Structure of D-Glucose and Configuration of D-Glucose https://youtu.be/m4PDNmnHyYI

1. Understanding
Structure & Configuration of
D-Glucose
Prof.Dr.P.Venkatesh
Jagan’s Institute of Pharmaceutical Sciences - Nellore
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2. Structure of D-Glucose
Elemental Analysis
and
Molecular weight determination
show that the molecular formula of Glucose is
C6H12O6
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3. Reduction of Glucose
Complete reduction with concentrated hydriodic acid in the presence of red
Phosphorous produces n-hexane as the major product.
Glucose n-hexane
 Indicates that the 6 carbon atom in the glucose molecule form a consecutive,
unbranched chain.
C-C-C-C-C-C
HI
Red P
C6H12O6 CH3CH2CH2CH2CH2CH3
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4. Glucose readily dissolves in water to give a neutral solution
 Indicates that the glucose molecule does not contain a
carboxyl group
NO
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O
O

5. Reaction with hydroxylamine & Hydrogen Cyanide
Glucose on reaction with hydroxylamine produces monoxime or adds one mole
of hydrogen cyanide to give a cyanohydrin.
Glucose Oxime Gluconitrile
 Indicates the presence of either an aldehyde or a ketone group, but not both.
C6H12O6
(CHOH) 4
H NOH
CH2OH
NH2OH HCN
(CHOH) 5
CH2OH
CN
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O
O
H

6. Oxidation
Mild oxidation of Glucose with bromine water gives Gluconic acid, a monocarboxylic acid
with molecular formula C6H12O7.
 Indicates the presence of an aldehyde group – since only the aldehyde group can be
oxidised to an acid by gaining one oxygen atom without losing any hydrogen atoms.
Glucose Gluconic acid
 Six carbon atoms in glucose forms a consecutive unbranched chain, so the aldehyde
group must occupy one end of this chain
C6H12O6
Br2/H2O
Mild Oxidation
(CHOH) 4
CH2OH
COOH
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7. Oxidation
Further oxidation of Gluconic acid with nitric acid gives Glucaric acid, a dicarboxylic acid with a
molecular formula C6H10O8.
 Indicates the presence of a primary alcohol group, since oxidation occurs with the loss of two
hydrogens and gain of one oxygen atom. i.e.,
Gluconic acid Glucaric acid
 Hence, - CHO & CH2OH occupies the two ends of the six carbon chain
CH2OH COOH
(CHOH) 4
CH2OH
COOH
HNO3
Strong Oxidation
(CHOH) 4
COOH
COOH
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8. Terminal aldehyde group
Glucose reduces ammoniacal solution of silver oxide [Tollen’s reagent] to metallic
silver or a basic solution of cupric ion [Fehling’s solution] to red cuprous oxide.
Silver mirror
Red Precipitate
 Confirms the presence of terminal aldehyde group.
C6H12O6
Ag(NH3)OH
Tollen's reagent
Ag ↓
C6H12O6
Cu(OH)2/NaOH
Fehling's Solution
Cu2O ↓
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9. Hydroxyl groups
Glucose on reaction with acetic anhydride in the presence of pyridine to form a pentaacetate.
Glucose Glucose pentaacetate
 Confirms five –OH groups in glucose molecule
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C6H12O6
(CH3CO)2O
C5H5N
(CHOCOCH 3)4
CHO
CH2OCOCH 3

10. Hydroxyl groups in a different carbon atom
Organic compounds with two hydroxyl groups attached to a single carbon atom usuallylose water to
produce a carbonyl group.
 This suggeststhat in Glucose molecule, each one of the five hydroxyl group is attached to a different
carbon atom.
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OH
OH
O
-H2O

11. Structure of Glucose
These evidence concludes that Glucose is pentahydroxyhexanal
It can be represented by the following gross structure
*C = asymmetric carbon
Glucose [2,3,4,5,6 – pentahydroxy hexanal]
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CHO
CH2OH
*CHOH
*CHOH
*CHOH
*CHOH

