Addition of dichlorocarbene: practice and mechanism
KOH deprotonates chloroform and CCl₃⁻ loses Cl⁻: dichlorocarbene (:CCl₂) forms. It adds to the double bond, giving a 1,1-dichlorocyclopropane.
Example
The base deprotonates chloroform and the anion loses chloride, giving dichlorocarbene (:CCl₂). The carbene adds to the double bond in one step, giving a 1,1-dichlorocyclopropane.
Mechanism step by step
- 1
KOH deprotonates chloroform (CHCl₃ is fairly acidic thanks to the three Cl): CCl₃⁻ forms.
- 2
CCl₃⁻ loses a chloride (α-elimination): dichlorocarbene :CCl₂ forms, with a lone pair and an empty orbital.
- 3
The carbene adds to the double bond in one step: a 1,1-dichlorocyclopropane forms, stereospecifically.
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
Often confused with: Simmons–Smith cyclopropanation
Frequently asked questions
What is a carbene?
A neutral carbon with only two bonds and six valence electrons. It is electron-poor and very reactive.
How does it differ from Simmons–Smith?
Simmons–Smith transfers CH₂ (product without Cl), dichlorocarbene transfers CCl₂. Both form cyclopropanes and keep the alkene geometry.