WILEY-Klein-Third-edition_Chapter-Seven.pdf

Copyright © 2017 John Wiley & Sons, Inc. All rights reserved.
Organic Chemistry
Third Edition
Chapter 7
Alkyl Halides: Nucleophilic Substitution and
Elimination Reactions
David Klein
Klein, Organic Chemistry 3e

• Alkyl halides undergo substitution and elimination rxns
– What is the hybridization of each highlighted carbon in the
structures shown above?
– Which structure represents an alkyl halide?
7.1 Introduction to Alkyl Halides
Copyright © 2017 John Wiley & Sons, Inc. All rights reserved. 7-2 Klein, Organic Chemistry 3e

• Alkyl Halides can undergo a substitution reaction when reacted
with a nucleophile.
• Alkyl Halides can undergo an elimination reaction when reacted
with a base.
- what do “nucleophiles” and “bases” have in common?
7.1 Substitution and Elimination
Rxns

• When the reagent can act as a nucleophile or a base,
elimination and substitution will be competing reaction
pathways
• In the example above, which compound is the
substrate?
Rxns

Two main reasons why alkyl halides undergo substitution
and elimination reactions:
1. The halogen is electron-withdrawing, creating a partial
positive charge on the alpha carbon, making it susceptible
to nucleophilic attack.
2. The halogen acts as a leaving group, and for a substrate to
undergo a substitution/elimination reaction, it must
possess a good leaving group.
Rxns
Nuc:

7.1 Leaving Groups
Good leaving groups are the conjugate bases of strong
acids.

7.2 Alkyl Halides
• Alkyl halides are compounds where a carbon group
(alkyl) is bonded to a halide (F, Cl, Br, or I)
• Recall from section 4.2 the steps we use to name a
molecule
1. Identify and name the parent chain
2. Identify the name of the substituents
3. Assign a locant (number) to each substituents
4. Assemble the name alphabetically
• The halide group is the key substituent we will name
and locate

7.2 Alkyl Halide Nomenclature
• For each of these examples, convince yourself that
they are numbered in the most appropriate way.

• Some simple molecules are also recognized by their
common names.
– the alkyl group is named
as the substituent, and
the halide is treated as
the parent name
Methylene chloride is a
commonly used organic
solvent
Systematic Name Common Name
Cl Cl
Dihalo alkane
Cl Cl
alkyl dihalide
Dichloromethane Methylene Chloride

• Give reasonable names for the following molecules
• Try more examples with conceptual checkpoint 7.1

7.2 Alkyl Halide Structure
• Greek letters are often used to label the carbons of the
alkyl group attached to the halide
– Substitutions occur at the alpha carbon
WHY?
• The amount of branching at the alpha carbon affects
the reaction mechanism. There are three types of alkyl
halides

7.2 Alkyl Halide Structure
• Some alkyl halides are used as insecticides. For the
insecticides below…
– Label each halide as either primary, secondary, or tertiary
– For the circled atoms, label all of the alpha, beta, gamma,
and delta carbons.

7.2 Uses of Organohalides
• Halides appear in a wide variety of natural products
and synthetic compounds
• The structure of the molecule determines its function,
and functions include…
– Insecticides (DDT, etc.)
– Dyes (tyrian purple, etc.)
– Drugs (anticancer, antidepressants, antimicrobial, etc.)
– Food additives (Splenda, etc.)
– Many more

• A substitution reaction requires the loss of a leaving
group, and nucleophilic attack. There are two possible
mechanisms: (1) concerted, and (2) stepwise.
1. The concerted mechanism involves breaking of the bond to the
leaving group and making of the bond to the nucleophile at the
same time
7.3 SN2 Reactions

2. The stepwise mechanism the leaving groups leaves first, to give
a carbocation intermediate, followed by nucleophilic attack
(The stepwise mechanism is covered in section 7.10)
7.3 SN2 Reactions

7.3 SN2 – concerted substitution
• What do S, N, and 2 stand for in the SN2 name
• How might you write a rate law for this reaction?
• How would you design a laboratory experiment to confirm this
rate law? (Test yourself with conceptual checkpoint 7.2)

7.3 SN2 – stereochemistry
• How does the observed stereochemistry in the
following reaction support an SN2 mechanism?
• Practice drawing the products of SN2 reactions with
SkillBuilder 7.1

7.3 SN2 – backside attack
• The nucleophile attacks from the back-side
– Electron density repels the attacking nucleophile from the
front-side
– The nucleophile must approach the back-side to allow
electrons to flow from the HOMO of the nucleophile to the
LUMO of the electrophile.
– Proper orbital overlap cannot occur with front-side attack
because there is a node on the front-side of the LUMO

• Draw the transition state for the following reaction, which
explains why SN2 reactions proceed with inversion of
configuration
• Practice drawing transition states with SkillBuilder 7.2
7.3 SN2 – backside attack
Transition state symbol

