Chapter 18, Problem 18.47AP

Interpretation Introduction

(a)

Interpretation:

The product of reaction of $m-\text{cresol}$ with concentrated ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is to be stated.

Concept introduction:

The chemical reaction in which an electrophile group is replaced by another functional group is known as electrophilic substitution reaction. When the electrophilic substitution happens on an aromatic ring such as benzene then the reaction is known as electrophilic aromatic substitution.

Answer to Problem 18.47AP

The products formed on reaction of $m-\text{cresol}$ with concentrated ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ are shown below.

Explanation of Solution

The reaction of $m-\text{cresol}$ with concentrated ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is shown below.

Figure 1

This reaction is known as electrophilic substitution reaction. The ortho, para directing power of hydroxyl group is greater than the methyl group, therefore, the sulfonic group is directed according to the position of hydroxyl group. Due to the presence of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$, ${\text{SO}}_{\text{3}}\text{H}$ is added as an electrophile and two products are formed. Therefore, the products are shown below.

Figure 2

Conclusion

The products formed on reaction of $m-\text{cresol}$ with concentrated ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ are shown in Figure 2.

Interpretation Introduction

(b)

Interpretation:

The product on reaction of $m-\text{cresol}$ with ${\text{Br}}_{\text{2}}$ in ${\text{CCl}}_{\text{4}}$ (dark) is to be stated.

Concept introduction:

The chemical reaction in which an electrophile group is replaced by another functional group is known as electrophilic substitution reaction. When the electrophilic substitution happens on an aromatic ring such as benzene then the reaction is known as electrophilic aromatic substitution.

Answer to Problem 18.47AP

The products formed on reaction of $m-\text{cresol}$ with ${\text{Br}}_{\text{2}}$ in ${\text{CCl}}_{\text{4}}$ (dark) are shown below.

Explanation of Solution

The reaction of $m-\text{cresol}$ with ${\text{Br}}_{\text{2}}$ in ${\text{CCl}}_{\text{4}}$ (dark) is shown below.

Figure 3

Under dark conditions, there is electrophilic substitution reaction. In this reaction, bromine will get substituted at the benzene ring. The position of the bromine depends on the hydroxyl group as it has more ortho, para directing power than methyl group. Therefore, the products formed on reaction of $m-\text{cresol}$ with ${\text{Br}}_{\text{2}}$ in ${\text{CCl}}_{\text{4}}$ (dark) are shown below.

Figure 4

Conclusion

The products formed on reaction of $m-\text{cresol}$ with ${\text{Br}}_{\text{2}}$ in ${\text{CCl}}_{\text{4}}$ (dark) are shown in Figure 4.

Interpretation Introduction

(c)

Interpretation:

The product on reaction of $m-\text{cresol}$ with ${\text{Br}}_{\text{2}}$ (excess) in ${\text{CCl}}_{\text{4}}$, light is to be stated.

Concept introduction:

The chemical reaction in which an electrophile group is replaced by another functional group is known as electrophilic substitution reaction. When the electrophilic substitution happens on an aromatic ring such as benzene then the reaction is known as electrophilic aromatic substitution.

Answer to Problem 18.47AP

The product on reaction of $m-\text{cresol}$ with ${\text{Br}}_{\text{2}}$ (excess) in ${\text{CCl}}_{\text{4}}$, light is shown below.

Explanation of Solution

The reaction of $m-\text{cresol}$ with ${\text{Br}}_{\text{2}}$ (excess) in ${\text{CCl}}_{\text{4}}$, is shown below.

Figure 5

Under light conditions, there will be bromine radical formed and benzylic substitution will take place. Further, due to excess of bromine, bromine will get substituted at the benzene ring. The position of the bromine depends on the hydroxyl group as it has more ortho, para directing power than methyl group. Therefore, the products formed on reaction of $m-\text{cresol}$ with ${\text{Br}}_{\text{2}}$ (excess) in ${\text{CCl}}_{\text{4}}$ (light) are shown below.

