# 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