ChemActivity 29
Electrophilic Aromatic Substitution
1
29
ChemActivity
Part A: Electrophilic Aromatic Substitution
(What products are formed when a strong electrophile is added to benzene?)
Model 1: (review) Electrophilic Addition of HCl
H
H
Cl
Cl
H
Rxn 1
Cl
cyclohexene
carbocation intermediate
H
H
Rxn 2
H
Cl
Cl
benzene
carbocation intermediate
Cl
x
This product
DOES NOT form!
Critical Thinking Questions
1. For Rxn 1 (above) draw curved arrows showing the mechanism of electrophilic
addition of HCl. Include an appropriate carbocation intermediate in the box above.
Figure 1: Reaction Diagrams for Electrophilic Addition of HCl
H
Cl
V.E. (Potential Energy)
V.E.
carbocation
intermediate
H Cl
H
carbocation
intermediate
Reaction Progress (Rxn 1)
Cl
H Cl
Reaction Progress (Rxn 2)
2. Rxn 1 is slightly down-hill in terms of energy. Rxn 2 is very up-hill in terms of
energy (see Figure 1). Construct an explanation for the large difference in
energy between the reactants and the product in Rxn 2.
In Rxn 2 the product is not aromatic so it is much higher in energy than the aromatic reactant.
3. Draw the carbocation that would form in Rxn 2. Explain why this carbocation goes
back to the starting material (H–Cl and benzene) instead of forming the product.
ChemActivity 29
Electrophilic Aromatic Substitution
The elimination reaction shown (the reverse of the first step) restores aromaticity, and is therefore
favored over step two which results in a non-aromatic product.
2
ChemActivity 29
Electrophilic Aromatic Substitution
3
Model 2: Electrophilic Aromatic Substitution
• Recall that deuterium (D) has nearly identical reactivity as hydrogen (H).
• In the following reaction D–Cl reacts the same way as would H–Cl.
• When benzene is treated with D-Cl, an H is replaced with a D. (With excess D-Cl
this continues until all the H’s have been replaced and the product is C6D6.
H
H
D
C
H
C
C
C
C
D
Cl
H
Rxn 3
H
C
C
D
Cl
H
H
H
C
H
C
C
C
C
C
C
C
C
H
H
H
H
C
H
H
C
H
H
Critical Thinking Questions
4. Use curved arrows to show a reasonable mechanism for the reaction in Model 2.
(Hint: the first step is formation of a carbocation intermediate, just as in Rxn 2.)
5. The energy diagram for Rxn 2 (using DCl in place of HCl) is shown below using a
dotted line. On this same set of axes, draw a solid line to show an energy diagram
for Rxn 3 in Model 2. (Note: Rxns 2 and 3 have the same carbocation intermediate.)
D
Cl
carbocation
intermediate
for Rxns 2 & 3
V.E.
D
D Cl
Reaction Progress (Rxn 2 with DCl)
Reaction Progress (Rxn 3)
6. Construct an explanation for why Rxn 3 is much more likely to occur than Rxn 2.
(Note: Rxn 2 does not occur under normal circumstances.)
Rxn 3 yields a much lower energy product than Rxn 2, so the former is much more likely.
H
Cl
ChemActivity 29
Electrophilic Aromatic Substitution
4
7. Draw a generalized mechanism for EAS (Electrophilic Aromatic Substitution)
showing the substitution of an electrophile (E+) for one of the H’s of benzene in the
presence of a mild base (B). (Note: in Rxn 3, E+ = D+)
E
B
H
E
E
H
B
Information: Not All Electrophiles are Created Equal
• A very strong electrophile (E+) is required to break up an aromatic ring.
• The H of a strong acid will work since it exists essentially as H+.
Know the following strong acids.
δ
+
δ+
H
δ
+
δ–
H
Cl
δ
–
Br
δ
+
δ–
H
I
O
O
H
δ+
O
δ–
S
OH
hydrobromic acid
hydroiodic acid
N
O
δ
–
nitric acid
O
hydrochloric acid
H
sulfuric acid
O
• Suitable electrophiles (E ) for EAS (Electrophilic Aromatic Substitution) include
an H+ or D+ donated by a strong acid, or the electrophiles (E+) listed below.
