# Organic Chemistry I & II

Organic Chemistry I & II is designed for instructors who want an active, dynamic, and understandable approach to support their own efforts in the classroom. This ever-evolving textbook includes auto-graded questions, videos and approachable language in order to make difficult concepts easier to understand and implement.

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## Comparison of Organic Chemistry Textbooks

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### Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

### McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

### Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

### Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

### Pricing

Average price of textbook across most common format

#### $219 Hardcover print text only ####$301

Hardcover print text only

#### $301 Hardcover print text only ### Always up-to-date content, constantly revised by community of professors Content meets standard for Introduction to Organic Chemistry course, and is updated with the latest content ### In-Book Interactivity Includes embedded multi-media files and integrated software to enhance visual presentation of concepts directly in textbook Only available with supplementary resources at additional cost Only available with supplementary resources at additional cost Only available with supplementary resources at additional cost ### Customizable Ability to revise, adjust and adapt content to meet needs of course and instructor ### All-in-one Platform Access to additional questions, test banks, and slides available within one platform ## Pricing Average price of textbook across most common format ### Top Hat Steven Forsey, “Organic Chemistry”, Only one edition needed #### Up to40-60%more affordable Lifetime access on any device ### McGraw-Hill Carey & Giuliano, “Organic Chemistry”, 10th Edition ####$219

Hardcover print text only

### Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

#### $301 Hardcover print text only ### Wiley David R. Klein, “Organic Chemistry”, 3rd Edition ####$301

Hardcover print text only

## Always up-to-date content, constantly revised by community of professors

Constantly revised and updated by a community of professors with the latest content

### Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

### McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

### Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

### Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

## In-book Interactivity

Includes embedded multi-media files and integrated software to enhance visual presentation of concepts directly in textbook

### Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

### McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

### Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

### Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

## Customizable

Ability to revise, adjust and adapt content to meet needs of course and instructor

### Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

### McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

### Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

### Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

## All-in-one Platform

### Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

### McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

### Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

### Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

#### Dr. Steven Forsey, Ph.D.University of Waterloo

Steven Forsey is currently a Professor at University of Waterloo, teaching a variety of organic chemistry courses to Chemistry, Science, Chemical Engineering, Nanotechnology and distance education students. He received his Ph.D. (2004) for Synthetic Organic Chemistry from University of Waterloo, Ontario. He is a recipient of the Excellence of Science Teaching Award and has acted as the Teaching Fellow for the Department of Chemistry since 2016.

## Explore this textbook

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# Chapter 7: SN2 Reactions

Epoxy resins, a strong adhesive, are often used to build models. ​These adhesives make use of substitution reactions to stick to their base. [1]

## Learning Objectives

• Identify nucleophiles and electrophiles.
• Write the mechanism for an SN2 reaction and explain the term bimolecular nucleophilic substitution.
• Comprehend the factors that affect the rate of SN2 reactions: nucleophilicity, leaving group ability, solvent effects, and the effect of the electrophile's structure (steric interaction).
• Determine the stereochemical outcome of an SN2 reaction.
• Evaluate a series of nucleophiles and predict their relative rate of reaction towards an electrophile.
• Evaluate a series of electrophiles and predict their relative rate of reaction towards a nucleophile.
• Understand the effect of protic and aprotic solvents have on the rate of an SN2 reaction.

## 7.0 Introduction

Definitions:

Substitution Reaction: A reaction in which an atom or a group of atoms is replaced by another atom or group of atoms. In the above example, the nucleophile (Nuc) replaces the leaving group (LG).

Nucleophile: (“nucleus-loving”) Nucleophiles are Lewis bases and are electron-rich. They can be negatively charged or neutral and must have a lone pair of electrons to donate to the electrophile. Nucleophiles react with positively charged or slightly positively charged nuclei.

Electrophile: (“electron-loving”) Electrophiles are Lewis acids and are electron-deficient. They can be positively charged or neutral, with a slightly positively charged nucleus. Electrophiles accept electrons from the nucleophile. Note: to accept a pair of electrons an atom must have an empty antibonding orbital.

Leaving group: A leaving group must be able to leave as a relatively stable, weakly basic molecule or ion (stable meaning non-reactive).

Bimolecular: The rate of reaction is dependent on the concentration of the nucleophile and electrophile. The reaction rate is second-order.

