enter image description here

My thought-process:

According to me, this reaction is like the bromination of alkene but here instead of $\ce{CCl4}$ solvent, we have $\ce{H2O}$ as a solvent. For the first step a $3$-Membered Chlorinium Ion Intermediate is formed:

enter image description here

Now how do I proceed further? The solvent is $\ce{H2O}$ so it cannot act as a good nuclophile, whereas we still have the Cl- ion as a nucleophile available, so why doesn't the reaction proceed just like it would in $\ce{CCl4}$ medium, giving us a vicinal dichloride product.

According to the solution I have, a carbocation (the more stable out of the two possible carbocations) is formed and then $\ce{H2O}$ acts as a nucleophile and according to that the product is formed. But can a carbocation form when the intermediate is a 3 membered halonium ion? I don't know about this. I'm confused, isn't this a SN$1$ type reaction, so how can a carbocation form?

The answer given is:

enter image description here

P.S: This question was asked in JEE Mains $12$ April $2019$ Shift $1$

  • $\begingroup$ You underestimate how good a nucleophile water is. $\endgroup$
    – Waylander
    Mar 18, 2022 at 7:15
  • $\begingroup$ @Waylander Negatively charged species are better nucleophile than neutral species. In this case, we have bromide ion, so comparitively, isn't water a not so good nucleophile here? $\endgroup$ Mar 18, 2022 at 7:18
  • 3
    $\begingroup$ You are incorrect about how good a nucleophile Chloride ion is, and there is far more water present than chloride. $\endgroup$
    – Waylander
    Mar 18, 2022 at 9:21

1 Answer 1


Once you have the chloronium ion, you can think of it as having either of two carbocation structures, which are formed by breaking one or the other carbon-chlorine bond. Whichever carbocation is more stable forms with lower activation energy, so the nucleophile (water) will preferentially attack that carbocation center.

Here, if you break the carbon-chlorine bond on the right a primary carbocation is formed, whereas breaking the left bond gives a secondary cation. The secondary cation is more stable, so the water is directed accordingly.

  • $\begingroup$ Part of my question is why does bromide ion not attack? Is it becuase water is present in large quantity as compared to bromide ion? $\endgroup$ Mar 18, 2022 at 13:55
  • 1
    $\begingroup$ If you mean chlorine (rather than bromine), yes - assuming you are using chlorine water for the reaction there is much more water than chlorine. $\endgroup$
    – Ian Bush
    Mar 18, 2022 at 15:00

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