Table 2: Other Suitable Electrophiles (E+) for EAS
Reactant
E+ =
Reagents
Product/s
NO2
H2SO4
O
N
O
and
HNO3
SO3H
O
anhydrous
sulfuric
S
acid
O
O
Br
Cl
Br2 and FeBr3
or Cl
Br
or
or
*
Cl2 and FeCl3
R
R-X, AlX3
benzene
R
(X = Cl or Br)
or
*
R = alkyl group
other aromatic
+
O
O
ring
C
R
*
AlX3
C
R
X
(X = Cl or Br)
O
C
R
R = alkyl group
*The last three rows of electrophiles are generated only in the presence of a Lewis acid catalyst such as
FeX3 or AlX3. A supplementary activity on Lewis acid catalysts is assigned for homework.
ChemActivity 29
Electrophilic Aromatic Substitution
5
Part B: Substituent Effects
(How do substituents on an aromatic ring direct the placement of E+ and rate of EAS?)
Model 3: Electrophilic Aromatic Substitution with Toluene
H
O
C
C
C
C
C
C
O
H
C
N
C
H
C
C
C
C
H 3O
H
H
C
H
NO2
H
O2N
NO2
C
H 2O
C
H
H
H
H
C
C
O
C
C
CH3
C
C
H 3O
H
CH3
H
H
H
CH3
H
C
C
NO2
H
C
C
H2O
C
H
H
C
H
O
C
H
C
H
N
H
C
H
H
C
C
C
C
H
H
H
C
C
CH3
CH3
CH3
Critical Thinking Questions
8. Add curved arrows showing the movement of electrons in the two reactions above.
a) Draw any missing important resonance structures for each of the two
carbocation intermediates. (One from each set is re-drawn for you below.)
CH3
meta
CH3
NO2
CH3
NO2
H
NO2
H
CH3
H
CH3
CH3
para
O2N
H
O2N
H
O2N
H
b) In the meta set, none of the three resonance structures stand out as being most
important, but in the para set one resonance structure stands out as being most
important. Explain this statement and circle the most important res. structure.
All three resonance structures in the meta set are secondary. In the para set, the middle resonance
structure is tertiary, and therefore a larger contributor to the overall picture of electron distribution.
c) Of the two products at the top of the page, one forms and the other does not.
Circle the product that forms and explain your reasoning.
The para product is the only one that forms since the carbocation intermediate on this pathway is
lower in potential energy than the carbocation intermediate on the meta pathway.
ChemActivity 29
Electrophilic Aromatic Substitution
6
9. Draw the other two resonance structures of the intermediate that forms if the nitro
group adds to the ortho position (as shown in the mechanism below).
H
C
O
N
O
C
C
C
C
C
H
H
NO2
C
H
C
C
H
H
C
CH3
CH3
CH3
H
H
H
H
C
C
H
H
H
C
C
H
H
H
NO2
C
C
C
C
H
H
NO2
C
C
H 3O
H
CH3
H
NO2
C
C
H
H
C
C
C
C
C
C
H2O
H
C
C
CH3
H
a) The carbocation above is very close in potential energy to the intermediate in
the para set on the previous page. In what way are these two sets of resonance
structures similar?
Both sets of resonance structures consist of two secondary carbocation resonance structures and one
tertiary carbocation resonance structure.
b) Of the ortho intermediate and the para intermediate, one is slightly lower in
potential energy due to steric effects. Which one do you expect to be lower in
potential energy? ortho intermediate or para intermediate [circle one].
c) The following energy diagram shows pathways to the ortho, meta, and para
products. Label each pathway with the correct name.
= meta
= ortho
= para
CH3
CH3
CH3
V.E.
NO2
CH3
NO2
O
N
O
H3O
H2O (weak base)
ortho
meta
reaction progress
d) When toluene is mixed with nitric acid and sulfuric acid (giving +NO2) two of
the three products in the box above are formed. Circle these two and cross out
the other one. The ortho and para products form. The meta product does not form.
e) A methyl group, or any other alkyl group (R) is said to be an “ortho & para
director.” Explain why the R group "directs" the nitro to the o or p positions.
NO2
para
ChemActivity 29
Electrophilic Aromatic Substitution
7
The presence of the alkyl group appears to direct the electrophile to substitute for an H either ortho or
para to itself (the methyl group in this case).
Model 4: Steric vs. Electronic Effects in EAS
• For the reaction below, chemists say that the meta product is NOT formed
because of unfavorable electronic effects. (The electron arrangement in the
meta intermediate is not as favorable as in the ortho and para intermediates.)
• Generally, electronic effects are much more powerful than steric effects, but steric
effects can have an impact.
• With large E+ like nitro (+NO2), the para product is favored over ortho product.
R
R
O
N
R
R
O2N
O
NO2
H2O
NO2
R = -CH3
ortho (63 %)
meta (~3 %)
para (34 %)
ortho (12 %)
meta (~3 %)
para (85 %)
CH3
R=
CH
CH3
Critical Thinking Questions
10. Explain the very strong preference for the para product when R = isopropyl
(as compared to when R = methyl).
