Organic Chemistry (8th Edition)
8th Edition
ISBN: 9780134042282
Author: Paula Yurkanis Bruice
Publisher: PEARSON
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Chapter 20, Problem 53P
Interpretation Introduction
Interpretation:
The
Concept Introduction:
The Wohl degradation is the opposite of Kiliani Fischer synthesis which decreases the aldose chain by one carbon atom that is hexoses having ix carbons are converted to pentoses having five carbon and pentoses having five carbons are converted to tetroses having four carbons.
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Using the conditions of spontaneity to deduce the signs of AH and AS
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
This reaction is always spontaneous, but
proceeds slower at temperatures above
120. °C.
ΔΗ is
(pick one)
AS is
(pick one)
ΔΗ is
(pick one)
B
This reaction is spontaneous except above
117. °C.
AS is
(pick one)
ΔΗ is
(pick one)
This reaction is slower below 20. °C than
C
above.
AS is
|(pick one)
?
18
Ar
1
Calculating the pH at equivalence of a titration
Try Again
Your answer is incorrect.
0/5
a
A chemist titrates 70.0 mL of a 0.7089 M hydrocyanic acid (HCN) solution with 0.4574M KOH solution at 25 °C. Calculate the pH at equivalence. The pK of
hydrocyanic acid is 9.21.
Round your answer to 2 decimal places.
Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of KOH solution added.
pH
=
11.43]
G
00.
18
Ar
B•
Biological Macromolecules
Naming and drawing the products of aldose oxidation and reduction
aw a Fischer projection of the molecule that would produce L-ribonic acid if it were subjected to mildly oxidizing reaction conditions.
Click and drag to start drawing a
structure.
X
AP
‡
1/5
Naor
Explanation
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Chapter 20 Solutions
Organic Chemistry (8th Edition)
Ch. 20.1 - Prob. 1PCh. 20.2 - Prob. 2PCh. 20.2 - Prob. 3PCh. 20.3 - Prob. 4PCh. 20.3 - Prob. 5PCh. 20.3 - Prob. 6PCh. 20.4 - Prob. 7PCh. 20.4 - Prob. 8PCh. 20.5 - Prob. 9PCh. 20.5 - Prob. 10P
Ch. 20.5 - Prob. 11PCh. 20.6 - Prob. 12PCh. 20.6 - Prob. 13PCh. 20.6 - Prob. 14PCh. 20.7 - Prob. 15PCh. 20.8 - Prob. 16PCh. 20.9 - Prob. 18PCh. 20.10 - Prob. 20PCh. 20.10 - Prob. 21PCh. 20.10 - Prob. 22PCh. 20.11 - Prob. 23PCh. 20.11 - Prob. 24PCh. 20.12 - Prob. 25PCh. 20.12 - Prob. 26PCh. 20.14 - Prob. 28PCh. 20.15 - Prob. 29PCh. 20.15 - Prob. 30PCh. 20.16 - Prob. 31PCh. 20.17 - Prob. 32PCh. 20.18 - Refer to Figure 20.5 to answer the following...Ch. 20 - Prob. 34PCh. 20 - Prob. 35PCh. 20 - Prob. 36PCh. 20 - Prob. 37PCh. 20 - Prob. 38PCh. 20 - Prob. 39PCh. 20 - Prob. 40PCh. 20 - Prob. 41PCh. 20 - Prob. 42PCh. 20 - Prob. 43PCh. 20 - Prob. 44PCh. 20 - Prob. 45PCh. 20 - Prob. 46PCh. 20 - Prob. 47PCh. 20 - Prob. 48PCh. 20 - The 1H NMR spectrum of D-glucose in D2O exhibits...Ch. 20 - Prob. 50PCh. 20 - Prob. 51PCh. 20 - Prob. 52PCh. 20 - Prob. 53PCh. 20 - Prob. 54PCh. 20 - Prob. 55PCh. 20 - Prob. 56PCh. 20 - Prob. 57PCh. 20 - Prob. 58PCh. 20 - Prob. 59PCh. 20 - Prob. 60PCh. 20 - Prob. 61PCh. 20 - A hexose is obtained when the residue of a shrub...Ch. 20 - Prob. 63PCh. 20 - Prob. 64PCh. 20 - Prob. 65PCh. 20 - Prob. 66PCh. 20 - Prob. 67PCh. 20 - Prob. 68PCh. 20 - Prob. 69PCh. 20 - Prob. 70PCh. 20 - Prob. 71PCh. 20 - Prob. 72PCh. 20 - Prob. 73P
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