Part 8: SPECTROSCOPY

The following data is available from the question.

Note : Remember to cross reference things to confirm.... eg. IR may show C=C, use NMR to confirm that etc.

MS: M+ seen at 207 g/mol (odd MW suggests an odd number of N atoms, 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 69.54% C, 6.76% N and 8.27% H  (therefore there is 15.43% 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 partial molecular formula = C12H17N ?  (so that leaves 32). The missing piece is consistent with 2 oxygen atoms (because 2 x 16 = 32) .... hence  C12H17NO2
From here we get the IHD = 5

IR: Confirms the presence of oxygen. There is a C=O at about 1732 cm-1 (high for a ketone) and an NH band between 3400-3300cm-1.  C-O stretches can be seen around 1200cm-1.  There are no -OH nor triple bonds (CN or CC) absorbances.

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 above 160 ppm, four types of C=C at 120-140ppm (possibly ArC),  and 5 peaks between 0-60 ppm that are most likely from a hydrocarbon portion, those nearer 60ppm are deshielded by something electronegative such as O.

1H nmr: First of all we have 7 sets of peaks so we have 7 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 7.3 m 5 5 Ar-H = monosubstituted benzene,  C6H5- 4.2 q 2 CH2 coupled to 3H, deshielded (by O ?) 3.8 s 2 CH2 not coupled, slightly deshielded 2.9 t 2 CH2 coupled to 2H, slightly deshielded 2.5 t 2 CH2 coupled to 2H, slightly deshielded 1.9 s, brd 1 NH or OH 1.3 t 3 CH3 coupled to 2H, hydrocarbon portion

(m = multiplet, q = quartet, t = triplet, s = singlet, brd = broad)

The most significant structural information from this is :
• peak at 7.3 ppm indicates a monosubstituted aromatic C6H5-
• peaks at 4.2 and 1.3ppm indicate 2 pieces that have to be joined as a CH3CH2O- unit
• peaks at 2.9 and 2.5ppm indicate 2 pieces that have to be joined as a -CH2CH2- unit
• peak at 3.8ppm indicates an isolated (uncoupled) -CH2- unit
• peak at 1.9ppm corresponds to an NH (IR rules out OH)
• Lack of a peak above 9ppm rules out an aldehyde (RCHO) and an carboxylic acid (RCO2H)
Summary....
The MS indicated MW = 207
EA with MS gives the MF =
C12H17NO2
The IR showed the presence of C=O, NH and C-O
13C and H NMR gives  C6H5- , CH3CH2O-,  -CH2CH2-, and  -CH2-

Therefore the pieces we have are
C6H5-, CH3CH2O-,  -CH2CH2-, and  -CH2- , C=O and NH
This is in agreement with the molecular formula of C12H17NO2

Altogether...

 The pieces we have are C6H5-, CH3CH2O-,  -CH2CH2-, and  -CH2- , C=O and NH (2 end pieces, 4 middle pieces) The -CH2CH2-, and  -CH2- can not be connected to each other since that would give a different coupling pattern. Neither the -CH2CH2- can be connected to the CH3CH2O- because that would require a triplet near 4 ppm. The C=O .... not an aldehyde or ketone (13C nmr), nor an amide (IR too high), suggests an ester, so join C=O and CH3CH2O- to get  -CO2CH2CH3 Therefore the NH must come between the -CH2CH2- and  -CH2- to give -CH2CH2-NH-CH2- This gives two reasonable answers that are hard to distinguish : or

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 ?"

Major errors:

1. Missing the monosubstituted benzene
2. Incorrectly assigning OH instead of NH in the IR.
3. Using NH2 instead of NH as indicated by the IR and H nmr
4. Joining CH2 units in ways that gave different coupling patterns.