Ladies and gentlemen, I do not propose today to talk about flash photolysis, but rather

to refer to a subject which has always been an old love of mine, the study of combustion, and in particular, on this occasion, to speak about the kinetics of the slow combustion

of aliphatic hydrocarbons. First, such reactions are autocatalytic. They start from zero speed and gradually accelerate till they come to the end of the reaction, in some cases passing to an explosive process.

These reactions have been described and named by Professor Semenoff as degenerate explosive reactions. The first slide will show a classical diagram of Professor Bone and his colleagues for the

oxidation of ethane, mixtures of ethane and oxygen. And this shows very well the induction period, or time of incubation of the reaction, when there's apparently no reaction taking place. And then the acceleration. I think you can see the curves are labeled. Have we

got a pointer? No. The bottom curve moves over to the right, and there comes the end of the reaction. There's an acceleration measured in pressure terms, which finally becomes a constant when the reaction is completed. And there you see the falling off

of the reactants, the ethane, and the oxygen as the reaction proceeds. Also shown are the curves of the growth and decay of intermediate compounds. These compounds rise to a maximum

rate when the velocity reaches a maximum rate. And then they fall away as the reaction proceeds to completion. They include aldehydes, peroxides, particularly. Other things are seen in many of these reactions, particularly alcohols, like methanol in the case of ethane,

and ethylene oxide in the case of ethane also. Now the explanation, the pattern of this explanation, which was given so brilliantly by Professor Semenov many years ago now, explaining

the nature of this degenerate type of reaction, is to be seen, roughly speaking, in the next slide. Here you see his idea is that the velocity depends upon a branching chain reaction. V is given approximately by the top expression, where A is a nearly constant

but not quite. And it's exponential in terms of time, dependent upon two constants, F and G. F represents the rate of branching of the chains. And G represents the rate

of extinction. You might say that F is the birth rate, and G is the death rate. Now if the birth rate exceeds the death rate, we then have a positive value for the branching factor. Phi, the net branching factor, is equal to the difference of the birth rate

and the death rate. And you then have a reaction which develops exponentially with time. This shows the development of the reaction, shall we say, in terms of the first state and the final state. It starts from zero, accelerates to the passage from the one state to the

other, accelerates along an exponential curve, which finally falls away. And the end of the reaction, beginning and the end of the reaction, are achieved by this exponential

development. Now, in actual fact, Phi, the positive branching factor, becomes zero in cases of this kind. As the reaction proceeds and as the reactants, the oxygen and the fuel, are burned up, it becomes zero. And then Phi becomes negative, and the reaction

does not develop any further. It comes to an end by slow process. If Phi, however, is greater than a certain quantity, such that it does not fall to zero during the course of the reaction, we have the development of an explosion. And the passage from this

condition of reaction to that is always quite sharp. Now, the general explanation of this in terms of molecular chemistry has to be sought. This beautiful description, which

was originally given, as I say, by Professor Semenoff, completely gives a pattern for the requirements of any molecular theory of explosion. And several efforts have been made to account for it. Semenoff showed that you explain the case for hydrocarbons,

where Phi is curiously very small and gives rise to long induction periods, AB, gives rise to long induction periods measured in minutes and perhaps more. He explained that by postulating that there is a straight chain formed by the reaction of free radicals,

which operates with the fuel and the oxygen, a straight chain which produces an intermediate substance of longish life, moderate stability, but which is of life long enough to outlast

the length of the chain, but which itself also, reacting with the oxygen, can give rise to new centers, to new chains, and so you have a branching effect. The nature of this intermediate substance was not postulated at the time, nor has there been complete

agreement about it ever since. And that is what I want to talk about today. From the very early years when we tried to develop a theory of this explosion, I fixed my attention upon the growth of the aldehyde intermediate, which I always thought was the origin of the

delayed branching. On the other hand, many other people concluded that it was a peroxidic substance. And between the two of them, a tremendous controversy has arisen which perhaps

