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 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 from the University of Liverpool.

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  • $\begingroup$ That's a very nice use of JMol! $\endgroup$ – Klaus-Dieter Warzecha Feb 17 '14 at 8:28

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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