The hydride ion is not very stable. On the contrary a negative charge on a $\ce{sp^2}$ on the benzene ring would be much more stable. So why doesn't the benzene ring migrate instead of the hydride, as the benzene ring with a negative charge would be a better leaving group (as we know that less basic groups are better leaving groups)?

I've seen some examples where the benzene ring does migrate instead of the hydride for example when two nitro groups are attached at ortho positions. But I think the benzene migration should always happen instead of the hydride migration for reasons which I stated above.

P.S: Someone suggested me that the "real" mechanism is the transition mechanism. But even that does not explain why the transition state involving o-dinitro benzene would be more stable than the one involving $\ce{H}$ but the one involving just a benzene would be less stable than $\ce{H}$. How do we logically explain the discrepancy?

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    $\begingroup$ One possibility: hydrogen has better overlap for forming the delocalized bond in the transition state. You have to heavily withdraw electrons from the benzene ring (e.g. nitro groups) to make the hydrogen non-hydridic and force the benzene ring to migrate instead. $\endgroup$ – Oscar Lanzi May 12 '17 at 10:21
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    $\begingroup$ Actually I dont know why you think phenyl anion is very stable( not saying incorrect, just asking why?) . Its not resonance stabilized. If it had been Phenylacetaldehyde then maybe it would be stable anion but in that case the reactant may give Aldol condensation reaction. What am I missing? $\endgroup$ – Black Jack 21 May 12 '17 at 13:24
  • $\begingroup$ @AlphaRomeo Because it has a $sp^2$ carbon which stabilizes the negative charge. $\endgroup$ – user38977 May 12 '17 at 13:27
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    $\begingroup$ How? Pka of benzene is in 40s .. and I dont see much stabilization by inductive effect as resonance wont occur. Anyway, I will brush up my concepts :P $\endgroup$ – Black Jack 21 May 12 '17 at 13:28
  • $\begingroup$ @AlphaRomeo Who's talking about inductive effect here? I'm talking about $sp^2$ hybridization. $\endgroup$ – user38977 May 12 '17 at 13:29