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Lecture Comments (20)

1 answer

Last reply by: Professor Starkey
Mon May 12, 2014 11:00 PM

Post by Chris Kay on May 11, 2014

In the termination steps you show that the radicals are destroyed when they collide. How is this reaction possible when both radicals have a negative charge?

1 answer

Last reply by: Professor Starkey
Mon Feb 10, 2014 10:16 PM

Post by nneka igwemadu on February 10, 2014

Is there  a video on monochlorianation? or synthetic reaction?

1 answer

Last reply by: Professor Starkey
Thu Nov 7, 2013 9:27 AM

Post by Jude Nawlo on November 6, 2013

Hi Professor Starkey,

In order for free radicals to be produced, do we need both irradiation with light and high temperatures? Also, is everything neutral at the stage of initiation in free radical formation(as in, no free radicals necessary to begin)? I recently had a question on an exam asking what brought about free radical formation and I put irradiation with light, but the option of putting both irrad. with light AND high temps was on there too. I was just wondering if both were necessary in order to initiate this process. Your lectures are lifesavers - thank you so much for clarifying everything :)

1 answer

Last reply by: Professor Starkey
Sat Feb 2, 2013 2:13 PM

Post by marsha prytz on February 1, 2013

in the free radical halogenation the first step breaks the H from the CH4 to make CH3 and HCL, but in the second step you still have 2 CL atoms where does the first one come from or the last one that continues the radical mechanism?

1 answer

Last reply by: Professor Starkey
Tue Nov 27, 2012 11:25 PM

Post by Consolata Mogeni on November 26, 2012

Do you have any lectures on how to calculate the heats of hydrogenation for the overall reaction and the heats of hydrogenation for each propagation step?

1 answer

Last reply by: Professor Starkey
Thu Nov 15, 2012 1:03 AM

Post by Anthony Blair on November 13, 2012

I think I understand now, the chlorine is replacing one of the hydrogens, but still I don't know why the first CH3 in the chain does not show as being able to replace a hydrogen with a chlorine.

0 answers

Post by Anthony Blair on November 13, 2012


0 answers

Post by Anthony Blair on November 13, 2012

Why can you not put a chlorine on the first CH3 in the straight parent carbon chain? On the CH3 on the others, where exactly is the carbon attaching? Is a hydrogen leaving or does the carbon have 3 hydrogens and a chlorine attached?

1 answer

Last reply by: Professor Starkey
Tue Oct 2, 2012 11:27 PM

Post by Sara Valek on October 2, 2012

how about for benzenes? would 9 hydrogens on a tert-butyl side chain off of the aromatic ring react faster than the aromatic hydrogens when bromine and light are added?

1 answer

Last reply by: Professor Starkey
Sun Sep 30, 2012 11:41 AM

Post by sophia lin on September 29, 2012

how to calculate the percent for the structures?

1 answer

Last reply by: Professor Starkey
Sat Nov 5, 2011 3:16 PM

Post by Hannah Shin on November 1, 2011

For the allylic halogenation, why does Br not replace the H on the double bond instead of the other H next to it?

Free Radical Halogenation

Use bond dissociation energies (BDE) to calculate the ∆Ho of this reaction:
  • Bonds broken:
    For CH3CH2-H, the ∆Ho = +410 kJ/mol
    For Br-Br, the ∆Ho = +192 kJ/mol
    Total = +602 kJ/mol
  • Bonds formed:
    For CH3CH2-Br, the ∆Ho = -285 kJ/mol
    For H-Br, the ∆Ho = -368 kJ/mol
    Total = -653 kJ/mol
  • Overall ∆Ho = (+602 kJ/mol) + (-654 kJ/mol) = -51 kJ/mol
-51 kJ/mol
What is the rate equation for this reaction?
CH3Br + NaCN → CH3CN + NaBr
  • A + B → C + D
    rate = k[A]a[B]b
    k = rate constant at a given temperature.
rate = k[CH3Br][NaCN]
A. What happens to the rate of the reaction if [CH3Br] is doubled.
B. What happens to the rate of the reaction if [NaCN] is halved?
CH3Br + NaCN → CH3CN + NaBr
A. If [CH3Br] is doubled, rate doubles.
B. If [NaCN] is halved, rate halved.
Rank the following carbocations in order of increasing stability.
An increase in temperature will result in an increase or decrease of the reaction rate?
An increase in concentration will result in an increase or decrease of the reaction rate?

