Question

please explain .........​

Answer 1

Answer:

As we know, halogenation of alkene takes place by free radical mechanism.

Also, the increasing order of stability of free radical is :

methyl radical < 1° radical < 2° radical < 3° radical

As per the question, option (b) is correct because chlorine is attached to 3° carbon.

Whereas, in other options,

Option (a) → chlorine is attached to 2° carbon.

Option (c) → chlorine is attached to 1° carbon

Option (d) → chlorine is attached to 1° carbon

Hence, the major product formed will be (b), whose IUPAC name is 2-chloro-2-methylbutane.

Answer 2

The given reaction is free radical chlorination in presence of light energy.

The process involves substitution of hydrogen by chlorine. But removal of hydrogen results in the formation of a free radical.

The interesting point is that, the product depends on what type of free radical is formed by the the reactant.

So, the reactant would prefer forming a much more stable free radical for initiating the reaction.

The order of stability of free radicals is

tertiary (3°) > secondary (2°) > primary (1°)

The given reactant has 3°, 2° as well as 1° carbon atoms.

So obviously, the free radical that the reactant will prefer to form will be the tertiary free radical.

So, the C-H bond will homolytically cleave and the Cl₂ also homolytically cleave and free chlorine atom will attach to the 3° carbon.

The product will be 2-Chloro-2-methylbutane [option (2)]

HOPE THIS HELPS!!