What are the intermediate products involved in the following reaction?
$C H C l_{3} 4 O H^{-} - --- \to H C O O^{-} + 3 C l^{-} + 2 H_{2} O$

C C l_{3}^{-}

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C C l_{2}

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C H (O H)_{3}

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C O C l_{2}

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Solution

The correct options are
A $C C l_{3}^{-}$
C $C C l_{2}$
In this Reaction , CHCl3 reacts with hydroxide ion ( OH− )
CHCl3 + OH− ⟶ CCl3−
And when one chlorine is eliminated then Carbene is formed.
CCl3− ⟶ : CCl2 (carbene )
Here, two intermediates formed in this reaction , CCl3− and : CCl2 .

Since, Option A and B are correct.

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Similar questions

C H C l_{3} + 4 O H^{-} \to H C O O^{-} + 3 C l^{-} + 2 H_{2} O

C H C l_{3} + O H^{-} \to H C O O^{-}

R - C O O H

N_{3} H

R N H_{2}

R - C H_{2} - C C l_{2} - R ⟶ R - C H = C C l - R

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