Rate of reaction = k[nucleophile][electrophile], where k is equal to the rate constant.

Experimentally, this means if the nucleophile concentration is doubled, then the rate of reaction doubles. If the electrophile concentration is doubled, then the reaction rate doubles. If the nucleophile and electrophile concentrations are both doubled, then the rate quadruples.

Click on the video to see a bimolecular SN2 reaction in action. Make sure you take note of the red ball on the Potential Energy–Reaction Coordinate diagram and how the energy of the reaction relates to the animation.

What did the video show us?

• The reaction involved two molecules colliding into each other (bimolecular).
• The reaction is concerted, meaning that bond breaking and formation occurred in one step.
• For the reaction to occur, the molecules had to have a very specific orientation and the nucleophile and electrophile had to follow a very specific pathway. The nucleophile had to attack from the backside of the carbon-leaving group σ bond.
• For a reaction to occur, the electrophile and nucleophile had to collide with enough kinetic energy to surmount the energy barrier to the reaction (Ea).
• A high-energy transition state formed that involved both bond formation and breaking. In the transition state, the bonds are partially formed and broken.
• The products are at a lower potential energy than the reactants.
• The bonds of the products are completely formed; that is, they are not in the process of breaking or forming.

Activation Energy (Ea) is the minimum kinetic energy the molecules must have for a reaction to occur.

Transition State: A hypothetical species that exists between the reactants and products. Bonds are forming and breaking during this high-energy activated-complex phase. It is the transition point where the reactants form products or dissociate back to reactants.

Below is the mathematical description of the factors that affect the collision of molecules and the rate of reactions.

Steric Factor (p): Not all molecules will have the proper orientation or pathway for a reaction to occur; thus, the rate of reaction will be lower than if orientation did not matter. The steric factor will be different for every reaction.

Fraction of molecules with sufficient kinetic energy (f): Not all molecules will have sufficient energy to surmount the energy barrier, even if they have the correct orientation. The fraction of collisions equal to or greater than the activation energy needed to produce the product is given by the shaded areas under the curves below (Figure 7.3) and can be approximated by the equation f = e(−Ea/RT) where R is the gas constant (8.314 J/K.mol) and T is the temperature (Kelvin).

Note: ƒ is a very small number. If Ea = 75 kJ/mol and T = 298 K, then ƒ = 7 x 10−14 or 7 collisions in 100 trillion have sufficient energy to convert reactants to products.

As temperature increases, the number of molecules with sufficient energy to complete a reaction increases. For the above reaction, when T=308K (ƒ = 2 x 10−13). The value of f increases by a factor of 3. Collision theory, therefore, accounts nicely for the exponential dependence of reaction rates on reciprocal temperature. As T increases, ƒ increases exponentially. Collision theory also explains why reaction rates are so much lower than collision rates.

Collision rate Z[A][B] where Z is a constant related to the collision frequency. The collision rate can change dramatically with solvents that help or hinder the molecules from colliding with each other.

How does this relate to reaction rates in a bimolecular reaction?

The reaction rate is lower than the collision rate by the probability of the molecules having a favorable orientation when they collide (p) times the fraction of collisions equal to or greater than the activation energy needed to produce the products (f). Thus, the rate of reaction is equal to: Reaction rate = p x ƒ x collision rate.

Reaction ratepƒZ[A][B]

The experimental rate law for a bimolecular reaction is as follows.

Rate of reaction = k [A][B]

Now compare the experimental rate of reaction with that predicted by Collision Theory and you will see that the rate constant (k) must be equal to the collision rate times the probability of the molecules having a favorable orientation times the fraction of collisions that produce a product.

This is a very important concept to understand since the rate of an SN2 reaction is dependent on steric factors, kinetic energy, and solvent effects.

Q7.1 - Level 1

Which of the following is a reasonable definition of a concerted reaction?

A

It is a reaction which takes place in a series of steps

B

It is always an endothermic reaction

C

It is a reaction in which all bond-breaking and bond-forming occurs at the same time

D

It is a substitution reaction

Q7.2 - Level 2

Which of the following is the rate law for the given $S_N2$ reaction?

