[RedsAreRaw]

Had a question:
"(i) How do you account for the fact that when allyl bromide is treated with dilute H₂SO₄, there is obtained not only 1-bromo-2-propanol, but also 2-bromo-1-propanol? (ii)In contrast, allyl chloride yields only one product, 1-chloro-2-propanol. How do you account for this difference between the chloride and the bromide?"

I was thinking for (i) the intermediate would be a bromonium ion and attack from water from either side. So for (ii) I wasn't completely sure why chlorine wouldn't do the same. I was guessing that it wouldn't hold the chloronium ion in a three-ring structure like bromine would be able to.

Thanks!

[orthoformate]

Had a question:
"(i) How do you account for the fact that when allyl bromide is treated with dilute H₂SO₄, there is obtained not only 1-bromo-2-propanol, but also 2-bromo-1-propanol? (ii)In contrast, allyl chloride yields only one product, 1-chloro-2-propanol. How do you account for this difference between the chloride and the bromide?"

I was thinking for (i) the intermediate would be a bromonium ion and attack from water from either side. So for (ii) I wasn't completely sure why chlorine wouldn't do the same. I was guessing that it wouldn't hold the chloronium ion in a three-ring structure like bromine would be able to.

Thanks!

I think you are correct.