An isopropyl group is much larger than a methyl group so the former generates much larger steric
interactions. It makes sense, therefore, that para would dominate in the case of R= isopropyl.
11. "Without steric effects, you would expect the ortho product to be twice as abundant
as the para product." Explain this statement (see the Hint below).
R
R
R
R
O2N
H
H
H
H
H
H
H
H
NO2
H
H
H
NO2
H
H
R = alkyl group
H
H
NO2
ortho (66 %)
meta (~0 %)
para (33 %)
H
Hint: What is the difference
between this product and the
ortho structure in the box at left?
Expected Ratio if there were no Steric Effects
There are two different ortho H’s that can be substituted for, whereas there is only one para H. The
two ortho products shown above are identical to one another.
12. An alkyl group (R) is slightly electron donating. This means it donates electron
density into an attached aromatic ring. Based on this information, which of the
following EAS reactions do you expect to be faster? Explain your reasoning.
H2SO4/HNO3
R = alkyl group
R
(gives +NO2)
H2SO4/HNO3
(gives +NO2)
O2N
R
R
+
NO2
NO2
Hint: Which is better for the EAS reaction, to have a ring that more is electron rich or more electron poor?
ChemActivity 29
Electrophilic Aromatic Substitution
8
In an EAS reaction, the ring acts as a nucleophile so the more electron rich the ring, the faster the
reaction. For this reason an electron donating group such as an alkyl is expected to speed the rate of
EAS.
Part C: Resonance Donating and
Withdrawing Groups
(How do π donating and withdrawing groups affect rate and placement in EAS?)
Model 5: Second Order Resonance Structures
Second order resonance structures are like "assistant resonance structures." They are
not as important as regular (first order) resonance structures, but they can tell us some
information about the arrangement of electrons in a complicated molecule.
O
O
N
Nitrobenzene
O
O
O
N
O
O
N
O
N
Second Order Resonance Structures of Nitrobenzene
Critical Thinking Questions
13. Explain why the three second order resonance structures shown in Model 5 are
not full fledged resonance structures. That is, why are they "not as important" as a
regular, first order resonance structure such as the one at the far left, above?
The first order resonance structure above, left, has only two non-zero formal charges. Each of the
three second order resonance structures has four non-zero formal charges. In general, the fewer the
number of non-zero formal charges the more important the resonance structure.
14. This set of second order resonance structures tells us that the nitro group withdraws
electron density into its p orbitals very strongly from certain carbons on the
aromatic ring. Complete the composite drawing of nitrobenzene below by placing a
δ+ on appropriate carbons in the ring.
δ
O
Composite Drawing of Nitrobenzene
(showing a combination of all first and second order
resonance structures)
δ
N
O
δ
δ
δ
15. Add curved arrows to nitrobenzene at the top of the page showing how you would
change it into one of the second order resonance structures. Then use curved arrows
to generate each of the subsequent second order resonance structures.
ChemActivity 29
Electrophilic Aromatic Substitution
9
16. In the first step of an electrophilic aromatic substitution (EAS) reaction the
aromatic ring is acting as a nucleophile and reacting with the electrophile (E+).
Z
E
δ
δ
Z
Z
Base
E
Step One
δ
Step Two
E
H
a) Which is more likely to react with an electrophile, a carbon that is electron
rich, or a carbon that is electron poor and holds a δ+ charge [circle one]?
b) Assume that Z = nitro group and add δ+ to appropriate ring carbons on the
structure of the starting material above. (Hint: see Model 5.)
c) Construct an explanation for why, when "Z" is an electron withdrawing
group such as nitro, the meta product is the major product.
O
O
NO2
NO2
N
NO2
E
E
Base
E
E
minor products
major product
The electrophile is less attracted to the positions on the ring with a partial positive charge, and more
attracted to the other two positions (meta to a resonance electron withdrawing group such as nitro).
17. Any molecule with a lone pair next to the aromatic ring will have second order
resonance structures. The first in a set of three for aniline is shown below.
a) Add curved arrows to aniline showing how you would change it into the
second order resonance structure shown. Then use curved arrows to generate
the two missing second order resonance structures.
NH2
aniline
NH2
NH2
NH2
Second Order Resonance Structures for Aniline
b) In terms of electrons, does a NH2 group donate into or withdraw from [circle
one] the aromatic ring?
c) Complete the composite drawing of aniline below by placing a δ– on
appropriate carbons in the ring.