7.3 SN2- Kinetics
• Less sterically hindered electrophiles react more
readily under SN2 conditions. Tertiary halides are too
hindered to react via SN2 mechanism.
• To rationalize this trend, examine the
reaction coordinate diagram.
TABLE 7.2 EFFECT OF SUBSTITUENTS ON THE RATES OF SN2 REACTIONS
R B r
a c e to n e
R B r
+
⊝
⊝
omide is over one hundred times more reactive than ethyl bromide (a
hich is over one hundred times more reactive than isopropyl bromide
e). Notice that t-butyl bromide (a tertiary alkyl halide) is unreactive
rvations indicate that SN2 reactions are most ef ective for methyl halides
es, and SN2 reactions cannot be performed with tertiary alkyl halides.
X X
X
H3C X
Reactivity toward SN2
1° 2° 3°
Mos t
reactive
Unreactive
Methyl

7.3 SN2 - Kinetics
• Alkyl groups branching from the a and b carbons hinder the
backside attack of the nucleophile, resulting in a slower rate of
reaction.
To rationalize this trend, we must examine the energy diagram

7.3 SN2 – Rationalizing kinetic data
• What feature of the diagram is relevant to rationalize the
rate of reaction?
• What feature is relevant to rationalize the thermodynamics
of the reaction?

• Which reaction will have the fastest rate of reaction?
• WHY?
• 3° substrates react too slowly to measure.
X
H
H
H
Nuc:
X
H3C
H
H
Nuc:
X
H3C
H3C
H
Nuc:

• An example to consider: neopentyl bromide
• Draw the structure of neopentyl bromide
• Is neopentyl bromide a primary, secondary, or tertiary
alkyl bromide?
• Should neopentyl bromide react by an SN2 reaction
relatively quickly or relatively slowly? WHY?

7.4 SN2 – nucleophilicity
• What are the factors that contribute to the strength of
a nucleophile? (Review section 6.7)
• A strong nucleophile is needed for an SN2 rxn
• Be able to recognize a given nucleophile as being
strong or weak

7.4 SN2 – Nucleophilicity
Commonly nucleophiles used in substitution rxns:
• You should be able to determine if a nucleophile is strong or
weak. Which of the nucleophiles shown above would prefer
SN2?
• In general, anions are strong nucleophiles
• Polarizable atoms are good nucleophiles
• The solvent affects nucleophilicity

7.4 SN2 – solvent effects
• Need a polar aprotic solvent for SN2 rxns
• Protic solvents engage in H-bonding, and stabilize anionic species
(such as good nucleophiles).
• Aprotic solvents stabilize both cations and anions
• How do these factors play into the strength of a nucleophile in
protic versus aprotic solvent?

• Nucleophiles are less stable,
thus more reactive in aprotic
solvent.
• The activation energy will be
lower and the reaction faster
7.4 SN2 – solvent effects
Aprotic solvents are
best for SN2 reactions

• Halides are common leaving groups for laboratory use, but are not
common substrates in biological SN2 rxns.
• Both of these compounds are good methylating reagents: a good
nucleophile will attack the CH3 via an SN2 mechanism.
7.5 SN2 – biological alkylation
Good
leaving
group
Good
leaving
group

• When an alkyl halide is treated with a strong base, it can undergo
beta elimination (1,2-elimination) to form an alkene:
• A strong base will react in a concerted mechanism, called an E2
elimination.
7.6 introduction to E2 rxns

• E2 elimination is concerted, where the base removes a b-proton,
causing the loss of the leaving group and the formation of the C=C
bond. So, concerted elimination is bimolecular and follows
second-order kinetics:
• In what ways is E2 elimination similar to SN2 substitution? What
are the differences?
7.6 E2 - kinetics

• Consider a reagent such as NaOH, which is a strong nucleophile
(SN2) and a strong base (E2)…
… when the substrate is sterically hindered, E2 elimination will occur.
7.6 E2 – effect of the substrate

• Alkenes are given IUPAC names using the same
procedure to name alkanes, with minor modifications
1. Identify the parent chain, which includes the C=C double
bond
2. Identify and Name the substituents
3. Assign a locant (and prefix if necessary) to each substituent.
Give the C=C double bond the lowest number possible
4. List the numbered substituents before the parent name in
alphabetical order. Ignore prefixes (except iso) when ordering
alphabetically
5. The C=C double bond locant is placed either just before the
parent name or just before the -ene suffix
7.7 Alkene Nomenclature

1. Identify the parent chain, which should include the C=C double
bond
• The name of the parent chain should end in -ene rather than –
ane
• The parent chain should include the C=C double bond

2. Identify and Name the substituents
3. Assign a locant (and prefix if necessary) to each substituent. Give
the C=C double bond the lowest number possible
– The locant of the double bond is a single number, and is the
number indicating where the double bond starts. The alkene
above is located at the “2” carbon.

