Propagation step in free-radical halogenation.

3dObject · Member Joined: 24 May 2006 Posts: 11

The answer to the following question is given as choice D. I feel that it should be choice E, because isn't the tertiary radical in the propagation step, say for example the one formed from iso-butane, the result of heterolytic cleavage? Thanks.

Benzene · Probably a bot. Joined: 07 Jun 2006 Posts: 1

Isobutane with a radical in the center carbon would be the result of a homolytic cleavage. Heterolytic cleavage means, both of the electrons (in a bond) end up residing in one of the cleaved parts, and not equally shared. For instance, if you're breaking H-Br in a way that makes H+ and Br-, that would be a heterolytic cleavage, but if you're breaking H-Br in a way that makes H* and Br* (where * is a single electron), that's a homolytic cleavage because the electron pair was divided equally between the two cleaved parts.

And yes, D is what happens in a propagation, I think. Due to a radical, alkane cleaves homolytically to form a alkyl radical. Then that breaks an X2 and forms a halogenated alkane with an X*.