Chapter 13, Problem 33P

Write the structure of the organic product in each of the following reactions. If electrophilic

aromatic substitution occurs, assume only monosubstitution.

Interpretation Introduction

Interpretation:

Structures for the major organic product of an electrophilic aromatic monosubstituted reactions in each of the given reactions is to be written.

Concept introduction:

Ring activating groups such as hydroxide, amines, alkoxy groups etc. direct the incoming electrophile to ortho or para position with respect to their position.

Ring deactivating groups such as nitro, formyl, esters, acyl, etc. direct the incoming electrophile to meta position with respect to their position.

In electrophilic aromatic substitution, alkenes, which are converted to carbocations by protonation in the presence of a strong acid, can be used to alkylate benzene.

Aryl halides (halogens attached to the $\text{benzene}$ ring) undergo nucleophilic aromatic substitution reaction only if there are strong deactivating groups present either ortho and para to the halogen in the ring.

Heterocyclic aromatic compounds such as pyrrole, furan, thiophene have electron rich aromatic rings and are extremely reactive towards electrophilic aromatic reactions. The incoming electrophile attaches selectively to C2 carbon atom in case of heterocyclic aromatic compounds.

The reagent zinc amalgam and concentrated hydrochloric acid is used to convert a carbonyl group into methylene unit. This reaction is known as Clemmenson’s reduction.

Answer to Problem 33P

Solution:

Explanation of Solution

The given reactant molecule has a benzene ring with two carboxylic acid groups and one chlorine as substituents. The mixture of $HN{O}_3$ and $H_{2}S{O}_4$ forms a nitronium ion, which is the electrophile in this electrophilic aromatic substitution. A pair of $\pi$ electrons of benzene acts as a nucleophile.

The carboxylic acid groups are strong deactivating and meta directing groups. Chlorine atom slightly deactivates the ring but is ortho-para directing group. The incoming electrophile, nitronium ion, will be directed to the meta position with respect to the two carboxylic acid groups, which is also the para position with respect to the chlorine atom. Both the attached substituents reinforce electrophilic aromatic substitution at para position with respect to chlorine.

Thus, the reaction is as follows:

The given reactant molecule has a benzene ring attached to three substituents. The three substituents are amino group $(N{H}_2)$, trifluoromethyl $(C{F}_3)$, and nitro group $(N{O}_2)$. The two substituents $N{O}_{2}andC{F}_3$ are strong deactivators and are meta directors. $N{H}_2$ Substituent is a strong activator and is ortho-para director. Bromine is acetic acid used to add bromine atom on the ring. Bromine is the electrophile.

Due to the presence of two strong deactiving substituents, the incoming electrophile is directed to the meta position with respect to $N{O}_{2}andC{F}_3$, which is also the ortho position with respect to the $N{H}_2$ group.

Thus, the reaction is as follows:

For a biphenyl compound, two benzene rings are connected to each other by a single bond. If only one substituent is present in the biphenyl ring, then its position is designated as being ortho, meta, or para with respect to the other ring.

In biphenyl, ring B is considered as a substituent of ring A. Ring B is an aryl ring with a hydroxyl group attached to it. Hydroxyl group is an activating group and is ortho-para directing. Thus, monosubstitution takes place to para position with respect to ring B.

Thus, the reaction is as follows:

The given reactant molecule has a benzene ring attached to alkyl substituents. The one substituent is an isopropyl group while the other is tert-butyl group. $HN{O}_3$ in acetic acid is used to produce a nitronium ion which is the electrophile in this electrophilic aromatic substitution reaction. A pair of $\pi$ electrons of benzene acts as a nucleophile.

Both the substituents are strong activators and are ortho-para directors. Tert-butyl substituent is more sterically hindered than an isopropyl substituent. Thus, the electrophilic aromatic substitution takes place at ortho position with respect to the isopropy group, which is also the para position of tert-butyl group.

Thus, the reaction is as follows:

In the given reaction, $H_{2}S{O}_4$, which is a strong acid, will protonate the alkene to form a secondary carbocation, which behaves as an electrophile. As benzene in not substituted, the secondary octyl cation can be added through anywhere to the ring that yields the same product.

Thus, the reaction is as follows:

This is a Friedel Crafts acylation reaction. The benzene ring has two substituents attached –fluorine and methoxy groups. Acetic acid and aluminum chloride produce an acyl cation, which is an electrophile in this reaction.

The fluorine is a deactivator but ortho-para director. The methoxy group is a strong activator and ortho-para director. Thus, the strong activating methoxy group will direct the incoming acyl cation to its para position.

Thus, the reaction is as follows:

The given electrophilic aromatic substation reaction is nitration. The mixture of $HN{O}_3$ and $H_{2}S{O}_4$ forms a nitronium ion, which is the electrophile in this electrophilic aromatic substitution. A pair of $\pi$ electrons of benzene acts as a nucleophile.

The isopropyl group attached to the benzene ring is an activator and ortho-para director. The nitro group is a strong deactivator and meta director. When mononitration of this molecule takes place, the incoming electrophile is directed meta to nitro group, which is the para position with respect to the isopropyl group.

Thus, the reaction is as follows:

In the given reactant molecule, a benzene ring has two substituents attached. One substituent is the methoxy group while the other is methyl group. Both methyl and methoxy groups will activate the ring and are ortho-para directors. $H_{2}S{O}_4$ is a strong acid, and it is used to protonate the given alkene to produce a stable tertiary carbocation.