A

Rate = k[1-bromopropane]

B

Rate = k[NaCN]

C

Rate = k[NaCN][NaCN]

D

Rate = k[NaCN][1-bromopropane]

Q7.3 - Level 1

For the following $S_N2$ reaction, how would the rate of reaction change if the concentrations of both the hydrosulfide and butyl bromide concentrations were doubled? Assume no other changes.

A

Double the rate

B

Triple the rate

C

D

Sextuple the rate (6 times)

## 7.1 Stereochemistry of the Bimolecular Nucleophilic Substitution (SN2) Reaction

Notice how the hydrogens flip from one side of the molecule to the other. This is called inversion. For this particular reaction, stereochemistry is not important because no chiral centers are involved. However, observe how the hydrogens on the electrophilic carbon are flipped to the other side of the carbon after the nucleophilic substitution reaction.

### Question 7.4

What is the product formed in the following reaction? Think about the stereochemistry of the reactant and products.

Watch the video below for a step-by-step walk-through of the solution to this problem.

The above question illustrates the fact that when a substrate contains more than one stereocenter, inversion takes place only at the stereocenter being attacked by the nucleophile. You must also consider the overall stereochemistry of the molecule after the reaction since the chirality of the overall molecule may have changed.

Note: This does not mean that an electrophile with “R” stereochemistry will always produce a product with “S” stereochemistry since the R-S designation is based on nomenclature, however inversion always occurs.

Q7.5 - Level 3

When $(2S,4R)$-2-bromo-4-chloropentane is reacted with excess NaCN, both the bromine and chlorine atoms are substituted. What is the product formed?

A

a

B

b

C

c

Q7.6 - Level 2

Is the product formed in Question 7.5 chiral?

A

Yes

B

No

When (2S,4R)-2-bromo-4-methylchloropentane is reacted with excess NaCN, both the bromine and chlorine atoms are substituted. What is the product formed and is the product chiral? Click on the video for the answer.

Q7.7 - Level 1

Predict the major product formed in the following reaction.

A

1)

B

2)

C

3)

D

4)

Q7.8 - Level 3

Predict the major product formed in the following reaction.

A

1)

B

2)

C

3)

D

4)

Q7.9 - Level 3

Predict the major product formed when (S)-2-iodobutane is reacted with NaCN in DMSO.

A

1)

B

2)

C

3)

D

4)

Q7.10 - Level 2

Does the product of the given reaction have an R or S stereocenter?

A

$R$

B

$S$

Q7.11 - Level 2

Match the stereochemistry of the two substituted products formed.

Premise
Response
1

1)

A

$S$

2

2)

B

$R$

C

$S$

D

$R$

## 7.2 Nucleophiles, Electrophiles, and the Mechanisms

To solve organic chemistry questions, you must have a thought process or a methodology. However, before you can solve questions you must understand and learn the terminology. For substitution reactions, you must be able to recognize nucleophiles and electrophiles first, then you will be able to think about how (and where) the nucleophile attacks the electrophile.

Nucleophiles have lone pairs of electrons to donate: (“nucleus-loving”) Nucleophiles are Lewis bases and are electron-rich. They can be negatively charged or neutral. They must have a lone pair of electrons to donate to the electrophile. Nucleophiles react with positively charged, or slightly positively charged, nuclei. Examples of nucleophiles include the following.

Electrophiles have partial positive or fully positively charged atoms: (“electron-loving”) Electrophiles are Lewis acids and are electron-deficient. They can be positively charged or neutral with a slightly positively charged nucleus. Electrophiles accept electrons from the nucleophile. Examples of electrophiles include those found below.

Q7.12 - Level 2

Label the following atoms or groups as nucleophilic or electrophilic sites.

Premise
Response
1

1)

A

Electrophilic

2

2)

B

Electrophilic

3

3)

C

Nucleophilic

4

4)

D

Electrophilic

5

5)

E

Electrophilic

F

Nucleophilic

G

Nucleophilic

H

Nucleophilic

I

Nucleophilic

Watch the following video to see the explanation to Q7.12.

Q7.13 - Level 1

What is the nucleophile in the following S$_N$2 reaction?

A

1)

B

2)

C

3)

Q7.14 - Level 1

What is the electrophile in the following reaction?

A

1)

B

2)

C

3)

Q7.15 - Level 3

Click on the electrophilic carbon in the following molecule that would be attacked by a nucleophile in a bimolecular reaction. Click inside one of the circles.