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.


Free Radical Halogenation

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Free Radical Halogenation 0:06
    • Free Radical Halogenation
    • Mechanism: Initiation
    • Mechanism: Propagation Steps
  • Free Radical Halogenation 5:33
    • Termination Steps
    • Example 1: Terminations Steps
    • Example 2: Terminations Steps
    • Example 3: Terminations Steps
    • Example 4: Terminations Steps
  • Regiochemistry of Free Radical Halogenation 9:32
    • Which Site/Region Reacts and Why?
    • Bromination and Rate of Reaction
  • Regiochemistry of Free Radical Halogenation 14:30
    • Chlorination
    • Why the Difference in Selectivity?
  • Allylic Halogenation 20:53
    • Examples of Allylic Halogenation

Transcription: Free Radical Halogenation

Welcome back to Educator.0000

Next we are going to be talking about a reaction called a free radical halogenation.0002

This is a reaction of alkanes; and as the name implies, we are going to be doing a halogenation reaction; we are going to be adding a halogen to the alkane structure.0007

Here we have a very simple alkane such as methane; we are reacting with chlorine, Cl2; and this reaction is going to happen in the presence of either light or heat energy.0015

We could represent light by the symbol hν or heat with the symbol Δ; so we can get used to seeing perhaps either of those reaction conditions.0029

The products we are going to get here will be CH3Cl, so chloromethane, and HCl.0037

You can see that we have halogenated the alkane; you replace one of the hydrogens on the alkane with a halogen.0049

This is a very important reaction because it is a way that we can functionalize an alkane and convert them into an alkyl halide.0056

An alkyl halide is going to be very useful in organic synthesis; we could do lots of interesting reactions with alkyl halides.0064

Alkanes are readily available through the distillation of our crude oil from a refinery.0072

We have a lot of alkanes as natural raw materials; and this would be a great reaction as a way to convert them to alkyl halides.0081

Let's take a look at the mechanism of the free radical halogenation.0088

As the name implies, it is going to involve free radicals; and it is also an example of what is known as a chain reaction.0092

We are going to have two major steps that form the product; but before we can get to those two major steps, first we have to start the radical reaction.0099

That first step is called an initiation; initiation step is one that creates radicals.0107

The way a radical is created in these reaction conditions is that the chlorine reacts with either the hν or the Δ, the light or the heat.0116

We are going to break the chlorine-chlorine bond homolytically; one electron is going to go to each side; and we are going to form two chlorine atoms.0126

That is going to be our very first step in the mechanism--is to create a radical.0138

Once we have a radical started, now that radical is going to react with some other species; we will take the chlorine and we will react it with our alkane.0141

What is going to happen is that chlorine is unstable; it is very reactive; and what radicals like to do is they like to pluck an atom off of another species; we call that an atom abstraction.0150

It is going to see this hydrogen; it is going to grab that hydrogen; one electron is going to pair up with the radical; and the other electron is going to stay behind.0161

That is going to form a molecule of HCl, which we saw above is one of the products that we were expecting; and what else does it form?--it forms a methyl radical.0172

This type of a radical mechanism step is called a propagation step; because we are starting with one radical and the product we get still has a radical in the product.0187

One radical gets converted to a new radical; we call those types of steps: propagation steps.0199

We would describe this first step as a hydrogen atom abstraction; you abstracted or plucked off a hydrogen atom, a hydrogen with its one electron... with its one electron.0208

Now we have a methyl radical; what is that methyl radical going to want to do?0224

Again, it is unstable; it wants to stabilize itself; it is going to look around for an atom to steal; and what it is going to find is a molecule of Cl2.0227

We are going to have the exactly analogous mechanism where the carbon plucks a chlorine atom off of Cl2; that leaves a radical behind.0237

The product we get here now has a carbon-chlorine bond; and we've just made our product; we've just formed a molecule of chloromethane.0247

What else is formed in this step?--we also form Cl·, chlorine atom.0261

What can that chlorine atom do?--that can continue back and attack another equivalent of the alkyl halide; so this will continue the radical mechanism--the chain reaction.0269

It is called a chain reaction because as soon as you form just one radical, these propagation steps will go on and on and on; these happen many times.0287

These are the two main steps I described in how to form your product--is plucking off a hydrogen atom from the alkane and then the resulting alkyl radical plucking off a chlorine from Cl2.0299

