Alkene reactions with sodium tert-butoxide and CHBr₃

In a problem they gave me some reagents and they want me to figure out the end product.


The reagents given are:


2-butene reacting with sodium tert-butoxide and $\ce{CHBr3}$. 


Now, what I was thinking was that the oxygen would pull the hydrogen from $\ce{CHBr3}$ making a positive charge on the carbon. (I'm not quite sure if the positive charge would be strong since the net dipole moment on $\ce{CHBr3}$ would be zero). However, from there, I immediately thought of a carbene reaction, yet, it has 3 bonds with $\ce{Br}$. 


So, am I on the right track, and what are the next steps from here? Sorry if I didn't give you much to work with.




Strong bases react with chloroform $(\ce{CHCl3})$ and bromoform $(\ce{CHBr3})$ to form [dihalocarbenes](http://en.wikipedia.org/wiki/Dichlorocarbene) by alpha elimination. 


The net reaction is as follows:


$$\ce{RO- + CHCl3 -> ROH + Cl- + CCl2}$$


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


$$\ce{RO- + Cl3CH <<=> ROH + Cl3C:^{-}}$$


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


$$\ce{Cl3C:^{-} <=>> Cl- + Cl2C:}$$


The mechanism using arrow-pushing formalism can be found at [this website](http://www.chemtube3d.com/Carbenes-Formation%28AlphaElimination%29.html) from the University of Liverpool.


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Dibromo carbene will form in the reaction for sure because *tert*-butoxide is a very strong non nucleophilic base and here dibromo carbene will form after the alpha elemination. For the next step I think there would be an insertion reaction of the carbene in the double bond of 2-butene to form a 1,1-dibromo-2,3-dimethylcyclopropane.


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