# 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.

## What is a Top Hat Textbook?

Top Hat has reimagined the textbook – one that is designed to improve student readership through interactivity, is updated by a community of collaborating professors with the newest information, and accessed online from anywhere, at anytime.

• Top Hat Textbooks are built full of embedded videos, interactive timelines, charts, graphs, and video lessons from the authors themselves
• High-quality and affordable, at a significant fraction in cost vs traditional publisher textbooks

## Comparison of Organic Chemistry Textbooks

Consider adding Top Hat’s Organic Chemistry textbook to your upcoming course. We’ve put together a textbook comparison to make it easy for you in your upcoming evaluation.

### 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

Read the fully unlocked textbook below, and if you’re interested in learning more, get in touch to see how you can use this textbook in your course today.

# Chapter 8: SN1 Reactions and Distinguishing the Differences Between SN1 and SN2 Reactions

## Learning Objectives

• Comprehend the differences between SN1 and SN2 reactions.
• Write the mechanism for an SN1 reaction and explain the term unimolecular nucleophilic substitution.
• Comprehend the factors that affect the rate of an SN1 reaction: leaving group ability, solvent effects, and the effect of electrophile structure (carbocation stability).
• Predict the stereochemical outcome of an SN1 reaction.
• Evaluate a series of substrates and determine their relative rate of reaction in an SN1 reaction.
• Understand the effect of protic and aprotic solvents have on the rate of an SN1 reaction.
• Recognize when a carbocation rearrangement will occur and predict the products formed in the SN1 reaction.
• By examining the nucleophile, electrophile and solvent you will be able to predict whether an SN1 or an SN2 reaction will occur.

## 8.0 Introduction to SN1 Reactions

In chapter 7: SN2 Reactions with Alkyl Halides, we studied the SN2 reaction and the effect nucleophilicity, the structure of the electrophile, and solvents have on the rate of an SN2 reaction. In this chapter you will apply this knowledge to a mechanistically different substitution reaction, the SN1 reaction.

### Definitions: Review

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 that in order to accept a pair of electrons an atom must have an empty orbital.

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

## 8.1 Substitution Nucleophilic Unimolecular (SN1)

Definition: It is a nucleophilic substitution reaction in which the rate of reaction is dependent only on the concentration of the electrophile. The reaction rate is first order as shown by its rate equation:

Rate of reaction = k[electrophile]

where k is equal to the rate constant.

Experimentally, this means that if the electrophile concentration is doubled, the rate of reaction doubles. On the other hand, if the nucleophile concentration is doubled, the rate of the reaction does not change. The nucleophile is not involved in the rate-determining step.

The overall SN1 reaction occurs through a series of steps. During the course of these steps, intermediates are formed and the rate of reaction is determined by the slowest step. The slow step is called the rate-limiting step or the rate-determining step.

Make sure you understand the Potential Energy Reaction Coordinate diagram and how energy relates to reaction rates in the animation.

What did the video show us?

• The reaction is multi-step, and the first step involves a heterolytic cleavage of the carbon-leaving group σ bond and does not involve the nucleophile.
• Since there is only bond breaking in the first step, the activation energy is large and would be a slow endothermic reaction.
• The first step is the rate-determining step. This step is unimolecular as the rate of reaction is dependent on the concentration of only one of the reactants, the electrophile. Rate of reaction = k[electrophile]
• The reaction is multi-stepped and intermediates are formed before the final products are produced.
• Intermediates have fully formed bonds.
• The nucleophilic attack follows a specific pathway. The attack on the electrophile can only occur on either face of the carbocation and into the empty p orbital.
• The final products are at a lower potential energy than the reactants and intermediates.

### 8.1.1 Stereochemistry of an SN1 reaction

What is the stereochemistry of the products when the nucleophilic attack creates a chiral carbon? For example, what is the stereochemistry of the products when (R)-3-bromo-3-methylhexane undergoes an SN1 reaction? The first step in the mechanism is the loss of a leaving group. This produces an achiral carbocation. The carbocation is sp2 hybridized, trigonal planar, with an empty p orbital above and below the plane of the alkyl groups.

The nucleophile can only donate electrons to an empty orbital of the electrophile, since the filled orbital already has two electrons and cannot accept more. Thus, it is the empty p orbital that accepts the electrons from the nucleophile. Since the p orbital is on either side of the plane of the alkyl groups, the nucleophilic attack can occur from either face of the carbocation to form an equal mixture of enantiomers. Pathway A shows the nucleophilic attack from the top face of the drawing and pathway B shows it from the bottom face. Since each pathway has an equal probability of occurring, this step will produce an equal (50:50) mixture of R and S enantiomers, as shown in Figure 8.4.

A solution that contains an equal mixture of enantiomers is called a racemic mixture. A pair of enantiomers is called racemates.