We could describe this second step as a chlorine atom abstraction; so after our initiation step, all the steps of our free radical halogenation are propagations steps that are atom abstractions.0316

What else can happen in a radical reaction?--another thing that can happen are steps known as termination steps.0334

I put these separately because this is not part of the normal chain reaction that is used to form our akyl halide; but this can also happen--a termination step is something which destroys radicals.0341

If you had two radicals ever combining in a step, then these two electrons could come together to form a bond; and when two radicals collide, that would give a product that no longer has a radical.0361

So this would be a way of stopping a chain reaction; OK, now why did I put this separately?0374

Because although it is possible to have Cl· reacting with H3C·, and that would be a way to possibly make the chloromethane, this is going to be a very rare reaction.0381

Having two radicals collide is a rare reaction because any radical that you have exists in very small amounts.0396

You have a very very small concentration of any radical; it is a fleeting intermediate; and so what is it more likely to collide with?0414

Is it going to collide with a stable neutral molecule that you have in high concentration?--or is it randomly going to collide with some other radical that you have in a very small concentration?0423

It is obviously going to react with a stable molecule instead; so this particular mechanism accounts for a very small amount of product; this is not how you make products.0434

The propagation steps on the previous slide are the ways that you are going to do this reaction; but the fact that these termination steps occur means that it could give rise to some other side products.0453

For example, if we had two methyl groups, two methyl radicals reacting, that would be a way of forming a molecule of ethane; so that might be found as a very very minor product.0464

Very minor product; but you could see some traces of that and that is how it might happen.0480

Another reaction that would be a termination step is if you had some kind of inhibitor that could combine with a radical; that would form a bond and destroy a radical.0484

Inhibitor is something that can stop a chain reaction, a radical reaction.0498

An example of that is that oxygen gas is a diradical; the structure of oxygen... oh, sorry, I drew two bonds here.0504

The structure of O2 actually has, rather than a double bond, it has a single bond and two radicals.0521

So oxygen is an example of a molecule that you could have in relatively high concentrations; that would be a source of radicals that could scavage some existing radicals and quench a radical reaction.0532

The key here is that radical reactions are run in oxygen free atmospheres; the presence of oxygen would interfere with my radical reaction.0545

So we would want to do this reaction under an inert atmosphere such as nitrogen or argon or something like that.0565

Another thing to think about for the free radical halogenation other than the mechanism is the regiochemistry of the reaction.0574

The regiochemistry is when we ask the question: which site or which region of a molecule reacts?0580

If we reacted, instead of using methane, let's say we used this alkane to do a free radical halogenation.0587

If we were to do a bromination, that means we are going to replace one of these hydrogens with a bromine; we call it a monobromination reaction.0595

There are ten hydrogens in this structure; does that mean there are ten unique structures we can draw?--monobromination structures?0604

No, because these three methyl groups are all chemically equivalent; no matter which one I would react the bromine with, which of these nine hydrogens, they would all give the same product.0614

The other unique product would be if I replace this hydrogen with a bromine; so there is actually two possible bromination products that I can get.0627

It turns out that the major product by far is this first product; and we only get a trace if any of the second product.0636

This is a very selective reaction; we only get this product as the major product and we can ask why this is happening?0644

It looks like this hydrogen was the one that was abstracted in our radical mechanism.0655

Remember we are going to be forming a Br· and that Br· is now going to select a hydrogen to remove; what would be the advantage to removing this hydrogen.0661

It is on a tertiary carbon; so it turns out that the tertiary hydrogen is the easiest to remove.0671

If we consider the resulting radical that can form, we would have a tertiary radical, if we remove that hydrogen; if we removed any of the other nine hydrogens, we would end up with a primary radical.0682

Remember we talked about radical stabilities; and so we know there is a difference in those two stabilities; so what is going on here?0696

This reaction we are observing is not a question of thermodynamics; this is not going to be explained by thermodynamics.0705

It is not a question of which of these two products is more stable; they are both fine; they both have about the same product stability; bromine doesn't care where it goes.0713

The difference is a matter of kinetics; it is the path leading to the first product or the second product that varies.0723

When we have a lower energy intermediate, when we can form a tertiary radical, that is going to give us a lower energy transition state.0732

Which is going to give us a lower energy of activation and a faster reaction; and in this case and in many cases, the faster reaction is the one that leads to the major product.0740

