Organic chemistry 8th edition (2017) part 2

416  Chapter 9: Nucleophilic Substitution and b-Elimination

Because the solvent is polar protic, there could be a minor extent of SN1/E1. If the sodium
acetate were left out of the reaction and it were heated, the prediction would be SN1/E1.
Competition 4

The haloalkane is secondary. The Nuc/Base has a pKa of its conjugate acid near or
slightly below 11 (HCN, pKa 5 9.3) and hence is a moderate to weak base. However,
cyanide anion is an excellent nucleophile. Consequently, SN2 will dominate over E2.
Furthermore, the solvent DMF (dimethylformamide) is polar and aprotic and supports SN2 or E2, but it does not assist SN1 or E1. Because the reactant is chiral, the
SN2 inverts the configuration.
1

Competition 5

The haloalkane is tertiary; therefore, SN2 cannot occur. The Nuc/Base is a weak
base (pKa HN3 5 4.9); therefore, E2 is not obviously favored. However, the solvent is
not protic but is simply polar. Therefore, SN1 and E1 are not going to be favored. This
is a case that is difficult to predict using Figure 9.8 or Table 9.11.
However, the lack of a polar protic solvent means that E2 is most likely, even
with the weak base. The E2 occurs with an anti and coplanar arrangement of the Br
and H that are eliminated, giving an E alkene. Any substitution from an SN1 pathway
would lead to racemization of the chiral center that possessed the leaving group.

1

1

Example 9.8  SN1 or SN2, E1 or E2
Predict whether each reaction proceeds predominantly by substitution (SN1
or SN2) or elimination (E1 or E2) or whether the two compete. Write structural

formulas for the major organic product(s).
(a)

1

8

(b)

1

8

Solution
(a) A 3° haloalkane is heated with a strong base/good nucleophile. Elimination by
an E2 reaction predominates to give 2-methyl-2-butene as the major product.

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9.9  Analysis of Several Competitions Between Substitutions and Eliminations   417

8

1

1

1

(b) Reaction of a 1° haloalkane with this moderate nucleophile/weak base
gives substitution by an SN2 reaction.
8

1

1

2

Problem 9.8

Watch a video explanation

Predict whether each reaction
proceeds predominantly by substitution (SN1 or
2
1
(a)
1
SN2) or elimination (E1 or E2) or whether the two compete. Write structural
2
1
(a)
1
formulas for the major organic product(s).
(b)
(a)

1

1

(c)
(b)

1
1

(c)

1

1 2
2

1

1

1

1 2

2

(c)

1 2

1

(b)
  

1

1

2

2

MCAT Practice: Passage and Questions
Solvents and Solvation

Choosing the best solvent for a chemical reaction is an
extremely important aspect of organic chemistry. When
deciding upon a solvent, chemists consider the solubility of the reactants and products, as well as the mechanism of the reaction and the solvation of intermediates.
Further, for reactions that need heating to proceed in

a reasonable amount of time, the choice of solvent is
guided by its boiling point because this sets the temperature at which the reaction refluxes. Lastly, unless
the solvent is intentionally used as a reactant, such as
in a solvolysis, it must remain inert.

Questions
A. When performing an SN1 solvolysis, which of the
following solvents would be a poor choice for tertbutyl iodide (“dried” means that water has been
removed)?
1.

2.
3.
4.

80% water, 20% ethanol
Pure water
Dried acetonitrile
Dried acetic acid

B. When attempting to enhance the extent of SN2
substitution by the nucleophile ethylamine
(EtNH2), which of the following solvents would
be a poor choice for sec-butyl iodide?
1. Pure water
2. Acetonitrile
3. DMSO
4.tert-Butyl alcohol

C. When performing an SN2 reaction using NaCN
as the nucleophile reacting with n-butyl iodide,
which of the following solvents would be the
worst choice?
1.
2.
3.
4.

DMSO
DMF
Acetonitrile

Toluene

D. The reaction of diethylamine (Et2NH) and sec-butyl
iodide requires heating, but to optimize the extent of
SN2 over E2 the reaction cannot be too hot. Which
of the following solvents would best represent a
compromise solvent in which to reflux this reaction?
1.
2.
3.
4.

Diphenyl ether
Diethyl ether
THF
DMSO

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418  Chapter 9: Nucleophilic Substitution and b-Elimination

An important take-home lesson from this chapter is that understanding key
transition state or reactive intermediate geometries as well as relative transition
state energies allows the prediction of product stereochemistry and regiochemistry.
Backside attack in SN2 reactions, the anti and coplanar geometry of the H atom and
leaving group in E2 reactions, and the presence of carbocation intermediates in SN1
reactions are important examples of reaction geometries that dictate stereochemistry.
­Understanding the relative energies of alternative p
­ ossible transition states is also important. In the case of b-elimination reactions, relative transition state energies provide a rationale for Zaitsev’s rule of regiochemistry. As you go through the rest of this

book, try to learn key features of reaction mechanisms that dictate the stereochemistry and regiochemistry of reaction products. You should think of mechanisms as more
than just electron pushing: they involve three-dimensional molecular interactions
with associated relative energies that control the formation of products.

9.10  Neighboring Group Participation
So far, we have considered two limiting mechanisms for nucleophilic substitutions
that focus on the degree of covalent bonding between the nucleophile and the substitution center during departure of the leaving group. In an SN2 mechanism, the leaving
group is assisted in its departure by the nucleophile. In an SN1 mechanism, the leaving group is not assisted in this way. An essential criterion for distinguishing between
these two pathways is the order of reaction. Nucleophile-assisted substitutions are
second order: first order in RX and first order in nucleophile. Nucleophile-unassisted
substitutions are first order: first order in RX and zero order in nucleophile.
Chemists recognize that certain nucleophilic substitutions have the kinetic characteristics of first-order (SN1) substitution but, in fact, involve two successive displacement reactions. A characteristic feature of a great many of these reactions is the
presence of an internal nucleophile (most commonly sulfur, nitrogen, or oxygen) on
the carbon atom beta to the leaving group. This neighboring nucleophile participates
in the departure of the leaving group to give an intermediate, which then reacts with
an external nucleophile to complete the reaction.
The mustard gases are one group of compounds that react by participation of
a neighboring group. The characteristic structural feature of a mustard gas is a twocarbon chain, with a halogen on one carbon and a divalent sulfur or trivalent nitrogen
on the other carbon (S-C-C-Lv or N-C-C-Lv). An example
of a mustard gas is bis(2ClCH2CH2SCH2CH2Cl
chloroethyl)sulfide, a poison gas used extensively in World War I and at one time, at
bis(2-Chloroethyl)sulfide
least, manufactured by Iraq. This compound is a deadly
vesicant (blistering agent) and
quickly causes conjunctivitis and blindness.

