Question

Question: The product of the following reaction :
\[C{H_2} = CH - CC{l_3} + HBr\]
A. \[C{H_3} - CH(Br) - CC{l_3}\]
B. \[C{H_2}(Br) - C{H_2} - CC{l_3}\]
C. \[BrC{H_2} - CHCl - CHC{l_2}\]
D. \[C{H_3} - C{H_2} - CC{l_3}\]

Answer 1

Hint: The alkenes undergo an addition reaction with electrophiles. In the process of electrophilic addition generally carbocations are formed. The final product of the reaction depends on the type of carbocation stable in this reaction.

Complete step-by-step answer:
When \[C{H_2} = CH - CC{l_3}\] undergoes an addition reaction, there can be the possibility of two types of carbocation- \[C{H_3} - C^+H_2 - CC{l_3}\] and \[C^+{H_2} = CH_2 - CC{l_3}\]. Due to -I effect of Cl-group the carbocation \[C{H_3} - C^+H_2 - CC{l_3}\] will be destabilized in comparison to the carbocation \[C^+{H_2} = CH_2 - CC{l_3}\]. So, the $Br^-$ ion will attack on \[C^+{H_2} = CH_2 - CC{l_3}\] to form the final product. The complete reaction is shown below of the given alkene and HBr.

\[C{H_2} = CH - CC{l_3} + HBr \rightarrow C{H_2}(Br) - C{H_2} - CC{l_3}\]

Therefore the product of the reaction is \[C{H_2}(Br) - C{H_2} - CC{l_3}\].

Option ‘B’ is correct

Additional Information:For those reactions where the creation of a carbocation is not involved but the free radical is involved, anti-mechanisms markovnikov's apply. According to the anti-laws of Markovnikov's the carbon atom with the most hydrogen gets linked to the negative portion of the substrate. This is the proper Markovnikov addition reaction in reverse. In this reaction anti-Markovnikov's product is formed due to -I effect.

Note:If an electron withdrawing group such as a halo group is attached to any carbocation as it will pull electron density from that positively charged carbon making its electron density even less and as a result the carbocation will be less stable.