HW2solf10 - Homework 2 Solutions 1 a O C OCH CH O 2 2 O C b O H N CH2 C H N O CH2 C 4 6 c O OCH2CH2O C d O H N C 2 O H3CO O O O 3 The possible

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Unformatted text preview: Homework 2 Solutions 1. a. O C ( OCH CH O 2 2 O C ) b. O H N ( ( CH2 ) C H N O ( CH2 ) C 4 6 ) c. O ( OCH2CH2O C ) d. O ( H N ) C 2. ( ) ) ( ) ( O H3CO O O O 3. The possible polymers are: cis‐1,4‐polyisoprene ( ) trans‐1,4‐polyisoprene ) ( 1,2‐polyisoprene (may be atactic, syndiotactic, or isotactic) ) ( 3,4‐polyisoprene (may be atactic, syndiotactic, or isotactic) ) ( Thus there are a total of 8 different structures possible for polyisoprene. 4. a. First we need the DP so that we know the number of bonds: , , / 9615 / Note that there are 2 bonds for each repeat unit: 〈〉 2 9615 1.54 45,606 / 〈〉 213 b. For this we need to use the equation: / 〈〉 and since we are told this is a theta solvent, there are no excluded volume effects and what we are calculating is actually <r0>2. Also, we are given that: . 8 ∗ 10 Combining these equations we get: ∗ 〈〉 / . / ∗ , .∗ , . / 5.26 ∗ 10 〈〉 7.25 ∗ 10 725 c. The characteristic ratio is the ratio of the actual square end‐to‐end distance (answer to part b) divided by nl2 (the answer to part a). Doing this gives a characteristic ratio of 11.6 (7252/2132). ...
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This note was uploaded on 11/16/2011 for the course EMA 3066 taught by Professor Douglass during the Fall '10 term at University of Florida.

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