ClCH2CH2SCH2CH2Cl
ClCH2CH2SCH2CH2Cl
bis(2-Chloroethyl)sulfide
bis(2-Chloroethyl)sulfide

CH3
ClCH2CH2

N

CH2CH2Cl

bis(2-Chloroethyl)methylamine

Bis(2-chloroethyl)sulfide and bis(2-chloroethyl)methylamine are not gases at all. They
are oily liquids with a high vapor pressure, hence the designation “gas.” Nitrogen and
sulfur mustards react very rapidly with moisture in the air and in the mucous membranes of the eye, nose, and throat to produce HCl, which then burns and blisters
CH

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ClCH2CH2

N
CH3 CH2CH2Cl


9.10  Neighboring Group Participation   419

these sensitive tissues. What is unusual about the reactivity of the mustard gases is
that they react very rapidly with water, a very poor nucleophile.
1

1

Mustard gases also react rapidly with other nucleophiles, such as those in biological
molecules, which makes them particularly dangerous chemicals. Of the two steps in
the mechanism of the hydrolysis of a sulfur mustard, the first is the slower and is ratedetermining. As a result, the rate of reaction is proportional to the concentration of
the sulfur mustard but independent of the concentration of the external nucleophile.
Thus, although this reaction has the kinetic characteristics of an SN1 reaction, it actually involves two successive SN2 displacement reactions.

Mechanism 9.9
Hydrolysis of a Sulfur Mustard—Participation
by a Neighboring Group
Step 1:  Make a new bond between a nucleophile and an electrophile and ­simultaneously break a bond to
give stable molecules or ions.  The reason for the extremely rapid hydrolysis of the sulfur mustards is neighboring
group ­participation by sulfur in the ionization of the carbon-chlorine bond to form a cyclic sulfonium ion. This is the
rate-determining step of the reaction; although it is the slowest step, it is much faster than reaction of a typical primary
chloroalkane with water. At this point, you should review halogenation of alkenes (Sections 6.3D and 6.3F) and
compare the cyclic halonium ions formed there with the cyclic ­sulfonium ion formed here.

1

1

1

Step 2:  Make a new bond between a nucleophile and an electrophile.  The cyclic sulfonium ion contains
a highly strained three-membered ring and reacts rapidly with an external nucleophile to open the ring
followed by proton transfer to H2O to give H3O1. In this SN2 reaction, H2O is the nucleophile and sulfur is
the leaving group.
1

1

1

Step 3:  Take a proton away.  Proton transfer to water completes the reaction.

1

1

9

9

1

1

The net effect of these reactions is nucleophilic substitution of Cl by OH.
We continue to use the terms SN2 and SN1 to describe nucleophilic substitution
reactions. You should realize, however, that these designations do not adequately describe all nucleophilic substitution reactions.

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420  Chapter 9: Nucleophilic Substitution and b-Elimination

Example 9.9  Hydrolysis of Nitrogen Mustards
Write a mechanism for the hydrolysis of the nitrogen mustard bis(2-chloroethyl)methylamine.

Solution

Following is a three-step mechanism.
Step 1:  Make a new bond between a nucleophile and an electrophile and
simultaneously break a bond to give stable molecules or ions.  This is an
internal SN2 reaction in which ionization of the C!Cl bond is assisted by the
neighboring nitrogen atom to form a highly strained three-membered ring.
1

1

Step 2:  Make a new bond between a nucleophile and an electrophile. 
Reaction of the cyclic ammonium ion with water opens the three-membered ring.
In this SN2 reaction, H2O is the nucleophile and nitrogen is the leaving group.
1

1

9

1

Step 3:  Take a proton away.  Proton transfer to the basic nitrogen completes
the reaction.
1
1

Problem 9.9
Knowing what you do about the stereochemisry of SN2 reactions, predict the
product of hydrolysis of this compound.
1

Connections to Biological Chemistry
Mustard Gases and the Treatment
of Neoplastic Diseases
Autopsies of soldiers killed by sulfur mustards in
World War I revealed, among other things, very low
white blood cell counts and defects in bone marrow
development. From these observations, it was realized
that sulfur mustards have profound effects on rapidly

dividing cells. This became a lead observation in
the search for less toxic ­alkylating agents for use in
treatment of cancers, which have rapidly dividing
cells. Attention turned to the less reactive nitrogen
mustards. One of the first compounds tested was

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9.10  Neighboring Group Participation   421

Mechlorethamine

mechlorethamine. As with other mustards, the
reaction of mechlorethamine with nucleophiles is
rapid because of the formation of an aziridinium ion.

melphalan is chiral. It has been demonstrated that
the R and S enantiomers have approximately equal
therapeutic potency.

Mechlorethamine undergoes very rapid reaction with
­water (hydrolysis) and with other nucleophiles, so much
so that within minutes after injection into the body, it
has completely reacted. The problem for the chemist,
then, was to find a way to decrease the nucleophilicity
of nitrogen while maintaining reasonable water
solubility. Substitution of ­phenyl for methyl reduced the

The clinical value of the nitrogen mustards lies in
the fact that they undergo reaction with certain
nucleophilic sites on the heterocyclic aromatic amine
bases in DNA (see Chapter 28). For DNA, the most
reactive nucleophilic site is N-7 of guanine. Next
in reactivity is N-3 of adenine, followed by N-3 of
cytosine.

