# Preference of certain positions during chlorination of an alkene

I am currently doing the questions:

Write equations for the reaction of chlorine with propene and 2-butene using structural formulas.

So I know that propene is $$\ce{H3C-CH=CH2}$$ and chlorine is $$\ce{Cl2}$$.

So:

From this I observe that naturally, the double bond is removed in the presence of an addition reaction and that the chlorine atom bonds to the $$\ce{CH}$$ pair and the other $$\ce{Cl}$$ atom does not bond to any of them but attaches itself at the end.

• Why did the second $$\ce{Cl}$$ pair not attach itself to $$\ce{CH2}$$? Does this mean that atoms can only bond to $$\ce{CH}$$ pair and nothing more than that, e.g. not $$\ce{CH2}$$ or $$\ce{CH3}$$, etc.

Second question for the reaction of 2-butene with Chlorine. I know that 2-butene is $$\ce{H3C-CH=CH-CH3}$$

• I noticed, like what happened above, the chlorine atom bonds to only the $$\ce{CH}$$ pair. Here however there are two $$\ce{CH}$$ pairs so the Chlorine doesn't need to attach itself at the end of the chain, unlike the first picture.

Are my observations correct? My textbook has no mention of this.

• With respect to (1), I imagine that Cl bonding in those locations is energetically favorable given that there are no C-H bonds being broken. Of course this does NOT mean that atoms can only bond to the CH pair and nothing else but then we'd have to consider the C-H bond rupture which is an entirely separate process. – LordStryker Sep 29 '14 at 12:13

In your first example, propene plus chlorine, one of the Cl atoms bonded to a "CH" and the other, that you described as bonding to the end, bonded to a "CH2," so it's not just CH groups getting all of the chlorines. Instead, it's the carbons at the ends of the double bond, however many hydrogens they might be attached to.

I think part of your misunderstanding lies in the way that you're writing/imagining the alkene. The propene molecule has a shape, and the geometry of the molecule plays a major role in its reactions. The double bonded section is more or less flat, and the CH3 is spread out in space. The chlorine atoms attach to the ends of the double bond, wherever that may be. The mechanism for this attachment is beyond the scope of your question, but if you're curious, you can find more information here. At the very least, know that the chlorines do not attach to the carbons at the same time; there is an intermediate compound formed.

(If my drawing isn't clear, you can try building a model of the molecule.)

It would require the breaking of a C-H or C-C sigma bond in order for the chlorine to get to the CH3 carbon, and that's unlikely. As I mentioned above, the C=C section is planar, so these carbons are open on the sides for attack from a halogen. A bond still has to break for a new C-Cl bond to form, but in this case it is the weaker C-C pi bond that breaks. With easy physical access and a lower energy barrier, the reaction goes forward.