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

   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
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Introduction to Spectroscopy
Introduction to Spectroscopy
Donald L. Pavia,… 4th Edition
Chapter 5, Problem 7 ↓

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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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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
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