Part 8: SPECTROSCOPY
The following data is available from the
question.
Note
: Remember to cross
reference things to confirm.... e.g.
IR may show C=C, use NMR to
confirm that etc.
MS: M+
seen at 226 g/mol. There is a 1:1 m,m+2, indicating that 1 x Br is present.
IR:
There is a strong absorption at about 1680 cm-1 which is probably C=O (note that's low for a ketone) and at about 1600 cm-1 which is probably C=C. There are no other absorptions of note.... no OH, NH, C≡C or C≡N
13C NMR:
The proton decoupled spectrum shows a total of 8 peaks indicating 8 types of
C (don't ignore the expansion). By analysis of the chemical shifts, we have a peak at 197 ppm (C=O in the aldehyde or ketone range). There are 4 peaks between 155-110 ppm that are sp2 aromatic C, and then 3 peaks between 0-50 ppm that are
most likely from an sp3 hydrocarbon portion.
1H NMR:
The proton spectrum shows a total of 5 peaks indicating 5 types of H.
|
d/ppm
|
multiplicity
|
integration
|
Inference
|
| 7.8 |
d |
2 |
2 x ArH |
| 7.3 |
d |
2 |
2 x ArH |
|
3.7
|
t
|
2
|
CH2 coupled to 2H, deshielded |
| 3.1 |
t
|
2
|
CH2 coupled to 2H, slightly deshielded |
|
2.5
|
s
|
3
|
CH3 not coupled, slightly deshielded
|
(q = quartet, t = triplet, d = doublet, s = singlet)
The most significant structural information
from this are:
- adding integrals shows 11H (or a multiple of)
- since there are 4 ArH, it is a disubstituted benzene, C6H4
- there is a CH3 group
- there is a -CH2CH2- group
Summary....
The MS indicated MW = 226, Br.
The IR showed the presence of C=O
13C NMR shows 7 C types including C=O and 4 ArC.
H NMR gives disubstituted benzene, C6H4, -CH3 and -CH2CH2- groups
This information suggests a molecular formula = C10H11BrO
which matches the MW = 226 and has an IHD = 5 consistent with the C=O and the benzene system.
Altogether...
|
The fragments we have are:
The IR, 13C and H-NMR imply that the C=O carbonyl is a ketone. Since the IR frequency (1680cm-1) is a little low for a simple ketone (1715 cm-1), this suggests that it is a conjugated ketone which means it must be attached to the benzene ring.
The H-NMR data indicates that the benzene ring is disubstituted (since there are only 4 ArH present from the integration).
Since the 13C-NMR tells us that there are 4 types of ArC and the H-NMR tells us that there are 2 types of ArH, then the benzene must be para (i.e. 1,4-) substituted (ortho & meta would have 6 types of ArC).
NMR chemical shifts (and MS) are most consistent with the methyl ketone rather than the methyl on the benzene ring. |
|
The final step should always be to check what you
have drawn. The easiest thing to check is usually the coupling patterns you
would expect to see, and the chemical shifts of each unit. You should
be asking yourself : "Does my answer give me what the H-NMR shows ?"
Common errors:
MS incorrectly read the MW for the identified M+ (check the scale) and did not recongise the presence of the Br correctly due to the 1:1 m, m+2
IR missed the C=O, incorrectly assumed NH (peak at 3400 cm-1 is too weak) or C-O to be present (more care in the fingerprint region)
13C NMR ignored the specifics of the C=O range (i.e. subtype :197 ppm > 190 therefore it's an aldehyde or a ketone)
H NMR number of H in structure did not match integration, did not get the number of ArH (there are 2+2=4) from the integration, had hydrocarbion pieces that did not fit the integration or coupling patterns
General
- did not check MW with data from MS
- remember that when checking MW from MS, use integer masses of the lighter isotope
- wrong type of C=O (look at 13C NMR C=O range and H NMR to know that it is a ketone, check IR frequency)
- did not draw a disubstituted benzene
- wrong Ar substitution pattern (how many ArH? How many ArH/C types in NMR ?)
- confusing cyclohexane and benzene
- drawing structures that broke the rules of valence!