The methoxy group activates the ring stronger than the methyl group. Thus, the alkylation takes place at ortho position with respect to methoxy group.

Thus, the reaction is as follows:

The given reactant molecule shows two benzene rings connected by a $C{H}_2$ unit. One benzene ring (ring B) has two methyl substituents and one hydroxyl substituent. The other benzene ring (ring A) has the other substituted benzene ring as a substituent though a $C{H}_2$ unit. Bromine in $CHC{l}_3$ is used to add bromine to the aromatic ring.

When this compound undergoes electrophilic aromatic substitution, the incoming electrophile is directed to the para position with respect to the hydroxyl group in ring B, which is also the ortho position with respect to ring A.

The reaction is as follows:

The given reactant molecule has a benzene ring attached to one fluorine while the other reactant is an benzyl chloride.

The fluoro group is a deactivator, but it will direct the aryl group in the para position. In this Friedel-Craft alkylation reaction, the benzyl group will be directed at the para position with respect to the fluorine atom.

Thus, the reaction is as follows:

Aryl halides (halogens attached to the $\text{benzene}$ ring) undergo nucleophilic aromatic substitution reactions only if there are strong deactivating groups present either ortho and para to the halogen in the ring. In the given reactant molecule, a benzene ring has three substituents attached. There is a strong deactivating group (nitro group) present at ortho position to one of the bromine atoms in the reactant molecule. Because of this, that bromine will undergo nucleophilic substitution reaction. The nucleophile is piperidine ring, which will substitute the bromine ortho to the nitro group.

Thus, the reaction is as follows:

In the reactant molecule, a benzene ring is attached to two substituents. One substituent is acylamino while the other is ethyl substituent. Both these substituents are strong activators and ortho-para directors. This is a Friedel Crafts acylation reaction, in which the acyl group will add to the para position with respect to the acylamino group and ethyl group, which are strong activators.

Thus, the reaction is as follows:

The give reactant benzene ring has four substituents. Three substituents are methyl groups while one is an acyl group. The reagent zinc amalgam and concentrated hydrochloric acid is used to convert a carbonyl group into methylene unit. This reaction is known as Clemmenson reduction.

Thus, the reaction is as follows:

This is an example of a heterocyclic electrophilic aromatic substitution reaction. Thiophenes have electron rich aromatic rings and are extremely reactive towards electrophilic aromatic substitution, preferably at C2-C5 carbon atom in the ring. The thiophene ring has a carboxylic acid group as a substituent attached. Carboxylic acid group deactivates the ring and is a meta directing group. Thus, in bromination of substituted thiophene, the bromine will add to meta position with respect to the carboxylic acid group, which is also the C2 position of thiophene.

Thus, the reaction is as follows:

There is a strong deactivator group (nitro group) present at ortho position to the chlorine. Because of this, that chlorine will undergo nucleophilic substitution reaction. The nucleophile is sulfur, which will attack the chlorine and form the product.

Thus, the reaction is as follows:

This is an example of an electrophilic aromatic substitution followed by a nucleophilic aromatic substitution. In the given reactant molecule, a benzene ring has two chloro groups, which are deactivators but ortho-para directors. The mixture of $HN{O}_3$ and $H_{2}S{O}_4$ forms a nitronium ion, which is the electrophile in this electrophilic aromatic substitution. A pair of $\pi$ electrons of benzene acts as a nucleophile.

As a first step, nitration of the given reactant molecule takes place such that the nitro group will be attached to the ortho position with respect to one of the chlorine atoms and para with respect to the other chlorine atom. Thus, in the product for this step one, there is a benzene ring with three substituents, the two chlorine atoms meta to each other and one nitro group para to one chlorine and ortho to the other.

In this product, the nitro group is placed where it is ortho to one chlorine and para to the other one. Since there is a strong deactivator group (nitro group) present at ortho and para positions to both the chlorines, the aryl chloride will undergo nucleophilic substitution reaction. The nucleophile is ammonia. Both the chlorine atoms will be replaced by amino groups.

Thus, the reaction is as follows:

This is an example of an electrophilic aromatic substitution followed by a nucleophilic aromatic substitution. The mixture of $HN{O}_3$ and $H_{2}S{O}_4$ forms a nitronium ion, which is the electrophile in this electrophilic aromatic substitution. A pair of $\pi$ electrons of benzene acts as a nucleophile. In the given reactant molecule, a benzene ring has a chlorine atom and a trifluoromethyl group. Chlorine slightly deactivates the ring but is ortho-para directing. Trifluoromethyl group is a strong deactivator and a meta director.

If the reactant molecule undergoes a nitration reaction, then the nitro group is placed ortho with respect to the chlorine atom, which is also the meta position with respect to the trifluoromethyl group.

This first step produces a compound in which two strong deactivating groups are attached at ortho and para positions with respect to the chlorine atom. Thus, a nucleophilic aromatic substitution reaction will take place. The nucleophile is methoxide, which will attack the chlorine and form the product.

Thus, the reaction is as follows:

This is an example of an electrophilic aromatic substitution followed by a nucleophilic aromatic substitution. Here, NBS reagent adds bromine to benzylic carbon, so bromine will be added to the $C{H}_3$ group attached to the benzene. The product of this first step then easily undergoes nucleophilic substitution reaction. The nucleophile is the sulfur, which attacks at the bromine on benzylic carbon to give the product.

Thus, the reaction is as follows: