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I was wondering whether there is a trick in the following problem:

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Since both carbons are equally substituted, shouldn't the products be the same with or without the use of peroxides?

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    $\begingroup$ Equally substituted but not equal enviroments. The Ph group makes a significant difference to the stability of both radicals and cations. This will result in regioselectivity for both reactions. $\endgroup$ – Waylander Apr 15 '20 at 11:34
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The use of HBr with peroxide leads to an anti-Markonikov's product(known as the peroxide effect).The carbons are both substituted once, but not equally! One has a methyl group while the other side has a phenyl group. The chemical basis for the Markovnikov's rule is the formation of a a stable carbocation during the addition process.

So, with just HBr a nomal Markovnikov's addition takes place i.e the Br group gets attached adjacent to the phenyl group while the H on the next carbon(as a benzylic carbocation is more stable).Final product: Ph-CHBr-CH2-CH3. With the peroxide however , the opposite happens leading to the formation if Ph-CH2-CHBr-CH3.

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