has not yet been acceptably solved, although I have found no reason to change my opinion about the mechanism. It is agreed among everybody that at low temperatures, about say 100 degrees, when you oxidize the hydrocarbon, undoubtedly peroxides are formed. I had hoped

to be able to use the blackboard, but if we could have the lights a minute, I would like just to show something. I was not aware there was to be no pause between the last two lectures, and therefore I had no time to write up what I had hoped to show. But

the simple idea for low temperature oxidation of hydrocarbon, RH plus O2, that's the initiation process, giving R, a free radical, plus HO2, initiation, followed by chain reaction,

peroxidic radical, and the chain is continued. If you want to have a branching, it has

been suggested that there's a split in the peroxide molecule. Now nobody disagrees, as far as I know, with that peroxidic mechanism for the growth of peroxides at low temperature in hydrocarbons. They can be a source of danger sometimes when they collect at low

temperatures. And it is that scheme of reaction which so many people have endeavored to take over in order to explain what one might call the slow oxidation at higher temperatures. All the peroxidic theories of reactions of hydrocarbons between, say, 250 and 400 or

more following the same pattern, the degenerate branching type, have been either endeavored to be explained by an intermediate peroxide, ROOH, alternatively, as I have endeavored to show, by means of the aldehyde as the branching agent. The peroxidic theories have

involved the split of the peroxide, ROOH into RO and OH in order to produce the multiplication of radicals for branching. The aldehydic process has suggested that the aldehyde is the source

of branching and that the reaction of the aldehyde, as I shall show in a little later when we come to the discussion at the end, the reaction of the aldehyde also produces free radicals, R and RO2 and HO2, which can equally well account for the branching.

In fact, more so, because here we come to the big problem. No one has ever found any peroxides in the high temperature oxidation, with the exception of hydrogen peroxide and the addition of hydrogen peroxide to aldehydes. There have been no ROOH observed, and we

have been most careful to study it by gas chromatography and by paper chromatography. No one has ever found any peroxides at high temperature. Why? Because peroxides destroy themselves, they break down to form aldehydes. Now, the question therefore is to try to

find what is the true degenerate branching agent, aldehyde or peroxide. There is another point about these chain reactions. Reactions of hydrocarbons can be called chain thermal,

because as they develop, they also develop heat and they become self-heating. Under such circumstances, you get the explosion facilitated. But if the heat developed is not developed fast enough and can be conducted away, you then obtain an exponentially developing reaction,

which however goes to a final steady state. All these matters are not by any means purely academic. The true understanding of the anatomy of combustion of these compounds is of great importance to the discovery of how to control fire, how to control explosive processes,

in cases of engines and in many other cases. And of course, I need hardly refer to you to the dreadful explosion, which occurred in Yorkshire quite recently, when the reactor in which caprolactam is formed, which involves passing through a static reactor or

a static system, or rather a flowing system, high pressure mixture, cyclohexane, oxygen and ammonia. Here, the process was a steady state reaction, but into a break in the connection

tubes, a big drop of pressure occurred, and certain conditions were involved, which turned this steady state reaction, this degenerate reaction, if you like, into a sudden explosion. Now, it is important to know what the intermediate substances are, which are responsible for things of this kind. Before we can use our detective methods

for trying to find out which is the effective degenerate branching agent, peroxides or aldehydes, I will just state a few of the important criteria which must be taken account of and

explained in any theory. First of all, we have the induction period. Next slide, please. Here you see some curves for the development of the reaction, this is in terms of pressure change with time, of the reaction of ethane and oxygen. We have indeed put into the

mixture of ethane and oxygen certain possible aldehydic compounds. If you put in 0.2 millimetre of acid aldehyde, the induction period is a long one like that. If you put in none

at all, it is still longer. These smooth lines show the addition of aldehyde. If you put in 1 millimetre of aldehyde, the induction period is still less, but the velocity has not changed. Then if you put in 2 millimetres, the induction period is nearly removed and

