The document summarizes the process by which Emil Fischer determined the structure of glucose in the late 19th century. [1] Glucose has 4 chiral centers, allowing for 16 possible stereoisomers. [2] Through oxidation and degradation experiments, Fischer was able to determine the stereochemistry at carbons 3, 4, and 5, leaving only carbon 2 undefined. [3] An experiment interchanging end groups showed that glucose has structure A, with stereochemistry now fully characterized.

1. Structure of
Glucose
CHEM 81
Professor Hoover
Allyson Camitta
Ruby Davila
Tamra Fukumoto
Maria Calderon
1

2. 2
Structure of Glucose
Available 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
Structure 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
Possible Structures
1) 2) 3) 4) 5) 6) 7) 8)
 Focused on the D configuration only

5. Oxidation with HNO3
 Dilute nitric acid oxidizes both the
aldehyde and –OH groups of an aldose
5
to an aldaric acid

6. 6
Clue 1: Oxidation with HNO3
1) 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
Ruff 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
Clue 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
Clue 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
Possible Structures Left
 At this point Fischer knew the arrangement
of all atoms except C2

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 glucose
11
A) New product
B) Same product

12. Structure of Glucose
12