Optical activity: When a plane-polarized light passes through a solution containing a chiral molecule, the plane-polarized light will be deflected to the left or right. This is called optical activity. If the plane-polarized light is rotated clockwise, the rotation is positive (+) and the molecule is dextrorotatory. If the plane-polarized light is rotated counterclockwise, the rotation is negative (-) and the molecule is levorotatory.

In an SN1 reaction, both enantiomers are produced in an equal mixture. Thus, the optical activity of a solution of the products would be zero because the optical activity of one enantiomer cancels out the optical activity of the other enatiomer.

Q8.1 - Level 1

Which statements about S$_N$1 reactions are true and which statements are false?

Premise
Response
1

The rate of the reaction depends on the concentration of the nucleophile

A

False

2

The carbocation intermediate is achiral

B

True

3

The rate of the reaction depends on the concentration of the alkyl halide

C

False

4

The optical activity of a racemic mixture can either be positive or negative

D

False

E

True

F

True

Q8.2 - Level 2

Which of the following is the rate law for the given S$_N$1 reaction?

A

1)

B

2)

C

3)

D

4)

Q8.3 - Level 2

What are the major substitution products formed in the following reaction?

A

A mixture of 1) and 4)

B

A mixture of 3) and 1)

C

A mixture of 1) and 2)

D

A mixture of 3) and 4)

Q8.4 - Level 3

Two substitution products are formed in the following reaction. What relationship do they have with each other?

A

Enantiomers

B

Diastereomers

C

Constitutional isomers.

D

Meso compounds

Q8.5 - Need Level

Two substitution products are formed in the following reaction. What relationship do they have with each other?

A

Enantiomers

B

Diastereomers

C

Constitutional isomers

D

Meso compounds

E

Only one compound is produced

Q8.6 - Level 3

The given compounds were heated in methanol and the resultant S$_N$1 products were determined. Match the products with their observed stereochemistry.

Premise
Response
1

1)

A

Achiral product

2

2)

B

Enantiomers

3

3)

C

Achiral product

4

4)

D

Diastereomers formed

## 8.2 Factors that Affect the Rate of the SN1 Reaction

### 8.2.1 Introduction

When we think about the rate of an SN1 reaction, we must think about the mechanism and how the molecules climb the energy barrier to reach the transition state and then form the carbocation intermediate. The transition state of the rate-determining step is very important in understanding the rate of an SN1 reaction.

The rate determining step is endothermic, and in the transition state, the C-L σ bond lengthens and weakens (the C-Cl bond in Figure 8.6). The leaving group gains electrons from the σ bond and becomes negatively charged relative to its charge as a reactant. The carbon loses electrons from the σ bond and becomes positively charged. The key factor to this mechanism is the formation of the unstable intermediate. Thus, anything that helps stabilize the transition state and intermediates will increase the rate of an SN1 reaction by decreasing the activation energy of the rate-determining step.

The factors that affect the rate of an SN1 reaction are as follows:

Structure of Electrophile: Since the rate-determining step involves the formation of a carbocation, electron-donating groups help stabilize the forming carbocation and decrease the activation energy.

Solvents: Since the rate-determine step involves the formation of ions, protic solvents have a large effect on the rate of reaction due to solvation energies.

Leaving group: The leaving group gains electrons from the σ 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 σ bond.

### 8.2.2 Leaving Group

In the transition state, the leaving group gain 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. This was discussed in the SN2 module and only the tables are presented here.

Q8.7 - Level 3

Order the following compounds from most reactive to least reactive in an S$_N$1 reaction.

A

2)

B

1)

C

3)

Watch the following video to see the explanation to Q8.7.

Q8.8 - Level 1

Which of the following alkyl halides would react the slowest when heated in an aqueous solution of acetone?

A

1)

B

2)

C

3)

D

4)

### SN1 Substitution in the Biosynthesis of Terpenoids

Terpenes are the largest class of natural products (organic compounds synthesized by Mother Nature) and are comprised of more than 30,000 compounds from all kingdoms of life.

You may not have learned the lingo ‘terpene’ before, but you’ve probably heard of some famous terpene natural products like menthol, retinal (in vitamin A), and taxol.

When nature constructs terpene natural products, it uses a five-carbon building block called dimethylallyl pyrophosphate.

Step 1: From dimethylallyl pyrophosphate, dissociation occurs to form dimethylallyl carbocation with the release of pyrophosphate.

Step 2: Dimethylallyl carbocation is attacked by another fragment called isopentenyl pyrophosphate.

Step 3: The resulting carbocation is ultimately converted to geranyl pyrophosphate, which Nature can further utilize.

“Step 2” involving carbocation formation and a nucleophile attack, is reminiscent of an SN1 reaction.

