For "H"_3"CCH"_2"O(O=)CCH"_3 there should be 4 singlets in the ""^13"C"{""^1"H"} "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 d_1-"chloroform" signal appear as a triplet?
