Part 7: SPECTROSCOPY

The following data is available from the question:

MS: M = 113 (Odd, therefore an odd number of N, no isotope pattern for Cl or Br)

EA: The % data when used together with the MW data available from the MS, gives the molecular formula = C5H7NO2 . From here we get the IHD = 3

IR: There are significant absorptions at 1745cm-1 due to a C=O (higher than a ketone value) and at 2270cm-1 which could be either of the triple bond groups: an alkyne or a nitrile. There are also strong bands at about 1200cm-1 due to a C-O. It is also important to note the absence of bands for -OH, (so not an alcohol or carboxylic acid) especially with the amount of O in the formula..

13C nmr: The proton decoupled spectrum shows only 5 peaks indicating 5 types of C (reference TMS at 0 ppm). Since we know the formula, we know that all the C are different and there is no symmetry. By analysis of the chemical shifts, we get that 164 ppm is typical of O=C-O unit, 63 ppm is C-O and those at 24 ppm and 14 ppm are most likely just from hydrocarbon. This leaves the peak at 114 ppm which would correspond to a nitrile. It is way too high for an alkyne (65-90 ppm)

1H nmr: First of all we only have 3 types of H showing up. After this, it's a good idea to tabluate the information to make sure you get it all correctly matched up:
 

d/ppm 
multiplicity 
integration 
Inference 
4.3 
quartet 
2 
 CH2 coupled to 3H, deshielded, probably by O
3.4 
singlet 
2 
 isolated, uncoupled CH2 deshielded, too much to just be by a C=O
1.3 
triplet 
3 
 CH3 coupled to 2H

The simplest information we get from the H nmr here is from the integrals (total of 7H) which accounts for all those in the system. The most significant structural information in the H nmr data is that we have an ethyl group -CH2-CH3 . Since the -CH2- is also deshielded, this suggests it is actually an ethoxy group -O-CH2-CH3.

The IR frequency of the C=O and the 13C nmr both suggest that we have an ester system and a nitrile.

So with all this information we have the following pieces: O=C-O-CH2-CH3, -CH2- and -CN.

There is only one way to put this together as we have 2 end groups and 1 middle group.

Altogther...

Ethyl cyanoethanoate. This explains why the -CH2- is so deshielded since it is between a carbonyl and a nitrile.
 

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 ?"