ORG.CHEM W/TEXT+SOLU.MANUAL
ORG.CHEM W/TEXT+SOLU.MANUAL
15th Edition
ISBN: 9780393252125
Author: KARTY
Publisher: W.W.NORTON+CO.
Question
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Chapter 23, Problem 23.43P
Interpretation Introduction

(a)

Interpretation:

For the given pair of aromatic compounds, the compound that will undergo electrophilic aromatic substitution faster is to be determined.

Concept introduction:

Activating groups speed up an electrophilic aromatic substitution reaction because they stabilize the arenium ion intermediate that is produced. Deactivating groups destabilize the arenium ion intermediate.

Expert Solution
Check Mark

Answer to Problem 23.43P

The compound that will undergo electrophilic aromatic substitution faster is shown below:

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 23, Problem 23.43P , additional homework tip  1

Explanation of Solution

The given pair of compounds is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 23, Problem 23.43P , additional homework tip  2

By using table 23-3, the –OH group is strongly activating than CH3, i.e., alkyl group attached to the ring. The activating groups speed up an electrophilic aromatic substitution reaction. Therefore, the compound that will undergo electrophilic aromatic substitution faster is as shown below:

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 23, Problem 23.43P , additional homework tip  3

Conclusion

The compound that will undergo electrophilic aromatic substitution faster is determined by identifying substituents as a weak, moderate, or strong activator or deactivator.

Interpretation Introduction

(b)

Interpretation:

For the given pair of aromatic compounds, the compound that will undergo electrophilic aromatic substitution faster is to be determined.

Concept introduction:

Alkyl groups have electron donating inductive effect, thus stabilizing the positive charge on the adjacent carbon. The electrophilic aromatic substitution reaction rate increases with the increase in electron density around the ring. The groups which donate electrons, activates the aromatic ring. Electron-withdrawing group withdraws electron density from the ring, so the driving force for the aromatic ring to attack the electrophile is diminished relative to benzene; thus the rate of electrophilic aromatic substitution decreases.

Expert Solution
Check Mark

Answer to Problem 23.43P

The compound that will undergo electrophilic aromatic substitution faster is shown below:

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 23, Problem 23.43P , additional homework tip  4

Explanation of Solution

The given pair of compounds is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 23, Problem 23.43P , additional homework tip  5

In the given compounds, the CH3 substituent is the activating group having electron donating effect. The CF3 is a deactivating group and shows the electron-withdrawing effect. Therefore, the compound that will undergo electrophilic aromatic substitution faster is as shown below:

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 23, Problem 23.43P , additional homework tip  6

Conclusion

The compound that will undergo electrophilic aromatic substitution faster is determined by identifying substituents as a weak, moderate, or strong activator or deactivator.

Interpretation Introduction

(c)

Interpretation:

For the given pair of aromatic compounds, the compound that will undergo electrophilic aromatic substitution faster is to be determined.

Concept introduction:

Activating groups speed up an electrophilic aromatic substitution reaction because they stabilize the arenium ion intermediate that is produced. Deactivating groups destabilize the arenium ion intermediate. Activating groups stabilize the arenium ion intermediate by participation of lone pair on the atom of the substituent in resonance.

Expert Solution
Check Mark

Answer to Problem 23.43P

The compound that will undergo electrophilic aromatic substitution faster is shown below:

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 23, Problem 23.43P , additional homework tip  7

Explanation of Solution

The given pair of compounds is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 23, Problem 23.43P , additional homework tip  8

In the given pair of compounds, the CO2H substituent is a deactivating group. The OCH=O group has an atom with lone pairs, so it is an activating group. The lone pair on oxygen participates into the resonance structure of arenium ion intermediate and stabilizes it. Therefore, the compound that will undergo electrophilic aromatic substitution faster is as shown below:

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 23, Problem 23.43P , additional homework tip  9

Conclusion

The compound that will undergo electrophilic aromatic substitution faster is determined by identifying substituents as activator or deactivator.

Interpretation Introduction

(d)

Interpretation:

For the given pair of aromatic compounds, the compound that will undergo electrophilic aromatic substitution faster is to be determined.

Concept introduction:

Electron-withdrawing group withdraws electron density from the ring, so the driving force for the aromatic ring to attack the electrophile is diminished relative to hydrogen; thus, the rate of electrophilic aromatic substitution decreases. Electron-withdrawing group is a deactivator.

