Ethylene oxide is an important industrial chemical. Although most ethers are unreactive, ethylene oxide is quite reactive. It resembles $\mathrm{C}_{2} \mathrm{H}_{4}$ in its reactions in that addition reactions occur across the \(\mathrm{C}-\mathrm{O}\) bond in ethylene oxide. a. Why is ethylene oxide so reactive? (Hint: Consider the bond angles in ethylene oxide as compared with those predicted by the VSEPR model.) b. Ethylene oxide undergoes addition polymerization, forming a polymer used in many applications requiring a nonionic surfactant. Draw the structure of this polymer.

Short Answer

Expert verified
Ethylene oxide is highly reactive due to the significant strain in the \(\mathrm{C}-\mathrm{O}\) bond, resulting from the bond angles within the ring being forced to be around 60 degrees, smaller than the expected tetrahedral angle of 109.5 degrees. Upon addition polymerization, ethylene oxide forms poly(ethylene oxide) with a repeating unit structure given by \(\mathrm{[-CH_{2}CH_{2}O-]}_{n}\), where "n" represents the number of repeating units present in the polymer chain.

Step by step solution

01

Part a: Reactivity of Ethylene Oxide

To determine the reactivity of ethylene oxide, we need to consider its molecular structure and apply the VSEPR model. Ethylene oxide has the formula \(\mathrm{C}_{2}\mathrm{O}\), which corresponds to a ring structure with one oxygen and two carbons. According to the VSEPR model, the bond angles around an oxygen atom with two single bonds and two lone pairs should be approximately 109.5 degrees (tetrahedral molecular geometry). However, in ethylene oxide, the ring causes strain as the bond angles within the ring are forced to be around 60 degrees, much smaller than the expected tetrahedral angle. This makes the \(\mathrm{C}-\mathrm{O}\) bond highly strained and vulnerable to attacks from nucleophiles, leading to increased reactivity.
02

Part b: Structure of Polymer

Now, let's draw the structure of the polymer formed through the addition polymerization of ethylene oxide. During addition polymerization, the strained \(\mathrm{C}-\mathrm{O}\) bond of ethylene oxide breaks and the monomers join together. The polymer formed has a repeating unit structure derived from ethylene oxide: \(-\mathrm{C}\mathrm{H}_{2}\mathrm{C}\mathrm{H}_{2}\mathrm{O}-\) In the structure of the polymer, the bond angle around the oxygen atom approaches the expected value of the VSEPR model due to the opened ring, which reduces strain and creates a more stable configuration. To represent the polymer, we can use the following notation for the repeating unit: \(\mathrm{[-CH_{2}CH_{2}O-]}_{n}\) Here, "n" denotes the number of repeating units present in the polymer chain. The resulting polymer is called poly(ethylene oxide) and is widely used as a nonionic surfactant in various applications.

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

Estimate \(\Delta H\) for the following reactions using bond energies given in Table \(8.5 .\) $$ 3 \mathrm{CH}_{2}=\mathrm{CH}_{2}(g)+3 \mathrm{H}_{2}(g) \rightarrow 3 \mathrm{CH}_{3}-\mathrm{CH}_{3}(g) $$ The enthalpies of formation for \(\mathrm{C}_{6} \mathrm{H}_{6}(g)\) and \(\mathrm{C}_{6} \mathrm{H}_{12}(g)\) are 82.9 and $-90.3 \mathrm{kJ} / \mathrm{mol}\( , respectively. Calculate \)\Delta H^{\circ}$ for the two reactions using standard enthalpies of formation from Appendix \( 4 .\) Account for any differences between the results obtained from the two methods.

The codons (words) in DNA (that specify which amino acid should be at a particular point in a protein) are three bases long. How many such three- letter words can be made from the four bases adenine, cytosine, guanine, and thymine?

If one hydrogen in a hydrocarbon is replaced by a halogen atom, the number of isomers that exist for the substituted compound depends on the number of types of hydrogen in the original hydrocarbon. Thus there is only one form of chloroethane (all hydrogens in ethane are equivalent), but there are two isomers of propane that arise from the substitution of a methyl hydrogen or a methylene hydrogen. How many isomers can be obtained when one hydrogen in each of the compounds named below is replaced by a chlorine atom? $$ \begin{array}{ll}{\text { a. } n \text { -pentane }} & {\text { c. } 2,4 \text { -dimethylpentane }} \\ {\text { b. } 2 \text { -methylbutane }} & {\text { d. methylcyclobutane }}\end{array} $$

Draw the following incorrectly named compounds and name them correctly. a. 2-ethyl-3-methyl-5-isopropylhexane b. 2-ethyl-4-tert-butylpentane c. 3-methyl-4-isopropylpentane d. 2-ethyl-3-butyne

What types of interactions can occur between the side chains of the following amino acids that would help maintain the tertiary structure of a protein? $$ \begin{array}{ll}{\text { a. cysteine and cysteine }} & {\text { c. glutamic acid and lysine }} \\ {\text { b. glutamine and serine }} & {\text { d. proline and leucine }}\end{array} $$

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