electrons when undergoing backside attack.

55. Vinylic and Aryl Halides Cannot Undergo SN1 Reactions
Similarly vinylic and aryl halides do not undergo SN1 reactions. Vinylic and aryl carbocations are even more unstable than primary carbocations.
The positive charge on a vinylic or aryl cation would be on an sp carbons – sp carbon are more electronegative than the sp2 carbons that carry the positive charge of alkyl carbocations, so sp carbons are more resistant to becoming positively charged. In addition, a ring carbon cannot assume the 180o bond angles required for sp hybridization.

56. Comparing SN2 and SN1 Reaction
The characteristics of SN1 and SN2 reactions are compared in this table.
Remember that the “2” and “1” in SN2 and SN1 refer to the molecularity of the reaction (the number of molecules involved in the transition state of the rate-determining step), and not to the number of steps in the mechanism.
Indeed, the opposite is true, an SN2 reaction proceeds by a one-step concerted mechanism, whereas an SN1 reaction proceeds by a two-step mechanism with a carbocation intermediate.

57. Summary of Alkyl Halide Reactivity
This table summarizes the reactivity of the alkyl halides in SN2 and SN1 reactions.
Primary, secondary, and methyl halides undergo only SN2 reactions because of their relatively unstable carbocations.
All tertiary halides undergo only SN1 reactions, because steric hindrance makes them unreactive in SN2 reactions.
Primary and secondary allylic and benzylic halides undergo both SN1 and SN2 reactions.
Vinylic and aryl halides cannot undergo either SN1 or SN2 reactions.
Only primary and secondary allylic and benzylic halides undergo both SN1 and SN2 reactions. The conditions under which the reaction is carried out determine which reaction predominates.
What conditions favor an SN2 reaction? This is an important question because an SN2 reaction (a synthetic chemist’s friend) forms a single substitution product, whereas an SN1 reaction (a synthetic chemist’s nightmare) can form a pair of enantiomers if the halogen is attached to an asymmetric center.
When a compound can undergo both SN1 andSN2 reactions, three conditions determine which reaction will predominate:
1) the concentration of the nucleophile
2) the reactivity of the nucleophile
3) the solvent in which the reaction is carried out.

58. Rate Law for an SN2 Reaction + Rate Law for an SN1 Reaction
The rate law for a reaction that can undergo both SN2 and SN1 reactions.
To understand how the concentration and the reactivity of the nucleophile affect whether an SN1 or an SN2 reaction predominates, we must examine the rate laws for the two reactions. The rate law for the reaction of a compound – for example, an allylic halide – that can undergo both SN2 and SN1 reactions simultaneously is the sum of the individual rate laws.
From the rate law, one can see that increasing the concentration of the nucleophile increases the rate of the SN2 reaction but has no effect on the SN1 reaction. Therefore, when both reactions occur simultaneously, increasing the concentration of the nucleophile increases the fraction of the reaction that takes place by the SN2 pathway.
Similarly, increasing the strength of the nucleophile increases the rate of an SN2 reaction by increasing the value of the rate constant, because a more reactive nucleophile is better able to displace the leaving group.
Increasing the strength of the nucleophile has no effect on the rate of an SN1 reaction.
In summary:
A good nucleophile favors an SN2 reaction over an SN1 reaction.
A poor nucleophile favors an SN1 reaction, not by increasing the rate of the SN1 reaction itself, but by decreasing the a rate of the competing SN2 reaction.
An SN2 reaction is favored by a high concentration of a good nucleophile.
An SN1 reaction is favored by a poor nucleophile. (Note that a lot of SN1 reactions employ neutral molecules such as solvents as the nucleophile)
An SN2 reaction is favored by a high concentration of a good nucleophile.
An SN1 reaction is favored by a poor nucleophile.

59. Solvation
The solvent in which a nucleophilic substitution reaction is carried out also influences whether an SN1 or an SN2 reaction will predominate for compounds that can undergo both reactions (that is, allylic and benzylic halides, unless they are tertiary).
The dielectric constant of a solvent is a measure of how well the solvent can insulate opposite charges from one another. Solvent molecules can insulate a charge by clustering around it, so that the positive poles of solvent molecules surround negative charges while the negative poles of solvent molecules surround positive charges.
Polar solvents, such as water, have high dielectric constants and therefore are very good at insulating (solvating) charges. Non polar solvents, such as toluene and hexane, have low dielectric constants and insulate the charge around an ion poorly.
When an alkyl halide undergoes an SN1 reaction, the first step is dissociation of the carbon-halogen bond to form a carbocation and a halide ion. Energy is required to break the bond. If the reaction is carried out in a polar solvent, the ions that are produced are solvated. The collective ion-dipole interactions that take place when a solvent stabilizes a charged species represents a great deal of energy. So, the alkyl halide does not fall apart spontaneously in an SN1 reaction – polar solvent molecules pull it apart. An SN1 reaction, therefore, cannot take place in a nonpolar solvent.
The dielectric constant is a measure of how the solvent can insulate opposite charges from one another.