Chlorambucil

nucleophilicity, but the resulting compound was not
sufficiently soluble in water for intravenous injection.
The solubility problem was solved by adding a carboxyl
group. When the carboxyl group was added directly to
the aromatic ring, however, the resulting compound was
too stable and therefore not biologically active.
Adding a propyl bridge (chlorambucil) or an
aminoethyl bridge (melphalan) between the aromatic
ring and the carboxyl group solved both the solubility
problem and the reactivity problem. Note that

Melphalan

The nitrogen mustards are bifunctional alkylating
agents; one molecule of nitrogen mustard undergoes
reaction with two molecules of nucleophile. Guanine
alkylation leaves one free reactive alkylating group,
which can react with another base, giving cross links
that lead to miscoding during DNA replication. The
therapeutic value of the nitrogen mustards lies in their
ability to disrupt normal base pairing. This prevents
replication of the cells, and the rapidly dividing cancer
cells are more sensitive than normal cells.
(Continued)

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422  Chapter 9: Nucleophilic Substitution and b-Elimination

7

Guanine

Study Guide
9.1  Nucleophilic Substitution in Haloalkanes
●●

Nucleophilic substitution is any reaction in which a nucleophile replaces another electron-rich group
called a leaving group (Lv).
–A nucleophile (Nu:2) is an electron-rich molecule or ion that donates a pair of electrons to
another atom or ion to form a new covalent bond.

P

9.1

9.2 Mechanisms of Nucleophilic Aliphatic Substitution
●●

There are two limiting mechanisms for nucleophilic substitution, namely SN2 and SN1.
–In the SN2 reaction mechanism, bond forming and bond breaking occur simultaneously.
–SN2 reactions are bimolecular because both nucleophile and haloalkane concentrations
influence reaction rate.
–The nucleophile must approach the carbon-leaving group (C!Lv) bond from the backside in
order to populate the C!Lv antibonding orbital and allow reaction.
–In the SN1 mechanism, the leaving group departs first in the rate-determining step, leaving a
carbocation intermediate that reacts with the nucleophile in a second step.
–SN1 reactions are unimolecular because only the haloalkane concentration influences reaction rate.

9.3 Experimental Evidence for SN1 and SN2 Mechanisms
●●

The SN2 reaction can be identified based on kinetics of the reaction and stereochemistry of the
products.

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Study Guide   423

–Because an SN2 reaction is bimolecular, doubling the concentration of either haloalkane or

nucleophile will double the rate of the reaction.
–Because backside attack geometry is required, reaction at a chiral center results in inversion of
configuration.
–The SN1 reaction can also be identified based on kinetics of the reaction and stereochemistry of
the products.
–Because an SN1 reaction is unimolecular, doubling the concentration of only the haloalkane can
double the rate of the reaction.
–Because a planar and achiral carbocation intermediate is formed that can be attacked with roughly
equal probability from either face, reaction at a chiral center results in racemization of stereochemistry.
–The chiral center is often not completely racemized because the leaving group forms an ion pair
with the carbocation intermediate, partially blocking one face.
●●

The structure of the haloalkane influences the reaction rate and mechanism.
–Haloalkanes that can form more stable carbocations react faster if an SN1 mechanism occurs.
–Because SN1 reactions involve carbocations, rearrangements (1,2 shifts) can occur if they lead
to more stable carbocation intermediates.
–Steric hindrance on the backside of the C—Lv bond of a haloalkane slows down or possibly
prevents an SN2 mechanism.

●●

The more stable the anion produced upon reaction, the better the leaving group ability.

●●

Solvent properties can have an important influence on reaction mechanisms.
–Protic solvents are hydrogen-bond donors. The most common protic solvents are those
containing —OH groups.
–Aprotic solvents cannot serve as hydrogen-bond donors. Common aprotic solvents are acetone,

diethyl ether, dimethyl sulfoxide, and N,N-dimethylformamide.
–Polar solvents interact strongly with ions and polar molecules.
–Nonpolar solvents do not interact strongly with ions and polar molecules.
–The dielectric constant is the most commonly used measure of solvent polarity.
–Solvolysis is a nucleophilic substitution reaction in which the solvent is the nucleophile.
–Polar protic solvents accelerate SN1 reactions by stabilizing the charged carbocation
intermediate.
–Polar aprotic solvents accelerate SN2 reactions because they do not interact strongly with the
nucleophile.

●●

Nucleophiles are categorized as good, moderate, or poor.
–Good nucleophiles are generally anions. Moderate nucleophiles are generally neutral, with
one or more available lone pairs. Poor nucleophiles are generally polar protic solvents.
–All things being equal, the stronger the interaction of a nucleophile with solvent, the lower the
nucleophilicity.
–Small nucleophiles with very little steric hindrance are better nucleophiles for SN2 reactions.

P

9.2–9.4, 9.10–9.13, 9.15–9.22, 9.24–9.36

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424  Chapter 9: Nucleophilic Substitution and b-Elimination

Stereochemistry of an SN1 Reaction

Key Reactions
1. Nucleophilic Aliphatic Substitution: SN2 (Section 9.3) SN2 reactions occur in one step; departure
of the leaving group is assisted by the incoming nucleophile, and both nucleophile and leaving group
are involved in the transition state. The nucleophile may be negatively charged as in the first example or
neutral as in the second example.
2

1

9

1

9

1

9

1

2

1

9

2

SN2 reactions result in inversion of configuration at the reaction center. They are accelerated more in

polar aprotic solvents than in polar protic solvents. The relative rates of SN2 reactions are governed by
steric factors, namely the degree of crowding around the site of reaction.
2. Nucleophilic Aliphatic Substitution: SN1 (Section 9.3) An SN1 reaction occurs in two steps.
Step 1 is a slow, rate-determining ionization of the C—Lv bond to form a carbocation intermediate
followed in Step 2 by rapid reaction of the carbocation intermediate with a nucleophile to complete
the substitution. Reaction at a chiral center gives l­argely racemization, often accompanied with a
slight excess of inversion of configuration. Reactions often involve carbocation rearrangements and
are accelerated by polar protic solvents. SN1 reactions are governed by electronic factors, namely the
relative stabilities of carbocation intermediates. The following reaction involves an SN1 reaction
with a ­hydride shift.
1

