Introduction to Free Radical Substitution Reactions

by James

in Alkanes, Alkyl Halides, Organic Chemistry 1, Organic Reactions

The four posts on acid-basesubstitutionaddition, and elimination covered the 4 main reactions in organic chemistry I. In this second series of posts we go beyond these to introduce a few of the less common (but still important) reactions you learn in organic chemistry 1. We talked about rearrangements last time: today let’s talk about free-radical substitution, and leave cleavage (oxidative cleavage) for last.

As I’ve said with everything in this series, the point is not to understand why just yet, but to be able to see from the diagrams what bonds are broken and formed. You need to understand how to read  line diagrams and also the concept of isomers. But other than that no further skills are required. The point here is to be able to follow the plot – to see what is happening. A later series of posts will go into more detail as to why things happen.

Let’s take two really simple examples. If you take a simple hydrocarbon like ethane (CH3CH3) and keep it in the dark with chlorine gas (Cl2), absolutely nothing happens at all.

But when you open the cover and let the light shine in, something interesting happens. As seen in the first example below, one of the hydrogens on ethane is replaced by an atom of chlorine, forming CH3CH2Cl (ethyl chloride).  Along the way, we also form an equivalent of hydrochloric acid (HCl). In other words, we’re breaking C-H and forming C-Cl. (Why might this happen? Not going to talk about it right now. But don’t forget that light (photons) carry energy, and that energy can be enough to break certain types of bonds.)

If you recall, this is very similar to the pattern we saw for substitution reactions earlier: replacement of one group at carbon for another. What makes this substitution different from the previous example  is the fact that we need light or some other kind of initiator in order for the reaction to occur. That’s definitely not the case for the examples in the first substitution post. (Spoiler – that name “free radical substitution” is going to tell us the proposed path by which this substitution occurs, but for now you can just think of it as “light-dependent substitution”.)

The second reaction shows another example: treatment of cyclohexane with Cl2 (1 equivalent) gives cyclohexyl chloride and HCl. Same essential pattern: break C-H, form C-Cl. This reaction is also dependent on light being present (as will be all the reactions below).

You might notice that substitution of ethane or cyclohexane with chlorine can lead to only one possible product. So what happens if you use something a little different – like butane, for instance? Let’s look:

See that we have two possible products – 1-chlorobutane and 2-chlorobutane – and they’re formed in about a 3:2 ratio. Not coincidentally, this is also the ratio of CH3 hydrogens (there are 6) to CH2 hydrogens (4). So in essence, substitution on butane is pretty much random. This also holds true if we use a more complicated alkane like 2-methylbutane.

Note again that we’re always doing the same reaction: break a C-H bond, form a C-Cl bond. Same pattern. Different starting material, different product, but the same pattern. 

What happens if you use a large excess of chlorine gas? Well, it will start replacing all of the hydrogens on the hydrocarbon, like in the third example. This is the same reaction (substitution), just repeated multiple times. Nothing new here, actually!

By the way: we aren’t using theory to predict these results, these are results that experiments tell us. Nobody predicted these results in advance. They just did the experiments, and these are the results. That’s organic chemistry. Theory comes after we’ve had enough experimental results to start making hypotheses.

One last slide.  Chlorine isn’t the only halogen on the periodic table. What about fluorine (F2), bromine (Br2), and iodine (I2)?

Let’s look at bromine and iodine. Fluorine we’ll leave until the end.

Bromine is cool. When we take 2-methyl butane and treat it with bromine, we get one major product: 2-bromo-2-methyl butane. This is the major product even if we add a large excess of Br2. This is a useful property.

What about iodine? Well it doesn’t do much of anything, as you can see. Doesn’t matter how many equivalents of iodine you add, or how much light – nothing happens. Interesting!

So how can we explain all of these results? How can we come up with some hypotheses to explain all of these different experimental observations?

PS.OK, OK. I didn’t mention fluorine. Why haven’t we talked about fluorine yet? Because fluorine gas is the Honey Badger Of Chemistry  . It really doesn’t give a shit. It will react with anything you put in front of it, and unless you are 1) extremely technically competent 2) extremely safety conscious, and 3) a little crazy, fluorine gas is something you should never, ever work with.

[youtube]http://www.youtube.com/watch?v=vtWp45Eewtw[/youtube]

 

Next Post In The Series: Introduction To Oxidative Cleavage Reactions

 

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{ 5 comments… read them below or add one }

Sajeda June 5, 2013 at 8:34 pm

“Fluorine don’t care…fluorine don’t give a shit!”

that video was hilarious. Now ill remember flourine forever!

Reply

Eric October 22, 2013 at 11:04 pm

Are the carbons for 2-bromo-2-methyl butane numbered correctly?

Reply

Pam December 9, 2013 at 2:12 am

How do we know that a free radical halogenation reaction will have statistical distribution of products? would Cl more likely bond to a tertiary or secondary carbon rather than primary?

Reply

James Ashenhurst December 10, 2013 at 2:32 pm

It’s safe to assume that Cl will be able to replace primary, secondary and tertiary hydrogens in any molecule. This can lead to isomers. It will be more *selective* for tertiary over secondary over primary, but that does not mean that reactions at primary C-H bonds will not happen.

Reply

frank January 23, 2014 at 2:57 am

It is only possible for tertiary hydrogen

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