Question

Conversion Methanol to tertiary butyl cyanide?

Answer 1

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

https://upload.wikimedia.org/

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

• the nitrile (`"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 `"PBr"_3` dissolved in pyridine turns an alcohol into an alkyl bromide.
2. Taking acetylene (`"HC"-="CH"`) from a separate process, reacting it with sodium amide (`"NaNH"_2`) deprotonates one of the protons on acetylene, turning it into a nucleophile. This can backside-attack `"CH"_3"Br"` to generate propyne.
3. If you use `"HBr"` on a terminal alkyne or alkene (a hydrobromination), it will react in a Markovnikov fashion, which means the `"Br"` will add onto the more-substituted carbon---the one with less `"H"` atoms. So, both `"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. `"Li"` or `"MgBr"` are both low electronegativity, so the attached carbon is mostly negatively-charged in `"Li"^((+))` `""^((-)):"CH"_3`, for example.

   Despite the alkyl halide being secondary (`2^@`), which is generally borderline in preferring `"S"_N1` vs. `"S"_N2`, the strength of this nucleophile should favor `\mathbf("S"_N2)` displacement of one `\mathbf("Br")`.

6. Lastly, adding sodium cyanide would favorably give an `"S"_N1` reaction to form your product.

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