Ch 17 Reactions of Aromatic Compounds Flashcards
sigma complex
Like an alkene, benzene has clouds of pi electrons above and below its sigma bond framework. Although benzene’s pi electrons are in a stable aromatic system, they are available to attack a strong electrophile to give a carbocation. This resonance-stabilized carbocation is called a sigma complex because the electrophile is joined to the benzene ring by a new sigma bond.
electrophilic aromatic substitution
Electrophilic aromatic substitution
the substitution of an electrophile (E+) for a proton (H+) on the aromatic ring
Preliminary step (optional): Formation of the electrophile
Step 1: Attack on the electrophile forms the sigma complex.
Step 2: Loss of a proton regains aromaticity and gives the substitution product.
why doesn’t water act as a nucleophile and attack the carbocation at the end of electrophilic aromatic substitution?
Electrophilic aromatic substitution
Step 2 of the iodination of benzene shows water acting as a base and removing a proton from the sigma complex. We did not consider the possibility of water acting as a nucleophile and attacking the carbocation, as in an electrophilic addition to an alkene. This would remove aromaticity.
what is the first step of electrophilic aromatic substitution?
Electrophilic aromatic substitution Step 1: Attack on the electrophile forms the sigma complex.
May be preceeded by the Preliminary step: formation of the electrophile, E+
what is the second step of electrophilic aromatic substitution?
Electrophilic aromatic substitution Step 2: Loss of a proton regains aromaticity and gives the substitution product.
(Water may be used as the base because the arenium cations is stable in an aqueous solution)
Why doesn’t the double bond of benzene act like a typical alkene during bromination, and what can be done to facilitate the reaction?
Benzene is not as reactive as alkenes, which react rapidly with bromine at room temperature to give addition products. Because of this, bromine itself is not sufficiently electrophilic to react with benzene, and the formation of Br+ is difficult. Addition of a strong Lewis acid such as FeBr3 catalyzes the reaction by forming a complex with Br2 that reacts like Br+
what is the difference between a catalyst and a reagent?
a catalyst is regenerated in the reaction, and can theoretically work with an unlimited amount of reagent.
List the reagents/catalysts of bromination, chloronation, and iodination of benzene:
Br2 + FeBr3 - molecular bromine and ferric bromide complex, the latter acts as a Lewis acid catalyst.
Cl2 + AlCl3 - molecular chlorine and aluminum chloride complex, the latter acts as a Lewis acid catalyst.
½ I2 + HNO3 - molecular iodine and nitric acid reagents. Nitric acid is consumed in the reaction, so it is a reagent (an oxidant) rather than a catalyst. The iodine cation results from oxidation of iodine by nitric acid (shown).
Electrophilic aromatic substitution
In step 1 of iodination of benzene, HNO3 and H+ are converted to NO2 and H2O, using one electron from ½ I2 in the process, forming I+ needed for iodination of benzene. (net effect, an electron is transferred from ½ I2 to H+)
What is the first step in the halogenation of benzene?
Electrophilic aromatic substitution Step 1: Formation of a stronger electrophile, using a Lewis acid catalyst (FeBr3, AlCl3)/additional reagent (½ I2 + HNO3)
What is the second step in the halogenation of benzene?
Electrophilic aromatic substitution Step 2: Electrophilic attack and formation of the sigma complex.
What is the third step in the halogenation of benzene?
Electrophilic aromatic substitution Step 3: Loss of a proton gives the products.
What is the first step in the nitration of benzene?
Electrophilic aromatic substitution Step 1: Attack on the electrophile forms the sigma complex.
What is the second step in the nitration of benzene?
Electrophilic aromatic substitution Step 2: Loss of a proton gives nitrobenzene.
What is the preliminary step in the nitration of benzene?
Electrophilic aromatic substitution Preliminary steps: Formation of the nitronium ion, NO2+.
Nitric acid has a hydroxyl group that can become protonated and leave as water, similar to the dehydration of an alcohol.