ChemActivity 29
Electrophilic Aromatic Substitution
H
Composite Drawing of Aniline
(showing a combination of all first and second order
resonance structures)
10
H
Nδ
δ
δ
δ
18. Recall that, in the first step of an EAS reaction the aromatic ring is acting as a
nucleophile and reacting with the + charged electrophile (E+).
Z
Z
δ
δ
Z
E
Base
δ
Step One
Step Two
H
E
E
a) Assume that Z = an amino group (NH2) and add δ– to appropriate ring
carbons on the structure of the starting material above.
b) Construct an explanation for why, when Z = a strong electron donating
group such as NH2, the major products are ortho and para.
H
H
NH2
NH2
N
E
E
Base
NH2
E
major products E
minor product
The electrophile is attracted to positions on the ring with more negative charge. With a resonance
donating group such as an amino group, the ortho and para positions have the largest amount of
negative charge.
19. To say a group on the ring is a “meta director” means it directs an E+ (in the next
EAS reaction) to add the to the meta position on the ring. Complete the following:
a) A strong electron withdrawing group is…
an ortho/para director or a meta director [circle one].
b) A strong electron donating group (with a lone pair to donate into the ring) is…
an ortho/para director or a meta director [circle one].
Part D: Summary of Directing Effects
(Why are halogens “deactivators” but ortho/para directors in an EAS reaction?)
Model 6: Inductive Effects vs. Resonance Effects
• An inductive effect is the donation or withdrawal of electron density through
sigma bonds. (As shown in the diagram below, left.)
• A resonance effect is the donation or withdrawal of electron density as
demonstrated by 1st or 2nd order resonance structures (as shown below, right).
ChemActivity 29
Example of an Inductive Effect
Cl
Electrophilic Aromatic Substitution
Example of a Resonance Effect
Cl
Cl is more
electronegative
than C, so it steals
electron density
from the ring.
11
Cl
Resonance donating
effects places extra
electron density at
the ortho and para
positions on the ring.
Note: A halogen is an inductive withdrawing group and a resonance donating group.
ChemActivity 29
Electrophilic Aromatic Substitution
12
Table 6: Directing Effects of Various R Groups in EAS Reactions
R
R
R
E
ortho
Base
— NH2 , — NHR, — NR2
— OH, — OR
Phenols
Alkyl groups
–CH3, –CH2CH3 etc.
Benzene
–H
Halogens
— I
— Br
— Cl
O δC
Acyl groups
δ+ R'
O δ-
δ-
δ
+
S OH
Sulfate group
N
— F
Inductive
Effects
Resonance
Effects
weak e–
withdraw
moderate e–
withdraw
weak e–
donation*
strong e–
donation
strong e–
donation
weak e–
donation*
Product
Regiochemistry
ortho &
para
ortho &
para
ortho &
para
—
strong e–
withdraw
—
weak e–
donation
—
ortho &
para
slow
strong e–
withdraw
strong e–
withdraw
meta
very slow
strong e–
withdraw
strong e–
withdraw
meta
very slow
strong e–
withdraw
strong e–
withdraw
meta
very slow
meta
very slow
meta
very slow
Relative
Reaction
Rate
very very
fast
very fast
moderatel
y fast
1
O
Nitro group
δ+
Ammonium
meta
E
O δ-
O
Cyanide group
para
E
Identity of R
Amines
R
E
C
— NR3
δ-
N
strong e–
withdraw
strong e–
withdraw
strong e–
withdraw
strong e–
withdraw
*Electron donation by an alkyl group can be considered an inductive or a pseudo-resonance effect (see
hyperconjugation). Hyperconjugation cannot be demonstrated with 2nd order resonance structures.
Critical Thinking Questions
20. Summarize how Column 3 (Resonance Effects) is related to Column 4 (Product
Regiochemistry)?
Those groups with resonance donating effects are o,p directors. Those with resonance withdrawing
effects are meta directors.
21. Summarize how Column 5 (Reaction Rate Relative to Benzene) is related to the
overall level of electron withdrawal or donation? (consider both inductive and
resonance effects)
If a ring is more electron rich than benzene, it undergoes EAS faster than benzene and vice versa.
ChemActivity 29
Electrophilic Aromatic Substitution
13
22. According to Model 6, are halogens o/p directors or m directors? o/p directors
Br
δ
δ
δ
δ
δ
δ
δ
δ
δ
a) Halogens are powerful inductive withdrawing groups. This means that a
halogen removes electron density from all carbons of the ring. Put a δ+ next to
each ring carbon in the structure of bromobenzene drawn above.
b) But, a halogen has a lone pair to donate into the ring. This means they are
resonance donating groups. On the same drawing of bromobenzene, put a
small δ– next to the three carbons that receive electron donation from Br.
c) These two effects (inductive withdrawal and resonance donation) compete
with one another, but the inductive withdrawing effects win. Is this
consistent with the reaction rate for bromobenzene (on Table 6) relative to
plain benzene? Explain.
