Organic Chemistry, Binder Ready Version
Organic Chemistry, Binder Ready Version
2nd Edition
ISBN: 9781118454312
Author: David R. Klein
Publisher: WILEY
Question
Book Icon
Chapter 24.2, Problem 1CC

(a)

Interpretation Introduction

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 (a) based on whether it is aldose or ketose and the numbers of carbon atoms presented in it.

(b)

Interpretation Introduction

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 Introduction

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 Introduction

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 Introduction

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.

Blurred answer
Students have asked these similar questions
Use the observations about each chemical reaction in the table below to decide the sign (positive or negative) of the reaction enthalpy AH and reaction entropy AS. Note: if you have not been given enough information to decide a sign, select the "unknown" option. reaction observations conclusions A The reverse of this reaction is always spontaneous but proceeds slower at temperatures below 41. °C. ΔΗ is (pick one) AS is (pick one) ΔΗ is (pick one) B This reaction is spontaneous except above 94. °C. AS is (pick one) This reaction is always spontaneous, but ΔΗ is (pick one) C proceeds slower at temperatures below −14. °C. AS is (pick one) Х 00. 18 Ar 무ㅎ B 1 1
Draw the product of the reaction shown below. Ignore inorganic byproducts. + H CH3CH2OH HCI Drawing
please explain this in simple terms

Chapter 24 Solutions

Organic Chemistry, Binder Ready Version

Ch. 24.5 - Prob. 10ATSCh. 24.5 - Prob. 11ATSCh. 24.5 - Prob. 2LTSCh. 24.5 - Prob. 12LTSCh. 24.5 - Prob. 13ATSCh. 24.5 - Prob. 14ATSCh. 24.5 - Prob. 15ATSCh. 24.5 - Prob. 3LTSCh. 24.5 - Prob. 16PTSCh. 24.5 - Prob. 17ATSCh. 24.5 - Prob. 18ATSCh. 24.5 - Prob. 19CCCh. 24.5 - Prob. 20CCCh. 24.5 - Prob. 21CCCh. 24.5 - Prob. 22CCCh. 24.6 - Prob. 23CCCh. 24.6 - Prob. 24CCCh. 24.6 - Prob. 25CCCh. 24.6 - Prob. 26CCCh. 24.6 - Prob. 27CCCh. 24.6 - Prob. 28CCCh. 24.6 - Prob. 29CCCh. 24.6 - Prob. 30CCCh. 24.6 - Prob. 4LTSCh. 24.6 - Prob. 31PTSCh. 24.6 - Prob. 32ATSCh. 24.6 - Prob. 33ATSCh. 24.6 - Prob. 34CCCh. 24.6 - Prob. 35CCCh. 24.6 - Prob. 36CCCh. 24.6 - Prob. 37CCCh. 24.6 - Prob. 38CCCh. 24.7 - Prob. 5LTSCh. 24.7 - Prob. 39PTSCh. 24.7 - Prob. 40ATSCh. 24.7 - Prob. 41CCCh. 24 - Prob. 42PPCh. 24 - Prob. 43PPCh. 24 - Prob. 44PPCh. 24 - Prob. 45PPCh. 24 - Prob. 46PPCh. 24 - Prob. 47PPCh. 24 - Prob. 48PPCh. 24 - Prob. 49PPCh. 24 - Prob. 50PPCh. 24 - Prob. 51PPCh. 24 - Prob. 52PPCh. 24 - Prob. 53PPCh. 24 - Prob. 54PPCh. 24 - Prob. 55PPCh. 24 - Prob. 56PPCh. 24 - Prob. 57PPCh. 24 - Prob. 58PPCh. 24 - Prob. 59PPCh. 24 - Prob. 60PPCh. 24 - Prob. 61PPCh. 24 - Prob. 62PPCh. 24 - Prob. 63PPCh. 24 - Prob. 64PPCh. 24 - Prob. 65PPCh. 24 - Prob. 66PPCh. 24 - Prob. 67PPCh. 24 - Prob. 68PPCh. 24 - Prob. 69PPCh. 24 - Prob. 70PPCh. 24 - Prob. 71PPCh. 24 - Prob. 72PPCh. 24 - Prob. 73PPCh. 24 - Prob. 74PPCh. 24 - Prob. 75PPCh. 24 - Prob. 76PPCh. 24 - Prob. 77PPCh. 24 - Prob. 78PPCh. 24 - Prob. 79PPCh. 24 - Prob. 80PPCh. 24 - Prob. 81IPCh. 24 - Prob. 82IPCh. 24 - Prob. 83IPCh. 24 - Prob. 84IP
Knowledge Booster
Background pattern image
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY