I was reading about the bromination of CH4, and saw that since Br is "selective" in its bromination, when you brominate CH4 with Br2 you predominately get CH3Br, and not CH2Br2. What about CH3Br makes it less "preferable" for Br2 to brominate and create CH2Br2 as a product?
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1$\begingroup$ This paper is relevant: pubs.acs.org/doi/pdf/10.1021/jp061558h $\endgroup$– user55119Commented Jul 31, 2021 at 22:59
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Bromine will react with methane to give a kinetic or thermodynamic mix of products basically because the HBr bond is moderately weaker than HCl and the reaction with the bromine radical is more selective. This article seems to discuss this in detail: https://www.mdpi.com/2227-9717/8/4/443/htm. I did not find the relative reaction rates between CH4 and CH3Br[but didn't look too hard] but I suspect they are close to 4:3 at some temperature. To answer your question one would need to know the context in which the statement was made since halogenation reactions are generally controlled by reactant mixes and strong temperature control and can give very variable product mixes depending on temperature, inert gas concentration and relative reactant concentrations. If I had to optimize methylene bromide production there is probably a methane-bromine mix, photon flux and temperature that will maximize mono and di bromination and minimize bromoform production. This is why chemical engineers study so hard and charge the big bucks! to solve problems like this.
