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 146 g/mol (no Cl or Br isotope patterns - they require m and m+2 of 3:1 or 1:1 respectively).

EA: Elemental analysis data for C and H (i.e. standard analysis) showed 82.19% C and 6.85% H - no N is found (not reported), plus sine the MW is even (therefore there is 10.96% something else, most likely oxygen : remember oxygen can't be detected in a combustion analysis).  Use the MW from the mass spectra (above) to get the molecular formula = C10H10O
From here we get the IHD = 6

IR: Confirms the presence of oxygen. There is a C=O at about 1683 cm-1 (which is a little low for a simple ketone).  There are no -OH or triple bonds (CN or CC) absorbances. C=C possible at 1600 cm-1

13C nmr: The proton decoupled spectrum shows a total of 10 peaks indicating 10 types of C. By analysis of the chemical shifts, we have types in the region for 1 x C=O at 198 ppm (therefore aldehyde or ketone), six types of C=C at 125-150ppm (probably ArC),  and 3 peaks between 0-40 ppm that are most likely from a hydrocarbon portion.  

1H nmr: First of all we have 5 sets of peaks so we have 5 types of H showing up. After this, it's a good idea to tabulate the information to make sure you get it all correctly matched up:  

d/ppm
multiplicity
integration
Inference
8.0
 d
 1
 1 Ar-H 
7.3
 m
 3
 3 Ar-H
2.9
 t
 2
 CH2 coupled to 2H, slightly deshielded
2.6
 t
 2
 CH2 coupled to 2H, slightly deshielded
2.2
 p
 2
 CH2 coupled to 2H, hydrocarbon portion

(m = multiplet, p = pentet, t = triplet, d = doublet, s = singlet)

The most significant structural information from this is :
Summary....
The MS indicated MW = 146
EA with MS gives the MF = C10H10O
The IR showed the presence of C=O
13C and H NMR gives  C6H4- ,  -CH2CH2CH2-

Therefore the pieces we have are C6H4- ,  -CH2CH2CH2-  and C=O
This is in agreement with the molecular formula of C10H10O

Altogether...
C6H4- ,  -CH2CH2CH2-  and C=O

(only 3 middle pieces)
  • Note IHD = 6, therefore, need one more = ring
  • The C=O is a ketone (13C / H nmr)




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:

  1. Missing the disubstituted benzene
  2. Missing the ketone (incorrectly have an aldehyde instead)
  3. Incorrect use of elemental analysis data  (see laboratory manual, CAL, etc.)
  4. Joining CH2 units in ways that gave different coupling patterns.
  5. Including alkene C=C units (not seen in C or H NMR)