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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, 2018 at 7:43
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    $\begingroup$ H- does not act as a nucleophile, it is a strong base. $\endgroup$
    – Waylander
    Nov 28, 2018 at 8:25
  • $\begingroup$ I think that the deuterium isotope effect may have something to do with this $\endgroup$
    – Yusuf Hasan
    Nov 28, 2018 at 8:31
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    $\begingroup$ Nothing to do with the deuterium isotope effect. Consider cyclopentadienyl anion - is it aromatic? $\endgroup$
    – Waylander
    Nov 28, 2018 at 9:00
  • $\begingroup$ @Waylander Yes it is. But what does it mean? $\endgroup$
    – Yusuf Hasan
    Nov 28, 2018 at 9:12

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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

enter image description here

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  • $\begingroup$ I can't access that link.Can you please explain the mechanism of this reaction in detail? $\endgroup$
    – user70066
    Nov 28, 2018 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, 2018 at 10:11

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