Why when cyclopropane is mixed with bromine in the dark it undergoes ring-opening to form 1,3-dibromopropane?

Question

Why when cyclopropane is mixed with bromine in the dark it undergoes ring-opening to form 1,3-dibromopropane?

Answer 1

Cyclopropane is a fairly reactive alkane because of the sharp bond angles.  Bromination across one of the bonds breaks that highly stressed structure to form a straight chain molecule, which is under far less stress.

Answer 2

Bond in cyclopropane is not typical sigma-bond. The electron density is not between the nuclei, but on the side, just like the pi-bond. (because of the 120grad angles in the molecule, and sp3 hybridisation.) Thus the cyclopropane exhibits some of the properties of the alkenes. The ring opening lowers the energy a lot, due to the releasing of the strain - so it's very energetically favourable.

Answer 3

Actually

Bond in cyclopropane is not typical sigma-bond. The electron density is not between the nuclei, but on the side, just like the pi-bond. (because of the 120grad angles in the molecule, and sp3 hybridisation.) Thus the cyclopropane exhibits some of the properties of the alkenes. The ring opening lowers the energy a lot, due to the releasing of the strain - so it's very energetically favourable.