Q7.16 - Level 1

Click on the electrophilic carbon in the following molecule that would be attacked by a nucleophile in a bimolecular reaction. Click inside one of the circles.

Now that you can recognize a nucleophile and electrophile, what are the steps needed to determine the products that are formed? These are the steps I use.

Example: What is the product formed in the following reaction?

Step 1: Put charges and partial charges on atoms.

Step 2: Identify the electrophile and nucleophile

Step 3: Negative goes to positive. Show the flow of electrons with arrows. Arrows guide you to the products that are formed.

The arrows show the formation of a new oxygen–carbon bond and the breaking of the carbon–bromine bond. The arrows are a guide to the products that are formed.

Click on the video for a visual explanation.

If only all reactions were that simple. Now is the time to think about what factors influence the rate of a reaction.

## 7.3 Factors that Affect the Rate of Substitution Reactions

### 7.3.1 Introduction

When we think about the rate of reaction we must think about the mechanism and how the molecules climb the energy barrier to reach the transition state and then form products. The transition state is very important in understanding the concerted bimolecular SN2 reaction.

You may have noticed that the transition state has ten electrons around the carbon. How can this be, since it violates the octet rule? Molecular orbital theory nicely explains this and explains why the nucleophilic attack only occurs from the backside.

For an electrophile to accept electrons from the nucleophile it must possess an empty orbital for those electrons to go into. Molecular orbital theory shows us that the linear combination of two atomic orbitals will produce a bonding molecular orbital and an antibonding molecular orbital. An illustration of the orbitals is shown below. Notice that the empty orbital has a large lobe pointing to the back of the C–LG sigma bond, where overlap with a filled orbital from the nucleophile can occur.

How are the electrons transferred from the nucleophile to the electrophile? Electrons on the nucleophile, that are in a nonbonding filled molecular orbital (the lone pair), approach the rear of the C–LG sigma bond. The interaction between the filled and empty orbitals is stabilizing because the nonbonding electrons (lone pair) can now occupy a bonding orbital.

As the favorable overlap between the empty antibonding molecular orbital on the electrophile and nonbonding electrons on the nucleophile increases in the transition state, the C–LG sigma bond weakens. As the reaction progresses, the C–LG sigma bond continues to weaken and becomes longer. As the empty antibonding orbital becomes filled and forms a new Nu–C sigma bond, the C–LG sigma bond breaks. The other groups attached to the carbon are pushed to the other side of the carbon and inversion occurs.

Watch the following video to visualize the mechanism.

Now that you can visualize what is happening in the SN2 reaction, the next step is to think about the factors that affect the rate of reaction. Those factors are:

Nucleophilicity: Since the nucleophile attacks the electrophile the more reactive the nucleophile is the faster the reaction will be.

Structure of electrophile: Since the electrophile is being attacked by the nucleophile at an electrophilic atom, any steric interaction caused by alkyl groups bonded to, or close to, the electrophilic center will slow the reaction down. Thus, sterically unhindered electrophiles will undergo an SN2 reaction more readily.

Leaving group: The leaving group gains electrons from the sigma bond and becomes negatively charged relative to its charge in the reactant. Therefore, the reaction proceeds faster with leaving groups that can accept a negative charge. The leaving group may also be positively charged and become neutral when it gains electrons from the sigma bond.

Solvents: Since SN2 reactions involve polar or charged reagents, solvents can have a large effect on the rate of reaction due to solvation energies.

### 7.3.2 Nucleophilicity

In the transition state, the nucleophile forms a new sigma bond while the carbon-leaving group sigma bond is cleaved. One of the major factors in determining the rate of reaction is the energy-matching between the nonbonding electrons in the highest occupied molecular orbital of the nucleophile, and the lowest unoccupied molecular orbital (antibonding orbital) of the electrophile. Thus, a good nucleophile with respect to an electrophile may or may not be a good nucleophile with respect to another type of electrophile. It is difficult to do more than segregate nucleophiles into excellent, good, or poor; however, when reaction conditions are similar, trends are observed that differ based on the nucleophile’s basicity, polarizability, shape (steric factors), and solvation energies. Nonetheless, the following generalizations can be made regarding different factors.

