Alkene reactions with sodium tert-butoxide and CHBr₃
Strong bases react with chloroform $(\ce{CHCl3})$ and bromoform $(\ce{CHBr3})$ to form dihalocarbenes by alpha elimination. 

The net reaction is as follows:

$$\ce{RO- + CHCl3 -&gt; ROH + Cl- + CCl2}$$

The base removes the hydrogen atom from chloroform or bromoform to form an anion. Remember, when you deprotonate something, the electrons stay behind.

$$\ce{RO- + Cl3CH &lt;&lt;=&gt; ROH + Cl3C:^{-}}$$

The trihalomethyl anion is unstable and fragments to lose a halide ion and generate the carbene.

$$\ce{Cl3C:^{-} &lt;=&gt;&gt; Cl- + Cl2C:}$$

The mechanism using arrow-pushing formalism can be found at this website from the University of Liverpool.