# How Does This Reaction Proceed? [closed]

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

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

• – Jan Jun 14 '16 at 19:22

## 1 Answer

$\ce{CH_{3}C#CNa}$ 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:

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

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