Chemistry Reference
In-Depth Information
12.41
We must use Beer's law (eq. 12.5) to solve this problem.
A
=
ε
cl
, or
c
=
A
/
ε
l
c
= 0.65 / 224 c 1 = 2.90 x 10
-3
mol/L
To go further, 1 L methylcyclohexane will contain 2.90 x 10
-3
x 98 = 0.284 g toluene
as a contaminant. As you can see, ultraviolet spectroscopy can be a very sensitive
tool for detecting impurities.
12.42
Follow the example in eq. 12.7.
+
[
CH
3
CH
2
CH
2
CH
2
CH
2
O
H]
12.43
The molecular ion of 1-butanol will be:
2
1
+
[
CH
3
CH
2
CH
2
CH
2
OH]
Fragmentation between C-1 and C-2 would give a daughter ion with
m/z
= 31
(compare with eq. 12.8). A possible mechanism for this cleavage is:
+
+
H
H
CH
3
CH
2
CH
2
C
O
H
CH
3
CH
2
CH
2
+
C
O
H
H
H
m/z
= 31
12.44
The band at 1725 cm
-1
in the infrared spectrum is due to a carbonyl group, probably
a ketone. The quartet-triplet pattern in the NMR spectrum suggests an ethyl group.
The compound is 3-pentanone:
δ
2.7, quartet, area = 2 (or 4H)
O
H
3
C
CH
2
C
CH
2
CH
3
δ
0.9, triplet, area = 3 (or 6H)
12.45
The quartet-triplet pattern suggests that the ten protons are present as two ethyl groups.
This gives a partial structure of (CH
3
CH
2
)
2
CO
3
. The chemical shift of the CH
2
groups
(
δ
4.15) suggests that they are attached to the oxygen atoms. Finally, the infrared
band at 1745 cm
-1
suggests a carbonyl function. The structure is diethyl carbonate:
O
H
3
C
CH
2
O
C
O
CH
2
CH
3
12.46
The molecular formula, C
4
H
10
O, tells us that compound
A
is saturated. The fact that
A
can be oxidized with PCC to an aldehyde (note that
B
gives a positive Tollens'
test) tells us that
A
is a primary alcohol. We can draw only two structures for
A
that
are consistent with these data:
CH
3
CH
2
CH
2
CH
2
OH (1-butanol) and (CH
3
)
2
CHCH
2
OH (2-methyl-1-propanol)
Of these, only 2-methyl-1-propanol is consistent with the
1
H NMR spectrum shown in
Figure 12.1. The 6H signal at
δ
0.9 is due to the two equivalent methyl groups, the
heptet at
δ
1.7 is due to the CH, the 2H doublet at
δ
3.35 is due to the CH
2
adjacent
