Question

$ CH_3 - CH_2 - CH = CH_2 + HBr$ $ \xrightarrow{ROOR\text{(peroside)}} \underset{\text{Major}}{(X)} + \underset{\text{Minor}}{(Y)}$ $X$ and $Y$ respectively are

• A. $BrCH_2 - CH_2 - CH = CH_2$ and $C_2H_5 - CHBr - CH_3$
• B. $C_2H_5 - CH_2 - CH_2Br $ and $Br - CH_2 - CH_2 - CH =CH_2$
• C. $C_2H_5 - CH_2 - CH_2Br$ and $C_2H_5 - CHBr - CH_3$
• D. $C_2H_5CHBr - CH_3$ and $C_2H_5 - CH_2 - CH_2Br$

Answer 1

The Correct Option is C

$CH _{3}- CH _{2}- CH = CH _{2}+ HBr \xrightarrow[{\text { (peroxide) }}]{ ROOR }$ $ \underset{\text{(Major)}}{CH _{3}- CH _{2}- CH _{2}- CH _{2} Br }+ \underset{\text{(Minor)}}{CH_3CH_2 - CHBr - CH_3}$ The main responsible factor here is organic peroxide which dissociates butene into free radicals and modifies the electrophilic addition to free radical addition which passes through free radicals as their intermediates. i.e. $ R O O R \longrightarrow 2 R O^{\bullet} $ $R O^{\bullet}+ HBr \longrightarrow R- OH + Br ^{\bullet}$

Now, more stable free radical attacks $H - Br$ molecule to form anti-Markownikoff product.

The modified case of $HBr$ in presence of peroxide is called kharasck effect and considered as an example of anti Markounikoff off addition.