Identify the major product (99 %) formed when | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The reaction of $(CH_3)_3CH$ (isobutane) with $Br_2$ in the presence of $UV$ light is a free radical substitution reaction.Bromination is highly s

Vedclass · Accepted Answer

$(CH_3)_3CBr$

Vedclass · Answer

(B) The reaction of $(CH_3)_3CH$ (isobutane) with $Br_2$ in the presence of $UV$ light is a free radical substitution reaction.Bromination is highly selective compared to chlorination.The stability of the intermediate free radical determines the major product.The tertiary free radical $(CH_3)_3C^{\bullet}$ is more stable than the primary free radical $(CH_3)_2CHCH_2^{\bullet}$.Therefore,the bromine atom replaces the hydrogen atom at the tertiary carbon to form $2-$bromo$-2-$methylpropane as the major product $(99 \%)$.

Identify the major product $(99 \%)$ formed when $(CH_3)_3CH$ is treated with $Br_2$ in $UV$ light.

Similar Questions

The conversion of $CH_4$ to $CH_3Cl$ is an example of which of the following reactions?

Which alkane gives only one mono-chlorinated product?

Which of the following has the highest boiling point?

Pure methane can be produced by:

At $25,^oC$,Ethylene glycol is a:

Vedclass Test Series

Exam Paper Generator

Online Exam Module