Find the product formed in the following reaction

$\ce{NaH}$ breaks as $\ce{Na+}$ and $\ce{H-}$. Here $\ce{H-}$ acts as a nucleophile and attacks at a place where the electron density is less. But the $\ce{C-D}$ bond is stronger than the $\ce{C-H}$ bond, so the reaction can't be feasible in forward direction. The correct option given is '(d) all of these'. Where is the other $\ce{D}$ gone and how is $\ce{D}$ now attached at all the places?

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    $\begingroup$ C-D bond is the same as C-H. The difference is really, really small, much smaller than you might think after reading these words, and even smaller than that. $\endgroup$ – Ivan Neretin Nov 28 '18 at 7:43
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    $\begingroup$ H- does not act as a nucleophile, it is a strong base. $\endgroup$ – Waylander Nov 28 '18 at 8:25
  • $\begingroup$ I think that the deuterium isotope effect may have something to do with this $\endgroup$ – Yusuf Hasan Nov 28 '18 at 8:31
  • $\begingroup$ Nothing to do with the deuterium isotope effect. Consider cyclopentadienyl anion - is it aromatic? $\endgroup$ – Waylander Nov 28 '18 at 9:00
  • $\begingroup$ @Waylander Yes it is. But what does it mean? $\endgroup$ – Yusuf Hasan Nov 28 '18 at 9:12

An interesting thing to note here that prima facie, it seems that the deuterium is being substituted at different position in the molecule, but looking at the reaction mechanism we see that in fact the second deuterium is not moving at all after abstraction of the first, rather it is the resonance of the molecule and protonation at different sites which is responsible for the varied products

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  • $\begingroup$ I can't access that link.Can you please explain the mechanism of this reaction in detail? $\endgroup$ – Kaushki Nov 28 '18 at 9:41
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    $\begingroup$ @kaushki The key thing to realise is that H- deprotonates the deuterocyclpentadiene to give cyclopentadienyl anion. This is a stable and aromatic species so when it is quenched there is an equal probability of each carbon being protonated. $\endgroup$ – Waylander Nov 28 '18 at 10:11

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