1

9.4 Analysis of Several Nucleophilic Substitution Reactions
●●

●●

●●

P

Methyl or primary haloalkanes react through SN2 mechanisms because of an absence of steric
hindrance and lack of carbocation stability.
Secondary haloalkanes react through an SN2 mechanism in aprotic solvents with good nucleophiles, but
through an SN1 mechanism in protic solvents with poor nucleophiles.
Tertiary haloalkanes react through an SN1 mechanism because the steric hindrance disfavors SN2
backside attack, and the attached alkyl groups stabilize a carbocation.
9.5, 9.14, 9.23

Problem 9.5

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Study Guide   425

9.5  b-Elimination
●●

A b-elimination reaction involves removal of atoms or groups of atoms from adjacent carbon atoms.
–Dehydrohalogenation is a b-elimination reaction that involves loss of an H and a halogen atom
from adjacent carbons to create an alkene from a haloalkane.
–Zaitzev's rule predicts that b-elimination reactions give primarily the more highly substituted
alkene. Such reactions are called Zaitsev eliminations.

P

9.6

9.6  Mechanisms of b-Elimination
●●

The two limiting mechanisms for b-elimination reactions are the E1 and E2 mechanisms.
–In the E1 mechanism, the leaving group departs to give a carbocation; then a proton is taken off
an adjacent carbon atom by base to create the product alkene.
–E1 reactions are unimolecular because only the haloalkane concentration influences the rate of
the reaction.
–In the E2 mechanism, the halogen departs at the same time that an H atom is removed by base
from an adjacent carbon atom to create the product alkene.

–E2 reactions are bimolecular because both the haloalkane and base concentrations influence
the rate of the reaction.

9.7 Experimental Evidence for E1 and E2 Mechanisms
●●

E2 reactions are stereoselective in that the lowest energy transition state is the state in which the
leaving group and H atoms that depart are oriented anti and coplanar.
–This anti and coplanar requirement determines whether E or Z alkenes are produced. For
cyclohexane derivatives, both the leaving group and departing H atom must be axial.

●●

Both E1 and E2 reactions are regioselective, favoring formation of the more stable (Zaitsev) product
alkene (as long as Lv and H can be oriented anti and coplanar in the case of E2).
–The more stable alkene is generally the more highly substituted alkene.

P

9.7, 9.37–9.42
Problem 9.7

Key Reactions
3. b-Elimination: E1 (Sections 9.6, 9.7) An E1 reaction occurs in two steps: slow, rate-determining
breaking of the C—Lv bond to form a carbocation intermediate followed by rapid proton transfer to
solvent to form an alkene. An E1 reaction is first order in haloalkane and zero order in base. Skeletal
rearrangements are common.
1

(Continued)

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426  Chapter 9: Nucleophilic Substitution and b-Elimination

4. b-Elimination: E2 (Sections 9.6, 9.7) An E2 reaction occurs in one step: simultaneous reaction with
base to remove a hydrogen, formation of the alkene, and departure of the leaving group. Elimination is
stereoselective, requiring an anti and coplanar arrangement of the groups being eliminated.

2

1

1

1

2

9.8  Substitution Versus Elimination
●●

●●

When deciding which substitution or elimination mechanism dominates a reaction, analyze the structure
of the haloalkane, the choice of the solvent, and the relative base strength of the nucleophile.
Methyl or primary haloalkanes do not react through E1 or SN1 mechanisms.
–SN2 is favored for all nucleophiles except for exceptionally strong bases (H2N2) or sterically
hindered ones ( tert-butoxide), which cause E2 to predominate.

●●

Secondary haloalkanes can react through any of the mechanisms.
–If the nucleophile is a strong base (conjugate acids with pKa’s above 11, such as hydroxide,
alkoxides, acetylides, and H2N2), E2 predominates.
–Weak bases (conjugate acids with pKa’s below 11) that are good or moderate nucleophiles (see
Table 9.7) react predominantly by an SN2 mechanism.
–Poor nucleophiles (that are polar protic solvents) react through a combination of SN1/E1
pathways, the exact ratio of which is hard to predict.

●●

Tertiary haloalkanes cannot react by an SN2 mechanism.
–If the nucleophile is a strong base (conjugate acids with pKa’s above 11, such as hydroxide,
alkoxides, acetylides, and H2N2), E2 predominates.
–For other nucleophiles in a polar protic solvent, reaction is through a combination of SN1/E1
pathways, the exact ratio of which is hard to predict.

P

9.8, 9.43–9.62
Contrasting SN2, SN1, E2 and E1

9.9 Analysis of Several Competitions Between Substitutions
and Eliminations
●●

Predicting whether substitution or elimination reactions will dominate is a matter of following the logic
given in Section 9.8.

–Either Table 9.11 or the flowchart given in Figure 9.8 (summarized just above) will lead to a
successful analysis of the majority of reactions that organic chemists perform.
Problem 9.8

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Problems   427

9.10  Neighboring Group Participation
●●

Certain nucleophilic displacements that have the kinetic characteristic of SN1 reactions (first order in
haloalkane and zero order in nucleophile) involve two successive SN2 reactions.
–Many such reactions involve participation of a neighboring nucleophile.
–The mustard gases are one group of compounds whose nucleophilic substitution reactions involve
neighboring group participation.

P

9.9

Key Reactions
5. Neighboring Group Participation: (Section 9.10) Neighboring group participation is characterized
by first-order kinetics and participation of an internal nucleophile in departure of the leaving group,
as in hydrolysis of a sulfur or nitrogen mustard gas. The mechanism for their solvolysis involves two
successive nucleophilic displacements.
1

1

Problems
Red numbers indicate applied problems.