Halobenzenes undergo EAS more slowly than benzene. This is consistent with the hypothesis that they
are less electron rich than benzene, and that the overall affect of the halogen is to withdraw electron
density from the ring.
d) If you were an electophile (E+), which carbons on bromobenzene would you
be most attracted to? Explain your reasoning.
The ortho an para positions are most electron rich and are most likely to react with an electrophile.
e) Is your answer in part d) consistent with the data in Table 6?
Yes. The halogens are ortho/para directors, but overall deactivators with regard to EAS.
Information:
activating group = group that makes the rate of EAS faster than with benzene
deactivating group = group that makes the rate of EAS slower than with benzene
examples of activating groups
NH2
OH
OR
very strong activators
FAST
HN
O
C
R R
examples of deactivating groups
X
weak
activator
O
CH
O
C
OR
O
weak
deactivator
C
OH O
C
R
SO3H CN
NO2
NR3
very strong deactivators
SLOW
SPEED OF EAS REACTION
ChemActivity 29
Electrophilic Aromatic Substitution
14
Model 7: EAS Reactions with Di-Substituted Rings
CH3
CH3
CH3
CH3
Br
Br2/FeBr3
Br2/FeBr3
Rxn A
NO2
Rxn B
Br
OH
NO2
OH
Br
CH3
NO2
CH3
Br2/FeBr3
Br2/FeBr3
Rxn C
CH3
Rxn D
only trace
amounts of
products
NO2
CH3
Critical Thinking Questions
23. Consider the starting material in Rxn A, in Model 7.
a) Mark the position/s on the ring where the CH3 group would direct an E+.
b) Mark the position/s on the ring where the NO2 group would direct an E+.
c) Are these two directing effects opposed to one another or in agreement
[circle one]?
24. Consider the starting material in Rxn B, in Model 7.
a) Mark the position/s on the ring where the CH3 group would direct an E+.
b) Mark the position/s on the ring where the OH group would direct an E+.
c) Are these two directing effects opposed to one another or in agreement
[circle one]?
d) Label each group on the ring (OH and CH3) as a strong activator, weak
activator, weak deactivator or strong deactivator.
e) Construct an explanation for why the product shown is the major product.
In this case, the OH group is a strong activator and the methyl group is a weak activator. As expected,
the strong activator wins in terms of directing effects yielding the product shown.
25. Consider the starting material in Rxn C, in Model 7.
a) Mark the position/s on the ring where each CH3 group would direct an E+.
b) Construct an explanation for why the product below is formed in very small
amounts compared to the product shown above.
CH3
Br
Minor Product
CH3
Sterics direct the bromine to add to an activated position that is NOT between the two methyl groups.
26. Which reaction/s in Model 7 demonstrate the general rule that Friedel-Crafts
reactions are extremely slow with a deactivated aromatic ring.
ChemActivity 29
Electrophilic Aromatic Substitution
Rxn D. Even though the two nitro group direct to the same position, the effect of two strong
deactivating groups is to slow the rate of EAS to the point where no product is observed.
15
ChemActivity 29
Electrophilic Aromatic Substitution
16
Exercises for Part A
1. Show the mechanism and most likely products that result from the following
reactants. (Note: two weak bases, water and bisulfate ion are also in solution.)
O
HSO4
N
O
H2 O
2. Sulfuric acid with absolutely no water in it is called fuming sulfuric acid and
contains small amounts of the powerful electrophile SO3 (one resonance structure is
shown below). Construct a mechanism for the following reaction. Hint: the final
step is an intramolecular H atom transfter.
O
O
O
S
O
OH
S
O
3. Draw all possible resonance structures for the carbocation intermediate in Model 2.
4.
+
NO2 is formed when nitric acid (HNO3) and sulfuric acid (H2SO4) are mixed.
Draw the Lewis structure of each and construct a mechanism that explains
formation of +NO2. Hint: water and HSO4– are the other products formed in this
reaction.
5. When toluene is treated with sulfuric and nitric acids under special conditions, three
nitro (NO2) groups are substituted for hydrogens (at the 2, 4 and 6 positions on the
ring). The product is a highly explosive substance commonly known by a three
letter name. Draw the structure and write the common name and the chemical name
for this explosive substance.
