In the dehalogenation of vicinal dihalides why can't sodium be taken as a reagent?
I understand that zinc is taken because it is reactive and rate of reaction would be considerably high and it doesn't form complexes. Because of this many other metals can't be used including magnesium as it forms complexes. But I don't understand why sodium can't be taken.
You mean something like the following right $$ \mathrm{R-X ~+~ Na \longrightarrow R\cdot ~+ ~NaX} $$ and we add a proton source like an alcohole to transfer an H atom to $\mathrm{R\cdot}$: $$ \mathrm{R\cdot~ + ~R'-H} \longrightarrow \mathrm{R-H} $$ Well, its in principle possible. But a) you don't want an explosion right?, so be careful as this will probably be very exothermic b) it will probably be really messy as side products like $$ 2~\mathrm{R\cdot} \longrightarrow \mathrm{R-R} $$ can be formed. Also the whole approach is not really tolerant to pretty much any functional group and will just tear the substrate apart in many cases. The other alternatives (e.g. Zinc, $\mathrm{SmI_2}$) are just much milder, easier to use and much more versatile. But sure, try it out have fun (less than 0.5 g sodium should still be relatively harmless) ; )
Dehalogenation can be effected with many metals, e.g. zinc, magnesium, lithium and sodium in ammonia. Other reagents are sodium naphthalenide, phenyllithium, lithium aluminum hydride, chromium(II) chloride, sodium iodide, sodium sulfide and sodium selenide.
Usually good yields are obtained in these reactions; however, since the best route to the starting material is halogen addition to an alkene, the preparative value is limited. An advantage over dehydrohalogenation is the fixed position of the formed double bond.
The mechanisms of the dehalogenations are complex and depend on the particular conditions. The reactions are sometimes stereospecifically anti, but in other cases no stereospecificity has been observed.