Nucleophilic Aliphatic Substitution
9.10 Draw a structural formula for the most stable carbocation with each molecular formula.
(b)C3H71

(a)C4H91

(c)C5H111

(d)C3H7O1

9.11 The reaction of 1-bromopropane and sodium hydroxide in ethanol occurs by an SN2
mechanism. What happens to the rate of this reaction under the following conditions?
(a) The concentration of NaOH is doubled.
(b) The concentrations of both NaOH and 1-bromopropane are doubled.
(c) The volume of the solution in which the reaction is carried out is doubled.
9.12 From each pair, select the stronger nucleophile.
(a)H2O or OH2
(c)CH3SH or CH3S2
(e)Cl2 or I2 in methanol

(b)CH3COO2 or OH2
(d)Cl2 or I2 in DMSO
(f)CH3OCH3 or CH3SCH3

9.13 Draw a structural formula for the product of each S N2 reaction. Where configuration of
the starting material is given, show the configuration of the product.

(a)
(a)

(a)

(c)
(c)

(c)

(e)
(e)

(e)

(g)
(g)

1
1

1
1

1
1

1
11
1

#
1#

22

221

11 2

1

(b)

1
1

1
1
1

2

22 #
11 2

(b)
(b)

(d)

(d)

1

122

11 2

1

(d)

(f)
(f)

(f)

1
1

1

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11 in whole
221 or in2part. WCN 02-200-203

(g)

1
1

1

(h)
(h)

(h)

1
1

1


(a)
(a)
(a)

2 2 21 1 1

111

(g)
(g)
(g)

2 2 21 1 1

(d)
(d)

(d)

22 2
11 1
###

111

111
111

(c)
(c) and b-Elimination
(c)
111 1 1 1 2 2 2
428  Chapter 9: Nucleophilic Substitution

(e)
(e)
(e)

(b)
(b)
(b)

(f)(f)
(f)

111

111 1 1 1 2 2 2

(h)
(h)
(h)

111

9.14 Suppose you are told that each reaction in Problem 9.13 is a substitution reaction but
are not told the mechanism. Describe how you could conclude from the structure of the
haloalkane, the nucleophile, and the solvent that each reaction is an SN2 reaction.
9.15 Treatment of 1,3-dichloropropane with potassium cyanide results in the formation of
pentanedinitrile. The rate of this reaction is about 1000 times greater in DMSO than in
ethanol. Account for this difference in rate.
1

1

1,3-Dichloropropane

Pentanedinitrile

9.16 Treatment of 1-aminoadamantane, C10H17N, with methyl 2,4-dibromobutanoate in the
presence of a nonnucleophilic base, R3N, involves two successive SN2 reactions and gives
compound A. Propose a structural formula for compound A.

1

1-Aminoadamantane

1

Methyl
2,4-dibromobutanoate

1

2

A

9.17 Select the member of each pair that shows the greater rate of SN2 reaction with KI in
acetone.
(a)

(b)

(c)

(d)

9.18 Select the member of each pair that shows the greater rate of SN2 reaction with KN3 in
acetone.



(a)

(b)

9.19 What hybridization best describes the reacting carbon in the SN2 transition state?
9.20 Each carbocation is capable of rearranging to a more stable carbocation. Limiting yourself
to a single 1,2-shift, suggest a structure for the rearranged carbocation.
1



(a)

(b)

(c)
1

1
1

(d)

(e)

1

1

(f)

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Problems   429

9.21 Attempts to prepare optically active iodides by nucleophilic displacement on optically
active bromides using I2 normally produce racemic iodoalkanes. Why are the product
iodoalkanes racemic?
9.22 Draw a structural formula for the product of each S N1 reaction. Where configuration of
the starting material is given, show the configuration of the product.



(a)

(b)

1

1

S

(c)

(d)

1

1

9.23 Suppose you were told that each reaction in Problem 9.22 is a substitution reaction, but
you were not told the mechanism. Describe how you could conclude from the structure

of the haloalkane or cycloalkene, the nucleophile, and the solvent that each reaction is an
SN1 reaction.
9.24 Alkenyl halides such as vinyl bromide, CH2"CHBr, undergo neither SN1 nor SN2 reactions. What factors account for this lack of reactivity?
9.25 Select the member of each pair that undergoes S N1 solvolysis in aqueous ethanol
more rapidly.
(a)

(b)

(c)

(d)

(e)

(f)

9.26 Account for the following relative rates of solvolysis under experimental conditions
favoring SN1 reaction.

9.27 Not all tertiary haloalkanes undergo SN1 reactions readily. For example, the bicyclic compound shown below is very unreactive under SN1 conditions. What feature of this molecule is responsible for such lack of reactivity? You will find it helpful to examine a model
of this compound.

1-Iodobicyclooctane

9.28 Show how you might synthesize the following compounds from a haloalkane and a
nucleophile.
(a)
(e)

(b)

(c)
(f)

(d)
(g)

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430  Chapter 9: Nucleophilic Substitution and b-Elimination

9.29 3-Chloro-1-butene reacts with sodium ethoxide in ethanol to produce 3-ethoxy-1-­
butene. The reaction is second order, first order in 3-chloro-1-butene, and first order in
sodium ethoxide. In the absence of sodium ethoxide, 3-chloro-1-butene reacts with ethanol to produce both 3-ethoxy-1-butene and 1-ethoxy-2-butene. Explain these results.
9.30 1-Chloro-2-butene undergoes hydrolysis in warm water to give a mixture of these allylic
alcohols. Propose a mechanism for their formation.

"

1

"

1-Chloro-2-butene

2-Buten-1-ol

"

3-Buten-2-ol

9.31 The following nucleophilic substitution occurs with rearrangement. Suggest a mechanism for formation of the observed product. If the starting material has the S configuration, what is the configuration of the stereocenter in the product?

