Discuss the open-chain structure of glucose with the reactions that support it. Give the evidence that glucose also has a cyclic structure.

1. 1 · State the structure. Glucose is an aldohexose, CHO(CHOH)₄CH₂OH, C₆H₁₂O₆.
2. 2 · Give the evidence with equations.
   • HI, prolonged heating: n-hexane, so a straight chain
   • NH₂OH: oxime; HCN: cyanohydrin, so a carbonyl group
   • Bromine water: gluconic acid, so an aldehyde group
   • Acetic anhydride: pentaacetate, so five OH groups
   • Nitric acid: saccharic acid, so a CH₂OH group
3. 3 · Draw the Fischer projection. CHO at the top; OH right, left, right, right on C2 to C5; CH₂OH at the bottom.
4. 4 · Give the limits of the open chain.
   • No Schiff's test, 2,4-DNP test or NaHSO₃ product
   • The pentaacetate does not react with NH₂OH
   • Two crystalline forms: α (419 K) and β (423 K)
5. 5 · Explain the cyclic form. The C5 OH adds to CHO, giving a six-membered cyclic hemiacetal (pyranose). C1 becomes the anomeric carbon, giving α and β anomers.
6. 6 · Link to real life. Dextrose drips in hospitals are solutions of D-glucose.
7. 7 · Close in one line. So glucose reacts as an open-chain aldehyde in some reactions but exists mainly in the cyclic form.

Scoring tip: Write one equation and one conclusion per line. The conclusion ('so a straight chain') is what earns the mark.