Question

\[C{{H}_{2}}=CHCl\] reacts with \[HCl\] to form
A) \[C{{H}_{2}}Cl-C{{H}_{2}}Cl\]
B) \[C{{H}_{3}}-C{{H}_{2}}C{{l}_{2}}\]
C) \[C{{H}_{2}}=CHCl.HCl\]
D) None of these

Answer 1

Hint: In the electrophilic addition of hydrogen halide to alkene according to markownikoff’s rule most stable carbocation gives the major product. Again the carbocation which contains any electron donating group or +R effect group is most stable.

Complete Step by Step Answer:
When \[C{{H}_{2}}=CHCl\]is treated with hydrogen chloride an electrophilic addition reaction takes place.
In electrophilic addition reaction hydrogen halide gets attached to the alkene. In this reaction in the first step the alkene gets protonated to give the stable carbocation at the carbon atom where chlorine atom is attached. In the next step, that chlorine atom attacks the carbocation to form the desired alkyl halide.

Thus the reaction takes place as follows:
\[C{{H}_{2}}=CHCl+HCl\to C{{H}_{3}}-CHC{{l}_{2}}\]
Thus the hydrogen atom gets attached to the methilinic carbon atom and the chlorine atom gets attached to the carbon atom which contains one chlorine atom.
Thus the correct option is B.

Note: If this alkene is treated with a peroxide with alkyl halide then a different product will be formed. Then the reaction will proceed through a free radical mechanism. Then this will follow This anti markovnikov's rule. According to anti markovnikov's rule in electrophilic addition reaction when the alkene is treated with alkyl halide in presence of peroxide then the least stable carbocation forms the major product.