If we look at our energy versus progress of reaction diagram, we look at our starting materials, our alkane and our bromine, going to our products.0751

The question is would this starting material rather go to a primary radical... let me do this on the bottom... a tertiary radical?--or would it rather go to a higher energy primary radical?0764

The path leading to the tertiary radical is going to be a lower energy path; this energy of activation is going to be lower.0781

Therefore it is going to be a faster reaction than if I had to go all the way up to this primary radical and down to product; that is going to be a slower reaction.0794

This is slower and this is faster; so we are talking about a difference in the rates of reactions; so we describe this as a kinetic effect.0805

The lower energy intermediate will have the lower energy transition state and therefore will be the faster reaction.0816

This again is going to be a very common theme we see throughout organic mechanisms--is if we ever have a choice between two competing intermediates, it is going to be the lower energy intermediate that is favored.0824

Because we have a vast majority of the product is this tertiary alkyl halide being formed, we describe bromination as being very selective.0838

The rates of reaction are orders of magnitude different; primary going to secondary is 82 times faster; going to a tertiary center is 1640 times faster.0847

With a huge difference like this, we see a huge selectivity, a significant selectivity for the more substituted hydrogen being replaced.0858

If we look at this same reaction, but with chlorine--doing a chlorination reaction, we find that that is going to be a less selective reaction.0868

We still find a difference; but the secondary position only reacts 2.5 times faster; and the tertiary only reacts 4 times faster; so it is very small differences in rates.0877

What happens is we start to see statistics playing a role in terms of how many hydrogens are available to be abstracted when we look at these major products or these product mixtures.0890

Let's take a look at this reaction; we have this alkane reacting with chlorine and hν.0904

We are going to do a monochlorination; we are going to replace one hydrogen on this structure with a chlorine.0910

Once again we have many hydrogens; but they are all not going to lead to unique products; let's see if we can describe our various products here.0916

If call this carbon type A, has hydrogens type A, reaction with any of these three hydrogens would all lead to the same chloride product.0928

Hopefully you can see that this methyl, because you can rotate and interchange these two, this methyl is also chemically equivalent; and so this would also give the same alkyl halide product.0944

We describe these two... you can put that term up here... as being chemically equivalent; because they will lead to the same product if we react at either site.0958

But if we react it at this carbon, replace that hydrogen, that would lead to a unique product; this CH2 is unique as well so that could be product C.0972

How about this methyl?--do you think this would be the same as the first two methyls we saw?--or is this unique?0982

Can you imagine putting a chlorine on here?--would that give you the same structure as if you put a chlorine here?0989

No, they would actually be different compounds; because here you would have a chlorine on the first carbon and then a methyl on the carbon right next to it.0996

Here if you put a chlorine on this first carbon, you would have a CH2 right next to it; so those would be different compounds.1002

There is actually four unique products that we can form here; if we draw... let me draw this as line drawings.1009

Here is our carbon backbone; so let's call this A... see what B looks like... and C... and D; and what do these products look like?1016

If we react with either of these two methyls, if we attach a chlorine, those would all be the same product.1034

Again we don't add a chlorine right here; there is a carbon here; so we are making a bond from that carbon to a chlorine; so that is product A.1040

Product B is from this middle carbon; product C is from this CH2; and product D is from this end carbon; so yes, there are four unique products that we can form here.1049

If we look at the product ratios here, we find that we get A is formed with a 34% yield; B is 22%; C is 28%; and D is 16%; so what we get here is actually a huge mess of products.1063

In fact, that is what we typically see for a chlorination--is we get a mixture of all the possible monochlorination products; so we no longer have one major product like we would expect in bromination.1086

Not only would we not be able to predict really which is the major product, but let's see if we can figure out why we have such a diverse mixture here.1101

A is... the largest proportion of any product is A; even though it is on a primary carbon, it would have come from a primary less stable radical, why do we get so much A?1111

Simply because there is six hydrogens; it is a primary radical but there is six different hydrogens that could be abstracted; they would all lead to product A.1123

This is what I'm talking about when I am saying that statistics is starting to have a play in who is going to be your major product.1133

Where B... B only has one hydrogen; but abstracting that leads to a tertiary radical; so that would be very stable.1142

So even though there is just a single hydrogen that is leading that path, it is a tertiary intermediate so it is a very fast reaction; so that is where we are getting this balance.1150

Again this is a secondary radical which is pretty good; but we have only two of them; and this one, he is primary and he only has three so that is why we get the smallest number here.1162

