The patient thrives on a diet of fruit jam, bread, pasta, and coffee. She exercises intermittently. The healthcare provider recommends that she take the Benedict's test. Assume that the glucose...

The patient thrives on a diet of fruit jam, bread, pasta, and coffee. She exercises intermittently. The healthcare provider recommends that she take the Benedict's test. Assume that the glucose levels of the patient are high. Based on your understanding about this scenario, respond to the following: State the results that the test would indicate (specify the color of the solution). State the composition and the properties of the ketohexose derived from fruit jam. Describe the manner in which ketohexose acts as a reducing sugar in the test.

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llltkl | College Teacher | (Level 3) Valedictorian

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The Benedict's test allows us to detect the presence of reducing sugars (sugars with a free aldehyde or ketone group). The copper sulfate (CuSO4) present in Benedict's solution reacts with electrons from the aldehyde or ketone group of the reducing sugar to form cuprous oxide (Cu2O), a red-brown precipitate, which is insoluble in water.

CuSO4 → Cu^(2+) + SO4^(2-)

2 Cu^(2+)+Reducing Sugar → Cu^(+)
        (electron donor)

Cu^(+) → Cu2O (precipitate)

The final color of the solution depends on how much of this precipitate was formed, and therefore the color gives an indication of how much reducing sugar was present. The color of the solution changes in the order of blue (with no glucose present), green, yellow, orange, red, and then brick red or brown (with high glucose present).

Here, the patient has high glucose, so the test would indicate a high level of reducing sugar in the urine, with the solution turning orange or red/brown.

A ketohexose is a ketone-containing hexose (a six-carbon monosaccharide). In other words, it is a monosaccharide which has a 6 carbon backbone with a ketone group on C2. They have 3 chiral centers, so 8 different stereo isomers are possible.

When the hemi-acetal or ketal hydroxyl group is free, i.e. it is not locked, not linked to another (sugar) molecule, the aldehyde (or keto-) form (i.e. the chain-form) is available for reducing copper (II) ions. When a sugar is oxidized, its carbonyl group (i.e. aldehyde or ketone group) is converted to a carboxyl group.

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