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Chapter 13 443
Copyright © 2017 Pearson Education, Inc.
35.
A
=
c
l
e
c
=
A
l
e
e
=
0.40
4.0
*
10
-
5
=
10,000 M
-
1
cm
-
1
36.
The compound has the same chromophore as methyl vinyl ketone. So it must have approximately the same
value for its
l
max
, which is 219 nm.
37.
a.
>
CH CH
>
>
CH CH
2
b.
N(CH
3
)
2
>
N(CH
3
)
3
>
+
N(CH
3
)
2
>
N(CH
3
)
2
38.
a.
Blue results from absorption of light that has a longer wavelength than the light that produces purple
when absorbed. The compound on the right has two N(CH
3
)
2
auxochromes that cause it to absorb light
with a longer wavelength than the compound on the left, which has only one N(CH
3
)
2
auxochrome.
Therefore, the compound on the right is the blue compound.
b.
They will be the same color at pH
=
3 because the N(CH
3
)
2
groups will be protonated and, therefore,
will not possess the lone pair that causes the compound to absorb light of a longer wavelength.
39.
NADH is formed as a product; it absorbs light at 340 nm. Therefore, the rate of the oxidation reaction can
be determined by monitoring the increase in absorbance at 340 nm as a function of time.
CH
3
CH
2
OH
+
NAD
+
alcohol dehydrogenase
+
NADH
H
+
+
H
CH
3
C
O
40.
The Henderson–Hasselbalch equation (Section 2.10) shows that when the pH of the solution equals the p
K
a
of the compound, the concentration of the species compound in the acidic form is the same as the concen-
tration of the species compound in the basic form.
From the data given, we see that the absorbance of the acid is 0. We also see that the absorbance ceases
to increase with increasing pH after the absorbance reaches 1.60. That means that all of the compound is
in the basic form when the absorbance is 1.60. Therefore, when the absorbance is half of 1.60 (or 0.80),
half of the compound is in the basic form and half is in the acidic form; in other words, the concentration
in the acidic form is the same as the concentration in the basic form. We see that the absorbance is 0.80 at
pH
=
5.0. Therefore, the p
K
a
of the compound is 5.0.
41.
The molecular ion peak for these compounds is
m
>
z
=
86; the peak at
m
>
z
=
57 is due to loss of an ethyl
radical (86
-
29), and the peak at
m
>
z
=
71 is due to loss of a methyl radical (86
-
15).