I'm a little confused about this concept. So EK says for 3>2>1 we follow 1:4:5. This implies that to form a primary product, the substrate will require 5X as many collisions than a tertiary product? I don't understand this since the example they gave us 2-methyl butane has 9 primary, 2 secondary and 1 tertiary hydrogen. This applies for all types of reactions and is not limited to alkanes, right? Since bromine is more selective, should we ignore this rule?
You should go back and reread that section in depth.
1:4:5 is selectivity for radical halogenation of plain alkanes by chlorine ONLY
Also, the 1:4:5 ordering is for primary:secondary:tertiary
Bromine selectivity is around 1:82:1640
Fluorine is close to 1:1:1 IIRC
To solve these types of problems do the following:
Selectivity = 1(primary):4(secondary):5(tertiary)
2-methyl butane = 9 primary H, 2 secondary H, 1 tertiary H
(9 x 1) + (4 x 2) + (5 x 1) = 9(prim) + 8(sec) + 5(tert) = 22(total)
Therefore I would expect:
9/22 = 41% Primary chlorinated product
8/22 = 36% Secondary chlorinated product
5/22 = 23% Tertiary chlorinated product