60. Common Solvents
Listed here are the dielectric constants of some common solvents.
They are divided into two groups – protic solvents and aprotic solvents.
Note the dielectric constants are highest in polar or protic solvents.
How increasing the polarity of the solvent (that is, increasing its dielectric constant) will affect the rate of most chemical reactions depends only on whether or not a reactant in the rate-limiting step is charged:
If a reactant in the rate-determining step is charged, increasing the polarity of the solvent will decrease the rate of the reaction.
If none of the reactants in the rate-determining step is charged, increasing the polarity of the solvent will increase the rate of reaction.

61. The Reactants are Neutral
If the reactants are neutral, the charge on the reactants will be less than the charge on the transition state.
Increasing the polarity of the solvent will increase the rate of reaction
If none of the reactants in the rate-determining step is charged, increasing the polarity of the solvent will increase the rate of reaction.
Let’s begin by looking at an SN1 reaction. The alkyl halide, which is the only reactant in the rate-determining step of an SN1 reaction, is a neutral molecule with a small dipole moment. The rate-determining transition state has greater partial charges because as the carbon-halogen bond breaks, the carbon becomes more positive and the halogen becomes more negative. Since the partial charges on the transition state are greater than the partial charges on the reactant, increasing the polarity of the solvent will stabilize the transition state more than the reactant, which will increase the rate of the SN1 reaction.
If, however, the compound undergoing an SN1 reaction is charged, increasing the polarity of the solvent will decrease the rate of the reaction because the more polar solvent will stabilize the full charge on the reactant to a greater extent than it will stabilize the dispersed charge on the transition state.

62. If the Reactants are Neutral, Increasing the Polarity of the Solvent Increases the Rate
In graphical form, if the size of the charge on the transition state is greater than the size of the charge on the reactants, then a polar solvent will stabilize the transition state more than it will stabilize the reactants.
Increasing the polarity of the solvent will decrease the difference in energy (ΔG dagger), which will increase the rate of the reaction.

63. The Rate of an SN1 Reaction of an Alkyl Halide Increases as the Polarity of the Solvent Increases
Just a reiteration – The rate of an SN1 reaction of an alkyl halide increases as the polarity of the solvent increases.

64. One of the Reactants is Charged
If a reactant is charged, the charge on the reactants will be greater than the charge on the transition state.
Increasing the polarity of the solvent will decrease the rate of the reaction.
If a reactant in the rate-determining step is charged, increasing the polarity of the solvent will decrease the rate of the reaction.
How increasing the polarity of the solvent affects the rate of an SN2 reaction depends on whether the reactants are charged or neutral, just as it does in an SN1 reaction.
Most SN2 reactions of alkyl halides occur between a neutral alkyl halide and a charged nucleophile. Increasing the polarity of a solvent will have a strong stabilizing effect on the negatively charged nucleophile. The transition state also has a negative charge, but that charge is dispersed over two atoms. Consequently, the interactions between the solvent and the transition state are not as strong as those between the solvent and the fully charged nucleophile.
Therefore, increasing the polarity of the solvent will stabilize the nucleophile more than it will stabilize the transtion state, so the reaction will be slower.
If, however, the SN2 reaction occurs between an alkyl halide and a neutral nucleophile, such as an amine, the partial charges on the transition state will be larger than the partial charges on the reactants, so increasing the polarity of the solvent will increase the rate of the reaction.

65. If One of the Reactants is Charged, Increasing the Polarity of the Solvent Decreases the Rate
In graphical form, if the size of the charge on the reactants is greater than the size of the charge on the transition state, then a polar solvent will stabilize the reactants more than it will stabilize the transition state.
Increasing the polarity of the solvent will increase the difference in energy (ΔG dagger), which will decrease the rate of the reaction.
Summation - Because a polar solvent decreases the rate of an SN2 reaction when the nucleophile is negatively charged, we would like to carry out such a reaction in a nonpolar solvent. However, negatively charged nucleophiles will not dissolve in nonpolar solvents, such as toluene and hexane. Instead, an aprotic polar solvent is used. Because they are not hydrogen bond donors, they are less effective than polar protic solvents, such as water and alcohols, at solvating negative charges. Thus the rate of an SN2 reaction with a negatively charged nucleophile will be greater in an aprotic polar solvent that in a protic polar solvent. Consequently, an aprotic polar solvent is the solvent of choice for an SN2 reaction in which the nucleophile is negative charged, whereas a protic polar solvent is used if the nucleophile is a neutral molecule.
We have now seen that when an alkyl halide can undergo both SN2 and SN1 reactions, the SN2 reaction will be favored by a high concentration of a good, negatively charged, nucleophile in an aprotic polar solvent or by a high concentration of a good neutral nucleophile in a protic polar solvent, whereas the SN1 reaction will be favored by a poor, neutral, nucleophile in a protic polar solvent.

