
(a)
Interpretation:
Each of the given compounds should be classified based on whether it is aldose or ketose and the numbers of carbon atoms presented in it (example: aldopentose).
Concept introduction:
- Carbohydrates are classified into two based on the presence of either
aldehyde group orketone group, that is aldose and ketose. - Monosaccharides are simple sugars or carbohydrates.
- monosaccharides are classified on the basis of,
- Presents of aldehyde or ketone group such as aldo or keto.
- Number of carbon atoms presented in a carbohydrates, triose (3 carbon), tetrose (4 carbon), pentose (5 carbon), hexose (6 carbon) or heptose (7 carbon) (ose indicates the carbohydrates).
To classify: the given compound (a) based on whether it is aldose or ketose and the numbers of carbon atoms presented in it.
(b)
Interpretation:
Each of the given compounds should be classified based on whether it is aldose or ketose and the numbers of carbon atoms presented in it (example: aldopentose).
Concept introduction:
- Carbohydrates are classified into two based on the presence of either aldehyde group or ketone group, that is aldose and ketose.
- Monosaccharides are simple sugars or carbohydrates.
- monosaccharides are classified on the basis of,
- Presents of aldehyde or ketone group such as aldo or keto.
- Number of carbon atoms presented in a carbohydrates, triose (3 carbon), tetrose (4 carbon), pentose (5 carbon), hexose (6 carbon) or heptose (7 carbon) (ose indicates the carbohydrates).
To classify: the given compound (b) based on whether it is aldose or ketose and the numbers of carbon atoms presented in it.
(c)
Interpretation:
Each of the given compounds should be classified based on whether it is aldose or ketose and the numbers of carbon atoms presented in it (example: aldopentose).
Concept introduction:
- Carbohydrates are classified into two based on the presence of either aldehyde group or ketone group, that is aldose and ketose.
- Monosaccharides are simple sugars or carbohydrates.
- monosaccharides are classified on the basis of,
- Presents of aldehyde or ketone group such as aldo or keto.
- Number of carbon atoms presented in a carbohydrates, triose (3 carbon), tetrose (4 carbon), pentose (5 carbon), hexose (6 carbon) or heptose (7 carbon) (ose indicates the carbohydrates).
To classify: the given compound (c) based on whether it is aldose or ketose and the numbers of carbon atoms presented in it.
The given compound is Ketopentose.
(d)
Interpretation:
Each of the given compounds should be classified based on whether it is aldose or ketose and the numbers of carbon atoms presented in it (example: aldopentose).
Concept introduction:
- Carbohydrates are classified into two based on the presence of either aldehyde group or ketone group, that is aldose and ketose.
- Monosaccharides are simple sugars or carbohydrates.
- monosaccharides are classified on the basis of,
- Presents of aldehyde or ketone group such as aldo or keto.
- Number of carbon atoms presented in a carbohydrates, triose (3 carbon), tetrose (4 carbon), pentose (5 carbon), hexose (6 carbon) or heptose (7 carbon) (ose indicates the carbohydrates).
To classify: the given compound (d) based on whether it is aldose or ketose and the numbers of carbon atoms presented in it.
(e)
Interpretation:
Each of the given compounds should be classified based on whether it is aldose or ketose and the numbers of carbon atoms presented in it (example: aldopentose).
Concept introduction:
- Carbohydrates are classified into two based on the presence of either aldehyde group or ketone group, that is aldose and ketose.
- Monosaccharides are simple sugars or carbohydrates.
- monosaccharides are classified on the basis of,
- Presents of aldehyde or ketone group such as aldo or keto.
- Number of carbon atoms presented in a carbohydrates, triose (3 carbon), tetrose (4 carbon), pentose (5 carbon), hexose (6 carbon) or heptose (7 carbon) (ose indicates the carbohydrates).
To classify: the given compound (e) based on whether it is aldose or ketose and the numbers of carbon atoms presented in it.

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Chapter 24 Solutions
ORGANIC CHEMISTRY-PRINT (LL)-W/WILEY
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