The compound which reacts with $^{'} H B r^{'}$ obeying markovnikov's rule is :

H_{2} C = {C H}_{2}

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H_{3} C - C H = C H - {C H}_{3}

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{C H}_{3} {C H}_{2} C H = C H - {C H}_{3}

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({C H}_{3}) C H = {C H}_{2}

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Solution

The correct options are
C ${C H}_{3} {C H}_{2} C H = C H - {C H}_{3}$
D $({C H}_{3}) C H = {C H}_{2}$
As option A & B are symmetrical so it will make same product and there is no use of markovnikov's rule.
But in case of C & D,
Alkenes are unsymmetrical so it will make different product when it reacts with HBr in presence of peroxide according to anti-markovnikov's rule.

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H_{3} C - C H_{2} - C H_{3} | C = C H - C | H C_{2} H_{5} - C H_{3} | C H - C H_{2} - C H_{2} - C H_{3}

H_{2} C = C H - C H_{2} - + C H - C H_{3}

H_{3} C - C H_{3} | C | C H_{3} - + C H_{2}

C H_{3} - C H = C H_{2} + H B r p e r o x i d e \to ?

H_{2} C = C H - C H_{3} + H B r \to ?

A . C H_{3} - C H = C H_{2} + H C l ⟶ C H_{3} - C | C l H - C H_{3}

B . C H_{3} - C H = C H_{2} + H B r ⟶ C H_{3} - C H_{2} - C H_{2} B r

C . C H_{3} - C H = C H_{2} + H B r p e r o x i d e - ---- \to C H_{3} - C H_{2} - C H_{2} - B r

D . C H_{3} - C H = C H_{2} + H I p e r o x i d e - ---- \to C H_{3} - C | I H - C H_{3}

C H_{3} - C H_{2} - C H = C H - C H_{2} - C H_{3} + H B r \to

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