Question

$C{H}_3-\underset{C{H}_3}{\underset{|}{C}}=C{H}_2\underset{peroxide}{\overset{HBr}{\to }}C{H}_3-\underset{C{H}_3}{\underset{|}{C}}H-C{H}_{2}Br$

The intermediate of this reaction is?

• A. $C{H}_3-\stackrel{\bullet }{\underset{C{H}_3}{\underset{|}{C}}}-C{H}_{2}Br$
• B. $C{H}_3-\underset{C{H}_3}{\underset{|}{C}}H-\stackrel{\bullet }{C}{H}_2$
• C. $C{H}_3-\stackrel{Br}{\stackrel{|}{\underset{C{H}_3}{\underset{|}{C}}}}-\stackrel{\bullet }{C}{H}_2$
• D. None of the above

Answer 1

The correct option is A $C{H}_3-\stackrel{\bullet }{\underset{C{H}_3}{\underset{|}{C}}}-C{H}_{2}Br$

Anti Markownikoff's rule:- When an unsymmetrical alkene is treated with an unsymmetrical reagent $\left(HX\right)$, in the presence of peroxide, the negative part of the reagent $(-X)$ gets added to the unsaturated carbon of the double bond which contains more number of hydrogen atoms.

(Refer to Image)

Here $HBr$ in presence of peroxide gets added to unsymmetrical alkene according to Anti-MArkownikoff's rule.