Free Chemistry NEET Video Lectures for Haloalkanes and Haloarenes

00:00:21 [Music] Uh hello uh I am Dr. Romesh Ramapaniker. Uh I am an associate professor in chemistry uh at the department of chemistry at Indian Institute of Technology Kpur. So today I will be discussing the portions that are covered in unit 10 of the chemistry book by NCERTT for class 12 students and uh this particular chapter as you would know covers halo alkanes and haloarines. These class of compounds are also known as alkalhalides and arhalides. They are a class of organic compounds where a hydrogen atom that is present in

00:

00:56 a hydrocarbon is replaced by a halogen atom. Multiple hydrogen atoms can also be replaced by multiple number of hogen atoms and in those cases too they fall under the same category. Uh however their ideal names are halo alkanes and hoarines but you would find that the normally available and most common compounds are generally referred to as alkalhalides rather than halo alkanes. However, a systematic naming would require that we call them as ho alkanes. you should start with the name of the hogen that is attached uh to the

00:01:29 organic compound. So we will deal with the naming of these compounds and how these molecules are classified. Then we would go on to say how they can be prepared from simple static materials and finally we would talk about their properties and then look at a few examples of these class of compounds that are useful to mankind either synthetic or which are available naturally. Now to begin with I would like to say that a general representation of these molecules that is whenever you would like to show an alkalhalide that is a

00:02:00 hoalk alkine by a simple representation the term that we normally use is RX where R stands for the alkal group as you already know and X stands for the halogen atom that is attached to the alkal group and in organic molecules the most common hogen atoms that you would find are florine chlorine broine and iodine and we will not see the fifth fifth halogen atom the fifth atom that is present in the group 17 element aatin that is not normally found in organic molecules. So by halo alkanes we would

00:02:31 only mean compounds of florine chlorine bromine and iodine. Similarly hoarines are represented by a rx where a r stands for the ar group or an aromatic group and x is the hogen atom in this case too. Here x can be florine chlorine broine and hyden just like in the previous case. Now when we look at these molecules, if we want to simply distinguish between an alkalhalide and an arhalide by the simplest of possible ways, the best method is to assume that an alkalhalide would have the halogen atom attached to an sp3 hybridized

00:03:04 carbon atom. That means the carbon atom would be completely saturated. It will not be part of any double or multiple or triple bonds. But a saturated carbon atom that is sp3 hybridized attached to a halogen atom is often referred to as an alkalhalide or a halo alkan. Uh then in arids they should be attached to an sp2 hybridized carbon but most uh by the term an arhalide or a hoarin we would mean that they are attached to an aromatic ring or a higher order aromatic compound. So we will look into these

00:03:35 classifications and names in the coming slides. Okay. So before I go further, I would also like to say that these kind of molecules are widely used by chemists and also by non-chemists even without knowing what these compounds are. They one of the most common applications of these molecules are is that the lower members of this class of compounds are liquids. So they are used as solvents and they can dissolve most of the other organic compounds. So organic compounds unlike most of the inorganic compounds

00:04:05 that we come across for example sodium chloride potassium iodide and so on they are mostly non-polar compounds they are not ionic so such compounds needs to be dissolved in suitable solvents and halo alkanes are normally used as solvents for dissolving these non-polar organic compounds they also are highly useful as synthetic starting materials so whenever we would like to prepare organic compounds that contain multiple functional groups. One of the starting points for making these kinds of compounds are organo halogen compounds

00:04:39 or halo alkanes or haloarines. So I would also like to say that normally these com organic compounds that contain a hogen atom are called organo halogen compounds or organo halo compounds and so on. So you would find these terms being used interchangeably in between or just to say that halogen containing organic compounds and so on. So I have a few examples listed here. So you would find that these are also some of the examples that you would find in your book as uh being mentioned as organic compounds that contain halogen atoms. So

00:05:12 first of them here is chlorophyll. So this is an antibiotic. This particular molecule is an antibiotic. You can see that it has two chlorine atoms attached to this. So by antibiotic it is something that has antibacterial properties. So this can be used against u bacterial infections that can cause us various diseases. So this particular compound chloromicol is used against typhoid. It is also a good against plague and so on. But it is one that is most commonly used against typhoid. So it has chlorine. It has two chlorine

