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 147 g/mol but also an m+2 at 149. The 1:1 ratio of m : m+2 implies a Br isotope patterns.
Since the MW is odd, the N rule implies an odd number of N atoms are present.

IR:  There is a strong absorption at 2266 cm-1 which is probably C≡C or C≡N. There are no other absorptions of note.... no OH, NH, C=O or C=C.

13C nmr: The proton decoupled spectrum shows a total of 4 peaks indicating 4 types of C. By analysis of the chemical shifts, we have a peak at 117ppm (which must relate to the sp C) and then 3 peaks between 0-40 ppm that are most likely from an sp3 hydrocarbon portion. 

1H nmr: The proton spectrum shows a total of 3 peaks indicating 3 types of H. 

d/ppm
multiplicity
integration
Inference
4.3
6 lines
1
CH coupled to 5H, deshielded
3.0
d
2
CH2 coupled to 1H, slightly deshielded
1.8
d
3
CH3 coupled to 1H

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

The most significant structural information from this are:

Summary....
The MS indicated MW = 147, Br and N.
 The IR showed the presence of C≡C or C≡N
13C NMR shows 4 C types
H NMR gives  CH3CHCH2-

This information suggests a molecular formula = C4H6BrN which matches the MW = 147 and has an IHD = 2.

Altogether...

The fragments we have are:

CH3CHCH2-, Br, C, N

The IR, C-NMR and IHD data suggest that a C≡N is present

The H-NMR data indicates that the CH is the most desheilded at 4.3ppm which is most consistent with the Br being attached at that C.

3-bromobutanonitrile

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 ?"