4. List the numbered substituents before the parent name in
alphabetical order. Ignore prefixes (except iso) when ordering
alphabetically
5. The C=C double bond locant is placed either just before the
parent name or just before the -ene suffix
Note: This alkene has the E configuration, which must be indicated in
the name, in parentheses: (E)-5,5,6-trimethylhept-2-ene

7.7 Alkene isomerism
Recall how to assign E or Z to alkene stereoisomers…
• First, prioritize the groups attached to the C=C double bond
based on atomic number

7.7 Alkene isomerism
• If the top priority groups are cis to each other, it is the
Z isomer
• If the top priority groups are trans to each other, it is the
E isomer

7.7 Alkene Stability
• Because of steric strain, cis isomers are generally less
stable than trans

• The difference in stability can be quantified by
comparing the heats of combustion
• Think about how the heats of combustion of the cis and
trans isomers reveal their relative stability…

• Alkyl groups stabilize the C=C pi bond via hyperconjugation.
• Practice with Conceptual Checkpoint 7.16
More alkyl groups = more stable alkene

7.7 Cycloalkene stability
• In cyclic alkenes with less than 7 carbons in the ring, only cis
alkenes are stable. WHY?
• So we do not need to indicate if the alkene is cis or trans unless
the ring contains 8 carbons or more.
• When applied to bridged bicycloakenes, this rule is called
Bredt’s rule

7.7 Alkene Isomerism
• Apply Bredt’s rule to the compounds below
• The bridgehead carbon cannot have a trans pi bond unless one of
the rings has at least 8 carbons (otherwise the geometry of the
bridgehead prevents parallel overlap of the p-orbitals)
• Try building a handheld model of each compound shown above,
and see, first-hand, the relative geometric strain of each

7.8 E2 regioselectivity
• It is common for a substrate to have more than one b-carbon
that can be deprotonated by a strong base, and so E2 elimination
results in more than one alkene product.
• When ethoxide is used as the strong base, 2-methyl-2-
bromobutane gives two E2 products (shown above), with the
more stable alkene (the Zaitsev product) produced as the major
product, and the less substituted alkene (the Hofmann product)
is the minor product.
Hofmann product
Zaitsev product

• E2 elimination is a regioselective:
• When constitutional isomers are formed as the products of a
reaction, with one of them as the major product, the reaction is
regioselective
• The regioselectivity of an E2 reaction can be controlled by carefully
choosing the strong base used.
Hofmann product
Zaitsev product

• Experimental data indicates that a bulky, sterically hindered base
will favor the formation of the Hofmann product, but an
unhindered base (like ethoxide) will favor the Zaitsev product:

• Why does a sterically hindered base favor the Hofmann product?
• Sterically hindered bases (also called non-nucleophilic bases) are
useful in many reactions
• Practice with SkillBuilder 7.3

Practice the Skill 7.18 predict the major and minor products for the
following E2 reactions:
7.8 Regioselectivity of E2 rxns

• Consider the dehydrohalogenation of 3-bromopentane, where
two stereoisomers are possible products:
7.8 E2 stereoselectivity
Use this energy diagram and
the Hammond postulate to
explain why the trans isomer is
formed stereoselectively

• Consider dehydrohalogenation of the alkyl halide below:
7.8 Stereospecificity of E2
• You might imagine that it would be possible to form both the E
and Z alkene products, but only the E isomer is formed
only product
this isomer is
not formed
There is only one
b-hydrogen to be removed
For E2 elimination to occur

• To rationalize the stereospecificity of the reaction, consider the
transition state for the reaction
• In the transition state, the C-H and C-Br bonds that are breaking
must be rotated into the same plane as the pi bond that is
forming
• In other words, the b-hydrogen and the leaving group must be
co-planar.

• There are two rotamers where the b-hydrogen and the leaving
group are coplanar:
• Since we are comparing two different rotamers, the Newman
projections are a good tool to compare them
ophilic Substitution and Elimination Reactions
t the C− C single bond is free to rotate before the reaction occurs. If we imagine
we see that there are two ways to achieve a coplanar arrangement:
Br
H
Ph
H
Me Rotate the C C bond
Br
H
Ph
H
Me
onformation iscalled anti-coplanar, whilethesecond conformation iscalled syn-c
ntext, the terms anti and syn refer to the relative positions of the proton and the
ich can be seen more clearly with Newman projections:
Br Br
H
H
t-Bu
t-Bu
Ph
Ph
H
H
Me
Me
Anti-coplanar
(s taggered)
Syn-coplanar
(eclips ed)
Nucleophilic Substitution and Elimination Reactions
l that the C− C single bond is free to rotate before the reaction occurs. If we imagine rota
ond, we see that there are two ways to achieve a coplanar arrangement:
Br
H
Ph
H
Br
H
Ph
H
Me
rst conformation iscalled anti-coplanar, whilethesecond conformation iscalled syn-copla
is context, the terms anti and syn refer to the relative positions of the proton and the lea
p, which can be seen more clearly with Newman projections:
Br Br
H
H
t-Bu
t-Bu
Ph
Ph
H
H
Me
Me
Anti-coplanar
(s taggered)
Syn-coplanar
(eclips ed)
n viewed in this way, we can see that the anti-coplanar conformation is staggered, while the