00:05:44 atoms. And you would also find that these chlorine atoms are attached to an sp3 hybridized carbon atom. So this is an example of a halo alkan. Uh the second example that I have is thyroxine. So thyroxine is a pro hormone. So this is not exactly the hormone that is functional. You would find that it has four iodine atoms and all four iodine atoms are attached to aromatic rings. Now in the active form, one of the iodine atom is lost from it and it would be a triodo compound that is the active enzyme and this particular enzyme is

00:06:16 formed in our body and if you have a deficiency of this this is taken as a medicine. This is taken orally. So this is just an example to show that this polyhalogen compound, the compound that contains as many as four iodine atoms attached to an aromatic ring is a biologically active compound and something that we produce in our body. The third example is chloroquin. Uh this is used as a medication against malaria. It is also used it is taken in advance to prevent from getting affected by malaria or it can also be used as a

00:06:47 drug. So this has a chlorine atom that is attached to an aromatic ring. So although this as you see is a compound that has two rings attached but the chlorine is attached to one of the aromatic rings. So it is an example of a halo arene. So the first three compounds that I showed you are having rather complex structures which are difficult to comprehend. But you have a look at the fourth structure that I have here. This just has two carbon atoms and the first carbon is attached to three florine atoms whereas the second one is

00:07:15 attached to a chlorine and a bromine. This is called halo and this is a general anesthetic. So these are given to patients before they are subjected to a surgery. So this compound would be able to make you not feel the pain or the medical process that a person is supposed to undergo. So this class of compound simply gives you an idea that the structural diversity of these compounds are quite high and their applications are also quite a lot. However, while we are trying to use organo halogen compounds, we should be

00:07:47 aware of one particular fact that halogenated organic compounds degrade extremely slowly under natural conditions. That means if it happens to go into soil, if it happens to go into the environment, they generally take quite a lot of time to get disintegrated or decomposed. As a result, they continue to stay in the environment and sometimes they continue to stay in living organisms and most of them can cause problems. So therefore although initially when h hallogen compounds started to be introduced for industrial

00:08:16 applications they were widely accepted nowadays we would like to limit their usage because they could cause problems to the environment. Uh one of the reasons why they get degraded slowly is the is because the bacteria that are present in soil are generally not able to decompose them by their normal methods of decomposing organic compounds. So this retains these molecules in soil for a longer time. Okay. So now uh the next thing that we can do is to have a look at very simple examples of these class of compounds and

00:08:50 then try to see how we can classify them so that whenever we come across one of these compounds we are able to identify them and put them into a particular class and then able to understand them better u from our perspective. Okay. So the most simple classification that we can have is on the basis of number of halogen atoms. So if you take first of all when you see an organic compound that contains a halogen atom we can have a look at them and see how many halogen atoms are present. So I have ethane

00:09:17 molecule that is substituted with a halogen atom here. So this is a monohalo alalkkan by x as I have mentioned I mean either florine chlorine bromine or iodine. So this is a monohalo alkan. One h hallogen atom is attached to it. And the second one is a diehallo alkan. I have put one h hallogen atom each on the two carbon atoms of ethane. So this is a diehallo alkan. So they are normally called as di plus the name of the h hallogen atom and then the alkan that it corresponds to. And the third one is

00:09:46 trihaloalkkan where I have three h hallogen atoms attached to three different carbon atoms of a propane based molecule. uh the classification continues to be the same even in the case of aromatic compounds. So monohaloarene a compound that has only one h hallogen atom it can be dihaloarene when there are two h hallogen atoms present trihalorine and so on it has to be noted that these h hallogen atoms need not be the same. So whenever I write an X one of them can be a florine the other can be a florine the

00:10:17 third one can be an iodine and so on and all those compounds can be made and all the most of those compounds are known. Next we could go ahead and we can say that we should then once we know that once we have identified the number of halogen atoms present in a molecule the next thing to do is to look at the molecule and see where exactly is this hogen atom attached to the molecule. So based on that we could classify them as compounds that contain sp3 hybridized carbon atom that are those are saturated

00:10:47 carbon atoms attached to a h hallogen bond. So this normally all the simple alkalides or halo alkanes that we started discussing with. So most of these compounds fall into this category and they as I have already mentioned are shown as RX and a general representation for these class of compounds provided that there are no unsaturated bond within the molecule would be CN H2N + 1X that this so this is derived from the general representation for a saturated hydrocarbon which would be CN H2N plus2.

