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 177 g/mol. Since this is an odd number, it implies that an odd number of N atoms are present. No m+2, so no Cl or Br.

IR:  There is a strong absorption at about 1680 cm-1 which is probably C=O 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 7 peaks indicating 7 types of C. By analysis of the chemical shifts, we have a peak at 190 ppm (which must relate to the C=O and it is right on the border of the aldehyde/ketone versus carboxylic acid or derivative ranges. There are 4 peaks between 155-110 ppm that are aromatic C, and then 2 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
9.8 s 1 aldehyde H (i.e. O=C-H)
7.8 d 2 2 x ArH
6.7
d
2
2 x ArH
3.5
q
4
2 x CH2 coupled to 3H, deshielded
1.3
t
6
2 x CH3 coupled to 2H

(q = quartet, t = triplet, d = doublet, s = singlet)

The most significant structural information from this are:

Summary....
The MS indicated MW = 177, N.
 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, aldehyde -CHO and two equivalent CH3CH2- groups

This information suggests a molecular formula = C11H15NO which matches the MW = 177 and has an IHD = 5 consistent with the C=O and the benzene system.

Altogether...

The fragments we have are:

the fragments

The IR and H-NMR imply that the carbonyl is from an aldehyde.

In order to have two equivalent ethyl groups, they need to be attached to the N atom i.e. -N(CH2CH3)

The H-NMR data indicates that the benzene ring is disubstituted (since there are only 4 ArH present from the integration), therefore the aldehyde and the diethylamine must be attached to the benzene.

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.

4-N,N-diethylaminobenzaldehyde

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 ?"