The reaction of propene with $\mathrm{HOCl}\left(\mathrm{Cl}_{2}+\mathrm{H}_{2} \mathrm{O}\right)$ proceeds through the intermediate:

(A) $\mathrm{CH}_{3}-\mathrm{CH}^{+}-\mathrm{CH}_{2}-\mathrm{Cl}$

(B) $\mathrm{CH}_{3}-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}_{2}^{+}$

(C) $\mathrm{CH}_{3}-\mathrm{CHCl}-\mathrm{CH}_{2}^{+}$

(D) $\mathrm{CH}_{3}-\mathrm{CH}^{+}-\mathrm{CH}_{2}-\mathrm{OH}$

JEE Main · 2016 · 03 Apr 2016 · Offline · Hydrocarbons

The reaction of propene with

HOCl (Cl_{2} + H_{2} O)

proceeds through the intermediate:

Answer: (A)

Adding

HOCl

to propene is an electrophilic addition in which

Cl X^{+}

attaches first. It goes to the terminal carbon, leaving the positive charge on the middle carbon as the more stable secondary carbocation

CH X_{3} - CH X^{+} - CH X_{2} Cl

. Water then attacks this carbocation to finish the chlorohydrin, so the intermediate is

CH X_{3} - CH X^{+} - CH X_{2} Cl

JEE Main · 2016 · 03 Apr 2016 · Offline · Single correct option