Which reaction is this?

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?




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.


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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](https://i.stack.imgur.com/qtvsj.png)](https://i.stack.imgur.com/qtvsj.png)


(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.](https://i.stack.imgur.com/xFd1V.png)](https://i.stack.imgur.com/xFd1V.png)


This is called E1cb mechanism. (cb stands for intermediate **c**ar**b**anion)


**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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