Question

In which of the following will the Kharasch effect operate?
(A) $C{{H}_{3}}-C{{H}_{2}}-CH-C{{H}_{2}}+HCl$
(B) $C{{H}_{3}}-C{{H}_{2}}-CH=C{{H}_{2}}+HBr$
(C) $C{{H}_{3}}-CH=CH-C{{H}_{3}}+HBr$
(D) $C{{H}_{3}}-C{{H}_{2}}-CH=C{{H}_{2}}+HI$

Answer 1

Hint: Kharasch effect is observed when unsymmetrical alkene reacts with HBr in presence of peroxides, then the negative part of HBr i.e. $HB{{r}^{-}}$ gets attached to the carbon having a greater number of alpha hydrogens.

Complete step by step solution:
Kharasch effect also known as “peroxide effect” and termed as anti-markovnikov's rule is the effect shown when unsymmetrical alkenes react with alkyl halides (especially HBr) in the presence of peroxides (mostly of benzenes) and the reaction proceeds via a free radical mechanism. The effect acts contrary to the markovnikov's rule i.e. when an unsymmetrical reagent adds to an unsymmetrical alkene, the negative part of the reagent gets attached to the carbon atom having a smaller number of hydrogen atoms. While in Kharasch effect, the negative part of the reagent gets attached to the carbon atom having a greater number of hydrogen atoms.
The below reaction shows the markovnikov's rule
$C{{H}_{3}}-CH=C{{H}_{2}}+HCl\to C{{H}_{3}}-CHCl-C{{H}_{3}}$
While the below reaction shows the anti-markovnikov's rule,
$C{{H}_{3}}-CH=C{{H}_{2}}+HBr\to C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}Br$
Remember, the Kharasch effect mostly occurs with HBr and unsymmetrical alkenes and alkynes.

Hence, the correct option is (B).

Additional Information:
Alkenes belong to the group of unsaturated hydrocarbons i.e. one molecule of alkene contains at least one double bond. Due to the presence of pi electrons, they show additional reactions in which an electrophile attacks the carbon-carbon double bond to form the additional products. This reaction is called anti-markovnikov's addition or Kharasch effect. This reaction is only observed with HBr, not with HCl or HI as HBr follows a free radical mechanism which cannot be followed by HCl or HI.
$C{{H}_{3}}-CH=C{{H}_{2}}+HBr\to C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}Br$ (in presence of peroxide of benzene)

Note: Possibly you can get confused with option (C) as it has HBr but the alkene given is symmetrical in nature thus does not show Kharasch effect. It should be kept in mind that the Kharasch effect can be shown by unsymmetrical alkenes reacting with only HBr.