Nature uses the product from the sequence shown above, geranyl pyrophosphate, to make larger terpenes.

How does it work? The process shown earlier can be repeated to add another five-carbon surrogate to form farnesyl pyrophosphate. When two farnesyl pyrophosphate molecules join together (or ‘dimerize’) a molecule called squalene is formed.

This terpene can cyclize to form lanosterol, which is a precursor to cholesterol and all steroids in animals. Cholesterol is important for cell membrane permeability and also for the synthesis of vitamin D in our skin.

### Cholesterol in the movies

The famous terpene cholesterol comes up often on TV and the movies! For example, in the 2004 documentary film "Super Size Me", Morgan Spurlock eats McDonald’s food for 30 consecutive days. His cholesterol levels increased from 169 to 225 mg/dL. He also gained more than 24 pounds, while experiencing mood swings and sexual dysfunction. It took him 14 months to return to his normal weight.

### 8.2.3 Solvent Effects

The intermediates that are formed in the SN1 reaction are a carbocation plus a leaving group that is either negatively charged or neutral. Thus, the polar protic solvents would be expected to help stabilize the transition state and the intermediates by lowering the activation energy.​​

Below is a list of solvents and their relative rates of ionization of tert-butyl chloride. Given also is the dielectric constant (ε) which is a measure of a substance's ability to insulate charges from each other. This value can also be used to interpret a solvent's polarity; the higher ε means higher polarity and greater ability to stabilize charges.

Notice that the rate of ionization of tert-butyl chloride is greater for more polar protic solvents.

Q8.9 - Level 1

Which of the following solvolysis reactions would you expect to have the fastest rate of reaction?

A

1)

B

2)

C

3)

### 8.2.4 How do you Relate Thermodynamic Stability with a Kinetic Rate?

Kinetics refers to the rate of a reaction or how fast it will proceed. Kinetics does not tell you how much product is formed. Thermodynamics refers to the relative energies of the reactants and products or the concentrations of reactants and products. Ultimately thermodynamics tells you about the equilibrium constant. Thermodynamics will not tell you how fast the reaction is.

For the SN1 reaction, we can say that the rate of formation of the more stable intermediate will be faster than the formation of the less stable intermediate. This is because the more stable intermediate follows a pathway that has a lower energy transition state and thus has a smaller activation energy. A lower activation energy means that the reaction will proceed at a faster rate. (See Section 3.8 Hammond Postulate and Section 3.9 Kinetics vs. Thermodynamics)

### 8.2.5 Solvolysis Reactions: The Solvent as the Nucleophile

In many SN1 reactions, the solvent is also the nucleophile. These reactions are called solvolysis reactions. A solvolysis reaction is any substitution reaction where the nucleophile is the solvent. Therefore, either an SN1 or an SN2 reaction could be a solvolysis reaction, as long as the solvent is the nucleophile. For certain nucleophiles there are specific terms for the types of solvolysis reactions. For example, when water is the nucleophile the reaction is called a hydrolysis reaction, Figure 8.12. The second reaction shown in Figure 8.12 is an alcoholysis or more specifically an ethanolysis reaction because ethanol is the nucleophile. In the last reaction an acid is used in the decomposition of the alkyl halide to form the ester. Thus this reaction is called an acidolysis reaction.

### 8.2.6 Mixed Solvent Reactions

Many nonpolar or slightly polar organic compounds are not soluble in polar solvents. Likewise, many polar compounds are not soluble in nonpolar or slightly polar solvents. To dissolve such an organic compound, a mixed solvent system must be used.

For example, if we wish to perform an SN1 reaction with (S)-3-bromo-2,3-dimethylhexane and water, no reaction would occur because the starting material is not soluble in water. However, (S)-3-bromo-2,3-dimethylhexane is soluble in acetone, and acetone is miscible with water! Now, if the reaction is performed in a water/acetone solvent mixture, an SN1 reaction will occur to produce a racemic mixture of alcohols.

In the above example, acetone does not react with the carbocation intermediate. Water is the only nucleophile present. However, if there is more than one nucleophile present, a mixture of products will form.

### 8.2.7 The Influence of Solvents on the Stereochemistry of the Products

In the previous sections, it has been assumed that the leaving group and the cation (ion pairs) are separate ions that are completely dissociated and solvated. However, if a contact ion-pair forms, the ion pairs are close to each other and the leaving group will block one face of the carbocation. Nucleophilic attack from the side with the leaving group will be more difficult since the solvent has to displace the leaving group before a reaction can occur.

