azazel2113 Posted December 4, 2011 Share Posted December 4, 2011 Hi there, I have a question about a question. in the picture I have attached, the question says that the alkene undergoes hydrobromination. However, to me, the reaction given looks like a halogenation, not a hydrohalogenation. I was wandering if I am correct in thinking that or if I am missing something. Thank you Link to comment Share on other sites More sharing options...
Horza2002 Posted December 4, 2011 Share Posted December 4, 2011 Firstly, why don't you draw out the structures that you think you would get if it was a bromination. A hydrobromination is a specific type of halogenation (since bromine is a halogen). What do you get if you treat an alkene with bromine? Link to comment Share on other sites More sharing options...
azazel2113 Posted December 4, 2011 Author Share Posted December 4, 2011 Ok, so after I posted that question I worked it out as a hydrobromation. These are the two conformations of the product I got. How do they look? Link to comment Share on other sites More sharing options...
Horza2002 Posted December 4, 2011 Share Posted December 4, 2011 Lol, could you possible write any smaller They seem ok. So where did you hydrogen come from that you needed for the hydrobromination then? And which one did you decide was the lowest energy conformation? Link to comment Share on other sites More sharing options...
azazel2113 Posted December 4, 2011 Author Share Posted December 4, 2011 Sorry about the size of my writing haha. Took the picture with my camera phone. As far as where the H comes from, that is what is really confusing me. The next question actually asks to draw the mechanism of the most stable, which I believe is the first picture. However, I cannot figure out how to go from Br2 to HBr. Do you have any suggestions? Link to comment Share on other sites More sharing options...
Horza2002 Posted December 4, 2011 Share Posted December 4, 2011 If all you've put in is bromine, carbon tetrachloride and your start material, that would lead me to assume that it is actually a di-bromination....as you said, there is no hydrogen to come form anywhere. Link to comment Share on other sites More sharing options...
azazel2113 Posted December 4, 2011 Author Share Posted December 4, 2011 Ok, so if it is a di-bromination then br will be added to both sides of the alkene, correct? So in place of the H attatched to the C with a Cl, there would be another Br? Link to comment Share on other sites More sharing options...
hypervalent_iodine Posted December 5, 2011 Share Posted December 5, 2011 (edited) Correct. Are you familiar with the mechanism for di-brominations with Br2? It is said to go via a cyclic bromonium ion intermediate, like this: (Image from here) Ignore the image I got from there and that was sitting right here where you're reading, as it's been deemed inaccurate. See this post for an updated mechanism. Which means that you will really only get 1,2-dibromination. I am slightly confused as to why the questions states it is a hydrobromination, though. There's clearly no proton source in your reaction, so that isn't going to happen (unless I'm missing something). Edited December 5, 2011 by hypervalent_iodine Link to comment Share on other sites More sharing options...
azazel2113 Posted December 5, 2011 Author Share Posted December 5, 2011 Thank you very much. I think it was just an error on the professor's part, judging by his dismissive response. Link to comment Share on other sites More sharing options...
Horza2002 Posted December 5, 2011 Share Posted December 5, 2011 Hyper, the mechanism for the formation of the bromonium ion is wrong...you need another curly arrow from the bomrin thats being attacked lone pair to the other end of the alkene Link to comment Share on other sites More sharing options...
hypervalent_iodine Posted December 5, 2011 Share Posted December 5, 2011 That's what I get for not really looking at the picture, I guess. It's implicit Link to comment Share on other sites More sharing options...
Horza2002 Posted December 5, 2011 Share Posted December 5, 2011 Tut tut, I would expect better from someone like you to be honest... Link to comment Share on other sites More sharing options...
mississippichem Posted December 5, 2011 Share Posted December 5, 2011 That's what I get for not really looking at the picture, I guess. It's implicit Everyone and their mother knows where those electrons are going anyway . Strange that the diagram doesn't show them though. Link to comment Share on other sites More sharing options...
hypervalent_iodine Posted December 5, 2011 Share Posted December 5, 2011 I reckon I'll go back to my policy of drawing them all myself. And yes, damn Horza and his pedantic mechanism drawing Link to comment Share on other sites More sharing options...
Horza2002 Posted December 5, 2011 Share Posted December 5, 2011 No, I wouldn't say everyone knows where there going...loads of undergrads here when you ask where that bond comes from have no idea...they just draw it. And its not really about being pedantic...its about being accurate Link to comment Share on other sites More sharing options...
hypervalent_iodine Posted December 5, 2011 Share Posted December 5, 2011 (edited) Alright, sunshine, we get it and we were just poking fun. I happen to agree with you, I just missed it when going through and getting pictures. Anyway, problem solved and I'm sure the OP understands that there is an arrow in there that wasn't drawn in by the author. But heck, for the sake of accuracy: I even included the partial charges; just for you, Horza. Can't have you being any more disappointed in me, now can we? Edited December 6, 2011 by hypervalent_iodine Link to comment Share on other sites More sharing options...
Horza2002 Posted December 5, 2011 Share Posted December 5, 2011 (edited) You've still got it wrong. I did appreciate the partial charges Edited December 5, 2011 by Horza2002 Link to comment Share on other sites More sharing options...
mississippichem Posted December 5, 2011 Share Posted December 5, 2011 You've still got it wrong. Yeah thats it. Otherwise the cyclic bromonium intermediate never happens. Without that cyclic flow of electrons there is also a parity violation I believe. We just had to give you some hell Horza for leaving us Horza-less for so long. Good to have you back at the metaphorical hood. Link to comment Share on other sites More sharing options...
hypervalent_iodine Posted December 6, 2011 Share Posted December 6, 2011 You know it's bad when you wake up in the morning and the first sinking feeling you think is, 'Ah, crap, I forgot an arrow and why the hell did I post that at 2.23am'? Welcome back, Horza. Anyway, I have no idea what you're talking about, if you'll see, my mechanism is totally correct and I absolutely did not go and fix it and edit my post accordingly. Link to comment Share on other sites More sharing options...
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