Expert Solution
Check Mark

Answer to Problem 23.43P

The compound that will undergo electrophilic aromatic substitution faster is shown below:

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 23, Problem 23.43P , additional homework tip  10

Explanation of Solution

The given pair of compounds is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 23, Problem 23.43P , additional homework tip  11

In the given pair of compounds, the NO2 group is an electron-withdrawing group relative to H. Thus, it is a deactivating group. Therefore, the compound that will undergo electrophilic aromatic substitution faster is as shown below:

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 23, Problem 23.43P , additional homework tip  12

Conclusion

The compound that will undergo electrophilic aromatic substitution faster is determined by identifying substituents as activator or deactivator.

Interpretation Introduction

(e)

Interpretation:

For the given pair of aromatic compounds, the compound that will undergo electrophilic aromatic substitution faster is to be determined.

Concept introduction:

Electron-withdrawing group withdraws electron density from the ring, so the driving force for the aromatic ring to attack the electrophile is diminished; thus the rate of electrophilic aromatic substitution decreases. Electron-withdrawing group is a deactivator.

Expert Solution
Check Mark

Answer to Problem 23.43P

The compound that will undergo electrophilic aromatic substitution faster is shown below:

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 23, Problem 23.43P , additional homework tip  13

Explanation of Solution

The given pair of compounds is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 23, Problem 23.43P , additional homework tip  14

The Cl is a deactivating group. In the compound having two Cl atoms, the reaction is slower than the reaction in that with one atom. Therefore, the compound that will undergo electrophilic aromatic substitution faster is as shown below:

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 23, Problem 23.43P , additional homework tip  15

Conclusion

Presence of two deactivating groups decreases the rate of electrophilic aromatic substitution than that of one group.

Interpretation Introduction

(f)

Interpretation:

For the given pair of aromatic compounds, the compound that will undergo electrophilic aromatic substitution faster is to be determined.

Concept introduction:

Activating groups speed up an electrophilic aromatic substitution reaction because they stabilize the arenium ion intermediate that is produced. Activating groups stabilize the arenium ion intermediate by participation of lone pair on the atom of the substituent in resonance. Deactivating groups destabilize the arenium ion intermediate.

Expert Solution
Check Mark

Answer to Problem 23.43P

The compound that will undergo electrophilic aromatic substitution faster is shown below:

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 23, Problem 23.43P , additional homework tip  16

Explanation of Solution

The given pair of compounds is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 23, Problem 23.43P , additional homework tip  17

The SO3H is a deactivating group. The S atom of SCH3 substituent has lone pair of electrons. This lone pair takes part in resonance and stabilizes the arenium ion intermediate. Thus, SCH3 is an activating group, and it will speed up the electrophilic aromatic substitution reaction. The compound that will undergo electrophilic aromatic substitution faster is as shown below:

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 23, Problem 23.43P , additional homework tip  18

Conclusion

The compound that will undergo electrophilic aromatic substitution faster is determined by identifying substituents as activator or deactivator.

Interpretation Introduction

(g)

Interpretation:

For the given pair of aromatic compounds, the compound that will undergo electrophilic aromatic substitution faster is to be determined.

Concept introduction:

Activating groups speed up an electrophilic aromatic substitution reaction because they stabilize the arenium ion intermediate that is produced. The rate of electrophilic aromatic substitution reaction is increased by strong activating group.

Expert Solution
Check Mark

Answer to Problem 23.43P

The compound that will undergo electrophilic aromatic substitution faster is shown below:

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 23, Problem 23.43P , additional homework tip  19

Explanation of Solution

The given pair of compounds is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 23, Problem 23.43P , additional homework tip  20

The compound on the left has OH groups that are strong activating groups. The CH3 group is a weak activator. The rate of electrophilic aromatic substitution reaction is increased by strong activating group. Therefore, the compound that will undergo electrophilic aromatic substitution faster is as shown below:

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 23, Problem 23.43P , additional homework tip  21

Conclusion

The compound that will undergo electrophilic aromatic substitution faster is determined by identifying substituents as a weak, moderate, or strong activator or deactivator.

Interpretation Introduction

(h)

Interpretation:

For the given pair of aromatic compounds, the compound that will undergo electrophilic aromatic substitution faster is to be determined.

Concept introduction:

Activating groups speed up an electrophilic aromatic substitution reaction because they stabilize the arenium ion intermediate that is produced. As the number of activating group to the ring increases, the rate of electrophilic aromatic substitution reaction increases.

