In methyl alcohol solution, bromine reacts with ethylene to yield $B r C H_{2} C H_{2} O C H_{3}$ in addition to 1, 2-dibromoethane because

The ion formed initially may react with

B r^{-}

C H_{3} O H

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The methyl alcohol solvates the bromine

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The reaction follows Markownikoff's rule

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This is a free-radical mechanism

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Solution

The correct option is A The ion formed initially may react with $B r^{-}$ or $C H_{3} O H$
$C H_{2} = C H_{2} + B r_{2} C H_{3} O H - ---- \to$
$C H_{2} - C H_{2} + B r - C H_{2} - C H_{2} - O - C H_{3} | | B r B r$

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Similar questions

B r C H_{2} C H_{2} O C H_{3}

B r C H_{2} C H_{2} O C H_{3}

B r^{-} C H_{3} O H

In methyl alcohol solution, bromine reacts with ethylene to yield in addition to 1, 2-dibromoethane because

B r C H_{2} C H_{2} O C H_{3}

B r_{2}

1, 2 - dibromoethane

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