The major product of the following chemical reaction is CH3−CH|CH3−CH=CH2+HBr→C6H5CO2O2

C H 3 − C H | B r − C H | C H 3 − C H 3

C H 3 − C H | B r − C H | C H 3 − C H 3

C H 3 − C H | C H 3 − C H | B r − C H 3

C H 3 − C H | C H 3 − C H 2 − C H 2 B r

C H 3 − C H | B r − C H | C H 3 − C H 3

C H 3 − C | C H 3 | B r − C H 2 − C H 3

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The major product of the following chemical reaction is
$C H_{3} - \overset{\overset{C H_{3}}{|}}{C H} - C H = C H_{2} + H B r \overset{{(C_{6} H_{5} C O)}_{2} O_{2}}{\to}$

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$C H_{3} - \underset{C H_{3}}{\underset{|}{C H}} - \underset{B r}{\underset{|}{C H}} - C H_{3}$

$C H_{3} - \underset{C H_{3}}{\underset{|}{C H}} - C H_{2} - C H_{2} B r$

$C H_{3} - \underset{B r}{\underset{|}{C H}} - \underset{C H_{3}}{\underset{|}{C H}} - C H_{3}$

$C H_{3} - \underset{B r}{\underset{|}{\overset{\overset{C H_{3}}{|}}{C}}} - C H_{2} - C H_{3}$

answer is C.

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$C H_{3} - \underset{C H_{3}}{\underset{|}{C H}} - C H_{2} - C H_{2} B r$

Follows anti markornikoff rule

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