• The staggered rotamer, where the b-hydrogen and leaving group
are anti-coplanar, is much lower energy than the eclipsed
rotamer:
• The product resulting from the anti-coplanar rotamer is formed
H H
Ph
H
Me
rst conformation iscalled anti-coplanar, whilethesecond conformation iscalled syn-copla
s context, the terms anti and syn refer to the relative positions of the proton and the lea
, which can be seen more clearly with Newman projections:
Br Br
H
H
t-Bu
t-Bu
Ph
Ph
H
H
Me
Me
Anti-coplanar
(s taggered)
Syn-coplanar
(eclips ed)
n viewed in this way, we can see that the anti-coplanar conformation is staggered, while the
nar conformation is eclipsed. Elimination via the syn-coplanar conformation would invol
tion state ofhigher energyasaresult of the eclipsed geometry. T erefore, elimination occursm
y via the anti-coplanar conformation. In fact, in most cases, elimination is observed to o
sively via the anti-coplanar conformation, which leads to one speci f c stereoisomeric product
Me
Br
Ph
H
t-Bu
H
Me
t-Bu
Ph
H
Elimination
Ph
H
Me
t-Bu
Lower energy
rotamer
High energy
rotamer
ef rst conformation iscalled anti-coplanar, whilethesecond conformation iscalled syn-coplanar
this context, the terms anti and syn refer to the relative positions of the proton and the leavin
oup, which can be seen more clearly with Newman projections:
Br Br
H
H
t-Bu
t-Bu
Ph
Ph
H
H
Me
Me
Anti-coplanar
(s taggered)
Syn-coplanar
(eclips ed)
hen viewed in this way, we can see that the anti-coplanar conformation is staggered, while the syn
planar conformation is eclipsed. Elimination via the syn-coplanar conformation would involve
nsition state ofhigher energyasaresult of theeclipsed geometry. T erefore, elimination occursmor
pidly via the anti-coplanar conformation. In fact, in most cases, elimination is observed to occu
clusively via the anti-coplanar conformation, which leads to one speci f c stereoisomeric product:
Anti-coplanar
(s taggered)
Me
Br
Ph
H
t-Bu
H
Me
t-Bu
Ph
H
Elimination
Ph
H
Me
t-Bu
T e requirement for coplanarity isnot entirely absolute. T at is, small deviations from coplanar

• Evidence suggests that a strict 180° angle is not
necessary for E2 mechanisms.
• Similar angles (175–179°) are sufficient
• The term, anti-periplanar is generally used instead of
anti-coplanar to account for slight deviations from
coplanarity
• Although the E isomer is usually more stable because it
is less sterically hindered, the requirement for an anti-
periplanar transition state can often lead to the less
stable Z isomer

• Assuming they proceed through an anti-periplanar
transition state, predict the products for the following
reactions

• For the following substrate, the b-carbon has TWO b-hydrogens
• There are two different rotamers where a b-hydrogen is anti-
periplanar to the leaving group, and so two stereoisomers will be
formed.
• In this case, the reaction will be stereoselective, but not
stereospecific. (see Skillbuilder 7.4)
• E2 elimination will be stereospecific only when both the a and b
carbons are stereocenters

• It is very important to understand the difference between the
terms stereospecific and stereoselective.
• In a stereospecific rxn, the substrate is stereoisomeric and
results in one stereoisomer as the product
• In a stereoselective rxn, the substrate can produce two
stereoisomers as products, where one is the major product.
7.8 stereospecific vs. stereoselective

• Consider the dehydrohalogenation of a cyclohexane derivative,
where the leaving group is attached to the ring
• Given the anti-periplanar requirement, E2 elimination can only
occur when the leaving group is in the axial position.
CH3
3
CH3
CH3
CH3
3
CH3
C 3 H3C 3
Br Br
6
7
1
9
O
O
O
2
3
4
5
8
6
7
1
9
2
3
4
5
8
Base
Compound 1
Base
Compound 2
roblems 7.61, 7.62, 7.74
specif city of E2 Reactions on Substituted Cyclohexanes
vious section, we explored the requirement that an E2 reaction proceed via an an
rmation. T at requirement has special signif cance when dealing with substituted
all that a substituted cyclohexane ring can adopt two dif erent chair conformatio
Cl
Cl
air conformation, the leaving group occupiesan axial position. In the other chair con
g group occupies an equatorial position. T e requirement for an anti-periplanar co
hat an E2 reaction can onlyoccur from thechair conformation in which theleavingg
E2 elimination
can occur in this
chair conformation
E2 elimination
cannot occur

