Free radical reactions | Substitution and elimination reactions | Organic chemistry | Khan Academy
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Free radical reactions | Substitution and elimination reactions | Organic chemistry | Khan Academy

Let’s think about what type of
reaction we might be able to get going if we had
some methane and some molecular chlorine. So if we just let this be and we
didn’t heat it up or put in any UV light into this reaction,
pretty much nothing will happen. Both of these molecules are
reasonably happy being the way they are. But if we were to add heat into
it, if we were to start making all the atoms and
molecules vibrate more and bump into each other more, or
we were to add energy in the form of UV light, what we could
start doing is breaking some of these chlorine-chlorine
bonds. Out of all of the bonds here,
those are the weakest. That would be the most susceptible
to breakage. So let’s say we were to add some
heat, what would happen? So let’s see. Let me draw the valence
electrons of each of these chlorines. This chlorine has one, two,
three, four, five, six, seven valence electrons, and this
chlorine over here has one, two, three, four, five, six,
seven valence electrons. Now, when you add heat to this
reaction, enough for these guys to vibrant away from each
other, for this bond to break, what’s going to happen, and we
haven’t drawn an arrow like this just yet, but what’s going
to happen is that each of these chlorines, this
bond is going to break. Each of these chlorines are just
going to take their part of the bond. So this guy on the
left, he’s just going to take his electron. And notice, I draw it with
this half arrow. It looks like a fish hook. It’s just half an arrowhead. This means that this electron
is just going to go back to this chlorine, and this other
magenta electron is going to go back to the right chlorine,
so we can draw it like that. If it was up to me, I would have
drawn it more like this. I would have drawn it more like
this to show that that electron just goes back to the
chlorine, but the convention shows that you can show that
half of the bond is going back to the entire atom. Now, after this happens, what
will everything look like? Well, we’re still going to
have our methane here. It hasn’t really reacted. So we still have our methane. Let me draw it a little bit. So we still have our
methane here. And all that’s happened is,
because we’ve put energy into the system, we’ve been able
to break this bond. The molecular chlorine has
broken up into two chlorine atoms. So we have the one on the
left over here, and then we have the one on the right. And let me draw the left’s
valence electrons. It has one, two, three, four,
five, six, seven. I just flipped it over so that
the lone electron is on the left-hand side right here. And then you have the
guy on the right. He has one, two, three,
four, five, six, seven valence electrons. Now that each of these guys
have an unpaired electron, they’re actually very,
very, very reactive. And we actually call any
molecule that has an unpaired electron and is very reactive
a free radical. So both of these guys now
are free radicals. And actually, the whole
topic of this video is free radical reactions. Both of these guys are
free radicals. And you’ve probably heard the
word free radical before. In the context of nutrition,
that you don’t want free radicals running around. And it’s the exact same idea. It’s not necessarily chlorine
that they’re talking about, but they’re talking about
molecules that have unpaired electrons. They’ll react with some of your
cell’s machinery, maybe even with your DNA, maybe cause
mutations that might lead to things like cancer. So that’s why people think you
shouldn’t have free radicals in your body. But as soon as we form these
free radicals, in this step right here, where we put energy
in the system to break this bond, we call this
the initiation step. Let me put this. We used energy here. This was endothermic. We use energy. This right here is the
initiation step. And what we’re going to see in
general with free radical reactions is you need some
energy to get it started. But once it gets started,
it kind of starts this chain reaction. And as one free radical reacts
with something else, it creates another free radical,
and that keeps propagating until really everything
has reacted. And that’s why these can be
so dangerous or so bad for biological systems. So I’ve told you that
they react a lot. So how will they react now? Well, this guy wants to form
a pair with someone else. And maybe if he swipes by this
methane in just the right way, with just enough energy, what
will happen is he could take the hydrogen off of the carbon,
and not just the proton, the entire hydrogen. He will form a bond with the
hydrogen using the hydrogen’s electrons, so they’ll
get together and they’ll form a bond. The hydrogen will contribute
one electron. Notice, I’m drawing the
half-arrow again, so the hydrogen isn’t giving away the
electron to someone else. That would be a full arrow. The hydrogen is just
contributing its electron to half of a bond. And then the carbon, the carbon
would do the same. I’ll do that in blue. So the carbon, this valence
electron right here, could be contributed to half of a bond,
and then they will bond, and this bond over here
will break. And so the carbon over here on
the left, this carbon over here will take back
its electron. So what does it look like? What does everything look
like after that’s done? So our methane now, it’s
no longer methane. It is now, if you think
about it– so we have three hydrogens. It took its electron back. It is now a free radical. It now has an unpaired
reactive electron. The hydrogen and this chlorine
have bonded. So let me draw the chlorine. It has this electron
right over here. It has the other six valence
electrons: one, two, three, four, five, six. And we have the hydrogen with
its pink electron that it’s contributing to the bond. And so we have them
bonded now. This chlorine is no longer a
free radical, although this one out here is still
a free radical. Let me copy and paste it. So it’s hanging around. Copy and paste. And now, notice we had one free
radical react, but it formed another free radical. That’s why we call this
a propagation step. So this right here is
a propagation step. When one free radical
