An important use of radical-chain reactions is in the polymerization of ethylene and substituted ethylene monomers such as propene, vinyl chloride (the synthesis of which was discussed in Section \(7.6\) along with its use in the synthesis of poly(vinyl chloride), \((\mathrm{PVC}))\), and styrene. The reaction for the formation of PVC, where \(n\) is the number of repeating units and is very large, follows. (a) Give a mechanism for this reaction (see Chapter 29). (b) Give a similar mechanism for the formation of poly(styrene) from styrene. Which end of the styrene double bond would you expect R- to attack? Why?

Short Answer

Expert verified
Question: Explain the mechanisms for the formation of PVC and poly(styrene) through radical-chain reactions, and point out which carbon atom of the styrene double bond is attacked by the initiator radical. Answer: The mechanisms for the formation of PVC and poly(styrene) involve radical-chain reactions, consisting of initiation, propagation, and termination steps. For both PVC and poly(styrene), the initiation step involves the reaction between an initiator radical (R.) and the respective monomer to form a new radical intermediate. The propagation step involves the reaction of the growing radical intermediate with another monomer, leading to the extension of the polymer chain. Termination occurs when two radical intermediates react, forming the final polymer. In the case of poly(styrene), the initiator radical attacks the carbon atom of the styrene double bond that is bonded to the phenyl group. This is due to the stabilizing effect of the phenyl group's resonance on the generated radical, making that site more susceptible to attack by radicals.

Step by step solution

01

Mechanism for the formation of PVC

The formation of PVC involves a radical-chain reaction with the following steps: (a) Initiation: 1. The initiator molecule (R.) reacts with vinyl chloride to generate a new radical intermediate. \(\mathrm{R^. + CH_2=CHCl \rightarrow RCH_2CHCl}\). (b) Propagation: 2. The radical intermediate reacts with another vinyl chloride molecule to form a new intermediate with a longer chain. \(\mathrm{RCH_2CHCl} \cdotp \mathrm{ + CH_2=CHCl \rightarrow R(CH_2CHCl)_2}\). 3. The propagation step keeps repeating, extending the polymer chain until a termination event occurs. \(\mathrm{R(CH_2CHCl)_n + CH_2=CHCl \rightarrow R(CH_2CHCl)_{n+1}}\). (c) Termination: 4. Termination occurs when two radical intermediates react with each other to form the final PVC polymer. \(\mathrm{R(CH_2CHCl)_n + R'(CH_2CHCl)_m \rightarrow (CH_2CHCl)_nCH_2CHCl}_m\).
02

Mechanism for the formation of poly(styrene) from styrene

The formation of poly(styrene) also involves a radical-chain reaction similar to PVC, and it can be written as follows: (a) Initiation: 1. The initiator molecule (R.) reacts with styrene to generate a new radical intermediate. \(\mathrm{R^. + CH_2=CHPh \rightarrow RCH_2CHPh}\). (b) Propagation: 2. The radical intermediate reacts with another styrene molecule to form a new intermediate with a longer chain. \(\mathrm{RCH_2CHPh} \cdotp \mathrm{+ CH_2=CHPh \rightarrow R(CH_2CHPh)_2}\). 3. The propagation step keeps repeating, extending the poly(styrene) chain until a termination event occurs. \(\mathrm{R(CH_2CHPh)_n + CH_2=CHPh \rightarrow R(CH_2CHPh)_{n+1}}\). (c) Termination: 4. Termination occurs when two radical intermediates react with each other to form the final poly(styrene) polymer. \(\mathrm{R(CH_2CHPh)_n + R'(CH_2CHPh)_m \rightarrow (CH_2CHPh)_nCH_2CHPh}_m\).
03

Attack of R- on the styrene double bond

In the initiation step of the poly(styrene) formation reaction, the attacking radical R- is expected to attack the carbon atom of the double bond that is bonded to the phenyl group (-Ph). The reason for this specific attack is due to the stabilizing effect of the phenyl group on the generated radical intermediate. The resonance effect of the phenyl group helps in stabilizing the radical on the carbon atom attached to it, making this site more susceptible to attack by radicals.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The boiling point of methylcyclohexane \(\left(\mathrm{C}_{7} \mathrm{H}_{14}, \mathrm{MW} 98.2\right)\) is \(101^{\circ} \mathrm{C}\). The boiling point of perfluoromethylcyclohexane \(\left(\mathrm{C}_{7} \mathrm{~F}_{14}\right.\), MW 350) is \(76^{\circ} \mathrm{C}\). Account for the fact that although the molecular weight of perfluoromethylcyclohexane is over three times that of methylcyclohexane, its boiling point is lower than that of methylcyclohexane.

Account for the fact that, among the chlorinated derivatives of methane, chloromethane has the largest dipole moment and tetrachloromethane has the smallest dipole moment. $$ \begin{array}{llc} \hline \text { Name } & \begin{array}{l} \text { Molecular } \\ \text { Formula } \end{array} & \begin{array}{c} \text { Dipole Moment } \\ \text { (debyes, D) } \end{array} \\ \hline \text { Chloromethane } & \mathrm{CH}_{3} \mathrm{Cl} & 1.87 \\ \text { Dichloromethane } & \mathrm{CH}_{2} \mathrm{Cl}_{2} & 1.60 \\ \text { Trichloromethane } & \mathrm{CHCl}_{3} & 1.01 \\ \text { Tetrachloromethane } & \mathrm{CCl}_{4} & 0 \\ \hline \end{array} $$

Draw structural formulas for the following compounds. (a) 3-Iodo-1-propene (b) \((R)\)-2-Chlorobutane (c) meso-2,3-Dibromobutane (d) trans-1-Bromo-3-isopropylcyclohexane (e) 1 -Iodo-2, 2-dimethylpropane (f) Bromocyclobutane

Write a pair of chain propagation steps for the radical bromination of propane to give 1-bromopropane, and calculate \(\Delta H^{\circ}\) for each propagation step and for the overall reaction.

There are three constitutional isomers with the molecular formula \(\mathrm{C}_{5} \mathrm{H}_{12}\). When treated with chlorine at \(300^{\circ} \mathrm{C}\), isomer A gives a mixture of four monochlorination products. Under the same conditions, isomer B gives a mixture of three monochlorination products, and isomer \(\mathrm{C}\) gives only one monochlorination product. From this information, assign structural formulas to isomers \(A, B\), and \(C\).

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free