Conversion Methanol to tertiary butyl cyanide?
1 Answer
I will assume you mean t-butyl cyanide, but not t-butyl isocyanide.
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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.
- Using
PBr3 dissolved in pyridine turns an alcohol into an alkyl bromide. - Taking acetylene (
HC≡CH ) from a separate process, reacting it with sodium amide (NaNH2 ) deprotonates one of the protons on acetylene, turning it into a nucleophile. This can backside-attackCH3Br to generate propyne. - If you use
HBr on a terminal alkyne or alkene (a hydrobromination), it will react in a Markovnikov fashion, which means theBr will add onto the more-substituted carbon---the one with lessH atoms. So, bothBr will go onto the middle carbon. - Continue this to generate a geminal dibromide.
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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 orMgBr are both low electronegativity, so the attached carbon is mostly negatively-charged inLi(+) (−):CH3 , for example.Despite the alkyl halide being secondary (
2∘ ), which is generally borderline in preferringSN1 vs.SN2 , the strength of this nucleophile should favorSN2 displacement of oneBr . -
Lastly, adding sodium cyanide would favorably give an
SN1 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 theBr− falls off. It's a slow (rate-determining) step, but it happens, becauseCN− is a stronger base thanBr− , favoringBr− leaving andCN− adding.