• Which of the two molecules below will NOT be able to undergo
an E2 elimination reaction? WHY?
• It might be helpful to draw their chair structures and build a
model

• Rationalize the product(s) formed in the following two reactions
• One of the alkyl halides undergoes E2 elimination much faster
than it’s diastereomer. Why is there a difference in their rxn
rates?
(E2 rxn is slow)
(E2 rxn is fast)
Only one
product formed
Two products
formed

• There are many factors to consider in order to correctly predict
the product(s) of an E2 rxn and decide what the major product
will be.
– Will the substrate react stereospecifically? or will it be a
stereoselective E2 rxn?
– Will the substrate produce several regioisomeric alkenes? If
so, what will be the major product, given the steric hindrance
of the base that is used?
• The only way to master this material is to do lots of practice
problems. Start with Skillbuilder 7.5, then go from there.
7.8 drawing products of E2 rxns

• Consider the following reaction, where a substitution and
elimination products are formed when dissolving:
• The substrate is 3˚, so SN2 substitution is not possible. The
reagent is EtOH, which is not a strong base, and so E2 elimination
is unlikely.
• It turns out the formation of the substitution and elimination
products follow first-order kinetics, which confirms neither SN2
nor E2 is occurring.
7.9 Unimolecular Rxns (SN1 and E1)

• The mechanisms of substitution and elimination in this case both
start with the same step: ionization of the substrate
• After the carbocation is formed, it will either undergo
substitution or elimination, depending on how it reacts with the
solvent (EtOH).
The leaving group leaves to form
a carbocation intermediate
(SN1)
(E1)
7.9 Unimolecular Rxns (SN1 and E1)

• The substitution rxn of a 3˚ substrate, in an alcohol solvent like
EtOH, proceeds through a two-step (stepwise) mechanism
• The entire mechanism is actually 3 steps, but the last step is just
a proton transfer
• This is called “solvolysis” because the nucleophile is also the
solvent.
Loss of
leaving group
7.9 SN1 mechanism
Nucleophilic
attack

• The highest energy transition state, in SN1, is for the formation
of the carbocation intermediate. So, formation of the
carbocation is the rate determining step.
7.9 SN1 reaction coordinate

• Since the formation of the carbocation requires only ionization
of the substrate, the rate of it’s formation depends only on the
substrate, and so the rxn follows first-order kinetics
• Now it is clear that when substitution occurs via a carbocation
intermediate, it is called “SN1”
7.9 SN1 kinetics
The rate of SN1 substitution
depends only on the substrate

• A substitution reaction that occurs stepwise, where the leaving
group first leaves to form a carbocation intermediate, followed
by nucleophilic attack is called SN1 substitution.
• Remember that when the nucleophile is a neutral species, such
as an alcohol, there will be a proton transfer after nucleophilic
attack
7.9 SN1 mechanism

• The elimination rxn of a 3˚ substrate, in an alcohol solvent like
EtOH, proceeds through a two-step (stepwise) mechanism
• Here, EtOH is serving as a base (not as a nucleophile) to
deprotonate the carbocation and form an alkene
• Like SN1, the E1 mechanism is unimolecular, and follows the
same kinetics.
Loss of
leaving group
7.9 E1 mechanism
b-hydrogen
elimination

• The rate of E1 is the same as for SN1: in both cases, the rate
determining step is the formation of the carbocation
intermediate
Both E1 and SN1 share
The same rate determining step
7.9 E1 kinetics

• Because SN1 and E1 rxns proceed through a carbocation
intermediate, the carbocation may rearrange from 1,2-hydride or
methide shifts.
7.9 SN1/E1 rearrangements
SN1 product
SN1 product
after 1,2-methide
shift
SN1 product
SN1 product
after 1,2-hydride
shift

• When a 1˚ substrate is reacted under solvolysis conditions, only
the product resulting from rearrangement is observed
• Remember 1˚ carbocations are too unstable to form. So, in this
case the rearrangement occurs as the leaving group leaves
7.9 SN1/E1 rearrangements
This product is
not observed
Only
SN1 product
1,2-methide shift and
ionization is concerted,
3˚ carbocation is formed

• Experimental data indicates SN1 and E1 reactions are faster in a
polar protic solvent
7.9 SN1/E1 solvent effects

• Overall: For SN2 rxns, a polar aprotic solvent is best
• For SN1 rxns, a polar protic solvent is best
7.9 solvent effects on substitution
aprotic solvents raise the energy
of the Nu-
, which results in lower Ea
and a faster SN2 reaction
protic solvents stabilize the carbocation
Intermediate, which results in lower Ea
and a faster SN1 reaction

