Question

The reaction of propene with $HOCl$ 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 1

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.
$Acid + Base \rightleftharpoons Conjugate\,Base + Conjugate\,Acid$
In the case of hypochlorous and hypobromous acids $\left( {HOX} \right)$ 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.
$HOCl + HOCl\xrightarrow{{}}{H_2}O + OC{l^ - } + C{l^ + }$
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,
$C{H_2} = C{H_2}\xrightarrow{{HOCl}}{\text{ }}HO - C{H_2} - C{H_2} - Cl$