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Explain why protonation of thiophene by a strong acid leads to the formation of a polymer.

Short Answer

Expert verified
Protonation generates a reactive cation, leading to polymer formation through successive electrophilic attacks that restore aromaticity.

Step by step solution

01

Understanding Thiophene Structure

Thiophene is a five-membered aromatic heterocycle containing sulfur. Its aromaticity arises from the continuous overlap of p orbitals, forming a conjugated -electron system which is stable.
02

Effect of Protonation

When thiophene is exposed to a strong acid, it can become protonated. Protonation occurs on the carbon adjacent to the sulfur atom, disrupting the aromaticity by removing the equivalent electron from the -electron system.
03

Generation of a Reactive Cation

Protonation generates a positively charged ion or carbocation at the protonation site, making the thiophene ring a reactive cation. This reactive cation seeks to return to aromatic stability.
04

Electrophilic Attack and Polymer Formation

The positively charged thiophene can undergo electrophilic attack by another thiophene molecule, forming a bond and resulting in the loss of a proton to restore aromaticity. This process repeats, causing more thiophene units to link together, thus forming a polymer through a chain reaction.

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Key Concepts

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

Aromatic Heterocycles
Thiophene is an example of an aromatic heterocycle. Aromatic heterocycles are rings where one or more atoms within the ring are different from carbon. Thiophene contains sulfur, making it heterocyclic. Its aromaticity stems from a stable conjugated system, where electrons are shared across the entire ring structure due to the overlap of p orbitals. These overlapping p orbitals form what's known as a \( ext{pi} \)-electron system, providing stability to the molecule. The presence of a heteroatom, such as sulfur in thiophene, contributes to the chemical properties and reactivity of the molecule, impacting how it participates in reactions, such as protonation and polymerization.
Protonation Effects
Protonation involves the addition of a proton \( (H^+) \) to a molecule, and it drastically changes the molecule's properties. In the case of thiophene, protonation occurs at a carbon atom neighboring the sulfur. This specific point of attack results because the heterocyclic framework allows electron density from the conjugated system to enable easier protonation. However, crucially, this interrupts its aromaticity by temporarily reducing the electron count needed for stability. Without sufficient \( ext{pi} \)-electron delocalization, the molecule can't maintain its aromatic nature, thus destabilizing it and making it more reactive.
Electrophilic Attack
Once thiophene has been protonated, it becomes a reactive cation, forming a carbocation where positive charge is localized. This carbocation is highly unstable and seeks to regain stability by reverting to an aromatic arrangement. Consequently, an electrophilic attack occurs where this positively charged site targets the electron-rich system of another thiophene molecule. This interaction leads to the formation of a new bond between the two molecules. During this process, the structure promptly expels a proton to restore aromaticity, essentially self-healing in a sense. This incredibly reactive step is the precursor to polymer formation. Without this reaction step, the thiophene molecules would remain isolated rather than assembling into larger structures.
Polymers in Chemistry
Polymers are large molecules composed of repeated subunits. In the case of thiophene, polymerization begins when multiple units link together through the described electrophilic attack mechanism and go on to form a chain. These chains can continue growing, depending on the availability of thiophene molecules to react with.
  • This process results in a polymer—specifically polythiophene—a polymer material notable for its conductivity and stability.
  • Such polymers are behind many modern advancements in chemistry and materials science, including uses in electronic devices because of their electronic properties.
The whole sequence of forming these advanced materials starts from the fundamental understanding of reactions like protonation and the resulting electrophilic attacks within molecules like thiophene.

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Most popular questions from this chapter

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.

(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?

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

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