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$$ \ce{CH3OH + CH3CCNa+ -> ?} $$

The answer is $\ce{CH3CCH +CH3ONa+}$ but I don't understand why.

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closed as off-topic by Todd Minehardt, Jan, ron, Jon Custer, orthocresol Jun 14 '16 at 21:59

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$\ce{CH_{3}C#CNa}$ exists in a solution as a cation and anion shown below: $\ce{CH_{3}C#CNa <=> CH_{3}C#C- + Na+} $

$\ce{CH_{3}OH}$ has very slightly acidic hydrogen (thanks to electron pulling ability of oxygen), that can be removed by a strong base.

When put in the same solution, $\ce{CH_{3}C#C-}$ will pull out $\ce{H+}$ from $\ce{CH_{3}OH}$ like so: $\ce{CH_{3}C#C- + CH_{3}OH <=> CH_{3}C#CH + CH_{3}O-}$

$\ce{CH_{3}C#CH}$ is a weaker acid than $\ce{CH_{3}OH}$, which means that $\ce{CH_{3}C#C-}$ is a stronger base than $\ce{CH_{3}O-}$. Therefore, $\ce{CH_{3}C#C-}$ will completely pull out hydrogen from methanol.

And overall reaction looks like the following:

$\ce{CH_{3}C#CNa + CH_{3}OH <=> CH_{3}C#CH + CH_{3}ONa}$

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  • $\begingroup$ Propyne is much weaker acid so reaction will be complete $\endgroup$ – Mithoron Jun 14 '16 at 19:22
  • $\begingroup$ @Mithoron You're right. pKa of propyne is much bigger than pKa of methanol. I updated the answer. $\endgroup$ – Kairat Jun 14 '16 at 19:28

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