Chapter 26, Problem 26.27AP

Interpretation Introduction

(a)

Interpretation:

The product formed when $\text{2-methylpyridine}$ reacts with given reagent is to be shown.

Concept Introduction:

Pyridine is a heterocyclic compound which contains nitrogen atom. Pyridine nitrogen atom contain a lone pair. The lone pair of pyridine is not involved in the resonance with the aromatic ring. No delocation of electron takes place, hence pyridine acts as a base.

Answer to Problem 26.27AP

The product formed when $\text{2-methylpyridine}$ reacts with dilute $\text{HCl}$ is shown below.

Explanation of Solution

In the given compound, the lone pair of the pyridine nitrogen is not involved in the resonance with the ring. Pyridine acts as base in the presence of acid. The nitrogen abstracts the proton from acid. The product formed is $\text{2-methylpyridinium}\text{ion}$. The complete reaction is shown below.

Figure 1

Conclusion

The product formed in the given reaction is $\text{2-methylpyridinium}\text{ion}$.

Interpretation Introduction

(b)

Interpretation:

The product formed when $\text{2-methylpyridine}$ reacts with given reagent is to be shown.

Concept Introduction:

Pyridine is a heterocyclic compound which contains nitrogen atom. Pyridine nitrogen atom contain a lone pair. The lone pair of pyridine is not involved in the resonance with the aromatic ring. No delocation of electron takes place, hence pyridine acts as a base. The hydrogen atoms of pyridine ring are not highly acidic, so mild base can’t abstract proton from pyridine. It requires a strong base for abstraction of proton.

Answer to Problem 26.27AP

No reaction takes place when $\text{2-methylpyridine}$ reacts with dilute $\text{NaOH}$.

Explanation of Solution

Protons of pyridine are not highly acidic in nature. Sodium hydroxide cannot abstract proton from pyridine. It requires a very strong base for the removal of proton from pyridine to takes place. Hence, no reaction takes place between $\text{2-methylpyridine}$ and dilute $\text{NaOH}$.

Figure 2

Conclusion

No product is formed in the given reaction.

Interpretation Introduction

(c)

Interpretation:

The product formed when $\text{2-methylpyridine}$ reacts with given reagent is to be shown.

Concept Introduction:

Pyridine is a heterocyclic compound which contains nitrogen atom. Pyridine nitrogen atom contain a lone pair. The lone pair of pyridine is not involved in the resonance with the aromatic ring. No delocation of electron takes place, hence pyridine acts as a base. The hydrogen atoms of pyridine ring are not highly acidic, so mild base can’t abstract proton from pyridine. It requires a strong base for abstraction of proton.

Answer to Problem 26.27AP

The product formed when $\text{2-methylpyridine}$ reacts with $\text{C}{\text{H}}_{\text{3}}\text{C}{\text{H}}_{\text{2}}\text{C}{\text{H}}_{\text{2}}\text{C}{\text{H}}_{\text{2}}\text{-Li}$ is shown below.

Explanation of Solution

The butyl-lithium compound is a very strong base. It will abstract proton from the methyl group of $\text{2-methylpyridine}$. The carboanion intermediate formed is resonance stabilized. The lithiation of compound takes place and butane is formed as side product. The complete reaction is shown below.

Figure 3

Conclusion

The product formed in the given reaction is $\text{2-methyllithiumpyridine}$.

Interpretation Introduction

(d)

Interpretation:

The product formed when $\text{2-methylpyridine}$ reacts with given reagent is to be shown.

Concept Introduction:

Nitration reaction is aromatic electrophilic substitution reaction. The nitrating mixture contains nitric acid and sulfuric acid. The nitrosonium ion is formed as electrophile. The methyl group is an activating group, so it promotes electrophilic substitution reaction at ortho-para position.

Answer to Problem 26.27AP

The product formed when $\text{2-methylpyridine}$ reacts with $\text{HN}{\text{O}}_{\text{3}}\text{,}{\text{H}}_{\text{2}}\text{S}{\text{O}}_{\text{4}}$, heat; then $\text{O}\text{H}$ is shown below.

Explanation of Solution

The compound $\text{2-methylpyridine}$ reacts with $\text{HN}{\text{O}}_{\text{3}}\text{,}{\text{H}}_{\text{2}}\text{S}{\text{O}}_{\text{4}}$ to form nitro derivative of given compound. Electrophilic aromatic substitution takes place at position $-5$ of the pyridine ring. The methyl group is ortho-para directing group, so the electrophile attacks at position para to the methyl group. The product formed is $\text{2-methyl-5-nitropyridine}$. The complete reaction is shown below.

Figure 4

Conclusion

The product formed in the given reaction is $\text{2-methyl-5-nitropyridine}$.

Interpretation Introduction

(e)

Interpretation:

The product formed when $\text{2-methylpyridine}$ reacts with given reagent is to be shown.

