write the open chain structure of glucose.

Explanation

(1) Elemental analysis and molecular weight determination show that molecular formula of glucose is `C_(6)H_(12)O_(6)`.
(2) Complete reduction with Hydrogen Iodide and Red Phosphorus gives n-hexane,which indicates that carbon atoms in glucose are in straight chain.
(3) Glucose reacts with acetic anhydride to form penta acetyl derivative which shows the presence of 5-OH groups in glucose.
  (4) Glucose reacts with hydroxylamine to form an glucose oxime which indicates the presence of > C = 0 (keto) group in glucose.
(5) On mild oxidation with bromine water glucose is converted to gluconic acid, when reduced with excess of Hl yields n-hexanoic acid. `C_(5)H_(11)O_(5)CHOunderset(fol)to C_(5)H_(11)O_(5)COOHunderset(11)toCH_(3) (CH_(2))_(4) COOH`
gluconic acid n-hexanoic acid
This clearly shows that glucose contains 6 carbon atoms in a straight chain with -CHO at one end, which gives -COOH on oxidation.
(6) Based on the above reaction glucose open - chain structure can be given as follows:

From the above structure the IUPAC name of glucose is 2,3,4,5,6 Pentahydroxyhexanal.