Note: Nucleophilicity relates to how fast the reaction proceeds. This is a kinetic measurement. Thus, when it is said that one nucleophile is better than another it means that one nucleophile will react faster than the other.

Basicity:

1. A negatively charged nucleophile is stronger (reacts faster) than its neutral conjugate acid.

2. In a group of nucleophiles in which the nucleophilic atom is the same, nucleophilicity parallels basicity. The more basic the nucleophile, the more nucleophilic it is (reacts faster).

3. Moving across the periodic table, nucleophilicity parallels basicity.

Shape

Large, bulky nucleophiles are less able to reach the antibonding orbital of the electrophile than less sterically unhindered, small or “bullet-shaped” nucleophiles. Good examples of bullet-shaped nucleophiles are cyanide (NC¯) and azide (N3-), which are much more nucleophilic than would be predicted by their basicity.

An example of a bulky, hindered nucleophile versus an unhindered nucleophile is t-butoxide and methoxide. t-Butoxide is a stronger base than methoxide but because it is sterically hindered, t-butoxide is a much weaker nucleophile.

3D Molecule*: 2-methylpropan-2-ol

3D Molecule*: methanol

Polarizability: In polar protic solvents (solvents that can hydrogen bond), nucleophilicity increases down the periodic table, following the increase in size and polarizability.

I¯> Br¯> Cl¯> F¯
HSe¯> HS¯> HO¯
R3P > R3N

Most strong bases are good nucleophiles, but not all weak bases are poor nucleophiles. Why is iodide such a strong nucleophile compared with fluoride even though it is a very weak base? (Remember HI is a stronger acid than HF.) Iodide’s electrons are more loosely held than a small ion and can accommodate changes in shape more readily. Because a polarizable electron cloud can more readily change its shape, iodide’s electron cloud can move freely toward a positive charge and can engage in partial bonding as it attacks the electrophilic carbon atom.

This may explain why iodide is a good nucleophile but it does not explain why it is more nucleophilic, i.e., reacts faster than the more basic fluoride ion.

More basic ions form stronger hydrogen bonds with protic solvents than larger, polarizable ions. Thus, for ions like F¯, the solvent hinders its movement. The ion must shed or displace the solvent as it attacks the electrophilic carbon. Click on the video to see how the nucleophile must displace the solvent to reach the electrophile.

When the solvent is aprotic (cannot hydrogen bond with the nucleophile), nucleophilicity is as expected and follows basicity.

Below is a table of nucleophiles and their relative nucleophilicity in a polar protic solvent: NC¯ is the strongest and H2O is the poorest nucleophile in the table. Notice that weak polarizable bases like CN¯, HS¯, and I¯ are also excellent nucleophiles, but weak, slightly polarizable bases like H2O, CH3OH, and F¯are poorer nucleophiles.

Q7.17 - Level 2

Rank the following compounds from strongest to weakest base.

A

3)

B

2)

C

1)

Q7.18 - Level 2

Rank the compounds from strongest to weakest nucleophile in a protic solvent.

A

2)

B

1)

C

3)

Watch the following video to see the explanation to Q7.18.

### 7.3.3 Leaving Group

In the transition state, the leaving group gains electrons and becomes negatively charged relative to its charge in the reactant. Therefore, good leaving groups are those that form stable ions or molecules after they depart. There is a good correlation between leaving group ability and pKa values of acids. The conjugate bases of acids with low pKa values (strong acid) are good leaving groups. Or, if we just think about the base that is leaving, then the weaker the base the better the leaving group.

In the transition state, the nucleophile forms a new sigma bond while the electrophile-leaving group bond is cleaved. Thus, correlation for proton affinity (basicity) and affinity for a carbon is only an approximation. For example, the hydrogen halides (HI, HBr, HCl) are strong acids and have very weak conjugate bases. Sulfonic acid anions, such as the sulfonates from p-toluenesulfonic and methanesulfonic acids, like halides, are also very good leaving groups even though they are conjugate bases of relatively weaker acids. Another very good sulfonate leaving group is triflate (trifluoromethanesulfonate). Triflic acid is an extremely strong acid, which makes its conjugate base a very good leaving group.

Notice that halide anions like Brand I are both good nucleophiles and good leaving groups because they are polarizable weak bases. So how do you know if the halide is going to act as a nucleophile or a leaving group?

