Question

The reaction of propene with $HOCl\left( {C{l_2} + HCl} \right)$with ethene proceeds through the intermediate
A. $C{H_3} - CHCl - C{H_2}^ + $
B. $C{H_3} - C{H^ + } - C{H_2} - OH$
C. $C{H_3} - C{H^ + } - C{H_2} - Cl$
D. $C{H_3} - CH\left( {OH} \right) - C{H_2}^ + $

Answer 1

Hint:The chemical reaction where propene is treated with $HOCl\;$gives a single molecule since it is an addition reaction. Since the $HOCl\;$is an electrophile, the addition reaction here is an electrophilic addition reaction. By the mechanism of addition reaction we can the intermediate of the product.


Complete step by step answer:We all know that electrophiles are electron deficient species that are attracted to an electron-rich center. Here, we have an electrophile which will react with propene in an electrophilic addition reaction. Let us look in $HOCl\;$ to its mechanism so that we can get its intermediate.
Electrophilic addition reactions are the characteristic reactions of alkene where $\pi $-electron clouds act as an electron rich site for the attack of electrophiles. So the electrophile $HOCl\;$will attack the $\pi $-electron cloud of propene. The chemical formula of propene is$C{H_3} - CH = {\text{C}}{{\text{H}}_2}$. The electrophile $HOCl\;$due to its charge separation they form as$H{O^{\delta - }} - C{l^{\delta + }}$. There occurs a small negative charge for $O{H^ - }$and positive charge for$C{l^ + }$.
Now, let us look at what happens in propene. The $\pi $-electron will change the electron distribution by rearrangement and the positive charged $C{l^ + }$will attack them forming
     \[C{H_3} - CH = {\text{C}}{{\text{H}}_2}\mathop {\mathop {\xrightarrow[{{\text{electrophilic}}\;{\text{addition}}}]{{H{O^{\delta - }} - C{l^{\delta + }}}}{\text{C}}{{\text{H}}_3} - {\text{CH}} + - {\text{C}}{{\text{H}}_2}}\limits_{} }\limits^{} - {\text{Cl}}\]
The $C{H_3} - C{H^ + } - C{H_2} - Cl$ acts as an intermediate because the secondary carbocation is more stable in the process, that is why the positive charged $C{l^ + }$attacked at the end of the propene. Now this carbocation is attacked by $O{H^ - }$
     $C{H_3} - C{H^ + } - C{H_2} - Cl\mathop \to \limits^{O{H^ - }} C{H_3} - CH\left( {OH} \right) - C{H_2} - Cl$
The final product obtained is the propylene chlorohydrin and the intermediate formed is$C{H_3} - C{H^ + } - C{H_2} - Cl$.

Hence, option (C) is correct.


Note:We should always keep in mind that the electrophile attacks depend upon the degree of accessibility of -electron, any factors that increase the electron density of the double bond will increase its reactivity towards electrophilic reagent. Electrophiles are known as electron lovers as well.