Electrophilic aromatic substitution
Sulfuric acid (H2SO4) protonates the hydroxyl group of nitric acid (HNO3), allowing it to leave as water (H2O) and form a nitronium ion (NO2+). The nitronium ion reacts with benzene to form a sigma complex. Loss of a proton from the sigma complex gives nitrobenzene (PhNO2).
how do you reduce an aromatic nitro (PhNO2) group to an aromatic amino (PhNH2) group?
Aromatic nitro (- NO2) groups are easily reduced to amino (- NH2) groups by treatment with an active metal such as tin, zinc, or iron in dilute acid. Nitration followed by reduction is often the best method for adding an amino group to an aromatic ring.
what reagents are used in the sulfonation of benzene, what kind of reaction is it, and what is atypical about sulfonation relative to the standard process of this reaction type?
Sulfonation of benzene proceeds by electrophilic aromatic substitution, the reagent is “fuming sulfuric acid”, a mixture of 7% SO3 in H2SO4. This reaction has an optional (but likely in the presence of a strong acid such as H2SO4) third step in which the sulfonate group becomes protonated, forming benzenesulfonic acid and HSO4-.
addition of what to a benzenesulfonic acid forms a benzenesulfonate?
a base like NaOH
describe desulfonation
Sulfonation is reversible, and a sulfonic acid group may be removed from an aromatic ring by heating in dilute sulfuric acid. Desulfonation follows the same mechanistic path as sulfonation, except in the opposite order. A proton adds to a ring carbon to form a sigma complex, then loss of sulfur trioxide gives the unsubstituted aromatic ring. Excess water removes SO3 from the equilibrium by hydrating it to sulfuric acid (H2SO4).
what is the rate limiting step for electrophilic aromatic substitution?
formation of the sigma complex
explain the reaction-energy diagram for nitration of benzene and nitration of alkylbenzene.
which one of these will undergo nitration faster and why?
m-xylene undergoes nitration 100 times faster than p-xylene, because it has attack points that are both otho or meta to either methyl substituents simultaneously. All attack points on p-xylene are meta to one methyl group but para to the other, reducing favourability.
xylene
dimethylbenzene
styrene
vinylbenzene
anisole
methoxybenzene
toluene
methoxybenzene
explain the methoxy substituent in terms of resonance
Oxygen is a strongly electronegative group, yet it donates electron density to stabilize the transition state and the sigma complex. Recall that the nonbonding electrons of an oxygen atom adjacent to a carbocation stabilize the positive charge through resonance. The second resonance form puts the positive charge on the electronegative oxygen atom, but it has more covalent bonds, and it provides each atom with an octet in its valence shell. This type of stabilization is called resonance stabilization, and the oxygen atom is called resonance-donating or pi-donating because it donates electron density through a pi bond in one of the resonance structures. Like alkyl groups, the methoxy group of anisole preferentially activates the ortho and para positions.
explain the resonance structures of activating substituents on the benzene ring during electrophilic aromatic substitution
how could you brominate a benzene ring without a catalyst?
add an activating substituent, like methoxy or an amine (nitrogen with a nonbonding pair). These groups are so strongly activating that they quickly brominate in water without a catalyst. In the presence of excess bromine, this reaction proceeds to the tribromide.
list strongly activating compounds, any compounds you list must be in order from strongest to most weakly activating.
phenoxides, anilines, phenols, phenyl ethers, anilides, alkylbenzenes
explain meta-directors
An electron-donating substituent activates primarily the ortho and para positions, and an electron-withdrawing substituent (such as a nitro group) deactivates primarily the ortho and para positions. This selective deactivation leaves the meta positions the most reactive, and meta substitution is seen in the products. Meta-directors deactivate the meta position less than the ortho and para positions, allowing meta substitution.
:NR2 is a strong activator, but NO2 is a strong deactivator, explain why
The nitrogen in the niro group (NO2) has no lone pairs, and all resonance structures place a positive charge on the nitrogen, this causes it to withdraw electrons from the pi system via induction and deactivate the aromatic system. :NR2 has lone pairs, and is willing to donate them to the pi system of the aromatic ring, activating it instead.