• The better the leaving group, the faster the SN1 or E1 rxn
• Remember the rate-determining step for SN1 and E1 of alkyl
halides is the ionization step: the formation of a carbocation and
a halide ion
• So, the more stable the halide ion, the faster the ionization
7.9 SN1/E1 – the substrate

• The more stable the carbocation intermediate, the faster the SN1
and E1 reactions will be.
• Solvolysis rxns of 1˚ and 2˚ alkyl halides are often too slow to
observe the formation of SN1 and E1 products
• However, 3˚ alkyl halides, as well as benzylic and allylic halides
will undergo solvolysis at a practical rate thanks to the stability of
the carbocation intermediates:
Recall that benzylic and allylic carbocations
are resonance stabilized

• Be able to judge whether or not a given alkyl halide will undergo
a solvolysis (SN1 and/or E1) reaction.
• In general, a 1˚ or 2˚ alkyl halide will only undergo solvolysis if
rearrangement to a more stable carbocation is possible.
• 3˚, allylic and benzylic alkyl halides will undergo solvolysis to give
a mixture of SN1 and E1 products
• Practice with Conceptual Checkpoints 7.30, 7.31 and 7.32
benzylic substrate Mixture of SN1 and E1 products
is observed

• Like we saw with E2 eliminations, it is possible for E1 elimination
to yield more than one regioisomer, as in the following example:
• E1 reactions will always give the most stable alkene as the
major product, which will be the most substituted alkene
• So E1 reactions are regioselective, but we cannot control the
regioselectivity like we can with E2 reactions.
7.9 E1 – regioselectivity
Major product
of E1 elimination

• It is further possible to obtain several alkene stereoisomers in an
E1 reaction, as in the following example:
• It still holds true that the E1 reaction will give the most stable
alkene as the major product. When two stereoisomers are
obtained, the least sterically hindered isomer will be more stable.
• So E1 reactions are stereoselective. But realize a mixture of all
possible products is still obtained.
7.9 E1 – stereoselectivity
Major product
of E1 elimination

• When the a-carbon in an SN1 reaction is chiral, we obtain two
substitution products that have opposite configurations at the
reactive carbon:
• Recall that in an SN2 reaction, the Nuc does a backside attack,
and only the inversion of configuration product is obtained.
7.9 SN1 – stereoselectivity
The Nuc can attack
The carbocation from
either side

• Even though a mixture of configurations is obtained in SN1
substitution, typically more of the inversion product is observed
7.9 SN1 – stereoselectivity
The leaving group
will form an ion-pair
with the carbocation,
making it more difficult
for the nucleophile
to attack from the
same side

• As you learn all these mechanisms of substitution and
elimination, you should appreciate how we have come to know
how the mechanisms occur
• One way we to study a mechanism is to see how replacing a
hydrogen atom with it’s isotope, deuterium, affects the rate of a
reaction. If the rate is affected, it is likely that particular H atom
is involved in the rate determining step.
• 1
H is called hydrogen, abbreviated as “H”, and 2
H is called
deuterium, abbreviated as “D”
• Deuterium (D) has the same chemical reactivity as hydrogen (H)
7.10 Kinetic Isotope effects

• Consider the following reaction, where an alkyl halide undergoes
elimination with a strong base (which you already know is called
an E2 reaction).
• When we replace the b-hydrogens with deuteriums, the reaction
occurs at a slower rate. This is called the kinetic isotope effect

• C – D bonds are stronger than C – H bonds. So, if replacing the b-
hydrogens with deuteriums results in a slower rate by more than
a factor of 5, then we can conclude the breaking of the C – H
bond occurs during the rate determining step
• The reaction above is nearly 7 times slower with b-deuteriums
instead of b-hydrogens. This is one of the reasons why we know
that E2 elimination is a concerted elimination

• By now, it should be clear that a number of factors affect the
product(s) formed when reacting an alkyl halide with a
nucleophile and/or base (the substrate, the reagent, and the
solvent).
• It should also be clear that in many cases, a mixture of
substitution and/or elimination products will be obtained
7.11 Predicting Products

• It is also possible, for a given substrate, that only one mechanism
will occur
• In order to understand how to use these reactions, to transform
alkyl halides into a desired compound, one must be able to
predict ALL the products that will form in a given reaction, as well
as the major and minor product(s)

To successfully predict the product(s) formed in a given reactions, we
can follow a three-step analysis:
1. DETERMINE THE FUNCTION OF THE REAGENT
2. ANALYZE THE SUBSTRATE AND DETERMINE THE EXPECTED
MECHANISM(S)
3. CONSIDER ANY RELEVANT REGIOCHEMICAL AND
STEREOCHEMICAL REQUIREMENTS

• Remember what kind of reagents promote SN1, SN2, E1 and E2
SN2 = strong nucleophile E2 = strong base
SN1 = weak nucleophile E1 = weak base
• The following table is a good resource for categorizing reagents
and the mechanisms they will promote
7.11 Function of the reagent
E2 SN2 and E2 SN1 and SN2 SN1 and E1
STEP 1