Concept Introduction:

Pyridine is a heterocyclic compound which contains nitrogen atom. Pyridine nitrogen atom contain a lone pair. The lone pair of pyridine is not involved in the resonance with the aromatic ring. Hydrogen peroxide is a strong oxidizing agent. It will oxidize the nitrogen atom of pyridine to form $\text{N-oxide}$.

Answer to Problem 26.27AP

The product formed when $\text{2-methylpyridine}$ reacts with $\text{30\%}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ is shown below.

Explanation of Solution

Hydrogen peroxide is an oxidizing agent. It will oxidize the given compound. The oxidation reaction takes place at the nitrogen atom. $\text{N-oxide}$ is formed upon oxidation. The product formed upon oxidation of $\text{2-methylpyridine}$ is $\text{2-methylpyridine-N-oxide}$.

Figure 5

Conclusion

The product formed in the given reaction of $\text{2-methylpyridine}$ is $\text{2-methylpyridine-N-oxide}$.

Interpretation Introduction

(f)

Interpretation:

The product formed when $\text{2-methylpyridine}$ reacts with given reagent is to be shown.

Concept Introduction:

Pyridine is a heterocyclic compound which contains nitrogen atom. Pyridine nitrogen atom contain a lone pair. The lone pair of pyridine is not involved in the resonance with the aromatic ring. No delocation of electron takes place, hence pyridine acts as a base. The nitrogen atom lone pair attacks the methyl group of methyl iodide.

Answer to Problem 26.27AP

The product formed when $\text{2-methylpyridine}$ reacts with $\text{C}{\text{H}}_{\text{3}}\text{I}$ is shown below.

Explanation of Solution

The lone pair of nitrogen atom of pyridine is not involved in the resonance with the aromatic ring. The lone pair makes the pyridine compound basic in nature. When compound $\text{2-methylpyridine}$ reacts with methyl iodide, methylation of nitrogen atom takes place. The product formed in the given reaction is $N\text{, 2-dimethylpyridinium}\text{iodide}$. The complete reaction is shown below.

Figure 6

Conclusion

The product formed in the given reaction of $\text{2-methylpyridine}$ is $\text{N, 2-dimethylpyridinium}\text{iodide}$.

Interpretation Introduction

(g)

Interpretation:

The product formed when $\text{2-methylpyridine}$ reacts with given reagent is to be shown.

Concept Introduction:

Pyridine is a heterocyclic compound which contains nitrogen atom. Pyridine nitrogen atom contain a lone pair. The lone pair of pyridine is not involved in the resonance with the aromatic ring. No delocation of electron takes place, hence pyridine acts as a base. Lithiated pyridine compound reacts with aldehyde to form $\text{β-hydroxy}$ compound which upon $\text{β-elimination}$ forms $\alpha ,\beta -$ unsaturated compound.

Answer to Problem 26.27AP

The product formed when $\text{2-methylpyridine}$ reacts with product of part(c) + $\text{PhCH=O}$, then ${\text{H}}_{\text{2}}\text{O}$ is shown below.

Explanation of Solution

The lithiation product formed reacts with benzaldehyde, nucleophilic addition reaction takes place to form $\text{β-hydroxy}$ compound. The $\text{β-hydroxy}$ compound undergoes $\text{β-elimination}$ reaction to form $\alpha ,\beta -$ unsaturated compound. The complete given reaction is shown below.

Figure 7

Conclusion

The product formed when $\text{2-methylpyridine}$ reacts with product of part(c) + $\text{PhCH=O}$, then ${\text{H}}_{\text{2}}\text{O}$ is $\text{2-styrylpyridine}$.

Interpretation Introduction

(h)

Interpretation:

The product formed when $\text{2-methylpyridine}$ reacts with given reagent is to be shown.

Concept Introduction:

Pyridine is a heterocyclic compound which contains nitrogen atom. Pyridine nitrogen atom contain a lone pair. The lone pair of pyridine is not involved in the resonance with the aromatic ring. Hydrogen peroxide is a strong oxidizing agent. It will oxidize the nitrogen atom of pyridine to form $\text{N-oxide}$. The hydrogen gas in presence of a catalyst acts as reducing agent.

Answer to Problem 26.27AP

The product formed when $\text{2-methylpyridine}$ reacts with product of part (e) + ${\text{H}}_{\text{2}}$, catalyst is shown below.

Explanation of Solution

The compound $\text{2-methylpyridine-N-oxide}$ undergoes reduction reaction in presence of hydrogen and catalyst. The $\text{N-oxide}$ group reduction takes place to form $\text{2-methylpyridine}$. The complete reaction is shown below.

Figure 8

Conclusion

The product formed when $\text{2-methylpyridine}$ reacts with product of part (e) + ${\text{H}}_{\text{2}}$, catalyst is $\text{2-methylpyridine}$.