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Question: The reaction of propene with \(HOCl\left( {C{l_2} + HCl} \right)\)with ethene proceeds through the i...

The reaction of propene with HOCl(Cl2+HCl)HOCl\left( {C{l_2} + HCl} \right)with ethene proceeds through the intermediate
A. CH3−CHCl−CH2+C{H_3} - CHCl - C{H_2}^ +
B. CH3−CH+−CH2−OHC{H_3} - C{H^ + } - C{H_2} - OH
C. CH3−CH+−CH2−ClC{H_3} - C{H^ + } - C{H_2} - Cl
D. CH3−CH(OH)−CH2+C{H_3} - CH\left( {OH} \right) - C{H_2}^ +

Explanation

Solution

The chemical reaction where propene is treated with HOCl  HOCl\;gives a single molecule since it is an addition reaction. Since the HOCl  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  HOCl\; to its mechanism so that we can get its intermediate.
Electrophilic addition reactions are the characteristic reactions of alkene where π\pi -electron clouds act as an electron rich site for the attack of electrophiles. So the electrophile HOCl  HOCl\;will attack the π\pi -electron cloud of propene. The chemical formula of propene isCH3−CH=CH2C{H_3} - CH = {\text{C}}{{\text{H}}_2}. The electrophile HOCl  HOCl\;due to its charge separation they form asHOδ−−Clδ+H{O^{\delta - }} - C{l^{\delta + }}. There occurs a small negative charge for OH−O{H^ - }and positive charge forCl+C{l^ + }.
Now, let us look at what happens in propene. The π\pi -electron will change the electron distribution by rearrangement and the positive charged Cl+C{l^ + }will attack them forming
CH3−CH=CH2→electrophilic  additionHOδ−−Clδ+CH3−CH+−CH2−ClC{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 CH3−CH+−CH2−ClC{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 Cl+C{l^ + }attacked at the end of the propene. Now this carbocation is attacked by OH−O{H^ - }
CH3−CH+−CH2−Cl→OH−CH3−CH(OH)−CH2−ClC{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 isCH3−CH+−CH2−ClC{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.