Question

What product do you get when you react 3-bromocyclopentene with `NaOCH_2CH_3` dissolved in `HOCH_2CH_3`?

Answer 1

`"Sodium cyclopentadienide?"`

Explanation:

`C_5H_7Br + NaOEt rarr C_5H_6 +NaBrdarr +EtOH`

`C_5H_6 + NaOEt rarr [C_5H_5]^(-)Na^(+) +EtOH`

`"1 equiv"` gives the diene, `C_5H_6`; the 2nd equiv gives cyclopentadienyl anion. These days, in inorganic laboratories, you can prepare `100*g` quantities of base-free `NaCp` simply by boiling up sodium metal with dicyclopentadiene, `C_10H_12`,