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I believe that the first step will be the formation of a radical at the first carbon of the ethyl group because vinyllic radicals are quite stable. However, I'm not really sure what happens after that. The answer key has 'I' as being the correct answer.

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    $\begingroup$ You meant benzylic, rather than vinylic radical, isn't it? $\endgroup$ – Klaus-Dieter Warzecha Mar 9 '16 at 6:28
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You are correct in your prediction for this first reaction:

RXN1

The second is simply an $\mathrm{E}2$ $\beta$-elimination reaction, which forms product I:

RXN2

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  • $\begingroup$ With heat, would any IV formed just eliminate to the same product? Or do you think they're just doing the main product? Or would IV just not be very likely with a hard nucleophile like that? $\endgroup$ – SendersReagent Mar 9 '16 at 6:55
  • $\begingroup$ I don't think it would very easy for IV to form in basic conditions. In acidic conditions that could easily form, however. $\endgroup$ – ringo Mar 9 '16 at 6:59

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