Conversion Methanol to tertiary butyl cyanide?

Jul 6, 2016

I will assume you mean t-butyl cyanide, but not t-butyl isocyanide.

This was a difficult one, and I had to consider whether the original carbon on methanol became:

• the nitrile ($\text{C"-="N}$) carbon
• the central tert-butyl carbon
• one of the surrounding tert-butyl carbons

This is ultimately what I came up with:

I ended up choosing one of the surrounding tert-butyl carbons and building the molecule based on that.

1. Using ${\text{PBr}}_{3}$ dissolved in pyridine turns an alcohol into an alkyl bromide.
2. Taking acetylene ($\text{HC"-="CH}$) from a separate process, reacting it with sodium amide (${\text{NaNH}}_{2}$) deprotonates one of the protons on acetylene, turning it into a nucleophile. This can backside-attack $\text{CH"_3"Br}$ to generate propyne.
3. If you use $\text{HBr}$ on a terminal alkyne or alkene (a hydrobromination), it will react in a Markovnikov fashion, which means the $\text{Br}$ will add onto the more-substituted carbon---the one with less $\text{H}$ atoms. So, both $\text{Br}$ will go onto the middle carbon.
4. Continue this to generate a geminal dibromide.
5. Here is something you may not have heard of; using methyl lithium or methyl magnesium bromide means you are using an alkyl anion, which is one of the strongest nucleophiles there is. $\text{Li}$ or $\text{MgBr}$ are both low electronegativity, so the attached carbon is mostly negatively-charged in ${\text{Li}}^{\left(+\right)}$ ${\text{^((-)):"CH}}_{3}$, for example.

Despite the alkyl halide being secondary (${2}^{\circ}$), which is generally borderline in preferring ${\text{S}}_{N} 1$ vs. ${\text{S}}_{N} 2$, the strength of this nucleophile should favor $\setminus m a t h b f \left({\text{S}}_{N} 2\right)$ displacement of one $\setminus m a t h b f \left(\text{Br}\right)$.

6. Lastly, adding sodium cyanide would favorably give an ${\text{S}}_{N} 1$ reaction to form your product.

Since tert-butyl bromide is bulky, ${\text{CN}}^{-}$ can't just attack it directly, but it can coordinate with the alkyl halide and wait until the ${\text{Br}}^{-}$ falls off. It's a slow (rate-determining) step, but it happens, because ${\text{CN}}^{-}$ is a stronger base than ${\text{Br}}^{-}$, favoring ${\text{Br}}^{-}$ leaving and ${\text{CN}}^{-}$ adding.