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Chapter 22: Problem 80

Isoprene is the repeating unit in natural rubber. The structure of isoprene is a. Give a systematic name for isoprene. b. When isoprene is polymerized, two polymers of the form are possible. In natural rubber, the cis configuration is found. The polymer with the trans configuration about the double bond is called gutta percha and was once used in the manufacture of golf balls. Draw the structure of natural rubber and gutta percha showing three repeating units and the configuration about the carbon–carbon double bonds.

Short Answer

Expert verified
a. The systematic name for isoprene is 2-methyl-1,3-butadiene. b. The structures of natural rubber (cis configuration) and gutta percha (trans configuration) with three repeating units are: Natural rubber: \( \\ CH_2-CH=CH(CH_3)-CH_2-\\ | \\ CH_2-CH=CH(CH_3)-CH_2-\\ | \\ CH_2-CH=CH(CH_3)-CH_2 \) Gutta percha: \( \\ CH_2-CH=CH(CH_3)-CH_2-\\ | \\ CH_2-CH=CH(CH_3)-CH_2-\\ | \\ CH_2-CH=CH(CH_3)-CH_2 \)

Step by step solution

01

Identify the structure of isoprene

Isoprene is the simplest example of an isoprenoid and has the following chemical structure: \[CH_2=C(CH_3)CH=CH_2\]
02

Assign a systematic name to isoprene

To name isoprene systematically, according to the IUPAC system, we first need to identify the longest carbon chain containing the maximum number of double bonds. In this case, the longest carbon chain consists of all 5 carbons so it is a pentene. Since there is a methyl group at carbon 2, it is named 2-methylpentene. However, there are two double bonds in the structure, so it is a diene. The correct systematic IUPAC name for isoprene is 2-methyl-1,3-butadiene. a. The systematic name for isoprene is 2-methyl-1,3-butadiene.
03

Understand cis and trans configurations

Cis and trans isomers are geometric isomers that occur in alkenes with a carbon-carbon double bond. In a cis configuration, the similar groups or atoms are on the same side of the double bond, while in the trans configuration, they are on the opposite side of the double bond.
04

Draw the structures of natural rubber and gutta percha

b. When isoprene undergoes polymerization, it forms long chains of polymers. Natural rubber has the cis configuration around the carbon-carbon double bond while gutta percha has the trans configuration. Both natural rubber and gutta percha structures can be represented as: Natural rubber (cis configuration): \( \\ CH_2=C(CH_3)CH=CH_2 \rightarrow \\ CH_2-CH=CH(CH_3)-CH_2-\\ | \\ CH_2-CH=CH(CH_3)-CH_2-\\ | \\ CH_2-CH=CH(CH_3)-CH_2 \) Gutta percha (trans configuration): \( \\ CH_2=C(CH_3)CH=CH_2 \rightarrow \\ CH_2-CH=CH(CH_3)-CH_2-\\ | \\ CH_2-CH=CH(CH_3)-CH_2-\\ | \\ CH_2-CH=CH(CH_3)-CH_2 \)

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Most popular questions from this chapter

Structural, geometrical, and optical isomers can be drawn having the formula \(\mathrm{C}_{6} \mathrm{H}_{12}\) . Give examples to illustrate these types of isomerism for \(\mathrm{C}_{6} \mathrm{H}_{12}\)

Which of the following polymers would be stronger or more rigid? Explain your choices. a. The copolymer of ethylene glycol and terephthalic acid or the copolymer of \(1,2\) -diaminoethane and terephthalic acid $\left(1,2 \text { -diaminoethane }=\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\right)$ b. The polymer of \(\mathrm{HO}-\left(\mathrm{CH}_{2}\right)_{6}-\mathrm{CO}_{2} \mathrm{H}\) or that of c. Polyacetylene or polyethylene (The monomer in polyacetylene is ethyne.)

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? $$ \begin{array}{ll}{\text { a. } n \text { -pentane }} & {\text { c. } 2,4 \text { -dimethylpentane }} \\ {\text { b. } 2 \text { -methylbutane }} & {\text { d. methylcyclobutane }}\end{array} $$

Give the structure of each of the following aromatic hydrocarbons. $$ \begin{array}{ll}{\text { a. } o \text { -ethyltoluene }} & {\text { c. } m \text { -diethylbenzene }} \\ {\text { b. } p \text { -di-tert-butylbenzene }} & {\text { d. } 1 \text { -phenyl-2-butene }}\end{array} $$

Reagents such as \(\mathrm{HCl}\) , HBr, and $\mathrm{HOH}\left(\mathrm{H}_{2} \mathrm{O}\right)$ can add across carbon-carbon double and triple bonds, with \(\mathrm{H}\) forming a bond to one of the carbon atoms in the multiple bond and \(\mathrm{Cl}\) , Br, or OH forming a bond to the other carbon atom in the multiple bond. In some cases, two products are possible. For the major organic product, the addition occurs so that the hydrogen atom in the reagent attaches to the carbon atom in the multiple bond that already has the greater number of hydrogen atoms bonded to it. With this rule in mind, draw the structure of the major product in each of the following reactions.
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