9.32 Propose a mechanism for the formation of these products in the solvolysis of this
bromoalkane.

1

1

9.33 Solvolysis of the following bicyclic compound in acetic acid gives a mixture of products,
two of which are shown. The leaving group is the anion of a sulfonic acid, ArSO 3H.
A ­sulfonic acid is a strong acid, and its anion, ArSO32, is a weak base and a good leaving
group. Propose a mechanism for this reaction.

1

9.34 Which compound in each set undergoes more rapid solvolysis when refluxed in ethanol?
Show the major product formed from the more reactive compound.
(a)

(b)

(c)

(d)

9.35 Account for the relative rates of solvolysis of these compounds in aqueous acetic acid.

2

2

2

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Problems   431

9.36 A comparison of the rates of SN1 solvolysis of the bicyclic compounds (1) and (2) in acetic
acid shows that compound (1) reacts 1011 times faster than compound (2). Furthermore,
solvolysis of (1) occurs with complete retention of configuration: the nucleophile occupies the same position on the one-carbon bridge as did the leaving 2OSO2Ar group.

(a)  Draw structural formulas for the products of solvolysis of each compound.
(b)  Account for the difference in rate of solvolysis of (1) and (2).
(c)  Account for complete retention of configuration in the solvolysis of (1).

b-Eliminations
9.37 Draw structural formulas for the alkene(s) formed by treatment of each haloalkane or
halocycloalkane with sodium ethoxide in ethanol. Assume that elimination occurs by an
E2 mechanism.

(a)

(c)

(b)

(e)

(d)

(f)

9.38 Draw structural formulas of all chloroalkanes that undergo dehydrohalogenation when
treated with KOH to give each alkene as the major product. For some parts, only one
chloroalkane gives the desired alkene as the major product. For other parts, two chloroalkanes may work.
(a)

(b)

(c)

(d)

(e)

9.39 Following are diastereomers (A) and (B) of 3-bromo-3,4-dimethylhexane. On treatment with sodium ethoxide in ethanol, each gives 3,4-dimethyl-3-hexene as the major
­product. One diastereomer gives the E alkene, and the other gives the Z alkene. Which
diastereomer gives which alkene? Account for the stereoselectivity of each b-elimination.

9

9

9.40 Treatment of the following stereoisomer of 1-bromo-1,2-diphenylpropane with sodium
ethoxide in ethanol gives a single stereoisomer of 1,2-diphenylpropene. Predict whether
the product has the E configuration or the Z configuration.

2

1

"
1-Bromo-1,2-diphenylpropane

1,2-Diphenylpropene

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432  Chapter 9: Nucleophilic Substitution and b-Elimination

9.41 Elimination of HBr from 2-bromonorbornane gives only 2-norbornene and no
1-norbornene. How do you account for the regioselectivity of this dehydrohalogenation?
In answering this question, you will find it helpful to look at molecular models of both
1-norbornene and 2-norbornene and analyze the strain in each.

2-Bromonorbornane

2-Norbornene

1-Norbornene

9.42 Which isomer of 1-bromo-3-isopropylcyclohexane reacts faster when refluxed with potassium tert-butoxide, the cis isomer or the trans isomer? Draw the structure of the expected product from the faster-reacting compound.

Substitution Versus Elimination
9.43 Consider the following statements in reference to SN1, SN2, E1, and E2 reactions of haloalkanes. To which mechanism(s), if any, does each statement apply?
(a) Involves a carbocation intermediate.

(b) Is first order in haloalkane and first order in nucleophile.
(c) Involves inversion of configuration at the site of substitution.
(d) Involves retention of configuration at the site of substitution.
(e) Substitution at a stereocenter gives predominantly a racemic product.
(f) Is first order in haloalkane and zero order in base.
(g) Is first order in haloalkane and first order in base.
(h) Is greatly accelerated in protic solvents of increasing polarity.
(i) Rearrangements are common.
(j) Order of reactivity of haloalkanes is 3° . 2° . 1°.
(k) Order of reactivity of haloalkanes is methyl . 1° . 2° . 3°.
9.44 Arrange these haloalkanes in order of increasing ratio of E2 to SN2 products observed on
reaction of each with sodium ethoxide in ethanol.

(b)

(a)

(c)

(d)

9.45 Draw a structural formula for the major organic product of each reaction and specify the
most likely mechanism by which each is formed.
(a)

(b)

1

8

1

(c)
(R )-2-Chlorobutane

(d)

2

1

1

(e)

1
R

(f)

1

(g)

2

1

1

"

R
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Problems   433

9.46 When cis-4-chlorocyclohexanol is treated with sodium hydroxide in ethanol, it gives
mainly the substitution product trans-1,4-cyclohexanediol (1). Under the same reaction
conditions, trans-4-chlorocyclohexanol gives 3-cyclohexenol (2) and the bicyclic ether (3).

1

cis -4-Chlorocyclohexanol

trans -4-Chlorocyclohexanol

(a)  Propose a mechanism for formation of product (1), and account for its configuration.
(b)  Propose a mechanism for formation of product (2).
(c) 
Account for the fact that the bicyclic ether (3) is formed from the trans isomer but not
from the cis isomer.

Synthesis
9.47 Show how to convert the given starting material into the desired product. Note that some
syntheses require only one step, whereas others require two or more.

(a)

(b)

(c)

(d)

(e)

(f)

(g)

(h)

(i)

9.48 The Williamson ether synthesis involves treatment of a haloalkane with a metal alkoxide.
Following are two reactions intended to give benzyl tert-butyl ether. One reaction gives
the ether in good yield, and the other reaction does not. Which reaction gives the ether?
What is the product of the other reaction, and how do you account for its formation?

(a)

(b)

2 11

1

2 11

1

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434  Chapter 9: Nucleophilic Substitution and b-Elimination

9.49 The following ethers can, in principle, be synthesized by two different combinations of
haloalkane or halocycloalkane and metal alkoxide. Show one combination that forms
ether bond (1) and another that forms ether bond (2). Which combination gives the
higher yield of ether?