reacts, it created another free radical. Now, what’s that free radical
likely to do? You might be tempted to say,
hey, it’s going to just react with that other chlorine,
but think about it. These molecules, there’s a
gazillion of them in this solution, so the odds that
this guy’s going to react exactly with that other free
radical is actually very low, especially early on in the
reaction where most of the molecules are still either
methane or molecular chlorine. So this guy is much more likely
to bump into another molecular chlorine than he is
to bump into one of these original free radicals
that formed. So if he bumps into another
molecular chlorine in just the right way– so let me draw
another molecular chlorine. So that’s another molecular
chlorine. And each of these one, two,
three, four, five six, seven; one, two, three, four,
five, six, seven. There is a bond here. If they bump in just the right
way, this chlorine electron might get contributed, and this
free unpaired electron will be contributed and then
this CH3, I guess we could call it, this free radical, this
carbon free radical, or this methyl free radical,
will then form a bond with this chlorine. What’s everything going to
look like after that? Well, after that happens this
is now bonded to a chlorine. It’s now chloromethane. Let me draw it. So it’s carbon, hydrogen,
hydrogen, hydrogen. Now, it’s bonded
to a chlorine. Let me draw the electrons
so we can keep track of everything. We have that magenta electron
right over there. And then we have the chlorine
with its one, two, three, four, five, six, seven
valence electrons. They are now bonded. This is chloromethane. And now you have another free
radical because this guy– and I should have drawn it there. This guy, that bond
was broken, so he gets back his electrons. So he’s sitting over here. He is now a free radical. So this is another
propagation step. And we still have that original
free radical guy sitting out over here. So we keep forming more
and more free radicals as this happens. Now, eventually we’re going to
start running out of methanes and we’re going to start
running out of the molecular chlorines. So they’re going to be less
likely to react and you’re actually going to have more
free radicals around. So once the concentration of
free radicals gets high enough, then you might
start to see them reacting with each other. So when the concentration of
free radicals get high enough, you might see, instead of this
step happening– this will happen a long time until most
of the free radicals or most of the non-free radicals
disappear. But once we have a soup of
mainly free radicals, you’ll see things like this. You’ll see the methyl
free radical. So let me draw it like this. You’ll see him maybe reacting
with another methyl free radical, where they
both contribute an electron to form a bond. And then, once the bond forms,
you have ethane. I could just write
as CH3, H3C. So you might have something
like this. And so this type of a step where
two free radicals kind of cancel each other out, this
is a termination step because it’s starting to lower the
concentration of free radicals in the solution, but this is
only once the concentration of free radicals becomes
really high. You might also see some of the
chlorines cancel out with each other again, so a chlorine
free radical and another chlorine free radical. I’ll only draw the unpaired
electron. They can bond with each
other and form molecular chlorine again. That again is a termination
step. Or you could see something like
the methyl free radical. Just for shorthand, I’ll write
it like this: H3C. The methyl free radical and a
chlorine free radical might also just straight-up react and
form chloromethane, And form H3C-Cl. So this will all happen once
the concentration of free radicals gets really high. Now, another thing that might
happen once this reaction proceeds, and we have a lot of
the propagation steps, is that you might have a situation
where you already have a chloromethane, so it
looks like this. You already have a
chloromethane. And once you have enough of
these, it then becomes more likely that some free radical
chlorine might be able to react with this thing, so it
might actually add another chlorine to this molecule. And the way it would do it, this
chlorine over here– I’m just drawing the free
electron pairs. It would form a bond with this
hydrogen right over there. They would both contribute
their electrons. And then the carbon would
take back its electron. Notice, all of the
half-arrows. You’d be left with– the
hydrogen and the chlorine would have bonded. And now, this guy’s going to
be a free radical, but he’s going to be a chlorinated
free radical. So it’s going to
look like this. He has a free electron over
there: hydrogen, hydrogen. And then he might be
able to react with another chlorine molecule. He contributes an electron. Maybe this guy contributes
an electron. This guy– I don’t want to
draw a full arrow– he contributes an electron to a
bond, and then this guy takes his electron back and becomes
a free radical. And then we’re left with what? We’re left with a doubly
chlorinated methane. So then we have Cl,
Cl, and then a hydrogen and a hydrogen. And this could actually
keep happening. As the concentration of these
get higher, then it becomes more likely that this can react
with another chlorine. Of course, this chlorine
over here becomes another free radical. But the general idea here that
I wanted to show you is that once a free radical reaction
starts– the first step requires some energy to break
this chlorine-chlorine bond, but once it happens, these guys
are highly reactive, will start reacting with other
things, and as they react with other things, it causes more and
more free radicals, so it starts this chain reaction. And actually, all in all, this
required energy to occur. This step right here, this
propagation step, it requires a little bit of energy, but
it’s almost neutral. It requires energy to break
this bond, but it creates energy when this
bond is formed. It still requires a
little net energy. And then things like this start
to become exothermic. And especially once you start
getting to the termination steps, you start releasing
a lot of energy. So actually, all in all, this
reaction is actually going to release energy, but it needed
some energy to get started.

