Draw the major monobromination product formed by heating the following alkane with bromine.
Concepts and reason
Bromination of alkane takes place via free radical mechanism. It involves the formation of free radical. The stability of free radical has the following order:
Fundamentals
On Bromination, bromine will get substituted to more substituting carbon because its radical is more stable.
It involves three steps:
- Initiation step: In this step, bromine is released as bromine radical by action of heat on
. - Propagation step: In this step, bromine radical abstract hydrogen from more substituted carbon of alkane or carbon with less number of hydrogen to for stable radical.
- Termination step: In this step, 2 bromine radicals react to form again.
Answer:
By the action of heat, 
breaks into 2 Br radicals as shown below:
There are 8 possible sites for removal of hydrogen by bromine radical as shown below:
All the labeled sites contain one or more than one hydrogen groups.
The number of hydrogen attached to each site is as follows:
C-1 carbon has less number of hydrogen. Therefore, radical is form at C-1 carbon as shown below:
There is one methyl group at C-1 carbon and one methyl and one ethyl group at C-2 carbon of cyclopentane. After the removal of hydrogen, more substituted carbon forms stable radical but there must be hydrogen group present to form radical.
There are three possible shifts; one is methyl shift from C-2 to C-1 carbon that will result in following radical:
The suffix is ane
The IUPAC name of the given structure is 3-ethyl-2,2-dimethylhexane.