For H3CCH2O(O=)CCH3 there should be 4 singlets in the 13C{1H} NMR spectrum, and four only. It might be hard to see the carbonyl carbon, but this will occur well downfield (and in the spectrum pictured, the carbonyl carbon is quite distinguishable; of course with reagents like this, which are available in quantity, you make a solution of almost 100% sample with a drop or 2 of lock solvent).
In the DEPT-135 sequence, the methylene carbon of the ethyl residue should be negative.
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Why does the d1−chloroform signal appear as a triplet?
