Conversion Methanol to tertiary butyl cyanide?
1 Answer
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 (
"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
"PBr"_3 dissolved in pyridine turns an alcohol into an alkyl bromide. - 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. - 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. - Continue this to generate a geminal dibromide.
-
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") . -
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.