with 5 it is almost just a bit too much. The reaction rate continues at the same speed and the development period is removed, which we take to be the reason, to be the origin of the effect of the intermediate. Put in the intermediate which is formed during the

process and if you put in the right amount, which represents the stationary state when the reaction is going steady, you will be able to remove all the necessity for the exponential development. We use here also formaldehyde, dotted lines, and you see much

more formaldehyde has to be used at this temperature, which is about 310, much more formaldehyde to be used and even so you never get it down to zero. However, at very high temperatures, 450, 500 and beyond, acid aldehyde is no longer detectable. It has

been destroyed completely and formaldehyde has now taken over the role of the degenerate branching agent as we shall show. So you see there the induction period removed by the addition of the aldehydic intermediate. The next slide shows another interesting

effect on the rate of reaction as we change the temperature. Here again we are talking and I am going to refer mainly to ethane because there is not time to deal with the whole pattern of alkyl substances, but here we have in the case of ethane and similar

curves for other hydrocarbons, the effect of the rise of temperature on the velocity, temperature rise, the rate of reaction measured in terms of temperature rise, that is to say a sort of steady rise of temperature due to the stationary state of the reaction,

here is the maximum rate of pressure rise which measures the rate of reaction. They both go together of course. And you notice that the rate of reaction increases with rise of temperature up to a maximum of about 360 and then as the temperature rises

again it falls away and passes through a minimum until once more it starts to rise. This is erroneously, in my opinion, called the negative temperature coefficient of the process. Any theory will have to explain this change. The next curve shows characteristic pictures,

the next slide, of what are called cool flame phenomena. This curve is taken not by us, I forget the name of the author, it deals with the case of propane. I think it comes from the Moscow laboratory. And you see here as the development of the reaction

proceeds it starts to develop very rapidly as if it were going to pass to infinity, that is explosion. But at a certain pressure change it drops back suddenly and there is then a delay again while it goes up once more. You have this kind of periodic reaction

until finally it never does get to explosion in this case, it slows off and finishes. These pressure pulses or reaction pulses are accompanied by the emission of chemiluminescence.

When you take the spectrum of the chemiluminescence you find that it is the spectrum of formaldehyde, pure formaldehyde emission. When you take the spectrum and when you study also the oxidation of aldehydes you see the same phenomena, you see the same evolution of chemiluminescence,

the same formaldehyde picture comes out and we can therefore take the luminescence as indicative of the presence of aldehydes in the system. This build up of the intermediate proceeds until the temperature rise is great enough

to destroy the nature of the intermediate substance and so the reaction which was going to go on further suddenly drops back due to the denuding of the system of the intermediate. It then has to start again and build up once more and so it can do it several times.

Sometimes it will finish with an explosion, sometimes not. Now it's a curious fact that if you study the amount of light emitted there's only one quantum of light emitted per million molecules oxidized. It is not therefore the property of the reaction itself. It merely betrays the nature of

the intermediate substance. Both this curve and the one we had previously showing the drop and rise again of the velocity are indicative of the fact that the intermediate substance is destroyed and with rise of temperature. As the reaction proceeds here the temperature rises you see and destroys and starts again

and so on. Some people have suggested that it's the peroxides which are being produced and destroyed but I shall give pretty good evidence later that it's not in terms of peroxide but in terms of aldehyde that we must regard

this very remarkable phenomenon of cool flames. Now again it seemed to me that if aldehydes were the important intermediates we should be able to affect the rate of reaction, the kinetics of the reaction by means of ultraviolet light.