9
(a)

(b)

9

(c)

9

"

9

9

9.50 Propose a mechanism for this reaction.

9
9.51 Each of these compounds can be synthesized by an SN2 reaction. Suggest a combination
of haloalkane and nucleophile that will give each product.
(a) CH3OCH3

(b) CH3SH



(c) CH3CH2CH2PH2



(g)
(e) CH3SCH(g)
2C(CH3)3



(g)
(h) R

(d) CH3CH2CN
(g)
(f) (CH ) NH1 Cl2

3 3


(h) R
(i) CH2"CHCH2OCH(CH3)2
(h) R
(k)

1

(k)

2



1

(l)

(h) R
( j) CH2"CHCH2OCH2CH"CH2

2

(k)
(l)

1

2

(l)



1

(k)
Looking
Ahead

2

(l)

9.52OH2 is a very poor leaving group. However, many alcohols react with alkyl or aryl sulfonyl chlorides to give sulfonate esters.

9

1

9 9

9

9 9

1

(a)  Explain what this change does to the leaving group ability of the substituent.
(b)  Suggest the product of the following reaction.


9

9 9

1

2

1

9.53 Suggest a product of the following reaction. HI is a very strong acid.
1

Organic Chemistry Reaction Roadmap
Reaction
Roadmap

9.54 Use the reaction roadmap you made for Problems 6.54, 7.29, and 8.28 and update it to
contain the reactions in the “Study Guide” section as well as Table 9.1 of this chapter. Because of their highly specific nature, do not use reactions 3 and 5 or entry 7 of Table 9.1 on
your roadmap.

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Problems   435

9.55 Write the products of the following sequences of reactions. Refer to your reaction roadmap to see how the combined reactions allow you to “navigate” between the different
functional groups.
(a)

Reaction
Roadmap

(b)

   

An alkane

(c)

An alkene

(d)

 

        

An alkyne

An alkane

Multistep Synthesis Problems

Some reaction sequences are more useful than others in organic synthesis. Among the
reactions you have a learned thus far, a particularly useful sequence involves the combination of free radical halogenation of an alkane to give a haloalkane, which is then
subjected to an E2 elimination to give an alkene. The alkene is then converted to a variety
of possible functional groups. Note that free radical halogenation is the only reaction you
have seen that uses an alkane as a starting material.

9.56 Using your reaction roadmap as a guide, show how to convert butane into 2-butyne.
Show all reagents and all molecules synthesized along the way.

Butane

2-Butyne

9.57 Using your reaction roadmap as a guide, show how to convert 2-methylbutane into racemic 3-bromo-2-methyl-2-butanol. Show all reagents and all molecules synthesized
along the way.

2-Methylbutane

Reaction
Roadmap

Hexanedial

9.59 Using your reaction roadmap as a guide, show how to convert cyclohexane into racemic
3-­bromocyclohexene. Show all reagents and all molecules synthesized along the way.

Cyclohexane

Reaction
Roadmap

3-Bromo-2-methyl-2-butanol

9.58 Using your reaction roadmap as a guide, show how to convert cyclohexane into hexanedial. Show all reagents and all molecules synthesized along the way.

Cyclohexane

Reaction
Roadmap

Reaction
Roadmap

3-Bromocyclohexene

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436  Chapter 9: Nucleophilic Substitution and b-Elimination

Reaction
Roadmap

9.60 Another important pattern in organic synthesis is the construction of C!C bonds. Using your reaction roadmap as a guide, show how to convert propane into hex-1-en-4yne. Y
  ou must use propane as the source of all of the carbon atoms in the hex-1-en-4-yne
product. Show all reagents needed and all molecules synthesized along the way.

Propane
Reaction
Roadmap

Hex-1-en-4-yne

9.61 Using your reaction roadmap as a guide, show how to convert propane into butyronitrile.
You must use propane and sodium cyanide as the source of all of the carbon atoms in the
butyronitrile product. Show all reagents and all molecules synthesized along the way.

Propane

Sodium cyanide

Butyronitrile

Reactions in Context
9.62 Fluticasone is a glucocorticoid drug that has been used to treat asthma. In the synthesis
of fluticasone, the following transformation is used that involves a limiting amount of
sodium iodide. Analyze the structure using the chemistry you learned in this chapter and
draw the product of the reaction.

9.63 The following reaction sequence was used in the synthesis of several derivatives of prostaglandin C2. Analyze the structure using the chemistry you learned in this chapter and
draw the structures of the synthetic intermediates A and B.

A

B

9.64 The following reaction was used in the synthesis of various prostaglandin derivatives.
Analyze the structure using the chemistry you learned in this chapter and draw the product of the reaction.

t

t

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10
Alcohols

Fermentation vats
of wine grapes at
the Beaulieu Vineyards, California.
Inset: a model of
ethanol. (Ian Shaw/
Alamy Stock Photo)

Outline
10.1
10.2
10.3
10.4
10.5
10.6
10.7
10.8
10.9

Structure and Nomenclature of Alcohols
Physical Properties of Alcohols
Acidity and Basicity of Alcohols
Reaction of Alcohols with Active Metals

Conversion of Alcohols to Haloalkanes and Sulfonates
Acid-Catalyzed Dehydration of Alcohols
The Pinacol Rearrangement
Oxidation of Alcohols
Thiols

In this chapter, we study the physical and chemical properties of alcohols, a class
of compounds containing the !OH (hydroxyl) group. We also study thiols, a class of
compounds containing the !SH (sulfhydryl) group.

Ethanol

Ethanethiol

Ethanol is the additive in the fuel blend known as E85, the alcohol in alcoholic beverages, and an important industrial solvent. Ethanethiol, like all other low-molecularweight thiols, has a stench; such smells from skunks, rotten eggs, and sewage are caused
by thiols or H2S.