H 2SO4
HNO 3
toluene
common name = ___ ___ ___
chemical name = ___________________________________
6. Construct a reasonable mechanism for the following reaction called a Friedel-Crafts
alkylation.
Note: when R-X and AlX3 are mixed, you can assume the result is R and
assume this to be...
CH2CH3
AlCl3 (cat.)
Cl CH2CH3
CH2CH3
AlCl4
H
Cl
AlX4
ChemActivity 29
Electrophilic Aromatic Substitution
17
7. Draw the mechanism (use curved arrows) and most likely product/s that would
result from the following EAS reaction called a Friedel-Crafts acylation.
assume these species are present
O
O
Br
AlBr3
AlBr4
8. Draw a mechanism to explain the formation of each of the two Friedel-Crafts
products. Hint: think of (and draw) the Rδ+ group in the R–X–AlCl3 complex as a
carbocation (R+), then think about possible carbocation rearrangements.
CH3
CH2CH2CH3
Br
CH
CH3
CH2CH2CH3 AlCl3 (cat.)
MIXTURE of above two products
H
Br
9. Give an example (not appearing in this ChemActivity) of…
a) an alkyl halide (R–X) that will likely undergo rearrangement during a
Friedel-Crafts alkylation.
b) an alkyl halide (R–X) that will NOT undergo rearrangement during a
Friedel-Crafts alkylation.
10. Shown below are two ways of making the same target product starting from
benzene. Synthetic pathway b gives a higher % yield of the desired product. Explain
why.
Br
a
AlCl3 (cat.)
Br
HCl heated in
Zn/Hg amalgum
O
b
AlCl3
(you are NOT responsible
for this mechanism)
O
11. Read the assigned pages in your text and do the assigned problems.
12. Complete the mini-activity on Lewis acid catalysts found at the end of this
ChemActivity.
ChemActivity 29
Electrophilic Aromatic Substitution
18
Exercises for Part B
13. Of the choices in brackets, circle the word or phrase that makes the sentence true.
a) The pi system of the ring acts as [a nucleophile or an electrophile] in an
EAS reaction.
b) The [more or less] electron rich the pi system of the aromatic ring, the faster
the rate of EAS.
14. A nitro group is a very powerful electron withdrawing group. Which do you expect
will undergo EAS reaction faster: benzene or nitrobenzene [circle one], and
explain your reasoning.
15. Construct an explanation for the following finding: Even with the best electrophile
(E+), di-nitro benzene undergoes EAS extremely slowly. So slowly that not even
trace amounts of product are observed.
O2N
E
Base
NO PRODUCTS OBSERVED
NO 2
very very very slow
16. When a flask containing a mixture of 1 mole of nitrobenzene and 1 mole of toluene
is treated with one mole of D-Cl, deuterium is incorporated into the toluene ring,
but not the nitrobenzene ring. Explain. That is, why does the toluene “hog” all the
D-Cl, while the nitrobenzene does not get any?
This notation says that D is on the ring, but
does not specify which position on the ring.
CH3
NO2
1 mole D
Cl
D
D
CH3
1 mole toluene
NO2
1 mole nitrobenzene
observed
not observed
17. Read the assigned pages in your text and do the assigned problems.
Exercises for Part C
18. Draw three 2nd order resonance structures for phenol.
H
O
phenol
a)
Explain why the 2nd order resonance contributors are less important than 1st
order resonance structures, and contribute only a small amount to our
ChemActivity 29
Electrophilic Aromatic Substitution
19
overall understanding of phenol.
Explain the following statement: The 2nd order resonance structures help
explain why R groups with a lone pair (such as –OH) activate the ortho and
para positions toward electrophilic aromatic substitution (EAS).
b)
19. Draw three 2nd order resonance structures for benzoic acid.
HO
O
C
benzoic acid
a) Based on these 2nd order resonance structures, do you expect the carboxylic
acid group (COOH) to be a resonance donating group or a resonance
withdrawing group?
b) Explain the following statement: The 2nd order resonance structures in part c)
help explain why a carboxylic acid group deactivates the ortho and para
positions toward electrophilic aromatic substitution (EAS).
Information
So far we have argued that 2nd order resonance structures can be used to predict the
regiochemistry of an EAS reaction. The following questions ask you to consider the
potential energies of intermediates on the EAS reaction pathways. As with all reactions
with a small change in energy between reactant and product, it is the height of the
activation barrier that determines which product will form. The potential energy of the
intermediate is an excellent approximation (according to the Hammond Postulate) of the
height of the activation barrier. Simply put…
The pathway with the most favorable carbocation intermediate will likely dominate.