So we were able to make a reaction vessel with a quartz window at one end and to make it very powerful with light from a high pressure mercury lamp. Of course suitable filters were put in to cool the light, take out the infrared and also to fix the wavelengths

of the light, to fix values. And the idea at the back of my mind was that there would be an addition to the branching process affected by the fact that the light

itself is able to dissociate the aldehydic substance into two free radicals. In the case of acid aldehyde the molecule is broken down into a methyl radical and a radical CHO which itself can give rise to further new radicals. In other words

the additional effect of the light should be to increase the branching factor phi and so to reduce the induction period and to increase the velocity. The next curve shows similar cool flames which I have not mentioned in the case of

ethane which we found and one of the most difficult reactions to show. Next slide please. Yes, there are the combustion of ethane molecules. You have to do it in very large vessels to see it at all. And you see at different temperatures you begin to get these cool flame pulses. And in the case of temperature at 309 there are

one, two, three pulses before the reaction is completed. Here again are three pulses. Here we've gone down to two again. One. And then finally at the higher temperatures none at all. That's because the reaction is over before the heat rises enough to destroy

the aldehyde. And this development of the reaction in this way is one of the evidences of the importance of the nature of the intermediate. Now to get back to the photochemical reaction we took mixtures of ethylene and oxygen

and of methane and oxygen which of course oxidize around these temperatures a bit higher in the case of methane and irradiated these mixtures with light which is absorbed by aldehydes. That is to say light of wavelength longer than 3100 angstroms. Light which was not

absorbed by the peroxides, if there are any there, which are absorbent only at lower shorter wavelengths. The next slide shows the development of the oxidation of ethylene on irradiation.

The two bottom curves are light and dark. The black points are the rate of reaction in the dark. The curve above it shows the rate of reaction when the light is turned on. You notice that the velocity is greater and the induction period is reduced. There is a pure photochemical effect due to the increase in the branching factor phi.

And here again in the next two curves you have two more curves used at different pressures 3 and 4 at higher pressures of oxygen. Rate is greater for both the dark and light reactions but still you see the effect of the light process. Same for methane

which I will not show. And the next slide shows how you can actually turn these slow reactions into an explosion. Here these two curves are as before dark and light corresponding

to each other. No, yes, dark and light corresponding to each other. A slightly higher pressure of oxygen. Dark curve is a normal degenerate explosion and the light curve is turned into a true explosion. We can convert a normal degenerate explosion into a true explosion.

In fact we can explode the mixture by irradiating with ultraviolet light. I ask you to remember that the light used here was the light absorbed by the aldehydic intermediates. I said I wasn't

going to mention flash photolysis but still you can study these reactions by flash photolysis. You can observe the rate of growth and decay in the combustion reactions at very short intervals of time. And although I haven't got slide with me you can see the exponential rise of the hydroxyl radical during the short induction periods at the very high temperatures

of explosive processes and its decay. And it has also been shown in terms of methane I think it is, that the same rate of rise and decay of the methyl radical. In other words we see by flash photolysis the intermediate radicals expected in the

case of the particular hydrocarbon which we observe. And always of course the hydroxyl radical. Another curious effect is the need for maturing the vessel, the reaction reactor in which the experiments are carried out. If we start straight away with what you

might call nicely cleaned vessels, washed out with hydrofluoric or nitric acids, you get one set of curves. If then you don't wash the reaction vessel out but merely evacuate it and put in further charges, as we go from one run to another the reaction

changes. Until finally we get a consistent reproducible condition. This is called maturing. And there is something produced or some change produced on the surface as a consequence of the reaction which has been carried out in it. Markievicz found this to be so for

the oxidation of aldehydes. And other people, including ourselves, have found it to be so for the oxidation of hydrocarbons. So we restrict ourselves to a nitured vessel. And I may show you in a minute the reason why this change, or part of the reason

why this change in the nature of the reaction occurs as we pass from one vessel to another. In fact, I think there's a slide showing the maturing of a vessel. Next slide, please. This shows the maturing of a vessel in which the maturing takes place after one

run after the other. Here's the first run here, you see, corresponding to the points up there. But eventually, after you've gone on long enough, you find the products begin to steady themselves down. I don't know what happened out here, this is different. Anyway, it does steady itself down to a reproducible rate. The same is true for

the oxidation of hydrocarbons. Now, what was found later which seemed to throw a spanner into the works, as it were, was when Knox and I discovered when propane is oxidized,

