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 228 g/mol with m+2 at 230 g/mol. 1:1 ratio of m and m+2 indicates Br.

IR:  There is a strong absorption at about 1722 cm-1 which is probably C=O (note that's about normal for a ketone) and maybe at 1290 and 1250 cm-1 which is probably C-O. There is an absorption at 1570 cm-1 which is probably C=C but no other absorptions of note.... no OH, NH, C≡C or C≡N

13C NMR: The normal proton decoupled spectrum shows a total of 9 peaks indicating 9 types of C. By analysis of the chemical shifts, we have a peak at 166 ppm (C=O in the carboxylic acid derivative range), 6 peaks between 123-136ppm (ArC), 62ppm (maybe a deshielded sp3C) and a peak at 15 ppm that are most likely from another sp3 hydrocarbon. 

1H NMR: The proton spectrum shows a total of 6 sets of peaks indicating 6 types of H. The peaks at about 7 ppm are very close together, but the expansion provides the necessary clarity.

d/ppm
multiplicity
integration
Inference
8.15 t 1 Ar-H coupled to 2H (narrow J)
7.92
dt
1
Ar-H coupled to 1H and 2H
7.75
dt
1
Ar-H coupled to 1H and 2H
7.32
t
1
Ar-H coupled to 2H
4.31 q 2 CH2 coupled to 3H, deshielded
1.34 t 3 CH3 coupled to 2H

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

The most significant structural information from this are:

Summary....
The MS indicated MW = 228 and Br
 The IR showed the presence of C=O
13C NMR shows 9 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.
13C NMR shows 6 C types in the aromatic region and the H NMR 4 ArH therefore we have a disubstituted benzene ring.
H NMR gives  C6H4, -CH2-, and -CH3 groups.

This information suggests a molecular formula = C9H9OBr which gives MW = (9x12) + (9x1) + 16 +79 = 212 i.e. missing 16 = O.
Therefore, molecular formula = C9H9O2Br and has an IHD = 5 consistent with the C=O and a benzene ring.
Altogether...

The fragments we have are:

fragments

The IR, 13C and molecular formula imply that the C=O carbonyl is part of an ester. Note the IR C=O is too low to be an acyl halide.

The H-NMR 2H quarttet at 4.31 ppm and 3H triplet at 1.34 ppm indicate an ethyl group, -CH2CH3.

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

nearly there

Last step is the benzene substitution pattern. Since there are 4 types of ArC and 4 types of ArH, it can not be para (which has 4 Ar C and 2 ArH types). So it must be ortho- or meta-.

The observed H NMR coupling pattern is consistent with meta as the H at 8.15 ppm only shows longer range (narrower J coupling)... note how it is almost a singlet in the full scale H NMR spectrum.

ethyl m-bromobenzoate

ethyl 3-bromobenzoate

or

ethyl m-bromobenzoate

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). Identified the wrong halogen!

IR missed the C=O

13C NMR ignored the specifics of the C=O range (i.e. subtype :166 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