When it comes to monochlorination, your goal is typically not going to be which is the major product?--because that is not something we can really tell.1173

But this is a very nice exercise in determining chemical equivalents and being able to draw all the possible products and maybe naming them as well; it might be a nice exercise in testing your IUPAC rules.1181

One last thing, let's just ask quickly why would there be such a difference in the selectivity?1199

If they are both going through the same tertiary carbon radical, why would there be such a huge difference in the bromination reaction rate versus the chlorination reaction rate?1205

That is because remember we are not just forming this radical; we are breaking and forming other bonds.1217

We are breaking... we are forming... as we are abstracting this hydrogen, we are forming either a hydrogen-chlorine bond or a hydrogen-bromine bond.1223

The strength of those bonds is going to contribute to the overall ΔH for that rate determining step.1232

It turns out that for bromination that rate determining step is more endothermic; it is a larger climb; and in those cases, we see better selectivity.1242

Finally let's look at a case of doing an allylic halogenation; if we were to react cyclohexene with bromine and hν.1256

We look at these reaction conditions; and we should recognize that we are forming Br·; we are forming a bromine radical when we react bromine and light.1265

We are expecting to do a halogenation reaction; we are going to take one of the hydrogens from this structure and replace it with a bromine.1276

How many different hydrogens are there?--well, we could have this hydrogen, or one of these hydrogens, or one of these hydrogens; there is actually three possible products that we can draw here.1284

This hydrogen is on a double bond; that is called a vinyl hydrogen--on a π bond; and those have no reaction with halogenation reactions.1300

Remember free radical halogenation is a reaction of alkanes; it is for sp3 hybridized carbon--tetrahedral carbons; that is where we are going to do our atom abstraction.1314

We are not going to look at this position; but here we have a hydrogen and a hydrogen.1323

Let's take a look at the carbons bearing those hydrogens and see if we can imagine what the radical might look like that is going to result after atom abstraction.1330

This carbon has two carbons attached so we would describe it as secondary; and this carbon has two carbons attached; this is also secondary.1340

Are those radicals equal in their stability?--well no, because this carbon is not only secondary, it is also next to a π bond; we have a word to describe that; we call that being allylic.1351

We know that that radical would be the best one because it would have resonance stabilization compared to the other secondary radical.1363

So we would expect as our major product that we would replace one of the allylic hydrogens with a halogen.1374

The problem is that although we can have this free radical halogenation, bromine Br2, can also react with an alkene, a carbon-carbon double bond, in an addition reaction.1382

We would have a competition between the free radical halogenation and the addition; so it would be very useful if we had another source of Br· that was not Br2.1398

One that is used very commonly is this structure; this is called N-bromosuccinimide or NBS for short.1409

It could be used in the place of bromine; you can imagine this also being a source of Br·; it can be used to be a source of bromine radicals.1419

The mechanism is a little more complicated than just this; but do know that if you ever see NBS as one of your reaction conditions, you can take a look at that and know that we have Br·.1433

In fact NBS typically, again with light or heat or some way to initiate a radical reaction, is really what you would see.1445

Let's see if we could predict the product here if we took this alkene and react with Br·.1452

We would expect to do a free radical halogenation and not any other disruptive reaction where we add to the alkene.1459

Where would we remove a hydrogen?--once again we are going to ignore the hydrogens that are on the double bond; and we could take one of the hydrogens at either end.1469

This looks like a primary radical; but it is not just a primary radical; primary radical would be okay; but this is actually even more stable because it is resonance stabilized.1485

The thing to keep in mind here is because we have this resonance stabilized radical intermediate.1502

When this now reacts with Br2 to do a bromine atom abstraction, it has two different sites at which it can react and add the bromine.1509

In fact we are going to get two possible products here due to this allylic radical intermediate.1520

We are going to get a product where the bromine has added to the end carbon right where we had the original hydrogen; but we can also get a product where it is added to a middle carbon.1532

That could have happened up here as well; this was resonance stabilized; but because it was a symmetrical molecule, they would both lead to the same product.1549

But in certain cases, we want to look at both resonance forms of the resonance stabilized radical intermediate and be sure that we recognize that bromination can occur at either of those sites.1557

So for allylic bromination, we want to use NBS; and we want to draw out that allylic intermediate to see if maybe a surprise product might be formed at the same time.1569

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