Chapter 22, Problem 22.26P

Interpretation Introduction

(a)

Interpretation:

In the given reaction, the aromatic ring has just one chemically distinct, aromatic H, so a single electrophilic aromatic substitution will lead to a just a single product. The product of the given reaction is to be determined.

Concept introduction:

A Friedel-Craft’s alkylation reaction does not occur readily unless the halogen atom of the alkyl halide is bonded to a ${\text{sp}}^{\text{3}}$ hybridized carbon. Electrophiles in electrophilic aromatic substitution reactions typically must be generated in situ from more stable precursors that can be added as starting materials. Aluminum chloride acts as Lewis acid and forms a complex with chlorine. After heterolysis, electrophilic aromatic substitution reaction takes place. Arenium ion intermediate is a carbocation intermediate consisting of five ${\text{sp}}^{\text{2}}$ hybridized C atoms and one ${\text{sp}}^{\text{3}}$ hybridized C atom.

Answer to Problem 22.26P

The product of the given Friedel-Craft’s alkylation reaction is as shown:

Explanation of Solution

The given reaction is Friedal Craft’s alkylation reaction. In the first step, ${\text{AlCl}}_{\text{3}}$ acts as Lewis acid and forms a complex with ${\text{Cl}}_{\text{2}}$. Step 2 involves heterolysis reaction. Cl-C bond breaks slowly producing ethyl ion as carbocation electrophile and ${\text{AlCl}}_{\text{4}}{}^{-}$. Steps 3 and 4 are electrophilic aromatic substitution reactions in which the aromatic ring acts as nucleophile and attacks the electrophilic ethyl ion. The product of this step is called an arenium ion intermediate. In the next step electrophilic elimination takes place. A proton is eliminated and is assisted by the formation of a bond to a base that is present. Overall, electrophilic ethyl ion replaces ${\text{H}}^{\text{+}}$. The reaction is as shown below:

Conclusion

With the help of mechanism of electrophilic aromatic substitution reaction, the product of the given Friedel-Craft’s alkylation reaction was determined.

Interpretation Introduction

(b)

Interpretation:

In the given reaction, the aromatic ring has just one chemically distinct, aromatic H, so a single electrophilic aromatic substitution will lead to just a single product. The product of the given reaction is to be determined.

Concept introduction:

Electrophiles in electrophilic aromatic substation reactions typically must be generated in situ from more stable precursors that can be added as starting materials. Aluminum chloride acts as a Lewis acid and forms a complex with chlorine. After heterolysis, electrophilic aromatic substitution reaction takes place. The arenium ion intermediate is a carbocation intermediate consisting of five ${\text{sp}}^{\text{2}}$ hybridized C atoms and one ${\text{sp}}^{\text{3}}$ hybridized C atom.

Answer to Problem 22.26P

The product of the given halogenation reaction is as shown:

Explanation of Solution

The given reaction is Friedal Craft’s alkylation reaction. In the first step ${\text{FeCl}}_{\text{3}}$ acts as Lewis acid and forms a complex with ${\text{Cl}}_{\text{2}}$. In step 2, heterolysis of Cl-Cl bond takes place slowly producing ${\text{Cl}}^{\text{+}}$ and ${\text{AlCl}}_{\text{4}}{}^{-}$. Step 3 and 4 is electrophilic aromatic substitution reaction in which the aromatic ring acts as nucleophile and attacks the electrophilic ${\text{Cl}}^{\text{+}}$ ion. The product of this step is called an arenium ion intermediate. In the next step, electrophilic elimination takes place. A proton is eliminated and is assisted by the formation of a bond to a base that is present. Overall, the electrophilic ethyl ion replaces ${\text{H}}^{\text{+}}$. The reaction is as shown below:

Conclusion

With the help of mechanism of electrophilic aromatic substitution reaction, the product of the given reaction was determined.

Interpretation Introduction

(c)

Interpretation:

In the given reaction, the aromatic ring has just one chemically distinct, aromatic H, so a single electrophilic aromatic substitution will lead to a just a single product. The product of the given reaction is to be determined.

Concept introduction:

In Friedel Craft’s acylation reaction, the aromatic species is treated with acyl chloride. The product of Friedel Craft’s acylation reaction is ketone. Electrophiles in electrophilic aromatic substation reactions typically must be generated in situ from more stable precursors that can be added as starting materials. Aluminum chloride acts as Lewis acid and forms a complex with chlorine. After heterolysis, the electrophilic aromatic substitution reaction takes place. The arenium ion intermediate is a carbocation intermediate consisting of five ${\text{sp}}^{\text{2}}$ hybridized C atoms and one ${\text{sp}}^{\text{3}}$ hybridized C atom.

Answer to Problem 22.26P

The product of the given Friedel-Craft’s acylation reaction is as shown.

Explanation of Solution

The given reaction is Friedal Craft’s acylation reaction. In the first step, ${\text{AlCl}}_{\text{3}}$ acts as Lewis acid and forms a complex with acyl chloride. In step 2, heterolysis of Cl-C bond takes place slowly producing carbocation and ${\text{AlCl}}_{\text{4}}{}^{-}$. The step 3 and 4 is electrophilic aromatic substitution reaction in which the aromatic ring acts as nucleophile and attacks the electrophilic carbocation. The product of this step is called an arenium ion intermediate. In the next step, electrophilic elimination takes place. A proton is eliminated and is assisted by the formation of a bond to a base that is present. Overall, the electrophilic carbocation ion replaces ${\text{H}}^{\text{+}}$. The reaction is as shown below:

Conclusion

With the help of the mechanism of electrophilic aromatic substitution reaction, the product of the given Friedel-Craft’s acylation reaction was determined.