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Chapter 14: Problem 12

Sketch cis and trans structures for (a) butadiene, and (b) chloroprene. Use twodimensional schematics per footnote 11 of this chapter.

Short Answer

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Question: Sketch the cis and trans structures of (a) butadiene and (b) chloroprene using two-dimensional schematics.

Step by step solution

01

Drawing the backbone of butadiene and chloroprene

Start by drawing the backbone of butadiene and chloroprene. Butadiene (C4H6) has a structure of CH2=CH-CH=CH2, and chloroprene (C4H5Cl) has a structure of CH2=CCl-CH=CH2. Ensure that double bonds are used to represent the pi bonds.
02

Drawing cis and trans isomers for butadiene

For butadiene (C4H6), we have two structures: Cis-butadiene: Draw butadiene with the two double bonds having a cis configuration: both hydrogen atoms (H) on the left and right carbons should be located on the same side of the double bonds. Trans-butadiene: Draw butadiene with the two double bonds having a trans configuration: The hydrogen (H) atoms on the left and right carbons should be located on opposite sides of the double bonds.
03

Drawing cis and trans isomers for chloroprene

For chloroprene (C4H5Cl), we have two structures: Cis-chloroprene: Draw chloroprene with the double bond between C2 and C3 having a cis configuration: The hydrogen (H) atom on C2 and the chlorine (Cl) atom on C3 should be on the same side of the double bond. Trans-chloroprene: Draw chloroprene with the double bond between C2 and C3 having a trans configuration: The hydrogen (H) atom on C2 and the chlorine (Cl) atom on C3 should be on opposite sides of the double bond. The final step is to sketch these structures using two-dimensional schematics.

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Most popular questions from this chapter

For a linear freely rotating polymer molecule, the total extended chain length \(L\) depends on the bond length between chain atoms \(d\), the total number of bonds in the molecule \(N\), and the angle between adjacent backbone chain atoms \(\theta\), as follows: $$ L=N d \sin \left(\frac{\theta}{2}\right) $$ Furthermore, the average end-to-end distance \(r\) for a randomly winding polymer molecule in Figure \(14.6\) is equal to $$ r=d \sqrt{N} $$ A linear polytetrafluoroethylene has a numberaverage molecular weight of \(500,000 \mathrm{~g} / \mathrm{mol}\); compute average values of \(L\) and \(r\) for this material.

The following table lists molecular weight data for a polypropylene material. Compute (a) the number-average molecular weight, (b) the weight-average molecular weight, and (c) the degree of polymerization. \begin{tabular}{rcc} \hline Molecular Weight Range \((\mathrm{g} /\) mol \()\) & \(\boldsymbol{x}_{\boldsymbol{i}}\) & \(\boldsymbol{w}_{\boldsymbol{i}}\) \\ \hline \(8,000-16,000\) & \(0.05\) & \(0.02\) \\ \(16,000-24,000\) & \(0.16\) & \(0.10\) \\ \(24,000-32,000\) & \(0.24\) & \(0.20\) \\ \(32,000-40,000\) & \(0.28\) & \(0.30\) \\ \(40,000-48,000\) & \(0.20\) & \(0.27\) \\ \(48,000-56,000\) & \(0.07\) & \(0.11\) \\ \hline \end{tabular}

The number-average molecular weight of a poly(styrene-butadiene) alternating copolymer is \(1,350,000 \mathrm{~g} / \mathrm{mol}\); determine the average number of styrene and butadiene repeat units per molecule.

Using the definitions for total chain molecule length \(L\) (Equation 14.11) and average chain end-to-end distance \(r\) (Equation 14.12), for a linear polyethylene determine the following: (a) the number-average molecular weight for \(L=2500 \mathrm{~nm}\) (b) the number-average molecular weight for \(r=20 \mathrm{~nm}\)

Calculate the number-average molecular weight of a random nitrile rubber [poly (acrylonitrile-butadiene) copolymer] in which the fraction of butadiene repeat units is \(0.30\) assume that this concentration corresponds to a degree of polymerization of 2000 .
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