Alkene Reactions: Formation of Dichlorocyclopropanes using Dichlorocarbene on Alkenes

Chloroform (CHCl3) combined with a strong base will react with alkenes to create a dichlorocyclopropane functional group::

The reaction takes place on each double bonded carbon, forming a cyclopropane (three ring carbon or triangle) with chloroform, resulting in the addition of a dichlorocyclopropane functional group. The addition of this functional group may result in the molecule taking on a new stereochemistry conformation however, this reaction always forms a syn product: 

Syn Addition

The reaction mechanism is depicted below:

In the first step, the chloroform is deprotonated by the strong base (NaOH in this case) and a water molecule is formed and exits the reaction (not pictured). In the second step, the carbanion (carbon holding a negative charge) is stabilized by the sodium atom (Na+). This allows for the carbon-chlorine bond to be broken, causing carbon to have a lone pair of electrons and sodium and chloride to react and form NaCl (not pictured), which exits the reaction. The third reaction step involves the lone pair of electrons on the carbon atom to react with the two alkene carbons, forming the cyclopropane molecule.

The bases commonly used for this reaction (but not exclusive to) are: NaOH, KOH, and potassium tert-butoxide (KO tBu).