66. Intermolecular versus Intramolecular SN2 Reactions
Molecules that contain both a nucleophile and a leaving group in the same molecule can undergo intermolecular or intramolecular reactions.
The former reactions are used in the synthesis of polymers.
The latter reactions are used in the synthesis of cyclic molecules.
Which reaction is likely to occur depends on the concentration of the bifunctional molecule and the size of the ring that would be formed in the intramolecular reaction.
A low concentration of the reactant favors an intramolecular reaction because the two functional groups have a better chance of finding each other if they are in the same molecule.
A high concentration of reactant helps compensate for the advantage gained by tethering, thereby increasing the likelihood of an intermolecular reaction.

67. The Intramolecular Reaction is Favored When a Five- or Six-Membered Ring Can Be Formed
How much of an advantage an intramolecular reaction has over an intermolecular reaction also depends on the size of the ring that is formed.
An intramolecular reaction will be favored over an intermolecular reaction in the case where five and six membered rings can be formed since these rings are the most stable and therefore are the most easily formed.

68. 3-Membered and 4-Membered Rings are Less Stable
Three and four membered ring products are more strained, which makes them less stable.
However, even though it has more strain than the four-membered ring, the three membered ring can form more easily because only one bond must rotate. Thus, the three-membered ring is more apt to have its reacting groups in the conformation required for reaction.
Seven and larger membered ring forming reactions are more difficult.
Three-membered rings are generally formed faster than 4-membered rings.

69. Methylation by a Chemist
If an organic chemist wanted to put a methyl group on a nucleophile, methyl iodide would most likely be used. The iodide is the most easily displaced of the halides and methyl iodide is a liquid at room temperature.

70. Methylation by a Cell Methyl iodide is not found in living cells.
The cell uses S-adenosylmethionine, (SAM), a water-soluble compound, as a methylating agent, instead. SAM transfers a methyl group to a nucleophile.
Note the good leaving group – the positively charged sulfur - which can readily accept the electrons from the departing methyl group.

71. Nordrenaline to Adrenaline
Here is another example whereby SAM methylates noradrenaline to produce adrenaline in the body.

72. One Phospholipid is Converted to Another Phospholipid
Here SAM effects three methylations to produce a component of cell membranes.

73. The Leaving Group
An SN2 reaction with an alcohol requires conversion to an alkyl chloride or an alkyl bromide
Unless it is protonated, an alcohol is a very poor leaving group.
By converting it to a halide or a tosylate, the leaving ability of the OH group is markedly increased.

74. Poor Leaving Groups Generally, ethers do not undergo SN2 reactions
Epoxides are an exception as they are more reactive than ethers (due to strain)

75. Tosylate
Conversion to a tosylate is another way to increase the leaving ability of an OH group.

76. Worked Example
Rank the following compounds in order of their expected reactivity toward SN2 reaction:
CH3Br, CH3OTos, (CH3)2CHCl, (CH3)3CCl

77. Worked Example Solution:
With reference to effects of the substrate and the leaving group:
Reactivity of tertiary substrates is lesser than that of secondary substrates
Secondary substrates are less reactive than primary substrates

78. Worked Example
Predict whether the following substitution reaction is likely to be SN1 or SN2

79. Worked Example Solution:
This reaction probably occurs by an SN1 mechanism
HCl converts the poor –OH leaving group into an excellent –OH2+ leaving group, and the polar solvent stabilizes the carbocation intermediate

80. Worked Examples
Propose a mechanism for the biosynthesis of limonene from linalyl diphosphate

81. Worked Example Solution:
After dissociation of PPi, the cation cyclizes by attack of the double bond π electrons and –H is removed to yield limonene

82. Elimination Reactions : Zaitsev’s Rule
A nucleophile/Lewis base reacts with an alkyl halide resulting in a substitution or an elimination

