Sodium ethoxide has reacted with ethanoyl chloride. The compound that is produced in the above reaction is

diethyl ether

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2-butanon

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ethyl chloride

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ethyl ethanoate

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Solution

The correct option is D ethyl ethanoate
The given reaction is an example of $S N_{2}$ reaction, where the sodium ethoxide acts as a base and attack the ethanoyl chloride. The product formed is ethyl ethanoate with the formation of sodium chloride. The mechanism follows the formation of tetrahedral intermediate with the elimination of $C l^{-}$ ion.

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