Glucose: preparation, structure and D/L configuration

Preparing glucose

1. 1
   C₁₂H₂₂O₁₁ (sucrose) + H₂O → C₆H₁₂O₆ (glucose) + C₆H₁₂O₆ (fructose)

   Boil cane sugar with dilute HCl or dilute H₂SO₄ in alcoholic solution.
2. 2★
   (C₆H₁₀O₅)n (starch) + nH₂O → nC₆H₁₂O₆ (glucose)

   Commercial method: boil starch with dilute H₂SO₄ at 393 K under a pressure of 2 to 3 atm.

★ key step

Evidence for the open-chain structure

Each reaction proves one part of the structure CHO(CHOH)₄CH₂OH.

1. 1
   Molecular formula: C₆H₁₂O₆

   Found from elemental analysis and the molar mass.
2. 2
   C₆H₁₂O₆ + 14HI → CH₃(CH₂)₄CH₃ + 6H₂O + 7I₂ (prolonged heating)

   n-Hexane forms, so the six carbon atoms are in a straight chain.
3. 3
   CHO(CHOH)₄CH₂OH + NH₂OH → CH(=NOH)(CHOH)₄CH₂OH + H₂O

   An oxime forms, so a carbonyl group (C=O) is present.
4. 4
   CHO(CHOH)₄CH₂OH + HCN → CH(OH)(CN)(CHOH)₄CH₂OH

   A cyanohydrin forms, which again shows a carbonyl group.
5. 5
   CHO(CHOH)₄CH₂OH + Br₂ + H₂O → COOH(CHOH)₄CH₂OH + 2HBr

   Bromine water, a mild oxidising agent, gives gluconic acid with the same six carbons. So the carbonyl group is an aldehyde (CHO).
6. 6
   CHO(CHOH)₄CH₂OH + 5(CH₃CO)₂O → CHO[CH(OCOCH₃)]₄CH₂OCOCH₃ + 5CH₃COOH

   With acetic anhydride a stable pentaacetate forms, so there are five OH groups, each on a different carbon.
7. 7★
   CHO(CHOH)₄CH₂OH + 3[O] → COOH(CHOH)₄COOH + H₂O (nitric acid as oxidising agent)

   Saccharic acid, a six-carbon dicarboxylic acid, forms. Gluconic acid gives the same product, so the other end must be a primary alcohol (CH₂OH) group.

★ key step

D and L configuration

The prefix D or L shows the configuration, not the direction of rotation. It compares the sugar with glyceraldehyde. In D-(+)-glyceraldehyde, the OH on the asymmetric carbon is on the right in the Fischer projection; in L-(−)-glyceraldehyde it is on the left. For a sugar, look only at the lowest asymmetric carbon (C5 in glucose). In natural glucose its OH is on the right, so it is D-glucose. The sign (+) shows that it is dextrorotatory: it turns plane-polarised light to the right. Fructose is also D, but it is laevorotatory, so it is written D-(−)-fructose.

How to draw D-glucose: Fischer projection and Haworth ring

• Fischer projection: draw a vertical chain of six carbons with C1 at the top. C1 is CHO.
• C2: H on the left, OH on the right. C3: HO on the left, H on the right.
• C4: H on the left, OH on the right. C5: H on the left, OH on the right (this OH makes it D). C6 is CH₂OH.
• Haworth ring: draw a flat hexagon with the ring O at the back right corner. C1 is the right corner; C2 and C3 are at the front; C4 is at the left; C5 is at the back left, with CH₂OH (C6) pointing up.
• Groups on the right in the Fischer projection point down in the ring: OH down at C2 and C4, up at C3.
• At C1: OH down gives α-D-glucose; OH up gives β-D-glucose.

Why the open chain is not the whole story

• Although glucose has an aldehyde group, it does not give Schiff's test or the 2,4-DNP test, and it does not form an addition product with NaHSO₃.
• Glucose pentaacetate does not react with hydroxylamine, so it has no free CHO group.
• Glucose exists in two crystalline forms: α (m.p. 419 K), crystallised from a concentrated solution at 303 K, and β (m.p. 423 K), from a hot saturated solution at 371 K.
• Explanation: the OH at C5 adds to the CHO group, giving a cyclic hemiacetal. The ring has six atoms (five C and one O), like pyran, so this is called the pyranose structure.
• C1 becomes a new asymmetric carbon, the anomeric carbon. The α and β forms differ only in the position of the OH at C1; they are called anomers.