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 '18 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.