12. Configuration of D-Glucose
[Fisher’s proof]
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13. Relative configuration
 Glucose structure has 4 unlike asymmetric carbon atoms.
 This representation is incomplete – because it doesn’t give any idea about the spatial arrangement of
the hydroxyl groups and hydrogen atoms around these four asymmetric centres.
 Yet to determine the relative configuration of the asymmetric centres and absolute configuration of the
molecule.
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14. A key compound is D-arabinose, an aldopentose, which must have one of the
following structures
I II III IV
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H OH
H OH
H OH
CHO
CH2OH
O
H H
H OH
H OH
CHO
CH2OH
H OH
O
H H
H OH
CHO
CH2OH
O
H H
O
H H
H OH
CHO
CH2OH

15. Oxidation of D-arabinose with nitric acid gives an optically active dicarboxylic acid. Under
these conditions structure I & III would have given optically inactive meso diacids.
----plane of ----planeof
symmetry symmetry
I Optically Inactive III Optically Inactive
[meso compound] [meso compound]
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H OH
H OH
H OH
CHO
CH2OH
HNO3
H OH
H OH
H OH
COOH
COOH
H OH
O
H H
H OH
CHO
CH2OH
HNO3 H OH
O
H H
H OH
COOH
COOH

16. II Optically active
IV Optically active
 D-arabinose is therefore either II (or) IV and can be represented with configuration in doubt at C-3.
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O
H H
H OH
H OH
O
H H
H OH
H OH
HNO3
CH2OH COOH
COOH
CHO
3
O
H H
O
H H
H OH
O
H H
O
H H
H OH
HNO3
CHO
CH2OH
COOH
COOH
3

17. When D-arabinose subjected to Kiliani – Fischer synthesis, it gives Glucose & Mannose
D-arabinose V VI
[new asymmetric centre at C-2]
 These sugars differ only in configuration at C-2, which is the new asymmetric centre created in the chain
extension. Structure V & VI represent Glucose and Mannose.
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CHOH (?)
O
H H
H OH
CHO
CH2OH
i) HCN, ii) H2O/H+
iii) NaBH4
CHOH (?)
H OH
O
H H
H OH
CHOH (?)
O
H H
O
H H
H OH
+
CHO CHO
CH2OH CH2OH
3
4 4
2 2

18. Next step is to determine the configurationat C-4 and then to identify which is Glucose & Mannose
 Both Glucose & Mannose on oxidation with nitric acid give diacids which are optically active.
 This means that the hydroxyl group is on the right, as in structure VII and VIII.
 If it were on the left, VII would have yielded an optically inactive meso diacid, IX
VII VIII IX
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H OH
O
H H
H OH
H OH
CHO
CH2OH
O
H H
O
H H
H OH
H OH
CHO
CH2OH
H OH
O
H H
O
H H
H OH
COOH
COOH
---------------------- Plane of symmetry

19.  Structures VII & VIII represent D-Glucose & D-Mannose.
 Decide whether VII is Glucose & VIII is Mannose, or the other way around.
 To decide this, another aldohexose, L-Gulose (X) is used. When L-Gulose oxidised with nitric acid yields the same
dicarboxylic acid (XI) as that obtained from D-Glucose.
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O
H H
O
H H
H OH
HO H
CHO
CH2OH
__
__
__
H OH
O
H H
H OH
H OH
CH2OH
CHO
H
N
O
3
H OH
O
H H
H OH
H OH
COOH
COOH
H OH
O
H H
H OH
H OH
CHO
CH2OH
HNO3
1
6
6
1
X (L-Gulose)
Rotated 180°
XI
VII [D-Glucose]

20.  L-Gulose when turned upside down (180°) has the same configuration at the asymmetric centres as does
D-Glucose, except that the aldehyde& the primary alcohol groups are interchanged.
 Oxidation converts this groups to the carboxyl groups and thus the same diacid (XI) is obtained.
 Such a result is not possible with structure VIII, since the structure obtained by interchanging the ends
doesn’t represent a different sugar.
 Hence D-Glucose is represented by structure VII and D-Mannose is structure VIII.
D-Glucose
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H OH
O
H H
H OH
H OH
CHO
CH2OH