In the following reaction, I believe a polar aprotic solvent like propanone will be more favorable since the partial negative charge on the oxygen atom in propanone will attract the sodium ion from sodium hydroxide leaving the nucleophile, $\ce{OH-}$ to freely attack the carbonyl group:

$$\ce{CH3COOC2H5 + NaOH → CH3COONa + C2H5OH}$$

On the other hand, a polar protic solvent like ethanol will be less favorable because the partial charges on the oxygen and hydrogen atoms (created due to a difference in electronegativities) in the ethanol will surround the nucleophile, $\ce{OH-}$ and would prevent the attack on the carbonyl group slowing down the reaction.

However, while performing the experiment and obtaining conductances of the solutions at different time intervals using a conductivity probe to calculate rates and $E_\mathrm{a},$ it was observed that the rate of reaction was faster in ethanol and the activation energy in ethanol was also lower proving my argument incorrect.

Can someone help?

  • 2
    $\begingroup$ Consider the solublity of the reaction components in the solvent, also consider whether the choice of solvent influences alternative reaction pathways $\endgroup$ – Waylander Jul 9 at 6:52

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