Addition of Dichlorocarbene to alkenes to give dichlorocyclopropanes

by James

Description: Alkenes treated with chloroform (CHCl3) and strong base will form dichlorocarbenes, which add to alkenes to form cyclopropanes.

Notes: A strong base such as NaOH is required for this reaction.


Notes:Note that the syn product is always formed in these reactions (see example 2).

Mechanism: Deprotonation of chloroform by NaOH (Step 1, arrows A and B) leads to an anion, which then loses Cl in an alpha-elimination reaction (Step 2, arrow C). This forms the dichlorocarbene, which is extremely reactive. Reaction with the alkene (Step 3, arrows D and E) results in the cyclopropane. Note that the cyclopropanation step occurs through a concerted transition state (box) and thus arrows D and E occur at the same time.

Notes: Step 2 (alpha elimination) results in loss of chloride ion to give an empty orbital on carbon. The resulting species (divalent carbon, with one lone pair and one empty orbital) is called a carbene.

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