there is a large production of propylene. And this origin of this propylene began to be very difficult to understand in the first instance. However, we have now realized that there are in fact two chain reactions taking place. And that this propylene is

formed with the production at the same time of hydrogen peroxide. Now, in an unmatured vessel, the hydrogen peroxide, which is formed, is very rapidly destroyed. So you only measure as a product the ethylene. But in a carefully matured vessel, and in

the case that I'll show you, with a small amount of aldehyde in to reduce the induction period, you can show that there is an equal amount of, or nearly equal amount of formaldehyde, I mean hydrogen peroxide, formed as it is of ethylene. Next slide, please, will show

the growth of the products ethylene and formaldehyde. And, sorry, hydrogen peroxide. Now, this is one of the new results which has to be explained. And at first sight, it seems that it's difficult to understand what role the acetaldehyde is playing in this

process. However, it's not too difficult, eventually, to find that out. We'll miss the next slide, please, and pass on to the last one. Miss the next slide. That is right, thank you. The explanation of all these different phenomena, which has to be

achieved by any mechanism proposed, does require an extension of the theories of the oxidation process. Here we're dealing with ethane. Now, the process in ethane involves, as we suggest, the ethyl radical. That forms what might be called a peroxidic radical,

which is in fact a transition complex. It doesn't last in the length of time. It isn't responsible for the branching, because the branching must take place after a long interval, and this lasts very small fractions of seconds. But this peroxidic complex can break up

in three ways. The first one is to make ethylene and HO2 radical. The second one is to make acetaldehyde and the hydroxyl radical. And the third is to make formaldehyde and the methoxy radical. Here you see how it can break up. First reaction, HO2 and ethylene.

Second reaction, OH and acetaldehyde. Third reaction, methoxy and formaldehyde, H2CO. There are thus possible three different reactions following from the changes 1, 2 and 3. From

the reaction 1, the HO2 radical is produced as you see. This HO2 radical can then react

with more ethane, forming again the ethyl radical and hydrogen peroxide. And then this reaction can take place again, the ethyl radical, to form ethylene and HO2. In other words, there is a straight chain involving the formation of ethyl radicals and

hydrogen peroxide molecules. And in the limit, there ought to be the same amount of ethylene as hydrogen peroxide, which is what we have been able to demonstrate. It is, however, a straight chain. It does not give rise to any induction period or development. The second reaction is the one which produces

the aldehydic product, C2H6 plus OH. First of all, it can react with hydroxyl to make ethyl radicals. The aldehyde formed from 2 can react with oxygen to form ethylene. Intermediate CH3 plus CO plus HO2. And here you see the aldehyde operating in its function

as a branching agent, because we have already lost one radical in the manufacture of CH3-CHO, but we have produced two new ones. So there will be an increase in the number of HO2 centers, and therefore ethyl centers, as the reaction proceeds.

Another way of branching with the acid aldehyde, which would be of less importance, because we have got so much oxygen there, this will be the predominant reaction. However, this reaction can occur, again producing products CH3, CO, and H2O.

Finally, the third process, the CH3, you can react with oxygen to form formaldehyde and OH. That is to say, this CH3, which is formed here, can produce hydroxyl radicals and so contribute to the branching. The third process, which involves the destruction

of the peroxidic complex to form aldehyde and ethoxide, I mean, sorry, ethyl radicals and oh yes, taking three there, you see, the OCH3 can react by extracting hydrogen

either from the ethane or from any other hydrogen-carrying molecule, such as formaldehyde, formed in the process, to form ethyl alcohol, which is a major product of the process. So the major products, including also, which I haven't shown here, ethylene oxide, can

be accounted for by concluding that the first result is the breakdown of the peroxidic radical to give either ethylene on the one hand, acid aldehyde on the other, or methyl alcohol on the third. And it is that which we have been able to synthesize

together. Now, if you notice, of the three reactions, two and three are liable to cause branching. That is to say, to cause a multiplication of the ethyl radicals from a very low value