437
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438  Chapter 10: Alcohols

Alcohols are important because they can be converted into many other types of
compounds, including alkenes, haloalkanes, aldehydes, ketones, carboxylic acids, and
esters. Not only can alcohols be converted to these compounds, but these compounds
can also be converted to alcohols. Thus, alcohols play a central role in the interconversion of organic functional groups.
Hydroxyl groups are found in carbohydrates and certain amino acids. Following
are two representations for glucose, the most abundant organic compound in nature. On the left is a Fischer projection showing the configuration of all chiral centers.
On the right is a cyclic structure, the predominant form in which this molecule exists

in both the solid form and in solution. The amino acid l-serine is one of the 20 amino
acid building blocks of proteins.

2

1

D-Glucose

␤ -D-Glucopyranose

L-Serine

Because sulfur and oxygen are both Group 6 elements, thiols and alcohols undergo many of the same types of reactions. Sulfur, a third-row element, however, can
undergo some reactions that are not possible for alcohols. In addition, sulfur’s electronegativity and basicity are less than those of oxygen.

..
..

(a)

10.1 Structure and Nomenclature of Alcohols

(b)
º

º
Figure 10.1  Methanol,
CH3OH. (a) Lewis structure and
(b) ball- and-stick model.

A. Structure
The functional group of an alcohol is an !OH (hydroxyl) group (Section 1.3A)
bonded to an sp 3 hybridized carbon. The oxygen atom of an alcohol is also sp 3
­hybridized. Two sp 3 hybrid orbitals of oxygen form s bonds to atoms of carbon
and hydrogen, and the remaining two sp 3 hybrid orbitals each contain an unshared pair of electrons. Figure 10.1 shows a Lewis structure and a ball-and-stick
model of ­methanol, CH 3OH, the simplest alcohol. The measured C!O!H bond
angle in methanol is 108.9°, very close to the perfectly tetrahedral angle of 109.5°.

B. Nomenclature
In the IUPAC system, the longest chain of carbon atoms containing the !OH group
is selected as the parent alkane and numbered from the end closer to !OH. To show
that the compound is an alcohol, change the suffix -e of the parent alkane to -ol
(Section 2.3) and use a number to show the location of the !OH group. The location of the !OH group takes precedence over alkyl groups and halogen atoms in
numbering the parent chain. For cyclic alcohols, numbering begins with the carbon
bearing the !OH group. Because the !OH group is understood to be on carbon 1 of
the ring, there is no need to give its location a number. In complex alcohols, the number for the hydroxyl group is often placed between the infix and the suffix. Thus, for
example, both 2-methyl-1-propanol and 2-methylpropan-1-ol are acceptable names.

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10.1  Structure and Nomenclature of Alcohols   439

Common names for alcohols are derived by naming the alkyl group bonded
to !OH and then adding the word alcohol. Here are IUPAC names and, in
­parentheses, common names for several low-molecular-weight alcohols.

1-Propanol
(Propyl alcohol)

2-Propanol
(Isopropyl alcohol)

1-Butanol
(Butyl alcohol)

(S)-2-Butanol
((S)-sec-Butyl alcohol)

2-Methyl-1-propanol
(Isobutyl alcohol)

2-Methyl-2-propanol
(tert-Butyl alcohol)

Example 10.1  Alcohol Nomenclature I
Write IUPAC names for these alcohols.

(a)

(b)

Solution
(a)(R)-4-Methyl-2-pentanol.
(b)(1R,2R)-2-Methylcyclohexanol. Note that the designation of the
configuration as R,R specifies not only the absolute configuration of
each chiral center but also the fact that the !CH3 and !OH groups
are trans to each other on the ring. The alcohol can also be named
trans-2-methylcyclohexanol, and while this name specifies that the

hydroxyl and methyl groups are trans to each other, it does not specify
the absolute configuration of either group.

Problem 10.1
Write IUPAC names for these alcohols and include the configuration for (a).
(a)

(b)

We classify alcohols as primary (1°), secondary (2°), or tertiary (3°), depending on
whether the !OH group is on a primary, secondary, or tertiary carbon.
9

9

9 9

9 9

9 9
0

8

8

8

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440  Chapter 10: Alcohols

Example 10.2  Classification of Alcohols
Classify each alcohol as primary, secondary, or tertiary.
(a)

(b)

(c)

Solution
(a)  Secondary (2°)   (b)  Tertiary (3°)   (c)  Primary (1°)

Problem 10.2
Classify each alcohol as primary, secondary, or tertiary.
(a)

Diol
A compound containing two
hydroxyl groups.

Triol
A compound containing three
hydroxyl groups.

(b)

(c)

(d)

In the IUPAC system, a compound containing two hydroxyl groups is named as a
diol, one containing three hydroxyl groups as a triol, and so on. In IUPAC names for
diols, triols, and so on, the final -e (the suffix) of the parent alkane name is retained,
as, for example, in the name 1,2-ethanediol. As with many organic compounds,
common names for certain diols and triols have persisted. Compounds containing
hydroxyl groups on adjacent carbons are often referred to as glycols (Section 6.5).
Ethylene glycol and propylene glycol are synthesized from ethylene and propylene,
respectively, hence their common names.

1,2-Ethanediol
(Ethylene glycol)

1,2-Propanediol
(Propylene glycol)

1,3-Propanediol

1,2,3-Propanetriol
(Glycerol, glycerine)

Compounds containing !OH and C"C groups are often referred to as
­ nsaturated alcohols because of the presence of the carbon-carbon double bond. In
u
the IUPAC system, the double bond is shown by changing the infix of the parent alkane from -an- to -en- (Section 2.3) and the hydroxyl group is shown by changing the
suffix of the parent alkane from -e to -ol. Numbers must be used to show the location
of both the carbon-carbon double bond and the hydroxyl group. The parent alkane
is numbered to give the !OH group the lowest possible number; that is, the group
shown by a suffix (in this case, -ol) takes precedence over the group shown by an infix

(in this case, -en-).

Example 10.3  Alcohol Nomenclature II
Write IUPAC names for these unsaturated alcohols.
(a)

"

(b)

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