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 180 g/mol, with an m+2 pattern of equal intensity that implies Br.

IR:  There is a strong absorption at about 1739 cm-1 which is probably a C=O (note that's high for a ketone) and maybe around 1200 cm-1 which is probably C-O. There are no other absorptions of note.... no OH, NH, C=C, C≡C or C≡N.

13C NMR: The normal proton decoupled spectrum shows a total of 5 peaks indicating 5 types of C. By analysis of the chemical shifts, we have a peak at 173 ppm (C=O in the carboxylic acid derivative range), 60ppm (maybe a deshielded sp3C) and peaks at 36, 23 & 14ppm that are most likely from sp3 C hydrocarbon. 

1H NMR: The proton spectrum shows a total of 4 sets of peaks indicating 4 types of H. 

d/ppm
multiplicity
integration
Inference
4.15
q
2
CH2 coupled to 3H, deshielded
3.7
t
2
CH2 coupled to 2H, deshielded
2.75
t
2
CH2 coupled to 2H, slightly deshielded
1.2 t 3 CH3 coupled to 2H

(dq = doublet of quartets, dd = doublet of doublets, q = quartet, t = triplet, d = doublet, s = singlet)

The most significant structural information from this are:

Summary....
The MS indicated MW = 180 and Br.
 The IR showed the presence of C=O
13C NMR shows 5 C types including C=O that is most likely a carboxylic acid or derivative. Since there is no RCO2H in the HNMR (12 ppm) or -OH in the IR, it can not be a carboxylic acid.
H NMR gives 3 x  -CH2-, and a -CH3.

This information suggests an initial molecular formula = C5H9OBr which gives MW = (5x12) + (9x1) + 16 + 79 = 164 i.e. missing 16 = O.
Therefore, molecular formula = C5H9O2Br and has an IHD = 1 consistent with the C=O.
Altogether...

The fragments we have are:

fragments

The IR, 13C and molecular formula imply that the C=O carbonyl is part of an ester.

The H-NMR 2H quartet at 4.2 ppm and 3H triplet at 1.2 ppm indicate an ethyl group, -CH2CH3.

Given the chemical shift of the -CH2- it is likely attached to -O- so we have an ethoxy group : -OCH2CH3 in an ester.

partial solution

The other coupling patterns are two 2H triplets which implies that we have a CH2 next to a CH2. This means we have -CH2CH2- and the Br must be attached to the one end (hence the 3.7ppm)

almost there

From here, only one structure is possible...

molecule

ethyl 3-bromopropanoate

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:

MS incorrectly read the MW for the identified M+ (check the scale).

13C NMR ignored the specifics of the C=O range (i.e. subtype :173 ppm < 190 therefore it's a carboxylic acid or derivative)

H NMR had hydrocarbon pieces that did not fit the integration and coupling patterns

General

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