The reaction of propene with $H O C l$ proceeds via the addition of,
A) $C l^{+} & O H^{-}$ in a single step.
B) $C l^{+}$ in the first step.
C) $H^{+}$ in the first step.
D) $O H^{-}$ in the first step.

Answer

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Hint:We know that electrophilic addition is a reaction between an electrophile and nucleophile, adding to double or triple bonds. A molecule that tends to react with other molecules containing a donatable pair of electrons is called an electrophile.

Complete step by step answer:

The Bronsted Lowry theory is based on an acid and base reaction.
We know that an acid is a species which has the capacity of donating a proton that is hydrogen ion is called Bronsted-Lowry acid and a base is a species which has the capacity of accepting proton and a base it needs to have a lone pair of electrons on the base which bonds to the hydrogen ion.
Now, we see Conjugate acid.
According to Bronsted-Lowry theory, the conjugate acid is the chemical species which is formed after the base accepts the hydrogen atom is called conjugate acid.

A c i d + B a s e ⇌ C o n j u g a t e B a s e + C o n j u g a t e A c i d

In the case of hypochlorous and hypobromous acids

(H O X)

these weak Bronsted acids do not react as proton donors; and since oxygen is more electronegative than chlorine or bromine, the electrophile will be a halide cation.
Alkenes undergo Electrophilic addition reactions on self-ionization produces

C l^{+}

ion. This chloride ion attacks the propene first.

H O C l + H O C l \overset{}{\to} H_{2} O + O C l^{-} + C l^{+}

Therefore, option B is correct.

Note:
We must remember when ethylene on addition with hypochlorous acid forms a product where double bond breaks and electrophile and nucleophile addition take place.
The product is,