Figure 6

Conclusion

The product on reaction of $m-\text{cresol}$ with ${\text{Br}}_{\text{2}}$ (excess) in ${\text{CCl}}_{\text{4}}$, light is shown in Figure 6.

Interpretation Introduction

(d)

Interpretation:

The product on reaction of $m-\text{cresol}$ with dilute $\text{HCl}$ is to be stated.

Concept introduction:

The chemical reaction in which an electrophile group is replaced by another functional group is known as electrophilic substitution reaction. When the electrophilic substitution happens on an aromatic ring such as benzene then the reaction is known as electrophilic aromatic substitution.

Answer to Problem 18.47AP

No product is formed on reaction of $m-\text{cresol}$ with dilute $\text{HCl}$.

Explanation of Solution

The electron pair of oxygen of hydroxyl group is in conjugation with the phenyl ring. Due to this conjugation, it does not get protonated. Therefore, no reaction is taking place on adding dilute $\text{HCl}$ to $m-\text{cresol}$.

Conclusion

There is no product formed on reaction of $m-\text{cresol}$ with dilute $\text{HCl}$.

Interpretation Introduction

(e)

Interpretation:

The product on reaction of $m-\text{cresol}$ with $0.1M\text{NaOH}$ solution is to be stated.

Concept introduction:

The chemical reaction in which an electrophile group is replaced by another functional group is known as electrophilic substitution reaction. When the electrophilic substitution happens on an aromatic ring such as benzene then the reaction is known as electrophilic aromatic substitution.

Answer to Problem 18.47AP

The product on reaction of $m-\text{cresol}$ with $0.1M\text{NaOH}$ solution is shown below.

Explanation of Solution

The reaction of $m-\text{cresol}$ with $0.1M\text{NaOH}$ solution is shown below.

Figure 7

On adding sodium hydroxide, the hydroxyl group gets deprotonated to form a phenolate ion. Therefore, the product formed on reaction of $m-\text{cresol}$ with $0.1M\text{NaOH}$ is shown below.

Figure 8

Conclusion

The product on reaction of $m-\text{cresol}$ with $0.1M\text{NaOH}$ is shown in Figure 8.

Interpretation Introduction

(f)

Interpretation:

The product on reaction of $m-\text{cresol}$ with ${\text{HNO}}_3$ is to be stated.

Concept introduction:

The chemical reaction in which an electrophile group is replaced by another functional group is known as electrophilic substitution reaction. When the electrophilic substitution happens on an aromatic ring such as benzene then the reaction is known as electrophilic aromatic substitution.

Answer to Problem 18.47AP

The product on reaction of $m-\text{cresol}$ with ${\text{HNO}}_3$ is shown below.

Explanation of Solution

The reaction of $m-\text{cresol}$ with ${\text{HNO}}_3$ is shown below.

Figure 9

In this reaction, electrophilic substitution reaction occurs. Nitro will get substituted at the benzene ring. The position of the nitro group depends on the hydroxyl group as it has more ortho, para directing power than methyl group. Therefore, the products formed on reaction of $m-\text{cresol}$ with ${\text{HNO}}_3$ are shown below.

Figure 10

Conclusion

The product on reaction of $m-\text{cresol}$ with ${\text{HNO}}_3$ is shown in Figure 10.

Interpretation Introduction

(g)

Interpretation:

The product on reaction of $m-\text{cresol}$ with given acyl chloride in presence of ${\text{AlCl}}_{\text{3}}$ is to be stated.

Concept introduction:

The Friedel-Craft acylation is a type of electrophilic substitution reaction. ${\text{AlCl}}_{\text{3}}$ is used as Lewis acid which ionizes the carbon-halogen bond of the acid chloride and forms a positively charged carbon electrophile which is resonance stabilized. Then the electrophile reacts with benzene

Answer to Problem 18.47AP

The products on reaction of $m-\text{cresol}$ with given acyl chloride in presence of ${\text{AlCl}}_{\text{3}}$ are shown below.

Explanation of Solution

The reaction of $m-\text{cresol}$ with given acyl chloride in presence of ${\text{AlCl}}_{\text{3}}$ is shown below.

