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 but also an m+2 at 138. The 3:1 ratio of m : m+2 implies a Cl isotope patterns. MW is even, so N rule implies no N (or an even number of N atoms)
IR: There is a C=O at 1738 cm-1 which is a little high for a simple ketone. The only other absorbance of any note is at1207 cm-1 which could be C-O.
13C nmr:
The proton decoupled spectrum shows a total of 5 peaks indicating 5 types of
C. By analysis of the chemical shifts, we have a C=O at 170ppm, suggesting a
carboxylic acid derivative, 1 peak at about 62ppm (possibly a C with an
electronegative O attached) and then 3 peaks between 0-40 ppm that are
most likely from a hydrocarbon portion.
1H nmr:
The proton spectrum shows a total of 4 peaks indicating 4 types of H.
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CH2 coupled to 3H, deshielded
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CH2 coupled to 2H, deshielded
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CH2 coupled to 2H
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CH3 coupled to 2H
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(m
= multiplet, q = quartet, t = triplet, d = doublet, s = singlet)
The most significant structural information
deduced from this HNMR information are:
The fragments we have are:CH2CH2 and CH3CH2-, C=O, -O- and -Cl The chemical shift of the CH2 in the CH3CH2 group requires that it be attached to the O, so we have an ethoxy group. The IR and C-nmr data suggest that an ester is present. This implies that the CH2CH2 group is connected to the C=O and the Cl, the H-nmr chemical shift of the CH2 groups are consistent with this. |
ethyl 3-chloropropanoate
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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 ?"