ORGANIC CHEM. VOL.1+2-W/WILEYPLUS
ORGANIC CHEM. VOL.1+2-W/WILEYPLUS
12th Edition
ISBN: 9781119304241
Author: Solomons
Publisher: WILEY C
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Chapter 22, Problem 2LGP

Shikimic acid is a key biosynthetic intermediate in plants and microorganisms. In nature, shikimic acid is converted to chorismate, which is then converted to prephenate, ultimately leading to aromatic amino acids and other essential plant and microbial metabolites (see the Chapter 21 Learning Group problem). In the course of research on biosynthetic pathways involving shikimic acid, H. Floss (University of Washington) required shikimic acid labeled with 13 C to trace the destiny of the labeled carbon atoms in later biochemical transformations. To synthesize the labeled shikimic acid, Floss adapted a synthesis of optically active shikimic acid from D-mannose reported earlier by G. W. J. Fleet (Oxford University). This synthesis is a prime example of how natural sugars can be excellent chiral starting materials for the chemical synthesis of optically active target molecules. It is also an excellent example of classic reactions in carbohydrate chemistry. The Fleet–Floss synthesis of D - ( ) - [ 1 , 7 - 13 C ] -shikimic acid (1) from D-mannose is shown in Scheme 1.

(a) Comment on the several transformations that occur between d-mannose and 2. What new functional groups are formed?

(b) What is accomplished in the steps from 2 to 3, 3 to 4, and 4 to 5?

(c) Deduce the structure of compound 9 (a reagent used to convert 5 to 6), knowing that it was a carbanion that displaced the trifluoromethanesulfonate (triflate) group of 5. Note that it was compound 9 that brought with it the required 13 C atoms for the final product.

Chapter 22, Problem 2LGP, Shikimic acid is a key biosynthetic intermediate in plants and microorganisms. In nature, shikimic

SCHEME 1 The synthesis of D - ( ) - [ 1 , 7 - 13 C ] -shikimic acid (1) by H. G. Floss, based on the route of Fleet et al. Conditions:

(a) acetone, HA;

(b) BnCl, NaH;

(c) HCl, aq. MeOH;

(d) N a l O 4 ;

(e) N a B H 4 ;

(f) ( C F 3 S O 2 ) 2 O , pyridine;

(g) 9, NaH;

(h) H C O O N H 4 + P d / C ;

(i) NaH;

(j) 60% aq. C F 3 C O O H .

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1. Calculate the accurate monoisotopic mass (using all 1H, 12C, 14N, 160 and 35CI) for your product using the table in your lab manual. Don't include the Cl, since you should only have [M+H]*. Compare this to the value you see on the LC-MS printout. How much different are they? 2. There are four isotopic peaks for the [M+H]* ion at m/z 240, 241, 242 and 243. For one point of extra credit, explain what each of these is and why they are present. 3. There is a fragment ion at m/z 184. For one point of extra credit, identify this fragment and confirm by calculating the accurate monoisotopic mass. 4. The UV spectrum is also at the bottom of your printout. For one point of extra credit, look up the UV spectrum of bupropion on Google Images and compare to your spectrum. Do they match? Cite your source. 5. For most of you, there will be a second chromatographic peak whose m/z is 74 (to a round number). For one point of extra credit, see if you can identify this molecule as well and confirm by…
Please draw, not just describe!
can you draw each step on a piece of a paper please this is very confusing to me
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