Figure 11

The above reaction is an example of Friedel Crafts Acylation reaction. In the above reaction, ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{COCl}$ is an electrophile that gets substituted on the benzene ring. The position of the acyl group depends on the hydroxyl group as it has more ortho, directing power than the methyl group.

Figure 12

Conclusion

The products on reaction of $m-\text{cresol}$ with given acyl chloride in presence of ${\text{AlCl}}_{\text{3}}$ are shown in Figure 12.

Interpretation Introduction

(h)

Interpretation:

The product formed on reaction of $m-\text{cresol}$ with ${\text{Na}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ in ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is to be stated.

Concept introduction:

Loss of electrons is classified as an oxidation reaction. ${\text{Na}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ is an oxidising agent. Benzoquinone is formed on oxidation of phenols. It is also known as para-quinone and has the molecular formula ${\text{C}}_{\text{6}}{\text{H}}_{\text{4}}{\text{O}}_{\text{2}}$.

Answer to Problem 18.47AP

The product formed on reaction of $m-\text{cresol}$ with ${\text{Na}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ in ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is shown below.

Explanation of Solution

The reaction of $m-\text{cresol}$ with ${\text{Na}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ in ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is shown below.

Figure 13

The above reaction is an oxidation reaction. The phenol group that has no substitution at para group gets oxidized to $p-\text{quinone}$. In this reaction, ${\text{Na}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ is the oxidizing agent that oxidizes $m-\text{cresol}$. Therefore, the product formed on reaction of $m-\text{cresol}$ with ${\text{Na}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ in ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is shown below.

Figure 14

Conclusion

The product formed on reaction of $m-\text{cresol}$ with ${\text{Na}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ in ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is shown in Figure 14.

Interpretation Introduction

(i)

Interpretation:

The product on reaction of $m-\text{cresol}$ with triflic anhydride in pyridine at $0°\text{C}$ is to be stated.

Concept introduction:

Stille reaction is an example of coupling reaction. In Stille reaction, the triflate reacts with trimethylstannane in presence of $\text{Pd}$ catalyst and $\text{LiCl}$ to give a coupled product of hydrocarbon part of trimethylstannane and triflate.

Answer to Problem 18.47AP

The product formed on reaction of $m-\text{cresol}$ with triflic anhydride in pyridine at $0°\text{C}$ is shown below.

Explanation of Solution

The reaction of $m-\text{cresol}$ with triflic anhydride in pyridine at $0°\text{C}$ is shown below.

Figure 15

In the above reaction, the triflic anhydride combines with the hydroxyl group of cresol to form $m-\text{tolyltrifluoromethanesulfonate}$. The product formed is a useful reactant in Stille coupling reaction. Therefore, the product formed on reaction of $m-\text{cresol}$ with triflic anhydride in pyridine at $0°\text{C}$ is shown below.

Figure 16

Conclusion

The product on reaction of $m-\text{cresol}$ with triflic anhydride in pyridine at $0°\text{C}$ is in Figure 16.

Interpretation Introduction

(j)

Interpretation:

The product on reaction of product of part (i) with ${\left({\text{CH}}_{\text{3}}\right)}_4\text{Sn}$, excess $\text{LiCl}$, and $\text{Pd}{\left({\text{PPh}}_{\text{3}}\right)}_{\text{4}}$ catalyst in dioxane is to be stated.

Concept introduction:

Stille reaction is an example of coupling reaction. In Stille reaction, the triflate reacts with trimethylstannane in presence of $\text{Pd}$ catalyst and $\text{LiCl}$ to give a coupled product of hydrocarbon part of trimethylstannane and triflate.

Answer to Problem 18.47AP

The product on reaction of product of part (i) with ${\left({\text{CH}}_{\text{3}}\right)}_4\text{Sn}$, excess $\text{LiCl}$, and $\text{Pd}{\left({\text{PPh}}_{\text{3}}\right)}_{\text{4}}$ catalyst in dioxane is shown below.