• Once you determine what mechanism(s) will be favored by the
reagent, analyze the substrate (is it to see which mechanism(s)
will dominate… is the substrate 1˚, 2˚, or 3˚?
7.11 Analyze the substrate
STEP 2

• After analyzing the reagent and the substrate, you can say which
mechanism(s) will occur. Draw all the possible regio- and
stereoisomers, then choose the major, using the guidelines you
have learned.
• For SN2, you will observe a single product, which is inversion of
configuration at the a-carbon
7.11 regioselectivity/stereoselectivity
STEP 3

For E2, draw all the possible alkene isomers. Only alkenes which
result from a b-hydrogen anti-periplanar to the leaving group can
form
- if a bulky base is used, the Hofmann product is the major
- if a non-bulky base is used, the most stable alkene is the major
STEP 3

For SN1, draw the carbocation intermediate, consider if it will
rearrange. If not, then attach nucleophile to the carbocation. If it
rearranges, draw the resulting carbocation, then attach the
nucleophile to it.
- if a chiral carbon is formed by attack of the nucleophile, then
two products are formed (“R” and “S”). Draw them both.
STEP 3

For E1, draw the carbocation formed from loss of the leaving group.
If it will rearrange, draw the rearranged carbocation. Then, draw all
possible alkene isomers resulting from elimination of a b-hydrogen.
All possible alkene stereoisomers will form (E1 is not stereospecific).
- the major product will always be the most stable alkene.
PRACTICE THESE STEPS WITH SKILLBUILDER 7.7
STEP 3

Predict the product(s) of the following reaction, and label the major
product.
• STEP 1: ANALYZE THE REAGENT(S). NaOH is a strong base, and a
strong nucleophile, so SN2 and E2 will be favored
• STEP 2: LOOK AT THE SUBSTRATE. It is a 2˚ halide, so SN2 and E2
will occur, but E2 will dominate (because 2˚ substrates are
somewhat hindered, and backside attack is more difficult)
7.11 Predict the products

• STEP 3: consider the regio- and stereochemical requirements.
For the SN2 product, backside attack gives inversion of configuration
SN2 product

• STEP 3: consider the regiochemical and stereochemical
requirements.
For the E2 product(s), draw all the b-hydrogens that can be anti-
periplanar to the leaving group, then draw the resulting alkenes (use
Newman projections if necessary)
elimination
of Ha
elimination
of Hb
elimination
of Hc

• STEP 3: consider the regiochemical and stereochemical
requirements.
Now we have all the products resulting from SN2 and E2. Now label
the major product. E2 is major pathway, and the base is not
hindered, so the Zaitsev product is the major.
Do more examples with
Practice the Skill 7.37
most stable
alkene
Br NaOH
Br
MAJOR

• There are a variety of alternatives to alkyl halides for substitution
and elimination reactions, such as alkyl sulfonates
• Mesylates, tosylates, and triflates are excellent leaving groups.
They are also quite large, and so we usually use abbreviations
when drawing their structures (OMs, OTs, and OTf)
7.12 Other substrates
R
S
O
O
Sulfonate leaving groups are res onance-s tabilized
R
S
O
O R
S
O
O
⊝
⊝
Since sulfonate ions are such good leaving groups, alkyl sulfonates undergo substitution and elimina-
tion reactions, much like alkyl halides. Many sulfonates are commonly used, including mesylates,
tosylates, and trif ates:
S
O
O
O
mesylate
group
An alkyl mes ylate (ROMs )
S
O
O
O
tosylate
group
An alkyl tos ylate (ROTs )
S
O
O
O
triflate
group
An alkyl triflate (ROTf)
CH3 CF3
CH3
Tosylates are the most commonly used, although trif ates have the best leaving group. Recall that the
best leaving groups are the weakest bases. T e trif ate ion is one of the weakest known bases, because
it is the conjugate base of an especially strong acid. Compare the pKa values of the following sulfonic
acids (RSO3H):
Methanes ulfonic acid
(Ms OH)
CH3
S
O
O
O
H
Trifluoromethanes ulfonic acid
(TfOH)
CF3
S
O
O
O
H
p-Toluenes ulfonic acid
(Ts OH)
S
O
O
O
H CH3

