In the conversion of propene to 1-propanol, can we just use water in presence of peroxide instead of first converting it into 1-bromopropane and then reacting with $\ce{NaOH}$?
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$\begingroup$ Please keep in mind isomers when talking about organic compounds. your question is not immediately obvious that you are talking about 1-propanol until a reader reaches your discussion of 1-bromopropane. also knowing the specific/general peroxide you have in mind would be helpful. $\endgroup$– A.K.Jul 24, 2018 at 14:29
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No, you can't.
If you observe the mechanism of Kharash effect, an $\ce{HBr}$ molecule first undergoes homolytic bond cleavage to produce a $\ce{Br}$ free radical. To undergo that homolytic bond fission, the electronegativities of the two elements, attached by that bond, have to be very close, which is possible for $\ce{H}$ and $\ce{Br}$ ($\chi _ {\ce{H}} = 2.1; \chi_{\ce{Br}} = 2.96$). The benzoyl peroxide is just a free radical initiator.
Now if you come to water, electronegativity difference between $\ce{H}$ and $\ce{O}$ is huge $(\chi _ {\ce{H}} = 2.1; \chi_{\ce{O}} = 3.44)$. So, it is not possible at all to undergo homolytic bond cleavage of $\ce{H-O}$ bond even in presence of peroxides.
To do hydration according to anti-Markownikoff's rule (without any rearrangement), the most widely used method is Hydroboration-Oxidation. This is an alternative and better than doing first bromination in presence of peroxides and then hydrolysis, because in many other cases chances of rearrangement also remains there.
answered Jul 24, 2018 at 13:10