The reaction of propene with $H O C l$ proceeds via the addition of:

H^{+}

in first step

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C l^{+}

in first step

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O H^{-}

in first step

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C l^{+}

and

O H^{-}

in single step

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Solution

The correct option is B $C l^{+}$ in first step

In the case of hypochlorous and hypobromous acids (HOX), these weak Brønsted acids (pKa's ca. 8) do not react as proton donors; and since oxygen is more electronegative than chlorine or bromine, the electrophile will be a halide cation.

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Similar questions

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

H O C l (C l_{2} + H_{2} O)

H O C l ({C l}_{2} + H_{2} O)

3 C l O^{-} \to C l O_{3}^{-} + 2 C l

C l O^{-} + C l O^{-} K_{1} - \to C l O_{2}^{-} + C l^{-}

C l O_{2}^{+} C l O^{-} k_{2} - \to C l O_{3}^{-} + C l^{-}

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