About James Carlton

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100 thoughts on “Free radical reactions | Substitution and elimination reactions | Organic chemistry | Khan Academy

  1. Thanks a lot. You don't understand how much you are helping people from around the world. I can't thank Khan Academy enough for doing what they do. Keep it up!

  2. Thank you so much Mr Khan! After weeks of trying to understand this process it finally makes sense after watching your video. Thank you.

  3. why hasnt he added the HCL? what happened to it? isnt the product supposed to be chloromethane+hydrochloric acid?

  4. Do all free radicals reaction happen at very high temperatures? Like in the human body the temperature is not that high, so how could a radical reaction occur?

  5. I personally think that Khan Academy is the best source of online learning. When I first encountered this youtube channel i thought for a second. wait "Would I be able to understand,possibility are there might be few terms which might be used in the video which might not be explained or talked about in my reference book" But trust me, it covers every single term and concept. I sometime wander whether he had read my course book or what lol

  6. Thank you very much Khan Academy, you have really been a great help to me, very clear and easy to understand. Keep up the good work. 🙂

  7. I just like the way Sal repeats what he already said, when he is drawing! Anyway's you always save me during exam days. LOL

  8. Dumb question. I thought that heat cannot be used as energy, or at least thats what I was taught in Animal Phys. Can you clarify for me? Thanks in advance.

  9. Cl isnt the best example because it has a small electron cloud what makes it more difficult to split into radicals by heating them up, but in general all halogens are good, the Cl just happens to be the worst in this case.

  10. Can anybody help me out here? Khan said that a chloromethane radical can bond with another chlorine radical to add more chlorines to the carbon in place of its hydrogens. Can two chloromethane radicals combine to form dichloroethane? Or is it so unlikely that it is disregarded in the reaction's products?

  11. the overall reaction is CH4 + Cl2 > CH3Cl + HCl
    so if once through homolysis chlorine free radicals are formed after breaking the only Cl molecule..then where did another Chlorine molecule come from?

  12. how to know which molecule forms bonds with which molecule.
    11:26 .

    why chlorine formed bond with hydrogen only, why not other molecules??

  13. Did some high surfer just find out how many reactions the unpaired electrons cause and just call them radicals, that is how i would like it to have happened in my head

  14. 4:54 Small mistake?
    Shouldn't it be Chlorine? Heard Sal say Carbon… Got me confused for a second.
    Great video though.

