Ch 17 Reactions of Aromatic Compounds

Question 1

sigma complex

Answer 1

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.

Question 2

electrophilic aromatic substitution

Answer 2

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.

Question 3

why doesn’t water act as a nucleophile and attack the carbocation at the end of electrophilic aromatic substitution?

Answer 3

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.

Question 4

what is the first step of electrophilic aromatic substitution?

Answer 4

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⁺

Question 5

what is the second step of electrophilic aromatic substitution?

Answer 5

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)

Question 6

Why doesn’t the double bond of benzene act like a typical alkene during bromination, and what can be done to facilitate the reaction?

Answer 6

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⁺

Question 7

what is the difference between a catalyst and a reagent?

Answer 7

a catalyst is regenerated in the reaction, and can theoretically work with an unlimited amount of reagent.

Question 8

List the reagents/catalysts of bromination, chloronation, and iodination of benzene:

Answer 8

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).

Question 9

Answer 9

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⁺)

Question 10

What is the first step in the halogenation of benzene?

Answer 10

Electrophilic aromatic substitution Step 1: Formation of a stronger electrophile, using a Lewis acid catalyst (FeBr₃, AlCl₃)/additional reagent (½ I₂ + HNO₃)

Question 11

What is the second step in the halogenation of benzene?

Answer 11

Electrophilic aromatic substitution Step 2: Electrophilic attack and formation of the sigma complex.

Question 12

What is the third step in the halogenation of benzene?

Answer 12

Electrophilic aromatic substitution Step 3: Loss of a proton gives the products.

Question 13

What is the first step in the nitration of benzene?

Answer 13

Electrophilic aromatic substitution Step 1: Attack on the electrophile forms the sigma complex.

Question 14

What is the second step in the nitration of benzene?

Answer 14

Electrophilic aromatic substitution Step 2: Loss of a proton gives nitrobenzene.

Question 15

What is the preliminary step in the nitration of benzene?

Answer 15

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.

Question 16

Answer 16

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).

Question 17

how do you reduce an aromatic nitro (PhNO₂) group to an aromatic amino (PhNH₂) group?

Answer 17

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.

Question 18

Answer 18

Question 19

Answer 19

Question 20

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?

Answer 20

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^-.

Question 21

addition of what to a benzenesulfonic acid forms a benzenesulfonate?

Answer 21

a base like NaOH

Question 22

describe desulfonation

Answer 22

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).

Question 23

Answer 23

Question 24

Answer 24

Question 25

Answer 25

Question 26

what is the rate limiting step for electrophilic aromatic substitution?

Answer 26

formation of the sigma complex

Question 27

explain the reaction-energy diagram for nitration of benzene and nitration of alkylbenzene.

Answer 27

Question 28

which one of these will undergo nitration faster and why?

Answer 28

*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.

Question 29

xylene

Answer 29

dimethylbenzene

Question 30

styrene

Answer 30

vinylbenzene

Question 31

anisole

Answer 31

methoxybenzene

Question 32

toluene

Answer 32

methoxybenzene

Question 33

explain the methoxy substituent in terms of resonance

Answer 33

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.

Question 34

explain the resonance structures of activating substituents on the benzene ring during electrophilic aromatic substitution

Answer 34

Question 35

how could you brominate a benzene ring without a catalyst?

Answer 35

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.

Question 37

list strongly activating compounds, any compounds you list must be in order from strongest to most weakly activating.

Answer 37

phenoxides, anilines, phenols, phenyl ethers, anilides, alkylbenzenes

Question 38

explain *meta*-directors

Answer 38

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.

Question 39

:NR₂ is a strong activator, but NO₂ is a strong deactivator, explain why

Answer 39

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.