For the next question, think about the reactivity of the nucleophile and leaving group.

Q7.19 - Level 3

What is the product of the following reaction?

A

1)

B

2)

C

3)

D

4)

Watch the following video to see the explanation to Q7.19.

What would happen if sulphuric acid was added to NaBr and methanol?

The first step would be an acid–base reaction or a proton transfer.

Now the electrophile is the less stable protonated alcohol (alkyloxonium ion) and water is the leaving group. Bromide is a good nucleophile and a weak base, water is a poor nucleophile and a weak base. The equilibrium will shift to the less reactive species and form the product, methylbromide.

When you visualize the reaction, always think about energy changes. Protonation of the alcohol produces the alkoxonium ion. The reaction is slightly exothermic because sulfuric acid is a stronger acid than the alkoxonium ion. The alkoxonium ion is an intermediate that has fully formed bonds. The bromide ion would then attack to produce the more stable products as shown in the Potential Energy–Reaction Coordinate diagram.

Q7.20 - Level 2

Order the compounds from most reactive to least reactive towards a nucleophile, if the reaction was carried out in the same reaction conditions, with the same nucleophile.

A

1)

B

2)

C

3)

Watch the following video to see the explanation to Q7.20.

### 7.3.4 Solvent Effects

SN2 reactions generally occur in an ocean of solvent and, since SN2 reactions generally involve partially or fully charged atoms or molecules, it is expected that solvents will change the relative rate of an SN2 reaction.

Definitions

Polar protic solvents contain at least one hydrogen connected directly to an electronegative atom and can form hydrogen bonds with both reactants and products.

Polar aprotic solvents have a dipole moment but do not contain a hydrogen atom connected directly to an electronegative atom and cannot act as hydrogen bond donors with the reactants and products. Aprotic solvents can be H-bond acceptors because they possess lone pairs of electrons. The intermolecular forces involved in solvation of the reactants and products are dipole–dipole interactions and dispersion forces.

Non polar solvents have an overall dipole moment of zero. These include CCl4, hexane, cyclohexane, and benzene, which do not have a hydrogen bonded to an electronegative atom. These solvents are generally not used in ionic reactions.

Solvents can change the rate of the reaction by changing the solvation energy of the reactants, transition state, and products by making them more or less stable.

As stated in Section 4.2, iodide is a better nucleophile than fluoride in a protic solvent. The reactivity of the fluoride ion decreases, due to strong hydrogen bonding with the solvent, and inhibits its ability to reach the electrophile. Thermodynamically, we can also state that cations and anions are stabilized by polar protic solvents and, thus, these solvents decrease their reactivity.

What happens when we change the solvent to an aprotic solvent like acetone?

Aprotic solvents cannot stabilize ions through hydrogen bonding, but can stabilize the ions through electron donation by heteroatoms, dispersion and dipole-ion forces. In the case of acetone the oxygen atom stabilizes the cation through electron donation, and the anion through dipole-anion and dispersion intermolecular forces. The anion-dipole interaction is a weaker intermolecular force than hydrogen bonding, thus the nucleophile (anion) has a higher potential energy (less stable) and is more reactive in an aprotic solvent.

The solvent also affects the transition state. In an SN2 reaction the charge is dispersed between the nucleophile and the leaving group. Solvents that are slightly polar help stabilize the disperse transition state. These two factors increase the rate of an SN2 reaction by decreasing the magnitude of the activation energy.

Since polar aprotic solvents increase the rate of reaction, why not use nonpolar aprotic solvents? Nonpolar aprotic solvents might not increase the rate of reaction because the polar transition state would not be stabilized by the solvent; in addition, since the nucleophile is partially charged, or has a full negative charge, generally they will not dissolve in nonpolar solvents.

Crown ethers are very good co-solvents for SN2 reactions because they solvate cations very well and enhance the solubility of the anionic nucleophile in nonpolar solvents. Using these solvents creates an essentially “naked” nucleophile and extremely large reaction rates are observed. (For more information, google “crown ethers.”)

The rate of reaction will also change if the polarity of the protic or aprotic solvent is increased or decreased. A much more thorough discussion is needed than can be presented here. Look to other resources to see how solvents affect reaction rates.

Q7.21 - Level 1