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ORGANIC CHEMISTRY-WILEYPLUS NEXTGEN
4th Edition
ISBN: 9781119760924
Author: Klein
Publisher: WILEY
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Question
Chapter 24.7, Problem 39PTS
(a)
Interpretation Introduction
Interpretation:
The each given disaccharides are to be identified whether it is reducing sugar or not.
Concept introduction:
For the carbohydrates,
- Anomeric position is carried by an OH group for Hemiacetals.
- Anomeric position is carried by an alkoxy group for acetals.
- Hemiacetal sugars are in equilibrium with their open chain form.
- Acetal sugars are not in equilibrium with their open chain form.
- Reducing sugar: Sugars that contain
aldehyde or a freeketone groups that are oxidized tocarboxylic acids are classified as reducing sugars. So, the open chain form of sugars can reduce certain oxidizing agents (Tollens’, Fehling’s and Benedict’s) and become carboxylic acids.
To identify: the given compound (a) is a reducing sugar or not.
(b)
Interpretation Introduction
Interpretation:
The each given disaccharides are to be identified whether it is reducing sugar or not.
Concept introduction:
For the carbohydrates,
- Anomeric position is carried by an OH group for Hemiacetals.
- Anomeric position is carried by an alkoxy group for acetals.
- Hemiacetal sugars are in equilibrium with their open chain form.
- Acetal sugars are not in equilibrium with their open chain form.
- Reducing sugar: Sugars that contain aldehyde or a free ketone groups that are oxidized to carboxylic acids are classified as reducing sugars. So, the open chain form of sugars can reduce certain oxidizing agents (Tollens’, Fehling’s and Benedict’s) and become carboxylic acids.
(c)
Interpretation Introduction
Interpretation:
The each given disaccharides are to be identified whether it is reducing sugar or not.
Concept introduction:
For the carbohydrates,
- Anomeric position is carried by an OH group for Hemiacetals.
- Anomeric position is carried by an alkoxy group for acetals.
- Hemiacetal sugars are in equilibrium with their open chain form.
- Acetal sugars are not in equilibrium with their open chain form.
- Reducing sugar: Sugars that contain aldehyde or a free ketone groups that are oxidized to carboxylic acids are classified as reducing sugars. So, the open chain form of sugars can reduce certain oxidizing agents (Tollens’, Fehling’s and Benedict’s) and become carboxylic acids.
Expert Solution & Answer
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Students have asked these similar questions
From the given compound, choose the proton that best fits each given description.
a
CH2
CH 2
Cl
b
с
CH2
F
Most shielded:
(Choose one)
Least shielded:
(Choose one)
Highest chemical shift:
(Choose one)
Lowest chemical shift:
(Choose one)
×
Consider this molecule:
How many H atoms are in this molecule?
How many different signals could be found in its 1H NMR spectrum?
Note: A multiplet is considered one signal.
For each of the given mass spectrum data, identify whether the compound contains chlorine, bromine, or neither.
Compound
m/z of M* peak
m/z of M
+ 2 peak
ratio of M+ : M
+ 2 peak
Which element is present?
A
122
no M
+ 2 peak
not applicable
(Choose one)
B
78
80
3:1
(Choose one)
C
227
229
1:1
(Choose one)
Chapter 24 Solutions
ORGANIC CHEMISTRY-WILEYPLUS NEXTGEN
Ch. 24.2 - Prob. 1CCCh. 24.2 - Prob. 2CCCh. 24.2 - Prob. 3CCCh. 24.2 - Prob. 4CCCh. 24.2 - Prob. 5CCCh. 24.2 - Prob. 6CCCh. 24.4 - Prob. 7CCCh. 24.4 - Prob. 8CCCh. 24.5 - Prob. 1LTSCh. 24.5 - Prob. 9PTS
Ch. 24.5 - Prob. 2LTSCh. 24.5 - Prob. 12PTSCh. 24.5 - Prob. 13PTSCh. 24.5 - Prob. 3LTSCh. 24.5 - Prob. 16PTSCh. 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. 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. 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. 86IPCh. 24 - Prob. 87IPCh. 24 - Prob. 88IPCh. 24 - Prob. 89IP
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- Don'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_forwardPLEASE ANSWER BOTH i) and ii) !!!!arrow_forward
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