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Understanding
Structure & Configuration of
D-Glucose
Prof.Dr.P.Venkatesh
Jagan’s Institute of Pharmaceutical Sciences - Nellore
Chem Eazy
Structure of D-Glucose
Elemental Analysis
and
Molecular weight determination
show that the molecular formula of Glucose is
C6H12O6
Chem Eazy
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
Chem Eazy
Glucose readily dissolves in water to give a neutral solution
 Indicates that the glucose molecule does not contain a
carboxyl group
NO
Chem Eazy
O
O
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
Chem Eazy
O
O
H
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
Chem Eazy
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
Chem Eazy
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 ↓
Chem Eazy
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
Chem Eazy
C6H12O6
(CH3CO)2O
C5H5N
(CHOCOCH 3)4
CHO
CH2OCOCH 3
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.
Chem Eazy
OH
OH
O
-H2O
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]
Chem Eazy
CHO
CH2OH
*CHOH
*CHOH
*CHOH
*CHOH
Configuration of D-Glucose
[Fisher’s proof]
Chem Eazy
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.
Chem Eazy
A key compound is D-arabinose, an aldopentose, which must have one of the
following structures
I II III IV
Chem Eazy
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
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]
Chem Eazy
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
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.
Chem Eazy
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
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.
Chem Eazy
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
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
Chem Eazy
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
 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.
Chem Eazy
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]
 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
Chem Eazy
H OH
O
H H
H OH
H OH
CHO
CH2OH

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Structure and Configuration of D-glucose

  • 1. Understanding Structure & Configuration of D-Glucose Prof.Dr.P.Venkatesh Jagan’s Institute of Pharmaceutical Sciences - Nellore Chem Eazy
  • 2. Structure of D-Glucose Elemental Analysis and Molecular weight determination show that the molecular formula of Glucose is C6H12O6 Chem Eazy
  • 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 Chem Eazy
  • 4. Glucose readily dissolves in water to give a neutral solution  Indicates that the glucose molecule does not contain a carboxyl group NO Chem Eazy 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 Chem Eazy 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 Chem Eazy
  • 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 Chem Eazy
  • 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 ↓ Chem Eazy
  • 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 Chem Eazy 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. Chem Eazy 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] Chem Eazy CHO CH2OH *CHOH *CHOH *CHOH *CHOH
  • 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. Chem Eazy
  • 14. A key compound is D-arabinose, an aldopentose, which must have one of the following structures I II III IV Chem Eazy 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] Chem Eazy 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. Chem Eazy 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. Chem Eazy 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 Chem Eazy 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. Chem Eazy 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 Chem Eazy H OH O H H H OH H OH CHO CH2OH