Organic Chemistry Emil Fischer Glucose

1,481 views

Published on

Project for my sisters Organic Chemistry Class. I helped type and insert most of the animations.

Published in: Education, Business, Technology
0 Comments
2 Likes
Statistics
Notes
  • Be the first to comment

No Downloads
Views
Total views
1,481
On SlideShare
0
From Embeds
0
Number of Embeds
1
Actions
Shares
0
Downloads
0
Comments
0
Likes
2
Embeds 0
No embeds

No notes for slide

Organic Chemistry Emil Fischer Glucose

  1. 1. Structure of Glucose CHEM 81 Professor Hoover Allyson Camitta Ruby Davila Tamra Fukumoto Maria Calderon1
  2. 2. 2Structure of GlucoseAvailable data: Meso compounds are optically inactive and unsymmetrical compounds are optically active Aldose contains an aldehyde moiety 6 carbon atom (hex-)Therefore, glucose is an aldohexose. Emil Fischer
  3. 3. 3Structure of Glucose (cont.)  Four chiral center  24 = 16 stereoisomers possible  During this time a method to assign the stereochemistry to the chiral center did not exist D-glucose  Fischer designated the aldohexoses with the -OH group at C-5 projecting to the right as D sugars.
  4. 4. 4 Possible Structures1) 2) 3) 4) 5) 6) 7) 8)  Focused on the D configuration only
  5. 5. Oxidation with HNO3  Dilute nitric acid oxidizes both the aldehyde and –OH groups of an aldose5 to an aldaric acid
  6. 6. 6 Clue 1: Oxidation with HNO31) 2) 3) 4) 5) 6) 7) 8)  Glucose is oxidized and optically active  Therefore, it cannot be molecule 1 or 7 since both would give optically inactive aldaric acids
  7. 7. 7Ruff Degradation An aldose is shortened by one carbon Steps: 1. Oxidation of aldose to an aldonic acid with bromine and water 2. Oxidative decarboxylation of aldonic acid to an aldose with hydrogen peroxide and ferric sulfate
  8. 8. 8Clue 2: Ruff Degradation Degradation of glucose gives an aldopentose  Further degradation of the aldopentose gives an aldotetrose Possible Structures:1) 2) 3) 4) 5) 6) 7) 8) Followed by oxidation with HNO3 gives an optically inactive aldaric acid  Eliminates 5, 6, 7, 8  Establishes conformation at C4 of glucose
  9. 9. 9Clue 3: Ruff degradation Degradation of glucose gives an aldopentose Possible Structures:1) 2) 3) 4) 5) 6) 7) 8) Followed by oxidation with HNO3 gives an optically inactive aldaric acid  Eliminates 1, 2, 5, 6, 7  Establishes confirmation at C3 of glucose
  10. 10. 10 Possible Structures Left  At this point Fischer knew the arrangement of all atoms except C2
  11. 11. Clue 4: Interchange of Two End Groups  If the product was different then structure A was glucose.  If product after interchange was the same, then structure B was glucose11 A) New product B) Same product
  12. 12. Structure of Glucose12

×