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Give the suitable chemical tests that defines the open chain structure of glucose. |
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Answer» Solution :(i) The molecular formula of glucose is `C_(6)H_(12)O_(6)`. (ii) On prolonged reaction with HI, it forms n-hexane, suggesting that all the six CARBONS of glucose are linked in a straight chain. `{:(""CHO),("|"),(" "(CHOH)_(4)),("|"),(""CH_(2)OH):}overset(HI,Delta)rarrunderset("(n - Hexane)")(CH_(3)-CH_(2)-CH_(2)-CH_(2)-CH_(2)-CH_(3))` (iii) Glucose reacts with hydroxyl amine to form an oxime and adds a molecule of hydrogen cyanide to form cyanohydrin. These reactions confirm the presence of carbonyl  group in the glucose.  (iv) Glucose gets oxidized to six carbon carboxylic acid (gluconic acid) on reaction with a MILD oxidizing agent like bromine water. This indicates that carbonyl group present is the aldehydic group. `{:(""CHO),("|"),(" "(CHOH)_(4)),("|"),(""CH_(2)OH):}overset(Br_(4)" Water")rarr{:(""COOH),("|"),(" "(CHOH)_(4)),("|"),(underset("Gluconic acid")(""CH_(2)OH)):}` (v) Acetylation of glucose with acetic anhydride gives glucose pentaacetate with confirms the presence of five -OH groups. Since it exists as a stable COMPOUND, the five -OH groups should be attached to the different carbon atoms.  (vi) When oxidized by nitric acid, glucose as well as gluconic acid yields dicarboxylic acid, saccharic acid. This indicates the presence of primary alcoholic `(-OH)` group in glucose.  The exact spatial arrangement of different -OH group was given by Fischer after studying many other properties. Its configuration is correctly represented as I. So, the gluconic acid is represented as II and saccharic acid as III. 
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