Suppose we have an alkene(e.g. 2-butene). We try to subject it to allylic bromination using NBS/$\ce{CCl4}$ and then the mixture is separated using fractional distillation. According to my understanding, the reaction proceeds with a free radical pathway and thus we must get three products namely cis, trans 1-bromobut-2-ene and the rearranged product. However, the answer is given as 2 . Where am I going wrong?
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$\begingroup$ Is your but-2-ene starting material an E/Z mixture or single isomer? $\endgroup$– WaylanderCommented Jan 6, 2019 at 10:59
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$\begingroup$ Not specified and that is the main problem $\endgroup$– Aditya GargCommented Jan 6, 2019 at 11:03
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1 Answer
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You are quite correct that this is a radical bromination, know as the Wohl–Ziegler reaction. The mechanism is here Wohl–Ziegler. The key point is that the radical intermediate at the allylic position is stabilised by resonance with the double bond giving a 3-centred intermediate. This can brominate at either end (favouring the more substituted), but it does not isomerise the double bond. So I think the question setters are refering to 3-bromobut-1-ene and 1-bromobut-2-ene as the possible products without concerning themselves with the E/Z isomers.
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1$\begingroup$ The OP suggested (E)-"ex"-2-butene as a substrate (bp 0.8 oC). Given that the W-Z NBS reaction is normally done in refluxing CCl4 (bp 77 oC) to enhance the solubility of NBS. There may be difficulty in containing the substrate in the reaction vessel. Dry ice/actone condenser??? $\endgroup$ Commented Jan 6, 2019 at 23:30
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2$\begingroup$ A poorly chosen example substrate, perhaps. $\endgroup$ Commented Jan 7, 2019 at 8:49