• Sulfonates are such good leaving groups because they are very
stable (like halides, they are the conjugate bases of strong acids)
• Based on pKa values, which sulfonate is the best leaving group?
7.12 Alkyl sulfonates
lates, and trif ates:
S
O
O
O
mesylate
group
n alkyl mes ylate (ROMs )
S
O
O
O
tosylate
group
An alkyl tos ylate (ROTs )
S
O
O
O
triflat
grou
An alkyl triflate (R
CH3
CH3
ylates are the most commonly used, although trif ates have the best leaving group. Recall th
leaving groups are the weakest bases. T e trif ate ion is one of the weakest known bases, b
the conjugate base of an especially strong acid. Compare the pKa values of the following su
s (RSO3H):
Methanes ulfonic acid
(Ms OH)
pKa = –1.9
CH3
S
O
O
O
H
Trifluoromethanes ulfonic acid
(TfOH)
pKa = –14
CF3
S
O
O
O
H
p-Toluenes ulfonic acid
(Ts OH)
pKa = –2.8
S
O
O
O
H CH3
R
S
O
O
O
X (X = , Br, or Cl)
An alkyl s ulfonate
An alkyl halide
ulfonate, the leaving group is a sulfonate ion, which is highly stabilized by resonance:
R
S
O
O
O
Sulfonate leaving groups are res onance-s tabilized
R
S
O
O
O R
S
O
O
O
⊝
⊝
⊝
ions are such good leaving groups, alkyl sulfonates undergo substitution and elimina-
much like alkyl halides. Many sulfonates are commonly used, including mesylates,
if ates:
S
O
O
sylate
roup
te (ROMs )
S
O
O
O
tosylate
group
An alkyl tos ylate (ROTs)
S
O
O
O
triflate
group
An alkyl triflate (ROTf)
CH3 CF3
CH3
Sulfonate ions are
Resonance stabilized!

• Sulfonates are made from the corresponding alcohol
• Realize we are just strapping the “Ts” group to the oxygen of the
alcohol… no change in the carbon atom bearing the OH group
occurs
Alcohols are not
good leaving groups
now we have a good
leaving group!

• To envision the compounds that can be synthesized from an alkyl
tosylate, treat them the same as you would an alkyl halide
With a strong Nu/strong base, a 1˚ substrate
gives mostly SN2, with a little E2
With a strong Nu/strong base, a 2˚ substrate
gives mostly E2, with a little SN2

• Alcohols can also be used in substitution and elimination
reactions, and used as starting materials to make alkyl halides and
alkenes.
• We need strongly acidic conditions to do these reactions, because
OH is a bad leaving group, but H2O is a good leaving group
7.12 Alcohols
1˚ alcohol 1˚ bromide

• The mechanism will be either SN1 or SN2, depending on the
substrate. 1˚ alcohols react via SN2, but 2˚ and 3˚ alcohols react via
SN1
• Strongly acidic conditions are protic conditions, which would favor
SN1. But, since 1˚ carbocations are too unstable to form, 1˚
alcohols react via SN2 mechanism
7.12 Alcohols
3˚ alcohol 3˚ bromide

• Alcohols will undergo E1 elimination when reacted with H2SO4
• Again, the strongly acidic conditions are protic conditions, which
favors E1 for 2˚ and 3˚ substrates
7.12 Alcohols
E1 mechanism
3˚ alcohol
3˚ alcohol alkene

• The whole point to organic synthesis is to make valuable, complex
compounds from cheap and readily available starting materials
• You now know how to make a variety of compounds starting with
an alkyl halide.
7.13 Synthetic strategies
1˚ halide

• In order to envision how a desired compound can be made, you
need to be able to recall the reactions you can use (meaning you
have to remember these reactions!!!)
3˚ halide

• When thinking about how to make something, we FIRST think
about what the finished product will look.
• If we were building a brick house, we would first imagine what the
house will look like. THEN we would decide what bricks would be
used to make it.
• Organic synthesis is the same way: we first look at the desired
product, and from there we decide what substrates and reactants
we would need to use to make it
• This approach is called retrosynthetic analysis

• Suppose we need to synthesize the following ether:
STEP 1: identify a bond in the target molecule that can be made using
a reaction that you know.
STEP 2: draw the substrate and the nucleophile necessary to for the
reaction.
7.13 retrosynthetic analysis
We can make this bond by
SN2 reaction between an
alkyl halide and an alkoxide
The retrosynthetic arrow is used to show
we are “thinking backwards” with regards
to the reaction we could do

• There are two C – O bonds in an ether, so we could also envision
an alternative SN2 reaction to make it:
• You will find that when “thinking backwards” this way, more than
one reaction will often come to mind to make a target compound
Alternatively, we could
make this bond via SN2
These are the reactants we
would need

STEP 3: verify that the reaction you have proposed is reasonable
STEP 4: draw the reaction in the forward direction
1˚ halide
(good substrate for SN2)
Strong, unhindered Nu
(good for SN2)
YES! We expect
this reaction to work
We are trying to do an SN2 reaction,
So we might as well use an aprotic solvent, right?

What reactants would you need in order to make the following
compound as the product of a substitution reaction?
Try to do this on your own, and when you want to check your
answer, or if you get completely stuck, refer to SkillBuilder 7.8

WILEY-Klein-Third-edition_Chapter-Seven.pdf

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