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.


2 Answers 2


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.

  • $\begingroup$ That's a very nice use of JMol! $\endgroup$ Feb 17, 2014 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.