[vande060]

in my book they give two reactions in hexane as such

for one problem:

NH3 + CH2CH3- (from CH3CH2Li) ---> NH2- + CH3CH3

for the other

CH3CH2Li + D20 ---> CH3CH2D + LiOD

what i dont get is how in the top problem they presented ethylithium as dissociated on the left side, and on the bottom they did not

then on one of the right hand sides they presented LiOD, and in the other equation Li was ignored other than the parenthesis to indicate where the CH3CH2 base came from.

is it because LiOD is insoluble in hexane, so you cannot ignore Li, and in the top equation Li is only a spectator ion?

[Smrt guy]

alkyl lithiums are largely ionic but there is still significant covalent character.  I think in the top reaction they were merely pointing out that the counterion does not matter in this case (e.b. EtMgBr would also deprotonate NH3).  In hexane, LiOD would precipitate out.