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Propose mechanisms for the reaction of 2 -pyridone with a general electrophile, \(\mathrm{E}^{+},\) to give (a) the 3-substituted and (b) the 5-substituted products.

Short Answer

Expert verified
2-Pyridone reacts with an electrophile at the 3- or 5-position via resonance-stabilized carbocations, leading to substitution at these positions.

Step by step solution

01

Analyze the Structure of 2-Pyridone

2-Pyridone has a carbonyl group (C=O) at the 2-position and a five-membered ring that includes a nitrogen atom. This nitrogen can donate electrons and activate the aromatic ring towards electrophilic substitution.
02

Identify Reactivity of Positions

The most likely positions for electrophilic substitution are positions 3 (para to the nitrogen) and 5 (meta to the nitrogen). These positions are typically reactive due to the resonance stabilization offered by the electron-donating nitrogen.
03

Mechanism for 3-Substitution

In this step, the electrophile, \(E^+\), attacks the 3-position of the pyridone ring. This attack forms a carbocation at the 3-position, stabilized by resonance through the nitrogen. The resonance form shows the positive charge delocalized to the nitrogen, enhancing stability. Finally, deprotonation occurs, restoring aromaticity and yielding the 3-substituted product.
04

Mechanism for 5-Substitution

Similarly, the electrophile, \(E^+\), attacks the 5-position. The 5-position carbocation is stabilized by resonance as the positive charge can be delocalized across the aromatic system. This leads to another resonance stabilized intermediate. Deprotonation restores aromaticity and results in the 5-substituted product.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

2-Pyridone Reactivity
2-Pyridone is a fascinating compound when it comes to electrophilic aromatic substitution, often abbreviated as EAS. In simple terms, the reactivity of 2-pyridone can be attributed to its structural build-up. The foundational structure of 2-pyridone consists of a pyridine ring, which includes a carbonyl group at the 2-position, hence the name. This carbonyl presence is crucial as it influences electronic distribution in the molecule.
Nevertheless, what really sparks 2-pyridone's reactivity for EAS is the nitrogen atom in the pyridine ring. The nitrogen atom acts as an electron-donor because of its lone pair of electrons. This action modifies the electronic conditions of the ring. Specifically, the lone pair can partially delocalize into the ring, activating it towards electrophilic assault. Understanding this electron-donating capacity is vital for predicting the preferred positions for substitution, as we'll explore further.
Carbocation Stability
During electrophilic aromatic substitution, the arrival of an electrophile results in a carbocation intermediate. This is a fleeting but crucial step for the reaction pathway. In our 2-pyridone example, when an electrophile approaches, it targets the electron-rich spots in the ring. For the 3-substituted product, a carbocation forms as the electrophile attaches to the 3-position. The fascinating part of this process is the stability that the carbocation finds through resonance. Consider these points about carbocation stability:
  • Carbocations are more stable when the positive charge is spread out or delocalized over multiple atoms.

  • In 2-pyridone, the nitrogen's ability to share its lone pair extends this stability.
  • Graphically, you can often depict this by showing resonance structures where the positive charge migrates to the nitrogen momentarily.
The resonance effect dramatically stabilizes the intermediate carbocations, making the electrophilic attack feasible and efficient.
Resonance Stabilization
Resonance stabilization is at the heart of many organic reactions, and it's a keyword when discussing the chemistry of 2-pyridone. When we talk about resonance stabilization, we mean the capacity for a molecule to dissipate electronic charges over different atoms, stabilizing reactive intermediates like carbocations.
This is vital in 2-pyridone reactions. Let's see how:
  • Upon electrophilic attack, a charge distribution across the ring increases stability.

  • The lone pair on the pyridine nitrogen helps spread the charge, supporting the positive charge that temporarily arises during substitution.
  • Imagining the structure in terms of different resonance forms helps illustrate this sharing of charge.
By fostering a lowering of energy states in intermediates, resonance allows reactions to proceed more smoothly, ultimately resulting in the successful substitution at the desired positions.

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Most popular questions from this chapter

Draw the structures of (a) 4-methylpyridine, (b) 2 -chloropyrrole, (c) 1,4 -dihydropyridine, (d) tetrabromopyrrole, (e) nicotinic acid.

(a) Explain how thiophene achieves a \(6 \pi\) -aromatic system. (b) What experimental evidence is there for the aromatic character of thiophene? (c) Does thiophene possess more or less aromatic character than furan? Rationalize your answer.

Three isomeric chloro-derivatives of pyridine \((\mathbf{A}, \mathbf{B}\) and \(\mathbf{C}\) ) analyse as containing \(40.58 \%\) C \(, 2.04 \%\) H and \(9.46 \%\) N. The \(^{1}\) H NMR spectroscopic data for the compounds are as follows where \(\mathrm{d}=\) doublet, \(\mathrm{d} \mathrm{d}=\) doublet of doublets and \(\mathrm{t}=\) triplet: $$\begin{array}{ll} \hline \text { Compound } & ^{1} \mathbf{H} \text { NMR } \delta / \text { ppm } \\ \text { A } & 7.66(\mathrm{t}, J=7.6 \mathrm{Hz}) \\ & 7.31(\mathrm{d}, J=7.6 \mathrm{Hz}) \\ \text { B } & 8.64(\mathrm{d}, J=2.1 \mathrm{Hz}) \\ & 8.25(\mathrm{t}, J=2.1 \mathrm{Hz}) \\ \text { C } & 8.70(\mathrm{dd}, J=3.0 \text { and } 0.3 \mathrm{Hz}) \\ & 8.13(\mathrm{dd}, J=9.0 \text { and } 3.0 \mathrm{Hz}) \\ & 7.68(\mathrm{dd}, J=9.0 \text { and } 0.3 \mathrm{Hz}) \\ & \\ \hline \end{array}$$ In each isomer, \(\mathrm{Cl}\) atoms are in either the 2 - or \(3-\) position with respect to the \(\mathrm{N}\) atom. Suggest structures for \(\mathbf{A}, \mathbf{B}\) and \(\mathbf{C}\).

(a) Why is the acylation of pyrrole not carried out under Friedel-Crafts conditions? (b) With reference to the nitration of pyrrole, show how acetyl nitrate behaves as a nitrating agent. Which site in pyrrole is preferentially nitrated?

(a) How many isomers of dimethylpyridine do you expect? Draw their structures and give each a systematic name. (b) Draw diagrams to show the dipole moments in furan and pyrrole.

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