1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Read this doi.org/10.1039/JR9500001282 $\endgroup$
    – Waylander
    Jun 13, 2022 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, 2022 at 8:22

1 Answer 1

1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.