83. Zaitsev’s Rule for Elimination Reactions
In the elimination of HX from an alkyl halide, the more highly substituted alkene product predominates
Elimination reactions almost always give mixtures of alkene products.
Zaitsev’s Rule predicts the major product will, in most cases, be the more substituted alkene

84. Mechanisms of Elimination Reactions
E1 reaction
Breaking of C–X bond produces a carbocation intermediate that yields the alkene by base removal of a proton
Like substitution reactions, elimination reactions can take place by a variety of mechanisms, including E2, E1 and E1cB mechanisms which differ in the timing of the breaking of the C-H and C-X bonds.
In the E1 reaction, the C-X bond breaks first to give a carbocation intermediate which undergoes subsequent base abstraction of H+ to give the alkene.

85. Mechanisms of Elimination Reactions
E2 reaction
Simultaneous cleavage of C–H bond and C–X bond produces the alkene without intermediates
In the E2 mechanism, base induced C-H bond cleavage is simultaneous with C-X bond cleavage, giving the alkene in a single step.

86. Mechanisms of Elimination Reactions
E1cB reaction
Proton undergoes base abstraction, yielding a carbanion (R-) intermediate
Carbanion loses X-, yielding the alkene
cb stands for conjugate base.
In the E1cB mechanism, base abstraction of the proton occurs first, giving a carbanion intermediate. This anion, then undergoes loss of X- in a subsequent step to give the alkene.
All three mechanisms occur in the laboratory, but the E1cB mechanism predominates in biological pathways.

87. Worked Example
What is the alkyl halide source of the following alkene?

88. Worked Example Solution:
For maximum yield, the alkyl halide reactant should not give a mixture of products on elimination

89. The E2 Reaction and the Deuterium Isotope Effect
E2 reaction: Reaction involving treatment of an alkyl halide with a strong base
The most common elimination pathway
Abides by the rate law
Rate = k ×[RX] ×[Base]
Deuterium isotope effect: C–H bond is more easily broken than a corresponding C–D bond

90. Mechanism of the E2 Reaction with an Alkyl Halide
The E2 reaction takes place in one step without intermediates.
As the base begins to abstract H+ from a carbon next to the leaving group, the C-H bond begins to break, a C=C bond begins to form, and the leaving group begins to depart, taking with it the electron pair from the C-X bond.
Among the pieces of evidence supporting this mechanism is that E2 reactins show second-order kinetics and follow the rate law Rate = k ×[RX] × [Base]. Thus both the base and the alkyl halide take part in the rate-limiting-step.

91. Mechanism of the E2 Reaction with an Alkyl Halide
Since the reaction involving breakage of a C-H bond is faster than that involving breakage of a C-D bond, the breakage of this bond must occur in the rate determining step consistent with E2 reaction involving a one-step process.
A second piece of evidence in support of the E2 mechanism is provided by a phenomenon known as the deuterium isotope effect. The C-D bond is weaker than a C-D bond. Since elimination of D-X occurs more slowly than H-X, the C-D bond must be broken in the rate-limiting-step.

92. Geometry of Elimination - E2
Termed periplanar geometry
Hydrogen atom, the two carbons, and the leaving group lie in the same plane
Syn periplanar: H and X are on the same side of the molecule
E2 eliminations occur with periplanar geometry – all four reacting atoms – the hydrogen, the two carbons, and the leaving group – lie within the same plane.
In syn periplanar geometry, the H and the X are on the same side of the molecule.

93. Geometry of Elimination - E2
Anti periplanar: H and X are on the opposite sides of the molecule
In antiperiplanar geometry, the H and the X are on opposite sides of the molecule.
Antiperiplanar geometry is energetically preferred because it allows the substituents on the two carbons to adop t a staggered relationship; syn geometry requires them to be eclipsed.

94. Significance of Periplanar Geometry
When all orbitals are periplanar, overlap in the transition state can easily occur
In both the SN2 and E2 reactions an electron pair pushes out the leaving group on the opposite side of the molecule.

95. Geometry of Elimination - E2
E2 is stereospecific
Meso-1,2-dibromo-1,2-diphenylethane with base gives cis 1,2-diphenyl
RR or SS 1,2-dibromo-1,2-diphenylethane gives trans 1,2-diphenyl
Only the E alkene is afforded in this reaction. This requires that the reaction be highly stereospecific.
For the Z alkene to be formed in this instance, the transition state would have to have syn periplanar geometry. This would be higher in energy. As a consequence, none of the Z-alkene is formed.