Explanation of Solution

The product of part (i) is shown below.

Figure 16

The reaction of product of part (i) with ${\left({\text{CH}}_{\text{3}}\right)}_4\text{Sn}$, excess $\text{LiCl}$, and $\text{Pd}{\left({\text{PPh}}_{\text{3}}\right)}_{\text{4}}$ catalyst in dioxane is shown below.

Figure 17

In the above reaction, $m-\text{tolyl}\text{trifluoromethanesulfonate}$ reacts with ${\left({\text{CH}}_{\text{3}}\right)}_4\text{Sn}$ to form a stannane. The stannane then further couples with tetramethyl stannane to form $m-\text{xylene}$. Therefore, the product on reaction of product of part (i) with ${\left({\text{CH}}_{\text{3}}\right)}_4\text{Sn}$, excess $\text{LiCl}$, and $\text{Pd}{\left({\text{PPh}}_{\text{3}}\right)}_{\text{4}}$ catalyst in dioxane is shown below.

Figure 18

Conclusion

The product on reaction of product of part (i) with ${\left({\text{CH}}_{\text{3}}\right)}_4\text{Sn}$, excess $\text{LiCl}$, and $\text{Pd}{\left({\text{PPh}}_{\text{3}}\right)}_{\text{4}}$ catalyst in dioxane is shown in Figure 18.

Interpretation Introduction

(k)

Interpretation:

The product on reaction of product of part (i) with $\left(E\right)-{\text{CH}}_{\text{3}}\text{CH=CH}-\text{B}{\left(\text{OH}\right)}_{\text{2}}$, aqueous $\text{NaOH}$, and $\text{Pd}{\left({\text{PPh}}_{\text{3}}\right)}_{\text{4}}$ catalyst is to be stated.

Concept introduction:

The Suzuki coupling reaction is a reaction in which an aryl or vinylic boronic acid is coupled to an aryl or vinylic iodide or bromide. It is a $\text{Pd}\left(0\right)$ catalysed reaction. This reaction can be used to prepare biaryls, aryl-substituted alkenes, and conjugated alkenes.

Answer to Problem 18.47AP

The product on reaction of product of part (i) with $\left(E\right)-{\text{CH}}_{\text{3}}\text{CH=CH}-\text{B}{\left(\text{OH}\right)}_{\text{2}}$, aqueous $\text{NaOH}$, and $\text{Pd}{\left({\text{PPh}}_{\text{3}}\right)}_{\text{4}}$ catalyst is shown below.

Explanation of Solution

The product of part (i) is shown below.

Figure 16

The reaction of product of part (i) with $\left(E\right)-{\text{CH}}_{\text{3}}\text{CH=CH}-\text{B}{\left(\text{OH}\right)}_{\text{2}}$, aqueous $\text{NaOH}$, and $\text{Pd}{\left({\text{PPh}}_{\text{3}}\right)}_{\text{4}}$ catalyst is shown below.

Figure 19

The above reaction is Suzuki coupling reaction. In this reaction, $m-\text{tolyl}\text{trifluoromethanesulfonate}$ reacts with the boronic acid in presence of sodium hydroxide and $\text{Pd}{\left({\text{PPh}}_{\text{3}}\right)}_{\text{4}}$ catalyst to give a coupled product. Therefore, the product on reaction of product of part (i) with $\left(E\right)-{\text{CH}}_{\text{3}}\text{CH=CH}-\text{B}{\left(\text{OH}\right)}_{\text{2}}$, aqueous $\text{NaOH}$, and $\text{Pd}{\left({\text{PPh}}_{\text{3}}\right)}_{\text{4}}$ catalyst is

Figure 20

Conclusion

The product on reaction of product of part (i) with $\left(E\right)-{\text{CH}}_{\text{3}}\text{CH=CH}-\text{B}{\left(\text{OH}\right)}_{\text{2}}$, aqueous $\text{NaOH}$, and $\text{Pd}{\left({\text{PPh}}_{\text{3}}\right)}_{\text{4}}$ catalyst is shown in Figure 20.