at the beginning to a higher value at the end. And so the branching effects of reaction two, and also three, involve an increase in the ethyl radicals, and therefore an increase in reaction one. In other words, the ethylene will be formed as the reaction develops will

be formed faster and faster because of the steady increase in the pressure of the ethyl radicals formed from reaction B or C or A. So we can understand how the process of oxidation grows with the formation of ethylene in large quantities and hydrogen peroxide

without the need of a branching process of its own. It follows the branching reactions of reaction two. Now, I hope that I have tried to make a strong case for the actions of the aldehyde intermediate in creating the growth and development of the reaction

process of oxidation. There have been many other efforts in terms of peroxides. In fact, so many that even the authors appear to have become confused. It has not been necessary to make any modifications in the aldehydic theory with the exception

of this addition when we discovered that one of the first products of the reaction is ethylene and hydrogen peroxide. Now, a final proof, I think a final proof,

of the importance of aldehydes as compared with the importance of peroxides was provided by Stern. I'm not sure where he worked in Moscow or where he worked, but I've never had the pleasure of meeting him, and I should like to, because in my opinion, he has vindicated the aldehydic theory. If you take a reactor in which the oxidation

is proceeding and extract halfway through the reaction, extract all the contents by liquid air into a trap in which there is presence of mercury, mercury will destroy all peroxides if there are any there, but will not touch the aldehydes. He then lets it warm

up and puts the reactants back into a new reactor and start at the right temperature and finds that the reaction starts off without any induction period at all. In other words, the aldehyde which was there in the first instance has been preserved and continues

its reactive function without any necessity for induction period for its buildup. The peroxide, however, which has been removed, has had no effect on the process at all. He's chosen between aldehyde and peroxide by following the growth of a reaction

to halfway and then removing the whole lot and destroying one of the competitors, one of the candidates for the degenerate branching and not the other. He's been able to show that the aldehyde is the functioning intermediate. We have studied very carefully

I hope with an open mind, and we've had of course to take over some ideas of the peroxides, which of course we understand are there too, but we do wish to stress the point that there is no functioning peroxide in the course of the high-temperature

reactions. We have studied it by gas chromatography, we have studied it by paper chromatography and find no evidence at all which would be required for the peroxidic intermediate.

Finally, I would like to say again that these ideas are not academic. It is important to find out what the intermediate substances are in order that we can understand the mechanism of combustion and in order that we can proceed to influence the mechanism of combustion by the judicious use of additives or inhibitors.

In the final state now, when you remember after the negative temperature coefficient, the reaction rises again. We have the result of the production of formaldehyde. Formaldehyde is a much more stable aldehyde than any of the others such as acid aldehyde, propion

aldehyde and so on. Formaldehyde is the last substance formed by what we call aldehydic degradation. You see the aldehyde here reacting with hydroxyl has given rise to

a CH3 group and the CH3 group reacting with more oxygen has given formaldehyde. Now this formaldehyde remains in the reaction chamber, not very active at the lower temperatures although you saw some effect on the induction period. It remains in the reaction chamber

at high temperatures, say 400 to 550 and so on. And it is at that temperature that it takes over the role of the effective intermediate. In other words, the rise again of the velocity of reaction is in my opinion to be explained very largely at any rate

by the development of the reaction through the agency of the intermediate formaldehyde. There are other stories which must go on at another time. At very much higher temperatures, we get evidence that the hydrogen peroxide reaction is developed much more strongly.

This hydrogen peroxide reaction you see has a temperature coefficient and as the temperature rises the formation of ethylene increases so that at about, shall we say about 600,

500 or 600, it's nearly all ethylene that is formed and the branching process which is still going on is functioning through the formaldehyde. I don't expect you to swallow all that in one go and I hope I have tried to make out a case for the importance of the intermediate substance, whatever it is, and to make a case

out that the intermediate substance is the aldehyde. Many people are still sweating away with the peroxide theory but if they can explain all the phenomena which I have described, which have to be explained, I shall be prepared to change my mind but I'm not prepared at present to do so because they have not

been able to do it either. Thank you.