The expected product is 3-chloropropan-1,2-diol since secondary carbocation is more stable than primary but it is not formed.

I have surfed the internet a lot regarding this and none seems to be a simple, reasonable answer. Can anyone please explain this mechanism clearly.

  • $\begingroup$ Read this doi.org/10.1039/JR9500001282 $\endgroup$
    – Waylander
    Jun 13 at 7:39
  • $\begingroup$ Yes thank you Waylander, ive read articles 260 and 261. It seems to be very complicated though. Is the reason really a tricky one or is there a simpler one? I'm only a high school student so I would appreciate a simpler answer? Nevertheless, thank you. -GW $\endgroup$ Jun 13 at 8:22

1 Answer 1


The answer seems to be pretty simple, surprisingly.

I've asked my chemistry sir and this is what he said. see the underlined step

In the underlined step, 2° carbocation is supposed to be formed.

But there is -I effect from OH group which makes the carbocation close to it, that is, 2° carbocation less stable.

Hence to avoid that, 1° carbocation is formed away from OH group and then the nucleophile 'OH minus' from HOCl attacks the 1° carbocation so formed.

Thus, 2-chloropropan-1,3-diol is formed as an intermediate instead of 3-chloropropan-1,2 diol.


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