  15. i don't get how he says that one free radical can form multiple free radicals and increase the concentration, as i understand it, a FR reacts with a molecule so it becomes a FR but the intitial FR will turn to a normal molecule so the total amount of FR won't change. can anyone explain?

  16. when you realise even at alevel uk they dont go over basic uni notation… ie sn11/2 e1/2
    i mean we are taugh it but not explicitly told what sn2 is etc or the differences

  17. if we could have complete preperation of csir net chemistry on khan academy with question solving and theory ,it would be very very helpfull and very beneficial for us.

  18. One thing that confuses me is that multiple Hydrogen atoms can be substituted by halogen atoms what are the conditions Could somebody help please

  19. Thanks, this video helped a lot.
    But i wonder, what are the probabilities, that after iniciation the free chlorines will look for another partner. Isn´t it less energetically demanding to bond with each other again, that break bonds in other molecules?

  20. You should've seen my face when everything started making sense lol. Thanks for the amazing explanation

  21. what if the methane are change to cyclohexane . At the termination will the cyclohexane react with cyclohexane ?

  22. So with that logic, is it possible to form chloroETHANE as a chlorinated free radical methyl 'bump' with a free radical methyl group? Or even longer alkane chain (with regard to the level of concentration of course)?

  23. Instructor non-verbal cues purssued im a punk. Glad this video! Diction is good for academic colab commnication.

  24. The heat or uv light should be given all the time or just to break the first Cl2 molecule bond at initiation step?

  25. #YouCanLearnAnything except how Antioxidants work on a mechanical level, you just have to trust that they do what we're told "correct the free radical".

  26. During a UV free radical chlorination you say to use UV light. So just for fun I used the formula E=hf and figured out the charge in joules of a 400 nm UV light (6.626 X 10*-19 J) which is the WEAKEST UV wavelength  and I found out it is more than enough joules for  homolytic bond cleavage not only in the chlorine BUT ALSO in a Cl/C bond. I found the dissociation bond energies from charts on line. So let us say I have DCM and want to make tetrachloromethane. So if I use the weakest UV……..400 nm then I have enough energy in the UV to not only break Cl/Cl bonds but also the Cl/C bonds that I want to make. THAT DOES NOT MAKE SENSE ?????????????????? Let us say I want to make benzyl chloride. 400 nm UV light is SLIGHTLY more joules needed to break a benzyl C/H bond (6.26 X 10*-19 J) and WAY MORE THAN ENOUGH to break a benzyl C/Cl bond. Do you see my point. As I make the product the UV light is enough energy to destroy what I make. So how does it work??????????????????? Also the UV light at 400 nm is enough to turn the whole pot of toluene into benzyl radicals that combine with each other as much as the chlorine… seems like that would not be good??????????????????????????? Also I looked up the energy needed to break a Cl/Cl bond (4 X 10*-19 J) and that corresponds to a light source of ABOUT 500 nm………..WHICH IS GREEN VISABLE LIGHT. Although at this point I was really frustrated and started doing math in my head but it is close. So now I am really confused because that means you do not even need UV light to make Cl free radicals. SO is this true?????????? Can I make Cl free radicals with green visible light?????????? IF SO THEN WOULD IT NOT BE BETTER to use 450 nm blue visible light with NO UV light. That way the photons are slightly under the energy needed to break benzyl Cl/C bonds and benzyl C/H bonds and in the case of an ALKYL chloride blue light is not enough to break Cl/C bonds. BUT blue light is good enough to break Cl/Cl bonds so your product wont be destroyed as you make it by UV light. I thought I had a good understanding of free radical halogenation until I did the math. WHAT IS THE DEAL. HELP ME HELP ME HELP ME PLEASE

  27. 5:36 you said freeradicals are the ones with unpaired electrons and are neutral but CH3 • is not neutral it has 5 electrons please tell me how come that works?

  28. At first I was put off by the rough diagrams but you helped me understand this a lil more. Will surely be returning throughout my A-levels! 🙂

  29. 7-7:30 minutes in video. Cl. with H. is now HCl? And now stable? H. released from CH4 and Cl. Radical from Cl2? What is happened with HCl?

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