If a contact ion-pair forms, the rate of reaction would be slower from the face from which the leaving group left. Thus we would expect to see a greater percentage of the product with inversion of the chiral carbon and a lower percentage of the product with retention of the chiral carbon. This is indeed what we see in some reactions. In the example below, an enantiomeric excess (ee) of the (S)-1-phenylethanol is formed. That is, more product is formed from the nucleophilic attack on the opposite side of the leaving group (inversion of stereocenter) than is formed from the side that the leaving group was on (retention of stereocenter). If this occurs, the products will not form a racemic mixture and the solution would be optically active. For an explanation of enantiomer excess, review the stereochemistry module, Chapter 6.3.6.

## 8.3 Effect of Structure on the Rate of Reaction

### 8.3.1 Electronic Effects

The rate-determining step in the SN1 reaction is the formation of the carbocation. A carbocation does not have a full octet and is unstable. In previous sections, we have seen that by increasing the stability of the transition state and carbocation, the rate of reaction increases because the activation energy of the rate-determining step decreases. Thus, it is expected that any groups attached to the carbocation that stabilize a carbocation will also increase the rate of an SN1 reaction.

Carbocation stability

Experimentally, it has been shown that as the number of alkyl groups attached to the carbocation increases the stability of the carbocation increases.

This can be explained by electron donation of the adjacent C-C or C-H σ bonds into the empty carbocation’s p orbital. Alkyl groups donate electron density inductively through σ bond conjugation or hyperconjugation.

Increasing the number of alkyl groups attached to the carbocation increases the stability of the cation. This suggests that, in SN1 conditions, only secondary and tertiary carbocations will form.

Q8.10 - Level 1

Which of the following alkyl halides would be most reactive in a unimolecular reaction?

A

1)

B

2)

C

3)

D

4)

Q8.11 - Level 2

Sort the alkyl halides from fastest to slowest in a S$_N$1 reaction.

Premise
Response
1

Fastest reaction

A

3)

2

Second fastest

B

2)

3

Slowest reaction

C

1)

### 8.3.2 Reactivity of Allylic, Benzylic, Vinyl, and Aryl Alkyl Halides

Both allylic and benzylic alkyl halides are very reactive in SN1 conditions. They are more reactive than tertiary alkyl halides. This is because resonance stabilization of the carbocation intermediate spreads the charge onto different atoms.

Vinyl and aryl alkyl halides do not react under normal SN1 conditions because the positive charge is not resonance stabilized by the π bond.

Q8.12 - Level 2

Order the following compounds from most reactive to least reactive in a S$_N$1 reaction in methanol

A

1

B

4

C

2

D

3

Watch the following video to see the explanation to Q8.12.

Q8.13 - Level 2

Sort the alkyl halides from fastest to slowest in a S$_N$1 reaction.

Premise
Response
1

Fastest reaction

A

3)

2

Second fastest

B

4)

3

Third fastest

C

1)

4

Slowest reaction

D

2)

Q8.14 - Level 2

Sort the alkyl halides from fastest to slowest in an S$_N$1 reaction

Premise
Response
1

Fastest reaction

A

2)

2

Second fastest

B

4)

3

Third fastest

C

1)

4

Slowest reaction

D

3)

### 8.3.3 Carbocation Rearrangement

Any time a carbocation is formed, a rearrangement can occur to form a more stable carbocation. A hydrogen, methyl, or whole alkyl group can move. Shown below are methyl and hydride migrations.

The group that migrates must be involved in hyperconjugation prior to the shift or β to the carbocation.

The Potential Energy Reaction Coordinate Diagram shows the decrease in potential energy in forming the tertiary carbocation. However, carbocation rearrangement does not have to produce a more stable carbocation. For example, carbocation rearrangement between adjacent secondary carbons can take place to produce a mixture of secondary products.

Alkyl groups can also rearrange. For example, the cyclobutane ring rearranges to form the more stable cyclopentane ring in the following SN1 reaction.

Q8.15 - Level 1

Would the following carbocation undergo a hydride shift or a methyl shift?

A

Hydride shift

B

Methyl shift

Watch the following video to see the explanation to Q8.15.

Q8.16 - Level 3

Identify the major substitution product(s) formed in the following reaction.

A

1)

B

2)

C

3)

D

An equal mixture of 1) and 2)

E

An equal mixture of 1), 2) and 3)

Q8.17 - Level 3

Which of the following compounds is most likely to undergo rearrangement in a solvolysis reaction?

A

2-chloro-6-methyloctane

B

2-chloro-2-methyloctane

C

2-chlorooctane

D

2-chloro-3,3-dimethyloctane

Q8.18 - Level 2

What are the products produced in the given solvolysis reaction?

A

An equal mixture of 1) and 2)

B

An equal mixture of 1) and 3)

C

An equal mixture of 2) and 4)

D

An equal mixture of 2) and 3)

E

All of the substitution products will be produced equally

Q8.19 - Level 1

Sort the following carbocations from most stable to least stable.

Premise
Response
1

Most Stable