# In the mechanism of Cannizzaro reaction why does the hydride migrate rather than the benzene ring? [duplicate]

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?

## marked as duplicate by Jan, Todd Minehardt, airhuff, paracetamol, Pritt BalagopalJun 14 '17 at 4:53

• @AlphaRomeo Because it has a $sp^2$ carbon which stabilizes the negative charge. – user38977 May 12 '17 at 13:27
• @AlphaRomeo Who's talking about inductive effect here? I'm talking about $sp^2$ hybridization. – user38977 May 12 '17 at 13:29