# What is/are the product(s) of the reaction of (5,5-²H₂)cyclopenta-1,3 diene with NaH, and quenched with H₂O?

$$\ce{NaH}$$ 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?

• 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. – Ivan Neretin Nov 28 '18 at 7:43
• H- does not act as a nucleophile, it is a strong base. – Waylander Nov 28 '18 at 8:25
• I think that the deuterium isotope effect may have something to do with this – Yusuf Hasan Nov 28 '18 at 8:31
• Nothing to do with the deuterium isotope effect. Consider cyclopentadienyl anion - is it aromatic? – Waylander Nov 28 '18 at 9:00
• @Waylander Yes it is. But what does it mean? – Yusuf Hasan Nov 28 '18 at 9:12