Question

Identify the major product (99%) formed when (CH₃)₃ CH is treated with Br₂ in UV light.

• A. (CH₃)₃ CCH₂Br
• B. (CH₃)₃-C-Br
• C. CH₃-CH- Br CH₃
• D. CH₃-(CH₂)₂ - CH₂Br

Answer 1

Answer: (B)

(CH₃)₃-C-Br

Answer 2

When (CH₃)₃CH (a tertiary alkane) is irradiated with Br₂ in the presence of UV light, the bromine radical abstracts the most easily removed hydrogen (at the tertiary center) giving the most stable tertiary radical. Bromine then adds to this radical forming the tertiary bromide. Thus, the major product is

$$\text{(CH₃)₃C–Br}$$