One test for glucose involves Fehling's solution. Circle the part of the glucose molecule that reacts with Fehling's solution. In solution, sugar molecules exist in an equilibrium in straight-chain and ring forms. To change from the straight-chain form to the ring form, the oxygen of the hydroxyl on carbon number 5 joins to the carbonyl carbon. This is shown below for glucose. Draw the structure of a ring form for fructose.

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One test for glucose involves Fehling's solution - Scottish Highers Chemistry - Question 13 - 2015

Question 13

One test for glucose involves Fehling's solution. Circle the part of the glucose molecule that reacts with Fehling's solution. In solution, sugar molecules exist i... show full transcript

Worked Solution & Example Answer:One test for glucose involves Fehling's solution - Scottish Highers Chemistry - Question 13 - 2015

Step 1

Circle the part of the glucose molecule that reacts with Fehling's solution.

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Answer

The part of the glucose molecule that reacts with Fehling's solution is the aldehyde group, which is located at the first carbon (C1). This can be identified by the presence of the -CHO functional group.

Step 2

Draw the structure of a ring form for fructose.

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Answer

The ring form of fructose is derived from the straight-chain structure by the reaction of the hydroxyl group on carbon 5 with the carbonyl (C2) to form a five-membered ring, known as a furanose. The correct structure is as follows:

       H    OH
        \  /  
         C      
        / \    
       O   C
      /     \  
  HO-C      CH2OH
      | 
     CH2OH

Ensure the stereochemistry is correct by placing the hydroxyl (-OH) and the hydrogen (H) in the proper positions based on the original straight-chain structure.

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