Question

In the reaction, \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{CH}} = {\rm{C}}{{\rm{H}}_{\rm{2}}} + {\rm{H}} - {\rm{X}} \to {\rm{Product}}\] . What is the product?
A) \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}} - {\rm{C}}{{\rm{H}}_{\rm{3}}}\]
B) \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{C}}{{\rm{H}}_{\rm{2}}} - {\rm{C}}{{\rm{H}}_2}{\rm{X}}\]
C) \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}} - {\rm{CHX}} - {\rm{C}}{{\rm{H}}_{\rm{3}}}\]
D) \[{\rm{C}}{{\rm{H}}_3} - {\rm{C}}{{\rm{H}}_2}{\rm{X}} - {\rm{CH}} = {\rm{C}}{{\rm{H}}_2}\]

Answer 1

Hint: Here, the reaction of an alkene with a hydrogen halide (HI, HCl, HBr) is given. So, the Markovnikov rule is to be followed while answering this question.

Complete Step by Step Solution:
Let's first understand what the Markovnikov rule states. According to this rule, when an alkene undergoes a reaction with hydrogen halide, the double bond of the alkene breaks and the hydrogen atom forms a bond with that carbon atom of the double bind to which most numbers of atoms of hydrogen are bonded. And the halogen atoms get attached to another carbon atom of the double bond. For example, the reaction of propene and hydrogen chloride gives 2-iodo-propane.
\[{{\rm{H}}_{\rm{3}}}{\rm{C}} - {\rm{CH}} = {\rm{C}}{{\rm{H}}_{\rm{2}}} + {\rm{HCl}} \to {{\rm{H}}_{\rm{3}}}{\rm{C}} - {\rm{CH(Cl)}} - {\rm{C}}{{\rm{H}}_3}\]

In the same way, the given alkene undergoes a reaction with HX. Markovnikov's rule is to be followed here also.
\[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{CH}} = {\rm{C}}{{\rm{H}}_{\rm{2}}} + {\rm{H}} - {\rm{X}} \to {{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{CH(X)}} - {\rm{C}}{{\rm{H}}_3}\]
Therefore, option C is right.

Additional Information: Let's discuss what Markovnikov's rule is. This rule is the opposite of Markovnikv's rule. This rule applies to the reaction of an alkene with HBr in the medium of peroxide. According to this rule, the halogen atom forms a bond with that carbon atom of the double bond to which most numbers of hydrogen atoms are bonded, and the hydrogen atom forms a bond with another carbon atom of the double bond.
\[{{\rm{H}}_{\rm{3}}}{\rm{C}} - {\rm{CH}} = {\rm{C}}{{\rm{H}}_{\rm{2}}} + {\rm{HBr}} \overset{Peroxide}{\rightarrow} {{\rm{H}}_{\rm{3}}}{\rm{C}} - {\rm{C}}{{\rm{H}}_2} - {\rm{C}}{{\rm{H}}_2}{\rm{(Br)}}\]

Note: It is not to confuse the reaction of other hydrogen halides except HBr with alkene in presence of catalyst peroxide. Only, with HBr, the anti-Markovnikov product is obtained. With other hydrogen halides, Markovnikov product is obtained with peroxide.