Question

The intermediate during the addition of HCl to propene in the presence of peroxide is
A) \[{\rm{C}}{{\rm{H}}_{\rm{3}}}\mathop {\rm{C}}\limits^{\rm{ \bullet }} {\rm{HC}}{{\rm{H}}_{\rm{2}}}{\rm{Cl}}\]
B) \[{\rm{C}}{{\rm{H}}_{\rm{3}}}\mathop {\rm{C}}\limits^{\rm{ + }} {\rm{HC}}{{\rm{H}}_{\rm{3}}}\]
C) \[{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{H}}_{\rm{2}}}\mathop {\rm{C}}\limits^{\rm{ \bullet }} {{\rm{H}}_{\rm{2}}}\]
D) \[{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{H}}_2}\mathop {\rm{C}}\limits^{\rm{ + }} {{\rm{H}}_2}\]

Answer 1

Hint: The reaction intermediates are short-lived and highly reactive species formed due to heterolytic and homolytic fission of bonds. Some reaction intermediates are carbocations, nitrenes, benzyne, etc.

Complete Step by Step Answer:
Let's understand what a carbocation is. A carbocation defines a species carrying the positively charged atom of carbon. The positive charge on the carbon atom is due to the electron pair attraction by the electronegative atom bonded to that carbon atom.

Let's understand the peroxide effect in detail. When an alkene undergoes a reaction with HBr in peroxide, the reaction gives a product contrary to Markovnikov's rule. But, there is no effect of peroxide on HCl and HI.

Here, propene undergoes a reaction with HCl in the presence of peroxide. The reaction happens the following way:

Fig: Formation of carbocation

The above reaction shows that the H-Cl bond breaks and the formation of a stable carbocation take place. The secondary carbocation is the one in which the positively charged carbon atom is bonded to two other carbon atoms. And secondary carbocation is formed because of its more stability than primary carbocation.
Hence, option B is right.

Note: When an asymmetrical alkene undergoes a reaction with a hydrogen halide, the addition of H atoms takes place at the Carbon atom of the double bond to which more numbers of hydrogen atoms are boned. And the halogen atom adds to the other atoms of the double bond.