20. Consider electrophilic aromatic substitution (EAS) performed on nitrobenzene.
O
O
NO2
NO2
N
NO2
E
Base
E
major product
a)
E
E
minor products
How does the placement of E+ on the nitrobenzene ring differ from an EAS
reaction starting with toluene or aminobenzene (aniline)?
ChemActivity 29
b)
O
Electrophilic Aromatic Substitution
20
Draw the intermediate on the reaction pathway to the major (meta) product.
(Be sure to include all important resonance structures.)
O
O
O
N
N
Base
E
carbocation intermediate
c)
O
E
meta
ONLY product
Draw the intermediate on the reaction pathway to the para product. (Be sure
to include all important resonance structures.)
O
O
O
N
N
Base
E
E
carbocation intermediate
d)
e)
f)
g)
para (and ortho)
NOT FORMED
Any resonance structure in which two + charges are next to each other is
very unfavorable. Circle all unfavorable resonance structures above.
Construct an explanation for why the intermediate on the reaction pathway
to the meta product is lowest in potential energy.
Draw an energy diagram showing all three pathways (ortho, meta and para).
Construct an explanation for why the meta product is strongly favored in
this reaction over the para (and ortho) product.
21. Aniline gives only ortho and para products in an EAS reaction.
a) Draw the intermediate on the reaction pathway to the para product. Be sure to
draw all FOUR important resonance structures.
NH2
NH2
Base
E
aniline
b) Draw the intermediate on the pathway to the meta product.
c) Construct an explanation for why the intermediate on the pathway to the meta
product is higher in potential energy than the intermediate on the pathway to
the para product.
E
ChemActivity 29
Electrophilic Aromatic Substitution
21
d) Draw an energy diagram showing all three pathways (ortho, meta and para).
22. Aniline reacts with a given electrophile 100 times faster than toluene and 1000
faster than benzene.
a) Explain why an EAS intermediate for aniline such as the one in part a) above
is lower in potential energy than the intermediate you drew for parasubstitution of toluene in Model 3.
b) Construct an explanation for why aniline undergoes EAS much faster than
toluene.
23. Read the assigned pages in your text and do the assigned problems.
Exercises for Part D
24. Consider the following reactions:
NO2
NO2
FeCl 3
NO2
NO2
Cl2
NO2
Cl
Cl
I
Cl
a
c
Cl
Cl
II
Br
Br2
e
Br
f
f NOT formed
Construct an explanation for why Product c is not formed in Rxn I.
Construct an explanation for why Product d is not formed in Rxn I.
Construct an explanation for why Product f is not formed in Rxn II.
25. In the reaction below, two major products are observed.
a) Draw them and construct an explanation for why they are formed instead of
the other two possibilities.
b) Which of the two major products do you expect to dominate and why?
CH3
Cl2/FeCl3
C
O
d
c and d NOT formed
Cl
Cl
FeBr3
a)
b)
c)
b
OH
26. Mark each of the following statements True or False based on your current
understanding. (If false, cite an example of a substituent for which it is false.)
a) T or F: A strong activator overpowers the directing effects of a weak activator.
b) T or F: A weak activator overpowers the directing effects of a deactivator.
c) T or F: All activators are o/p directors.
ChemActivity 29
Electrophilic Aromatic Substitution
22
d) T or F: All deactivators are m directors.
27. Circle each of the following that help explain why EAS with fluorobenzene is
slower than EAS with phenol (hydroxybenzene)?
I.
F is more electronegative than O, generating stronger inductive effects.
II.
F holds it lone pairs very tightly, making it a weaker pi donator than O.
III.
F has more lone pairs than O.
IV.
F is smaller in size than O.
28. Consider the following reactions:
Cl2
I
FeCl 3
only product
Cl
OCH3
OCH3
Cl2
II
OCH3
FeCl 3
Cl
Cl
65 %
III
Cl2
35 %
FeCl 3
Cl
Cl
37 %
a)
b)
c)
63 %
Construct an explanation for why only the para product is observed in
reaction I.
Fact: Neglecting steric effects, the expected ratio of para:ortho is 1:2 for
reactions I, II and III. Construct an explanation for this fact.
Why is Reaction III closest to the expected 1:2 ratio?
29. According to the note at the end of Part D of this ChemActivity, which of the
following starting materials will yield detectable products in a Friedel-Crafts
alkylation or acylation.
benzene
bromobenzene
ortho-chloroaniline
phenol
nitrobenzene
2-bromo-6-chlorophenol
Note: for the last two compounds the ring is net activated since NH2 and OH are strong
enough activators to overpower even two weak deactivators.
