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I would assume options A and B are possible products due to the $\ce{I+}$ electrophile attacking the alkene, and hence are ruled out.

However, I am torn between options C and D. Both products are formed via the thermodynamically unfavored generation of $\ce{Br+}$, followed by an attack of either iodide or $\ce{H2O}$. While iodide might be a better nucleophile than water by virtue of the former being a charged ion, the concentration of water is likely to exceed that of the iodide ion manyfold. I am of the opinion that both C and D are possible products, and if one cannot be formed, I am unable to decide which one.

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  • $\begingroup$ You need to think about the mechanism and in particular whether you form a bromonium ion or an iodonium ion as an intermediate. If you form an iodonium ion, then one of the two things that adds has to be an iodine. Unlike the addition of HX, this isn't an electrophilic addition where you form a carbocation. $\endgroup$
    – orthocresol
    Commented Sep 29, 2018 at 11:55
  • $\begingroup$ @orthocresol if we're forming an iodonium ion, for the I to end up on the middle carbon then the attack by the nucleophile must be on the less substituted carbon, which I thought isn't the case for cationic 3-membered rings? Please correct me if I'm wrong. $\endgroup$
    – Jonathan Smith
    Commented Sep 29, 2018 at 14:25
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    $\begingroup$ Yes, you’re right. But I suppose there’s possibly a minor product arising from attack at the less substituted carbon. $\endgroup$
    – orthocresol
    Commented Sep 29, 2018 at 15:09
  • $\begingroup$ But is the probability of an attack at the less substituted carbon greater than the probability of the Br+ electrophile generated? $\endgroup$
    – Jonathan Smith
    Commented Sep 29, 2018 at 15:22
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    $\begingroup$ Good question - not one that I can definitively answer with rigorous proof, but based on experience, I would say yes. By the way, there is no discrete (i.e. free) Br+ or I+ being generated - the alkene reacts directly with I-Br and the selectivity in that step is determined by the inherent polarisation of the I-Br bond. $\endgroup$
    – orthocresol
    Commented Sep 29, 2018 at 15:45

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I would say C is unlikely to form because this requires the formation of a primary carbocation like carbon with no opportunity for a hydride shift.

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  • $\begingroup$ i think C this more or less requires a cation or attack by water on a less stable complex with an iodine passively sitting there. I would thing that a small number might react in reverse. [I despise this type of question. If the reaction was run they knew not the results before so why expect a student to know. If the prof found it give the data and discuss then ask the question.] $\endgroup$
    – jimchmst
    Commented Dec 14, 2022 at 22:14

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