[madscientist]

Hello fellow chem geeks :geek: (lol)

Can someone please help me with this question i am waiting on an organic chemistry book to arrive in the mail and just cant seem to find a definate answer to it online or in my general chemistry books.

the q states:

What is the product of the addition of Cl2 to propene?

a.)chloropropene   (not this one)

b.)1,3 dichloropropane

c.)dichloropropane  ( not this one)

d.)1,2 dichloropropyne  (not this one)

e.)1,3-dichloropropene

I have ruled out these three ( but could be wrong) and am left with (b.) and (e.) can somone point me in the right direction before im put in a stray jacket!!

cheers,

madscientist

[Levi]

Uhm... I'm really rusty at this on top of being new here, but I'll give it a shot.

Cl2 + C3H6 -->  ClCH2CHClCH3

Or 1,2 dichloropropane

Again, probably wrong, but seems right to me.

[movies]

Why don't you start by telling us why you ruled out the ones that you did and why you didn't rule out the others.

It might also help you to think about the mechanism for this process.

[madscientist]

a.)chloropropene  (not this one) ruled out due to their being 2 Cl atoms = dichlor, although now i think about it only one may be taken up

b.)1,3 dichloropropane

c.)dichloropropane  ( not this one)ruled out by help of fellow student

d.)1,2 dichloropropyne  (not this one)as above

e.)1,3-dichloropropene

[arnyk]

UV acts as a catalyst to break the Cl2 into two free radicals and also breaks the double bond in propene. (not sure about specifically UV on the double bond but it does break)

Do you know what a substitution rxn is?  Because that doesn't apply here .

Do you know what an addition rxn is?

[Dude]

What was the basis for your fellow student to rule out dichloropropane (or 1,2-dichloropropane)?

[arnyk]

Lol I love how we're all beating around the bush here and the only guy who's given a straight answer may or may not be right.  

[Levi]

I made a mistake. Corrected it.



This is the compound produced.  The addition reaction broke a double (unsaturated) bond that existed between the first two carbons.  The two chlorines then account for the radicals, allowing all carbons still to have four bonds.

I originally said 1,3 dichloropropane because I apparently don't know how to friggin count.  But if you, on the other hand, know how to count, then count the carbon atoms and notate the ones associated with chlorine.  Carbon atoms 1 and 2 are associated with chlorine.  1,2 dichloropropane = dichloropropane

[movies]

It's actually probably the Cl-Cl bond that breaks first.  The bond energy for the Cl-Cl bond is about 243 kJ/mol, while the pi bond in and alkene is around 266 kJ/mol (or if you prefer kcal/mol, 58 and 63, respectively).

The Cl radical then adds to the double bond to make a secondary carbon radical (more stable than a primary carbon radical), which then reacts with either a Cl₂ molecule (a radical chain propagation step) or another Cl radical (a radical chain termination step).

[Levi]

That's some excellent attention to detail.    I didn't think to look up bond energies. Kudos.

I've never heard of radical chain propagation or termination steps, but they sound interesting.  I was about to ask what the point of giving those two steps different names was, on account of both ending up with Cl radicals, but I figured propogation implies creating another Cl radical, whereas termination is using up the last Cl radical "propogated"?  Or am I talking out of my behind?

Should I have learned this in high school?  

[movies]

I was about to ask what the point of giving those two steps different names was, on account of both ending up with Cl radicals, but I figured propogation implies creating another Cl radical, whereas termination is using up the last Cl radical "propogated"?  Or am I talking out of my behind?

Nope, that's exactly correct!

I don't know if radical mechanism formalism is covered in all high school chem classes.  I think we covered it in my AP chem class in high school.  The three steps you should always have are an initiation, a propagation, and a termination.  There can, however, be several different initiation, propagation, or termination routes.