The reactant is nitrobenzene(only) and reagent is Na metal in solvent liquid ammonia along with ethanol (which i am assuming is playing the role of proton donor please correct me if i am wrong).

The first thing that comes in mind is "alkyne" but there is not any alkyne in nitrobenzene( not even in its resonance structures).How this reaction will proceed?

In answer, the DBE is just reduced by one.I was also trying what if it was toluene?Can i have a explanation based on mechanism?

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    $\begingroup$ Birch Reduction will occur with Nitrobenzene. $\endgroup$ – Soumik Das Oct 26 '18 at 8:46
  • $\begingroup$ The ethanol makes little sense, unless it is added in the end only, to destroy remaining Na metal. $\endgroup$ – Karl Oct 26 '18 at 9:06

Benzene Rings or Substituted Phenyl rings undergoes reduction in presence of $\ce{Na}$ in liq. $\ce{NH3}$ in presence of $\ce{EtOH}$ solvent, and generally forms substituted Cyclohexa-$1,4$-dienes.

This reduction generally follows Single Electron Transfer (SET) Mechanism.The single unpaired electron of Sodium is transferred to the Benzene ring forming a radical which subsequently forms a Carbanion. This carbanion eventually takes the Proton from $\ce{EtOH}$ and becomes Cyclohexa-$1,4$-diene.

But in the first step, the transfer of electron by $\ce{Na}$ occurs in that place, where the intermidiate formed after this step gets maximum stabilisation. In case of Electron-withdrawing group, such as $\ce{-NO2, - COOH, -SO3H, -CHO, -COOR, -COR }$ etc., that transfer occurs para to that group, as the negative charge generated is highly stabilised by that electron-withdrawing group. The mechanism thus in case of Nitrobenzene is as follows,

enter image description here

But if the group is electron donating such as $\ce{-CH3, -OH, -OMe, -NMe2 }$ etc. The transfer will occur in such a way to stabilise the generated negative charge by keeping it in meta to the group (where the effect of electron donation by the group is least). For those compounds, the mechanism and corresponding product will be as follows,
enter image description here

Thus, in your case, the product will be $\text{3-nitrocyclohexa-1,4-diene}$ as described in the first mechanism.


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