Chapter 18, Problem 18.52AP

Interpretation Introduction

(a)

Interpretation:

The compound $A$ with the help of the given data is to be identified.

Concept introduction:

The hydrogenation reaction is defined as a chemical reaction of the hydrogen molecule with a reactant molecule in the presence of catalyst like palladium, nickel, platinum. It is used to saturate the unsaturated compound. It is used to convert vegetable oils into solid fats.

Answer to Problem 18.52AP

The compound $A$ is $3,5-\text{dimethylphenol}$ and the reaction is shown below.

Explanation of Solution

The given compound has a formula ${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}\text{O}$. The given compound is insoluble in water but soluble in aqueous $\text{NaOH}$. It confirms the presence of alcohol which forms sodium salts in aqueous basic solution which make them soluble. Its unsaturation number is $4$ which is calculated by double bond equivalent number formula. The formula is shown below.

$DBE=\frac{2\text{C}+2+\text{N}-\text{H}-\text{X}}{2}$ …(1)

Where

• $\text{C}$ is the number of the carbon atom.
• $\text{N}$ is the number of the nitrogen atom.
• $\text{H}$ is the number of the hydrogen atom.
• $\text{X}$ is the number of the halogen atom.

Substitute the value of each atom in equation (1). Then the unsaturation number is calculated as shown below.

$\begin{array}{rll}DBE& =& \frac{2\text{C}+2+\text{N}-\text{H}-\text{X}}{2}\\ & =& \frac{2\times 8+2+0-10-0}{2}\\ & =& \frac{16+2-10}{2}\\ & =& 4\end{array}$

From the unsaturation number, the presence of an aromatic ring is predicted. The hydrogenation reaction with nickel catalyst at high pressure gives $3,5-\text{dimethylcyclohexanol}$ confirms the presence of an aromatic ring. Therefore, the reaction is shown below.

Figure 1

Therefore, the compound $A$ is $3,5-\text{dimethylphenol}$.

Conclusion

The compound $A$ is $3,5-\text{dimethylphenol}$ and the reaction is shown in Figure 1.

Interpretation Introduction

(b)

Interpretation:

The compound $B$ with the help of the given data is to be identified.

Concept introduction:

Alcohol gives nucleophilic substitution reactions with strong acids like $\text{HCl}$, $\text{HBr}$ and many more. It may be ${\text{S}}_{\text{N}}\text{1}$ or ${\text{S}}_{\text{N}}\text{2}$ reaction depending upon the type of alkyl group present. Also, halides are good leaving group as well as nucleophiles. When alkyl halide obtained after the reaction with hydrogen halide, it reacts with magnesium metal and forms Grignard reagent that are used in various organic reactions.

Answer to Problem 18.52AP

The compound $B$ is $p-\text{tolylmethanol}$ and the reaction is shown below.

Explanation of Solution

The given aromatic compound has a formula ${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}\text{O}$. The given compound is insoluble in water as well as in aqueous $\text{NaOH}$. It may be a nonalcohol compound or it may be alcohol whose proton is less acidic because alcohol compounds are soluble when reacted with aqueous $\text{NaOH}$. Its unsaturation number is $4$ which is calculated by double bond equivalent number formula. The formula is shown below.

$DBE=\frac{2\text{C}+2+\text{N}-\text{H}-\text{X}}{2}$ …(1)

Where

• $\text{C}$ is the number of the carbon atom.
• $\text{N}$ is the number of the nitrogen atom.
• $\text{H}$ is the number of the hydrogen atom.
• $\text{X}$ is the number of the halogen atom.

Substitute the value of each atom in equation (1). Then the unsaturation number is calculated as shown below.

$\begin{array}{rll}DBE& =& \frac{2\text{C}+2+\text{N}-\text{H}-\text{X}}{2}\\ & =& \frac{2\times 8+2+0-10-0}{2}\\ & =& \frac{16+2-10}{2}\\ & =& 4\end{array}$

When compound react with $\text{HBr}$ and form $1-\left(\text{bromomethyl}\right)-4-\text{methylbenzene}$ undergoes ${\text{S}}_{\text{N}}\text{1}$ mechanism. It reacts further with $\text{Mg}$ in $\text{THF}$ followed by water gives $p-\text{xylene}$. It is a general reaction mechanism with a Grignard reagent. The reaction is shown below.

Figure 2

Therefore, the compound $B$ is $p-\text{tolylmethanol}$.

Conclusion

The compound $B$ is $p-\text{tolylmethanol}$ and the reaction is shown in Figure 2.

Interpretation Introduction

(c)

Interpretation:

The compound $C$ with the help of the given data is to be identified.

Concept introduction:

Alky ether gives similar nucleophilic substitution reactions as alcohols with strong acids like $\text{HCl}$, $\text{HBr}$ and many more. It may be ${\text{S}}_{\text{N}}\text{1}$ or ${\text{S}}_{\text{N}}\text{2}$ reaction depending upon the type of alkyl group present. Also, halides are good leaving group as well as nucleophiles.

Answer to Problem 18.52AP

The compound $C$ is $\text{ethoxy}-3-\text{methylbenzene}$ and the reaction is shown below.

Explanation of Solution

The given aromatic compound has a formula ${\text{C}}_9{\text{H}}_{12}\text{O}$. The given compound is insoluble in water as well as in aqueous $\text{NaOH}$. It may be a nonalcohol compound or it may be an alcohol whose proton is less acidic because alcoholic compounds are soluble when reacted with aqueous $\text{NaOH}$. Its unsaturation number is $4$ which is calculated by double bond equivalent number formula. The formula is shown below.

$DBE=\frac{2\text{C}+2+\text{N}-\text{H}-\text{X}}{2}$ …(1)

Where

• $\text{C}$ is the number of the carbon atom.
• $\text{N}$ is the number of the nitrogen atom.
• $\text{H}$ is the number of the hydrogen atom.
• $\text{X}$ is the number of the halogen atom.

Substitute the value of each atom in equation (1). Then the unsaturation number is calculated as shown below.

$\begin{array}{rll}DBE& =& \frac{2\text{C}+2+\text{N}-\text{H}-\text{X}}{2}\\ & =& \frac{2\times 9+2+0-12-0}{2}\\ & =& \frac{18+2-12}{2}\\ & =& 4\end{array}$

The compound $C$ on reaction with $\text{HBr}$ gives $m-\text{cresol}$ and volatile alkyl bromide. This reaction is generally given by ether. The reaction is shown below.

Figure 3

Therefore the compound $C$ is $\text{ethoxy}-3-\text{methylbenzene}$.

Conclusion

The compound $C$ is $\text{ethoxy}-3-\text{methylbenzene}$ and the reaction is shown in Figure 3.