Question

Which of the following best describes the below molecule?

CH₂OH
|
C=O
|
OH-C-H
|
H -C- OH
|
H -C- OH
|
CH₂OH

A. L-Ketose
B. D-lactose
C. L-glucose
D. D-fructose
E. D-Aldose

Answer 1

The given molecule is identified as D-fructose due to its configuration of the hydroxyl group on the right side of the last chiral carbon in the Fisher projection and the presence of a ketone group at the second carbon. D. D-fructose

Step-by-step explanation:

The molecule in question is best described as D-fructose. When determining whether a sugar is a D- or L- isomer, we look at the orientation of the hydroxyl group (OH) on the last chiral carbon in the Fisher projection. In the case of this molecule, the OH group on the bottom-most chiral carbon (the penultimate carbon) is on the right side, which classifies it as a D-sugar.

Furthermore, the presence of the ketone group (C=O) at the second carbon (C2) identifies it as a ketose, rather than an aldose which would have an aldehyde group at the first carbon (C1). Therefore, since the molecule also contains six carbon atoms (hexose), it can be identified as D-fructose, a ketohexose where the C=O is at C2, differentiating it from D-glucose which has an aldehyde group at C1.

Answer 2

The correct classification for the molecule described by the student is D-fructose, a ketohexose, due to the ketone functional group at the second carbon and the orientation of the hydroxyl groups according to the given Fisher projection.

Step-by-step explanation:

The molecule described by the student is a monosaccharide, which can be classified based on the functional group and the arrangement of the hydroxyl (-OH) groups. The options given include various types of monosaccharides: L-Ketose, D-lactose, L-glucose, D-fructose, and D-Aldose. To determine the correct classification, we must analyze the placement of these groups in the molecule's structure, specifically the orientation of the -OH group on the penultimate (second-to-last) chiral carbon in the Fischer projection.

From the information provided, we know that if the -OH group is on the right for the penultimate carbon in a Fischer projection, the molecule is designated as a D-form. The molecule described has a ketone functional group (C=O) in the middle of the chain, which disqualifies it from being an aldose, thereby eliminating options A, C, and E. Considering that D-lactose is a disaccharide composed of glucose and galactose and not a monosaccharide, option B is also incorrect. This leaves us with D-fructose, which is a ketohexose and fits the description provided by the student, as it has the ketone functional group at the second carbon and the mentioned orientation of -OH groups.