Question

you are asked to prepared 1-bromo-2,2,3,3-tetramethlylbuane by exposure to 2,2,3,3-tetramethylbutane to Br2 in the presence of light. How can yield of this monobrominated product be maximized while minimazing the formation of polybrominated product?

Answer 1

Answer: Treatment of alkanes with Br₂ in presence of light is the radical substitution of R-H that generates the alkyl bromide and HBr, is that the desired product (in this case CH₃C(CH₃)₂C(CH3)₂CH₂Br) may be further brominated. The remaining H's that remain on 1-bromo-2,2,3,3-tetramethlylbuane are reactive toward Br. As a result, it is difficult to prevent the bromination of to give di,tri,terta,penta-bromo product and so on leading to polybromination of the product. We can minimize polybromination by using a large excess of the 2,2,3,3-tetramethylbutane reactant compared to Br2. We can also minimize polybromination yield by permitting the monobromination reaction to proceed only partly to completion. Each of these strategies causes the unreacted 2,2,3,3-tetramethylbutane to have a higher concentration than that of the monobromoalkane product. As a result, bromine atoms in the reaction mixture encounter and react with unbrominated molecules much more frequently than with 1-bromo-2,2,3,3-tetramethlylbuane molecules that are present in much lower concentration.