(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-STUD.SOLNS.MAN+SG(LL)
- How can you identify Birch reduction outcomes? What are you looking for to determine the finalproduct?arrow_forwardDon't used hand raiting and don't used Ai solutionarrow_forward2' P17E.6 The oxidation of NO to NO 2 2 NO(g) + O2(g) → 2NO2(g), proceeds by the following mechanism: NO + NO → N₂O₂ k₁ N2O2 NO NO K = N2O2 + O2 → NO2 + NO₂ Ко Verify that application of the steady-state approximation to the intermediate N2O2 results in the rate law d[NO₂] _ 2kk₁[NO][O₂] = dt k+k₁₂[O₂]arrow_forward
- PLEASE ANSWER BOTH i) and ii) !!!!arrow_forwardE17E.2(a) The following mechanism has been proposed for the decomposition of ozone in the atmosphere: 03 → 0₂+0 k₁ O₁₂+0 → 03 K →> 2 k₁ Show that if the third step is rate limiting, then the rate law for the decomposition of O3 is second-order in O3 and of order −1 in O̟.arrow_forward10.arrow_forward
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