Question
The reaction of dimethyl malonate with acetaldehyde (ethanal) under basic conditions yields a compound with formula $\mathrm{C}_7 \mathrm{H}_{10} \mathrm{O}_4$. The proton NMR is shown here. The normal carbon-13 and the DEPT experimental results are tabulated:$$\begin{array}{cll}\hline \text { Normal Carbon } & \text { DEPT-135 } & \text { DEPT-90 } \\\hline 16 \mathrm{ppm} & \text { Positive } & \text { No peak } \\52.2 & \text { Positive } & \text { No peak } \\52.3 & \text { Positive } & \text { No peak } \\129 & \text { No peak } & \text { No peak } \\146 & \text { Positive } & \text { Positive } \\164 & \text { No peak } & \text { No peak } \\166 & \text { No peak } & \text { No peak }\end{array}$$Determine the structure and assign the peaks in the proton NMR spectrum to the structure.DIAGRAM CANT COPY
Step 1
The compound has the formula $\mathrm{C}_7 \mathrm{H}_{10} \mathrm{O}_4$. This indicates that the compound contains 7 carbon atoms, 10 hydrogen atoms, and 4 oxygen atoms. Show more…
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