Answer
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Hint: The various tests given, cause the oxidation of the glucose compound having the aldehyde and the alcohol group. So, the reaction we can determine whether the oxidation of both the groups is possible or not.
Complete step by step answer:
-The glucose is a monosaccharide carbohydrate. In its Fischer projection, with the six carbon atoms aligned on the vertical line and aldehydic group on C-1 followed by hydroxyl group on the remaining five carbons such that in C-3 the hydroxyl group is on the other side with respect to the hydroxyl group on the other carbon atoms.
-In presence of nitric acid, the oxidation reaction of the glucose takes place. The $HN{{O}_{3}}$ being an oxidising agent. As the nitrogen atom present is in its highest oxidation state of (+5). It can be reduced to form nitrous acid, $HN{{O}_{2}}$ having the nitrogen atom in (+3) oxidation state. The nitric acid being a strong oxidising agent, it can oxidise the alcohol and the aldehyde. Thus, in the glucose the C-1 and C-6 have the aldehydic and the primary alcohol to oxidise into carboxylic acid forming the saccharic acid compound.
Note: This reaction indicates that the primary alcohol is present in the glucose compound. Also, the bromine water only oxidises the aldehydic group, being a mild oxidising agent. The Fehling’s solution used to detect the reducing sugars, in which the aldehydic group oxidised to form gluconic acid and cuprous oxide on reaction. Similarly, in case of the alkaline solution of iodine which oxidises the aldehydic group to carboxylic acid along with sodium iodide and water molecule during the reaction.
Complete step by step answer:
-The glucose is a monosaccharide carbohydrate. In its Fischer projection, with the six carbon atoms aligned on the vertical line and aldehydic group on C-1 followed by hydroxyl group on the remaining five carbons such that in C-3 the hydroxyl group is on the other side with respect to the hydroxyl group on the other carbon atoms.
-In presence of nitric acid, the oxidation reaction of the glucose takes place. The $HN{{O}_{3}}$ being an oxidising agent. As the nitrogen atom present is in its highest oxidation state of (+5). It can be reduced to form nitrous acid, $HN{{O}_{2}}$ having the nitrogen atom in (+3) oxidation state. The nitric acid being a strong oxidising agent, it can oxidise the alcohol and the aldehyde. Thus, in the glucose the C-1 and C-6 have the aldehydic and the primary alcohol to oxidise into carboxylic acid forming the saccharic acid compound.
Note: This reaction indicates that the primary alcohol is present in the glucose compound. Also, the bromine water only oxidises the aldehydic group, being a mild oxidising agent. The Fehling’s solution used to detect the reducing sugars, in which the aldehydic group oxidised to form gluconic acid and cuprous oxide on reaction. Similarly, in case of the alkaline solution of iodine which oxidises the aldehydic group to carboxylic acid along with sodium iodide and water molecule during the reaction.
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