In my book there is a reaction that is

$\ce{HCCl2-CH2F}$ in presence of $\ce{RO^- ROH}$ gives $\ce{CCl2=CH2}$

I could not understand my type of reaction it is?

And why does $\ce{F-}$ leave whe $\ce{Cl-}$ should leave as it is a better leaving group?

  • 3
    $\begingroup$ Which is the more acidic proton? ;-) $\endgroup$ Feb 7, 2017 at 19:36
  • $\begingroup$ @KlausWarzecha the hydrogen attached to the carbon which ha Cl $\endgroup$ Feb 7, 2017 at 19:41
  • $\begingroup$ @KlausWarzecha but the energy required for F- is more $\endgroup$ Feb 7, 2017 at 19:47
  • 2
    $\begingroup$ The two chlorine atoms attached to the other carbon have a very strong electron-withdrawing effect that weakens the C-F bond, making it even weaker than the C-Cl bond weakened by said fluorine. Thus, in this context the fluoride ion is a better leaving group than the chloride ion. $\endgroup$
    – Linus Choy
    Feb 8, 2017 at 8:24

2 Answers 2


It appears to be an e1cb mechanism, where formation of the carbanion on the dichloro end is the rate limiting step. This appears to be a far more acidic proton than the ones on the fluorine-bearing carbon. Once the carbanion is formed, it rapidly eliminates fluoride ion to give the stable dichloroethylene product, rather than eliminating chloride (admittedly a better leaving group) to give the carbene, a higher energy product.


Look at that $\ce{RO-}$. It is a base, right? A strong base. A strong base is likely to attack an acid. In this case, the most acidic hydrogen is the one which is the one in $\ce{CHCl2}$ (See explanation below). For convenience, let's call it $\ce{H-\alpha}$ and the other two $\ce{H-\beta}$.

Base attack most acidic proton

(The base shoud be $\ce{RO-}$, but the drawing tool I'm using does not support placeholder signs.)

Now we have this:

Carbanion is unstable, thus fluorine anion will eliminate and form a double bond.

This is called E1cb mechanism. (cb stands for intermediate carbanion)

Why is the hydrogen attacked

$\ce{H-\alpha}$ has totally three -I effect group: two chlorine and a fluoromethyl group. Meanwhile, $\ce{H-\beta}$ only has two -I effect group: a fluorine group and a dichloromethyl group. Therefore, the intermediate is more stable than it would be if $\ce{H-\beta}$ is eliminated.


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