Conversion Methanol to tertiary butyl cyanide?

1 Answer
Jul 6, 2016

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.