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 136 g/mol, m, m+2 pattern 3:1 implying that there is a Cl present..

IR:  There is a strong band near 1750cm-1 which is high for a simple ketone C=O. There are no absorptions 3500-3000cm-1 (rules out OH, NH, sp and sp2 CH). There are absorptions of near 1200 cm-1 that could be C-O.

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 1 peak at 170 ppm a carbonyl in a carboxylic acid derivative as it is below 190ppm. There are 2 peaks at 62 and 53 ppm that are most likely from sp3 C-eneg atom and two hydrocarbon fragements (22 and 14ppm). Note that there is NO C=O between 150-220ppm therefore it is not an aldehyde, ketone or any type of carboxylic acid or derivative.

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

d/ppm
multiplicity
integration
Inference
4.45
q
1

deshielded, CH coupled to 3H
4.18 q 2 deshielded, CH2 coupled to 3H
1.7 d 3 3H (CH3) coupled to 1H
1.23 t 3 3H (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 = 136 g/mol and Cl.
 The IR showed the presence of C=O in an acid derivative
13C NMR suggests a disubstituted benzene system with 5 types of C
H NMR gives -CH-, -CH- and 2 x -(CH3)
This information suggests an initial molecular formula = C5H9ClO which gives MW = (5x12) + (9x1) + 35 + 16 = 120 which means we are missing 16.
The missing 16 can be accounted for by another O (consistent with the carbonyl of an acid derivative).
Therefore, molecular formula = C5H9ClO2 which has an IHD = 1 consistent with the C=O double bond.

Altogether...

The fragments we have are:

initial fragments

The IR showed C=O at about 1750cm-1 so an ester rather than acyl chloride.

The H-NMR 2H quartet at 4.18 ppm and the 3H triplet at 1.23ppm indicate an ethyl group CH3CH2-

The H-NMR H quartet at 4.45 ppm and the 3H triplet at 1.7ppm indicate the group CH3CH-

Now we have:

revised fragments

To complete the structure, let's consider the ethyl group. It can only be attached to the eser group, but is it carbonyl side or alkosy side ?
The H-NMR shift of the methylene in the ethyl group (4.18ppm) tells us it is the alkoxy side... and then everything else falls into place...

 

solution

ethyl 2-chloropropanoate

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 missed the halogen in the MS

IR C=O is too low to be an acid chloride (@ 1800cm-1)

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

General

[Home]