# Conversion of benzenediazonium chloride to benzene using lithium aluminium hydride

• Why does benzenediazonium chloride get converted to benzene upon reacting with $$\ce{H3PO2}$$ and not $$\ce{LiAlH4}$$ ?
• Also, what is the reaction mechanism of the reaction with $$\ce{H3PO2}$$ ?

This Wikipedia page list all kind of nucleophilic substitution reaction on aromatic compound why can't $$\ce{LiAlH4}$$ / $$\ce{H-}$$ react like one of these to give benzene and expel nitrogen.

I just wanted to add that it's up to my knowledge that $$\ce{LiAlH4 / H-}$$ can't reduce the double bond of Benzene, because when styrene reacts with $$\ce{LiAlH4 / H-}$$ it gives ethyl benzene.

### Update

I was just researching (on google) for the mechanism where I came across this site suggesting the mechanism to be radical and as follow :

But still there is one problem the side products don't match this site states the side product to be $$\ce{H3PO3}$$ and $$\ce{HCl}$$ but as stated above the side products must be $$\ce{H2PXO2}$$.

• Consider the possibility that the PCl(OH)2 undegoes hydrolysis - how stable is PCl3 in water? – Waylander Mar 9 '19 at 20:30
• @Waylander I know PCl3 reacts with water vigourously to give H3PO4 and HCl – Advil Sell Mar 9 '19 at 20:34
• So do you think it likely that other P-Cl species mat hydrolyse easily? – Waylander Mar 9 '19 at 21:31
• @Waylander yeah they may ........ – Advil Sell Mar 9 '19 at 21:44

I just wanted to add that it's up to my knowledge that $$\ce{LiAlH4 / H^{−}}$$ can't reduce the double bond of Benzene, because when styrene reacts with $$\ce{LiAlH4 / H^{−}}$$ it gives ethyl benzene.
That answer is as good as any for the first part of your question. It demonstrates that LAH will react preferentially with aliphatic pi bonds rather than the aromatic ring. Similarly there are aliphatic pi bonds in $$\ce{Ar - N^+\equiv N}$$, and so the LAH will reduce the diazonium group rather than displacing it.