Expert Solution
Check Mark

Answer to Problem 23.43P

The compound that will undergo electrophilic aromatic substitution faster is shown below:

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 23, Problem 23.43P , additional homework tip  22

Explanation of Solution

The given pair of compounds is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 23, Problem 23.43P , additional homework tip  23

Both compounds above have activating groups. The compound on the right has two activating groups. Two activating groups to the benzene ring increase the rate of electrophilic aromatic substitution reaction. Therefore, the compound that will undergo electrophilic aromatic substitution faster is as shown below:

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 23, Problem 23.43P , additional homework tip  24

Conclusion

Presence of two deactivating groups decreases the rate of electrophilic aromatic substitution than that of one group.

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Chapter 23 Solutions

ORG.CHEM W/TEXT+SOLU.MANUAL

Ch. 23 - Prob. 23.11PCh. 23 - Prob. 23.12PCh. 23 - Prob. 23.13PCh. 23 - Prob. 23.14PCh. 23 - Prob. 23.15PCh. 23 - Prob. 23.16PCh. 23 - Prob. 23.17PCh. 23 - Prob. 23.18PCh. 23 - Prob. 23.19PCh. 23 - Prob. 23.20PCh. 23 - Prob. 23.21PCh. 23 - Prob. 23.22PCh. 23 - Prob. 23.23PCh. 23 - Prob. 23.24PCh. 23 - Prob. 23.25PCh. 23 - Prob. 23.26PCh. 23 - Prob. 23.27PCh. 23 - Prob. 23.28PCh. 23 - Prob. 23.29PCh. 23 - Prob. 23.30PCh. 23 - Prob. 23.31PCh. 23 - Prob. 23.32PCh. 23 - Prob. 23.33PCh. 23 - Prob. 23.34PCh. 23 - Prob. 23.35PCh. 23 - Prob. 23.36PCh. 23 - Prob. 23.37PCh. 23 - Prob. 23.38PCh. 23 - Prob. 23.39PCh. 23 - Prob. 23.40PCh. 23 - Prob. 23.41PCh. 23 - Prob. 23.42PCh. 23 - Prob. 23.43PCh. 23 - Prob. 23.44PCh. 23 - Prob. 23.45PCh. 23 - Prob. 23.46PCh. 23 - Prob. 23.47PCh. 23 - Prob. 23.48PCh. 23 - Prob. 23.49PCh. 23 - Prob. 23.50PCh. 23 - Prob. 23.51PCh. 23 - Prob. 23.52PCh. 23 - Prob. 23.53PCh. 23 - Prob. 23.54PCh. 23 - Prob. 23.55PCh. 23 - Prob. 23.56PCh. 23 - Prob. 23.57PCh. 23 - Prob. 23.58PCh. 23 - Prob. 23.59PCh. 23 - Prob. 23.60PCh. 23 - Prob. 23.61PCh. 23 - Prob. 23.62PCh. 23 - Prob. 23.63PCh. 23 - Prob. 23.64PCh. 23 - Prob. 23.65PCh. 23 - Prob. 23.66PCh. 23 - Prob. 23.67PCh. 23 - Prob. 23.68PCh. 23 - Prob. 23.69PCh. 23 - Prob. 23.70PCh. 23 - Prob. 23.71PCh. 23 - Prob. 23.72PCh. 23 - Prob. 23.73PCh. 23 - Prob. 23.74PCh. 23 - Prob. 23.75PCh. 23 - Prob. 23.76PCh. 23 - Prob. 23.77PCh. 23 - Prob. 23.78PCh. 23 - Prob. 23.79PCh. 23 - Prob. 23.80PCh. 23 - Prob. 23.81PCh. 23 - Prob. 23.82PCh. 23 - Prob. 23.83PCh. 23 - Prob. 23.84PCh. 23 - Prob. 23.85PCh. 23 - Prob. 23.86PCh. 23 - Prob. 23.87PCh. 23 - Prob. 23.88PCh. 23 - Prob. 23.89PCh. 23 - Prob. 23.90PCh. 23 - Prob. 23.91PCh. 23 - Prob. 23.92PCh. 23 - Prob. 23.93PCh. 23 - Prob. 23.94PCh. 23 - Prob. 23.95PCh. 23 - Prob. 23.96PCh. 23 - Prob. 23.97PCh. 23 - Prob. 23.1YTCh. 23 - Prob. 23.2YTCh. 23 - Prob. 23.3YTCh. 23 - Prob. 23.4YTCh. 23 - Prob. 23.5YTCh. 23 - Prob. 23.6YTCh. 23 - Prob. 23.7YTCh. 23 - Prob. 23.8YTCh. 23 - Prob. 23.9YTCh. 23 - Prob. 23.10YTCh. 23 - Prob. 23.11YTCh. 23 - Prob. 23.12YTCh. 23 - Prob. 23.13YTCh. 23 - Prob. 23.14YT
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