Question

Write the formulas of the seven doubly brominated isomers formed when 2,2-dimethylbutane reacts with Br2 in the presence of light

Answer 1

Answer : 2,2-dimethylbutane actually has got 3 methyl radicals and 1 ethyl radical attached to the 2nd carbon.

The pairs of positions for the Br atoms are  in these ways

(a) both on the far end of the ethyl  

$(H_{3}C)_{3} - C - CH_{2} - CHBr_{2}$  

(b) both on the near C of the ethyl  

$(H_{3}C)_{3} - C - CBr_{2} - CH_{3}$  

(c) both on the ethyl but not on the same C  

$(H_{3}C)_{3} - C - CHBr - CH_{2}Br  $

(d) both on one of the methyls  

$ (H_{3}C)_{2} (Br_{2}HC) - C - CH_{2} - CH_{3}$  

(e) both on methyls, but not the same methyl  

$(BrH_{2}C) (H_{3}C)_{2} - C - CH_{2} - CH_{3}$  

(f) one on a methyl and one on far end of ethyl  

$(H_{3}C)_{2} (BrHC) - C - CH_{2} - CH_{2}Br  $

(g) one on a methyl and one on near C of ethyl.  

$ (H_{3}C)_{2} (BrHC) - C - CHBr - CH_{3}$