Chemistry Reference
In-Depth Information
c
x
is the concentration of substance
x
(%w/v)
c
y
is the concentration of substance
y
(%w/v)
l
is the path length in cm
Using the data obtained above and since, in this example, path
length
1 cm,
Total absorbance at 252 nm, 0.733
959
c
x
570
c
y
(equation 1)
Total absorbance at 280 nm, 0.340
136
c
x
449
c
y
(equation 2)
Multiplying equation (1) by 136 and equation (2) by 959 yields
99.688
130 424
c
x
77 520
c
y
(equation 3)
326.06
130 424
c
x
430 591
c
y
(equation 4)
Subtraction of equation (3) from equation (4) gives
226.372
353 071
c
y
Therefore
c
y
0.00064% w/v
Substituting this value of
c
y
into equation (1) gives
c
x
0.00039% w/v
Q
4 Determine the structure of the unknown compound
from the spectral data shown in Figure 7.27.
A
4 There are many ways to approach a structure elucidation
problem of this type and as long as the student solves the structure
correctly (and explains the assignments made) marks will be
awarded. This example makes greatest use of NMR data to solve
the structure but it is equally acceptable to use MS or IR data to
solve the unknown.
The IR spectrum is fairly straightforward (suggesting a
simple molecule) with a strong peak at 1725 cm
-1
. This peak is due
to a carbonyl group.
