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I am asked to give the type of reaction (SN1, SN2 or acid-base reaction) for the following reactions. I know that both $\ce{MeOH}$ and $\ce{EtOH}$ polar protic solvents, so because the first is an SN2 reaction shouldn't the second reaction be that as well? It says that it is an acid-base reaction as $\ce{MeO-}$ reacts with the $\ce{EtOH}$ to from 1-bromopropane, $\ce{MeOH}$ and $\ce{EtO-}$ and then goes on to from a SN2 reaction.

I would appreciate if someone could explain why the second reaction would be an acid-base reaction at first instead of a SN2 reaction.

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  • $\begingroup$ EtO- would be more basic than MeO-...the proton xfer is fast $\endgroup$ – Eashaan Godbole Apr 17 '18 at 15:38
  • $\begingroup$ Note: $\ce{MeO-}$ is a strong base while, $\ce{EtOH}$ is a weak base. My question: how can you perform an acid-base neutralisation reaction then? + acid-base neutralistion also produces water as a product! $\endgroup$ – rv7 Jun 16 '18 at 15:42
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In both cases the alkoxide will be in an equilibrium with the alcohol. This will be occurring on a faster timescale than the substitution but in the first case it's irrelevant as the same species are present on either side of the equilibrium.

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Methanol is more acidic than ethanol, however, so the reaction lies in favour of methoxide and ethanol.

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    $\begingroup$ But EtOH is in vast excess so you are going to end up with EtO- $\endgroup$ – Waylander Jul 16 '18 at 17:00

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