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I need some serious help with the mechanism of ionic addition, both my textbook and my teacher are at odds with internet videos (khan academy) which both seem to be at odds with quote, "recent evidence" on the topic.

I don't need to know much of this for my exam, but basically I want to understand it properly because I hate the idea that my foundation is built up badly from the start.

Here's the deal:

During the ionic addition of say, Bromine-Bromine to ethene, my textbook says that 'recent evidence shows' that there is a 'bridged structure' formed when the diatomic bromine breaks up and attacks the double bond.

I get this, the bromine becomes polarised when approaching the ethene at a right angle, leading to one bromine atom losing an electron to the other, more negative bromine.

This positively charged bromine which is two electrons short of a full shell now needs two bonds, right? So it attacks the ethene double bond, forming the 'bridged structure' my book mentions by bonding with each Carbon on the ethene molecule. Fine. Cool. Then eventually the other negatively charged bromine atom can form a bond on the opposite side of the molecule from the first bromine, leading to a lovely looking (1,2)-dibromoethane:

This seems logical, right?

Except now, my book goes on to completely ignore what it just did during the exact same process with Chlorine gas, which makes no sense. No bridged structure, the pictures even say that the Chlorines end up on the same side of the molecule as seen on wikipedia and on any quick google image search of 'addition reaction':

This is what you get when googling 'addition reaction', but surely this is wrong because the chlorines must be on opposite sides of the molecule?

I've seen this in different places on the web too... Bromine somehow undergoes a different ionic addition than Chlorine which makes no sense, it basically bonds to one carbon even though it needs two (like the bromine did), so no bridged structure.

I don't get it folks. Bromine and Chlorine have the same valency, and they're more electronegative than Carbon or Hydrogen. Why are these two processes different, and why do they end up with significantly different products?

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  • $\begingroup$ The product is the same in both cases, so you are only really asking about the intermediate. Do not read too much into the chlorine atoms appearing on the same side. It does not mean anything because in this case there is free rotation about the C-C bond. Anyway, I was taught that both bromine and chlorine form a halonium ion (i.e. the three-membered ring), and that fluorine doesn't because it's too small. As such, addition of either Br2 or Cl2 proceeds with anti stereospecificity (which means they are added to opposite faces of the double bond). $\endgroup$
    – orthocresol
    Commented Nov 23, 2015 at 0:07
  • $\begingroup$ ...which also means that, in a sense, the diagrams that show the chlorine atoms on the same side are a little misleading - not wrong, because free rotation about C-C, but a little misleading. If the rotation was not allowed, for example in the addition to cyclohexene, you would exclusively get the trans diastereomer (halogens are on different sides of the ring). By the way, what textbook are you using? $\endgroup$
    – orthocresol
    Commented Nov 23, 2015 at 0:09
  • $\begingroup$ That's extremely helpful. I understand it a lot better now, thank you. Out of interest, what effect does the size of the atom have on the ability to form a halonium ion? That's interesting, and it's mentioned nowhere in my notes/textbook. $\endgroup$
    – Malnurturedㅤㅤㅤㅤ
    Commented Nov 23, 2015 at 0:13
  • $\begingroup$ Sort of a handwavy argument - the electron cloud of fluorine isn't large enough to reach over to both carbons at once. There is a bit in Carey and Sundberg that mentions that it is unstable relative to the corresponding carbocation - I don't have the book with me now but maybe someone else can find it. $\endgroup$
    – orthocresol
    Commented Nov 23, 2015 at 0:19
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    $\begingroup$ @orthocresol looks like an answer to me $\endgroup$
    – jerepierre
    Commented Nov 23, 2015 at 16:02

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