The product when molecular bromine reacts with cyclohexene at room temperature?

Question

aaronq · Answer

hm not quite. First of all, you drew benzene (not cyclohexane) and it would brominate twice, in a trans configuration|dw:1378576020454:dw| I found this website a while back, they give you synthesis problems and it's very interactive, look into it http://highered.mcgraw-hill.com/classware/ala.do?alaid=ala_946617

aaronq · Answer

ps the synthesis problems start at chapter 6, but you should start from chapter 1

abb0t · Answer

HAHA! Gotta work your way from Chapter 1 to finish this! LOL

anonymous · Answer

:o kay..

But that was given as one of the option I had to choose from..
So I chose that and other options are also similar to that U_U

not like that you drew :I

but anyways thanks.

abb0t · Answer

By the way, benzene is planar @aaronq I don't think it can be drawn in chair conformation.

abb0t · Answer

Or, wait. I think you have drawn benzene @Ryaan . Is it bromobenzene or what?

aaronq · Answer

benzene sure is planar, but he wrote "molecular bromine reacts with cyclohexene", then he drew a benzene  lol

Ryan, were they all benzene rings? if so, that can't be right. you can't turn cyclohexane into benzene by brominating it.

anonymous · Answer

:I

aaronq · Answer

If Br was present as a radical, it would be A, though i'm not sure if the Br-Br bond is cleaved homolytically at 25 degrees without uv light.

maybe @abb0t can weigh in on this?

Normally, bromination proceeds like this

abb0t · Answer

The photo that @aaronq has provided is very nice! Follow that. Just want to say that it's not a radical reaction. Yes, without UV light. It's still an electrophilic acditiong

abb0t · Answer

The only plausible answer would be $D$, but I don't see why the bromines are 2 carbons away from each other.

aaronq · Answer

Yeah, i also thought D was the most plausible answer, but by the way the mechanism proceeds, it's impossible to have them not adjacent to each other.

abb0t · Answer

Yeah, a hydride shift of the $^+$that forms would be energetically unfavorable. But I would just go with $D$ and if you get it wrong, try and argue it or get an explanation from a professor as to why it is wrong. Best of luck!