Question

The addition of $HBr$ to propylene takes place opposite to Markovnikoff’s rule in presence of:
A. Sunlight
B. Hydrogen peroxide
C. Platinum catalyst
D. None of these

Answer 1

Hint: The Markovnikoff’s rule states that with the addition of a protic acid \[HX\] to an asymmetric alkene, the acid hydrogen (\[H\] ) or electropositive part gets attached to the carbon with more hydrogen substituents, and the halide (\[X\] ) group or electronegative part gets attached to the carbon with more alkyl substituents. Alternatively, the rule can be stated that the hydrogen atom is added to the carbon with the greatest number of hydrogen atoms while the \[X\] component is added to the carbon with the fewest hydrogen atoms.

Complete step by step answer:
Mechanisms that do not involve a carbocation intermediate may react through other mechanisms that have other regioselectivity not dictated by Markovnikoff’s rule, such as free radical addition. Such reactions are said to be anti-Markovnikov, since the halogen adds to the less substituted carbon, the opposite of a Markovnikov reaction. Similar to a positive charged species, the radical species is most stable when the unpaired electron is in the more substituted position. The anti-Markovnikov rule can be illustrated using the addition of hydrogen bromide to propene in the presence of hydrogen peroxide (${H_2}{O_2}$ ). The reaction of HBr with substituted alkenes was prototypical in the study of free-radical additions. The addition of $HBr$ to propylene, under anti-Markovnikov's rule can be demonstrated from the following reaction:
$C{H_3} - CH = C{H_2}\xrightarrow{{{H_2}{O_2}}}C{H_3} - CH(Br) - C{H_3}$
Thus, the correct option is B. Hydrogen peroxide.

Note:
Early chemists discovered that the reason for the variability in the ratio of Markovnikov to anti-Markovnikov reaction products was due to the unexpected presence of free radical ionizing substances such as peroxides. The explanation is that \[HBr\] produces a \[Br \bullet \] radical, which then reacts with the double bond.