Author Topic: Ritter reaction, amination ter-alcohols with R-CN  (Read 1553 times)

0 Members and 1 Guest are viewing this topic.

acx01b

  • Guest
Ritter reaction, amination ter-alcohols with R-CN
« on: October 05, 2004, 05:58:00 PM »
Hi,

i didn't find many posts on the RITTER reaction (R-CN with tertiaries alcohols for example PCOH phenyl-1-cyclohexanol, to get a primary amine, through a way that avoids the deshydratation reaction that would happen by reacting tertiary halide with primary amine, or hexamine)


the nitrile: NaCN can be used too, Ph-CN, Et-CN
the alcohol must be acidified to get the "free" carbocation...
H2SO4 can be used for it..
I guess all strong acids can ?

do the yields depend of the nitrile used ?

but my biggest problem is not the yields:
NaCN + strong concentrated acid --> free HCN gaz
NaCN + diluted strong acid --> H+ CN-
same with R-CN...

proposed method:
0.1mol R3-OH is mixed with water and polar organic solvent
(ether, toluene?) and 0.15 mol H2SO4

THEN organic phase is separated
and reacted with 0.1mol R-CN or NaCN

finally the amide is hydrolized and no free HCN, no CN- has been formed at all...

will the alcohol sulfuric ester be formed or will the alkene be formed by deshydratation?

will the organic phase separation work? will the acidified alcohol leave the water phase?

tks for your comments

lutesium

  • Guest
You cannot create a protonated alcohol then...
« Reply #1 on: October 08, 2004, 02:20:00 PM »
You cannot create a protonated alcohol then seperate it for further reaction. The medium must be acidic for the alcohol to reamain as a cation as its unstable.

You want the oh group to leave (thats why you protonate it) so you may think other possiblities like replacing the oh group with a halogen or sulfonyl ester.

Or add the cyanide to the protonated oh group and the medium will also protonate the -CN of NaCN to release HCN as ritter does. The choice is yours  ::)