d) \text{O}\text{O}\newline \text{EtO} \text{C}(\text{Me})\text{C}(\text{H})\text{O} \xrightarrow{\text{NaOEt}, \Delta} \text{O}\newline \text{C}=\text{C}(\text{Me})\text{C}(\text{Me})=\text{O}
Added by Joaquin H.
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Step 1
The first part of the reaction is "(p NaOEt;", which means that sodium ethoxide (NaOEt) is being used as a base. This suggests that the reaction is likely a deprotonation or elimination reaction. Show more…
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