lecture 15 given

lecture 15 given - Critically Important Announcement! Exam...

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Critically Important Announcement! Exam 7 pm Thursday Night: Last Name Starts with Letter A-M, Go to Wel 2.224 Last Name Starts with Letter N-Z, Go to UTC 2.102A Dr. Sessler’s Office Hours will be in Welch 5.330 today (1:30 - 3:15 pm)
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Organic Chemistry I 310/318M Pre-Health Professionals Unique numbers: 54410, 54435, 54440, 54445, and 54655 Prof. Jonathan L. Sessler Lecture 15 Notes: Homework 3 due today; homework 4 due next Wednesday. Exam 1 is Thurs evening; it will cover chapts. 1-3, inclusive; it will be in 2 rooms. Dr. Sessler’s Office Hours will be in Welch 5.330 today (1:30-3:15 pm).
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Optical Purity Optical purity: A way of describing the composition of a mixture of enantiomers. Enantiomeric Enantiomeric excess: excess: The difference between the percentage of two enantiomers in a mixture. – optical purity is numerically equal to enantiomeric excess, but is experimentally determined. x 100 [ α ] sample Percent optical purity = [ α ] pure enantiomer Enantiomeric excess (ee) = %R - %S Review from last time…
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Calculation of e.e. • If we know for a pure enantiomer we can use the polarimeter to figure out the relative ratios of the two enantiomers in an arbitrary mixture, or more useful, the excess of one enantiomer over the other. This is the enantiomeric excess (e.e.). Example #1 We know that (+)2-iodobutane has = 15.9 o . We have a solution of pure 2-iodobutane but it has = 5.30 o . What is the enantiomeric excess? measured specific rot . rot . = 5.30 15.9 = 33.3% Ans: e.e. α [ ] D 24 ° [] D t [ ] D 24 °
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How much of the (-) enantiomer is present? Let x = minor (-) x + 0.33 = y y = major (+) y + x = 1.00 2x + 0.33 = 1.00 Thus, x = 0.33 or 33% (-) and y = 0.67% (+) Answer checks out: 67% - 33% = 33% (ignoring rounding error) 0.66 x (15.9 o ) + 0.33 (-15.9 o ) = 5.30 o Example continued
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Example #2 : suppose you are making up a lab unknown to test 118 or 210 lab students on e.e. and need to prepare said solution. If we want to generate a solution of 2-iodobutane with an = 3.98 o , how much of the pure (-) and (+) enantiomers must we mix together? Answer: 1 st we must solve the students’ problem and find e.e. measured specific rot . rot . = 3.98 15.9 = 25% e.e. Now repeat calculation of example #1 Let x = minor (-) x + 0.25 = y y = major (+) y+ x = 1.00 2x + 0.25 = 1.00 x = 0.375 or 37.5% (-) y = 0.625 or 62.5% (+) α [ ] D t
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So we must mix 37.5 ml of (-)-2-iodobutane with 62.5 ml of (+)- 2-iodobutane. Î For those with calculators. Please figure out what happens if T.A. is sloppy and uses 40 ml and 60 ml. How far from 3.98 o will the lab unknown be? (0.6 x 15.9) + (0.4 x –15.9) = 3.18 = 20% error! α [] D 24 ° [ ] D 24 ° Calculation continued
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Enantiomeric Excess Practice: Practice: a commercial synthesis of naproxen, a nonsteroidal anti-inflammatory drug (NSAID), gives the S enantiomer in 97% ee.
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This note was uploaded on 03/23/2010 for the course CH 54410 taught by Professor Sessler during the Spring '10 term at University of Texas.

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lecture 15 given - Critically Important Announcement! Exam...

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