30. Read the assigned pages in your text and do the assigned problems.
ChemActivity 29
Electrophilic Aromatic Substitution
23
Mini-Activity on Lewis Acid Catalysis
(What catalyst is needed to generate F+ Cl+ Br+ I+ and C+ electrophiles?)
Information: Lewis Acid Catalysts
• Any molecule or atom that wants more electrons (lacks electrons) is a Lewis acid.
• Any molecule or atom that wants to share its electrons (excess e-) is a Lewis base.
• A catalyst helps increase the reaction rate, but is not consumed in the reaction.
Figure A: Common Lewis Acids
Cl
δ–
δ–
Br
+
Fe δ
Cl
Cl δ–
δ–
Br
δ–
Cl
+
Fe δ
δ–
Cl
Al
Br δ–
iron chloride
δ–
Br
δ+
δ–
Br
Cl δ–
iron bromide
δ–
+
Al δ
Br δ–
aluminum chloride
aluminum bromide
In each molecule above, the electronegative halogens steal electron density from the
central metal, leaving the Fe or Al with a lack of electrons and almost a full + charge.
Critical Thinking Questions
31. Circle the most Lewis acidic atom in each molecule in Figure A.
Information
• The electron cloud around bromine is normally symmetrical (below, left).
• A passing molecule with a dipole moment will polarize this cloud (below, middle)
• An even stronger effect is generated when a Lewis acid catalyst such as FeBr3
binds to one of the Br atoms of Br2 (below, right)
Outline of Undisturbed
Electron Cloud of Br2
Outline of Electron Cloud of Br2
near transient + charge disturbance
Br
δ
+
Br
Br
Br2 with partial bond to FeBr3
(outline of electron cloud not shown)
Br
δ
–
Br
δ
+
δ
–
Br
Br
Fe
Br
Br
passing molecule
with a small dipole
32. Consider the picture above, right, showing Br2 bound to FeBr3.
a) Add to this drawing a depiction of the shape of the electron cloud of Br2 when
it is bound to FeBr3 (above right). Don’t include the electron cloud of FeBr3 in
your drawing.
b) Add δ+ and δ– where appropriate to the Br2 portion of this complex (above
right).
ChemActivity 29
Electrophilic Aromatic Substitution
24
33. Consider the reactions below:
Br
Br
Br
I.
Br
Br
II.
Br
Br
+
NO REACTION
H
III.
Br
Br
FeBr3 (cat.)
Br
H
Br
Br
a) Why does Br2 react with cyclohexene but not with benzene?
(see Rxns I & II, above)
b) Construct a reasonable mechanism for Reaction III that shows the role of the
catalyst. For simplicity, in your mechanism show…
Br
Br
FeBr3
as
Br
and
Br
FeBr3
Note: in the last step FeBr4_ acts as a base, generating FeBr4H, which decomposes into
HBr and FeBr3, regenerating the Lewis acid catalyst (as shown below).
H
Br
FeBr3
H
Br
FeBr3
ChemActivity 29
Electrophilic Aromatic Substitution
25
Information: Friedel-Crafts Alkylation and Acylation
• One of the most difficult and important objectives in organic synthesis is the
formation of new carbon-carbon sigma bonds.
• Electrophilic aromatic substitution (EAS) is one of the few ways to do this.
• As with the EAS bromination in Model 4, a Lewis acid catalyst (usually AlCl3) is
required to make a sufficiently strong electrophile (see H3Cδ+ below left).
• For simplicity, you can think of this methyl electrophile as a methyl carbocation
(see below right), though 1o and methyl carbocations do not exist.
δ–
Br
H
δ–
H
C
δ+
Br
Al
Br
≈
δ–
H
Br
H
Br
H
C
Br
Al
Br
δ–
Br δ–
H
Note: Al should
have a formal charge but this
representation is
more accurate.
This technique for making carbon-carbon sigma bonds was discovered accidentally by
Charles Friedel (1832-1899) and James Crafts (1839-1917). The two were carrying out
reactions involving AlCl3 using benzene as a solvent (which they erroneously thought be
totally inert). Friedel-Crafts Reactions work with two different types of carbon groups,
below designated as R.
X
X
Al
δ
–
X
δ
+
R
X = Cl or Br
R = alkyl or acyl group
X
R = alkyl group
CH3
CH3
methyl
CH
CH2CH3 ethyl
R = acyl group
isopropyl
O
O
C
C
CH3
etc.
CH3
ethanoyl (acetyl)
CH2CH3
propanoyl
34. What reagents would you use to carry out the following reactions?
O
etc.
ChemActivity 29
Electrophilic Aromatic Substitution
26