96. Worked Example
Exhibit the stereochemistry of the alkene obtained by E2 elimination of (1R,2R)-1,2-dibromo-1,2-diphenylethane
Draw a Newman projection of the reacting confirmation

97. Worked Example Solution: The reactant with correct stereochemistry
The drawing converted into a Newman projection

98. Worked Example
The alkene resulting from E2 elimination is (Z)-1-bromo-1,2-diphenylethylene

99. The E2 Reaction and Cyclohexene Formation
Cyclohexane rings need to possess anti-planar geometry in order to undergo E2 reactions
Hydrogen and leaving group in cyclohexanes need to be transdiaxial
The anti-periplanar requirement for E2 reactions requires a trans-diaxial relationship between the H and X groups being eliminated in cyclohexane ring systems.
If either the leaving group or the hydrogen is equatorial, E2 elimination can’t occur.
The lack of an anti-periplanar geometry may override the energetics favoring adherence to Zaitsev’s Rule.

100. The E2 Reaction and Cyclohexane Formation
The trans-diaxial requirement is met in isomeric menthyl and neomenthyl chlorides by the removal of HCl
HCl is removed from neomenthyl chloride upon reaction with ethoxide ion
200 times faster than menthyl chloride
Conformations of menthyl chloride and neomenthyl chloride are responsible for the difference in reactivity
The E2 Reaction and cyclohexane conformation

101. Dehydrochlorination of Menthyl and Neo-Menthyl Chlorides
The elimination for neomenthyl chloride is 200 times faster than the elimination for menthyl chloride because the latter requires a ring flip.
Note also in the latter case the Zaitsev’s rule is violated. The more stable more highly substituted alkene cannot form because the H and X cannot assume an anti-periplanar relationship that would afford the product predicted from Zaitsev’s Rule.

102. Worked Example
Between trans-1-bromo-4-tert-butylcyclohexane or cis-1-bromo-4-tert-butylcyclohexane, identify the isomer that undergoes E2 elimination at a faster rate
Draw each molecule in its more stable chair configuration and provide an explanation

103. Worked Example Solution:
The larger tert-butyl group is always equatorial in the more stable conformation
The cis isomer reacts faster under E2 reactions because –Br and –H are in the anti periplanar arrangement that favors E2 elimination

104. The E1 and E1cB Reactions
E1 reaction: Comprises two steps and involves a carbocation
The E1 elimination involves a unimolecular dissociation to give a carbocation, just as is observed in the SN1 case.
Indeed the best E1 substrates are also the best SN1 substrates.
In the E1 case, the dissociation is followed by loss of H+ from the adjacent carbon rather than by substitution.

105. The E1 and E1cB Reactions
E1 reactions start out along the same lines as SN1 reactions
Dissociation leads to loss of H+ from the neighboring carbon rather than substitution in SN1 reactions
Substrates optimal for SN1 reactions also work well for E1 reactions
Typically mixtures of substitution and elimination products are obtained.

106. Evidence Supporting E1 Mechanism
E1 reactions exhibit first-order kinetics that are consistent with a rate-limiting, unimolecular dissociation process
E1 reactions do not exhibit the deuterium isotope effect
Rate difference between a deuterated and nondeuterated substrate cannot be quantified
E1 reactions do not have specific geometric requirements like E2 reactions
The lack of the deuterium isotope effect is because the rupture of the C-D (C-H) bond occurs after the rate-limiting-step.

107. Eliminations of Menthyl Chloride
Since a planar carbocation is obtained as an intermediate in the E1 elimination, the requirement for anti-periplanarity does not exist. As a result Zaitsev’s rule is invariably obeyed.

108. The E1cB Reaction Takes place through a carbanion intermediate
Base promoted abstraction of a proton to afford an anion is the rate-determining-step in an E1cB reaction. This anion expels a leaving group on an adjacent carbon.
This reaction is particularly common in substrates that have a poor leaving group, such as an OH group that is two carbons removed from a carbonyl group.
The poor leaving group disfavors the alternative E1 and E2 possibilities, and the carbonyl group makes the adjacent hydrogen unusually acidic by resonance stabilization of the anion intermediate.

109. Biological Elimination Reactions
Elimination reactions occur in biological pathways
E1cB is a common reaction
Eliminations convert 3-hydroxyl carbonyl compounds to unsaturated carbonyl compounds on a regular basis

110. Summary of the Products Obtained
From Substitution and Elimination Reactions

111. Summary of Stereochemistry

112. Worked Example
Classify the following reaction as an SN1, Sn2, E1, E1cB, or E2 reaction
Solution:
This is an E1cB reaction as the leaving group is two carbons away from a carbonyl group
A summary of reactivity: SN1, SN2, E1, E1cB, and E2