Question: As we will learn in Chapter 30, styrene derivatives such as A can be polymerized by way of cationic rather than radical intermediates. Cationic polymerization is an example of electrophilic addition to an alkene involving carbocations.

a. Draw a short segment of the polymer formed by the polymerization of A.

b. Why does A react faster than styrene (C6H5CH=CH2)in a cationic polymerization?

Short Answer

Expert verified

Answer

a.

Polymerization of A

b. The methoxy group in A stabilizes an intermediate carbocation by resonance. Therefore, A has more reactivity than styrene for polymerization.

Step by step solution

01

Cationic polymerization

Cationic polymerization is a type of chain-growth polymerization which involves electrophilic addition to an alkene.

The polymerization is initiated by a lewis acid-base complex which adds to the monomer. The carbocation thus formed propagates the polymerization reaction.

Monomers having electron-donating groups that can stabilize the intermediate carbocation undergo cationic polymerization.

02

Mechanism of Cationic polymerization

Cationic polymerization involves three steps- initiation, propagation and termination.

In the initiation step, lewis acid and base react to form a lewis-acid base complex which reacts with monomer to form the carbocation.

In the propagation step, the carbocation formed reacts with the alkene monomer to form a new C-C bond. An addition reaction occurs forming another carbocation. This reaction is repeated to form long-chain polymers. The addition reaction follows Markonikov’s rule.

In the termination step, a new pi bond is formed by the loss of a proton from the carbon adjacent to the carbocation. This terminates the reaction.

03

 Step 3: Cationic polymerization of compound A

a. Compound A undergoes cationic polymerization due to the presence of an electron-donating methoxy group.

Polymerization of A

b. Compound A has a methoxy substituent which is an electron-donating group. Therefore, the carbocation intermediately formed during the reaction can be stabilized by resonance. This stabilization is absent in styrene. Therefore, compound A can react faster than styrene in cationic polymerization.

Resonance structures of the intermediate carbocation

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

Question: (a)Hard contact lenses, which first became popular in the 1960s, were made by polymerizing methyl methacrylate [CH2=C(CH3)CO2CH3] to form poly(methyl methacrylate) (PMMA). Draw the structure of PMMA. (b) More comfortable softer contact lenses introduced in the 1970swere made by polymerizing hydroxyethyl methacrylate [CH2=C(CH3)CO2CH2CH2OH] to form

poly(hydroxyethyl methacrylate) (poly-HEMA). Draw the structure of poly-HEMA. Since neither polymer allows oxygen from the air to pass through to the retina, newer contact lenses that are both comfortable and oxygen-permeable have now been developed.

Question:Draw the products formed in each reaction and include the stereochemistry around any stereogenic centers.

a.

b.

c.

d.

Consider the following Bromination:

(a) Calculate ΔH° for this reaction by using the bond dissociation energies in Table 6.2.

(b) Draw out a stepwise mechanism for the reaction, including the initiation, propagation, and termination steps.

(c) Calculate ΔH° for each propagation step.

(d) Draw an energy diagram for the propagation steps.

(e) Draw the structure of the transition state of each propagation step

Question: When HBr adds to (CH3)2C=CH2 under radical conditions, two radicals are possible products in the first step of chain propagation. Draw the structure of both radicals and indicate which one is formed. Then draw the preferred product from HBr addition under radical conditions.

Question: What alkane is needed to make each alkyl halide by radical halogenation?

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