00:11:21 So we have removed one of the hydrogen atoms and put a hogen atom. The simplest example is a methile halate or a halome methane. So I have methile group here attached to an X. Now I have shown the carbon atom in in blue color. This is just because I would like you to concentrate on those particular atom that is attached to the halogen here. And you can see that this C atom is attached to three hydrogen atoms. That is when we call it as a methile group. So it's a halomethane. And the next example, I have almost the same

00:11:52 structure except that one of the hydrogen atom is replaced with an alkal group. Then in the third structure, I have replaced two of the hydrogen atoms with two alkal groups and I have replaced all the hydrogen atoms with alkal groups. So based on the number of alkal groups that are attached to the carbon to which the halogen atom is attached, these molecules can now be called as primary, secondary or tertiary alkalhalidates or halo alkanes. So once again these classifications are based on the number of substitutions that are

00:12:24 available on the carbon atom to which the halogen is attached. So a primary hoalkkan is one in which the carbon attached to halogen is bonded only to one alkal groups and when there are two hydrogen atoms present. We call them as secondary when two alkal groups are present and tertiary when three alkal groups are present. Uh the normal notation, a simple notation if you do not like to write primary, secondary and tertiary in all cases would be to use one with the degree sign on top. But this should not

00:12:55 be read as uh one degree and this should not be read as two degree, 3°ree and so on. Instead although you write them as one with a degree on top, they have to be read as primary, secondary and tertiary. I quite often come across people who mention to this compound as a onederee alkal chloride, two degree alkal chloride and so on which is not the correct way of uh calling them. Now even while I have said that these are compounds that do not have unsaturated carbon atoms attached to halogens, we could also have these class of compounds

00:13:27 which are called alilic helides where the hogen atom as you can see is attached to a CH2. So I would uh show you this particular carbon here which I now I have highlighted. So this is a CH2 group. It is attached to a double bond and an X. So this comes under the hoalk alkan class or this is an alkalhalide. And here again the halogen atom is attached to a carbon that is sp3 hybridized. However the same carbon is also attached to a double bond. So these kind of compounds where the carbon atom attached to the hogen atom is near is

00:14:01 adjacent to a double bond are called alilic helides or alilh helides in general. So I have three such examples. So here you can see this is an open chain compound a linear compound where I have a double bond CH2 and the second example is where I have a double bond and on the so it's a six membered carbon ring. So it's a cycllohexane ring. So in this cycllohexane I have a double bond. So it's a cycllohexene. And on the carbon that is immediately after the double bond has a halogen atom. So this

00:14:31 again is an alilic helid. And in the third example I have a cyclopentine ring but now the CH2 that is attached to the double bond is going out of the ring but it is attached to a hogen atom. So this again is an alilicide. So alilicides are those class of compounds where you have a double bond that is attached to another CH2 followed by an X. So this all these uh out of these three structures you can see that this carbon is attached to one alkal groups and two hydrogen. So this is a primary alhalide

00:15:00 whereas this one this carbon is attached to two different groups one double bond and this and there is only one hydrogen present in this. So this is a secondary alhalide whereas this again is a primary alhalate. So these are the classifications for allehides. They are all halo alkanes because the halogen is not attached to a double bond. Instead, it is attached to a saturated carbon atom. Now the other class is benilicide. So these are compounds where the carbon attached to halogen is attached to an

00:15:28 aromatic ring to be precise to a phenile ring. So this is a primary benilicial helide or a benilehalide where the aromatic ring is attached to a CH2 and an X. And in the second example, this carbon is also attached to another R group. That R group can also be an aromatic ring. It can be an alkal ring too. So then we call it as a secondary benilehide. And if there are two R groupoups and a phenail ring, we call it as a tertiary benzilhalide. Now again out of these two R groupoups either one or both of them can also be aromatic

00:15:59 rings. So that doesn't matter. We still call them as benilehide and they would be tertiary because there are no hydrogens's attached to the carbon to which halogen is attached. Now we would go ahead and then talk about compound compounds containing sp2 CX bond. So these are compounds where the carbon that is attached to the halogen is sp2 hybridized. We started off our discussion by saying that aromatic halogen compounds or haloarines or arhalidates are those compounds where the halogen atom is attached to an sp2

00:16:29 hybridized carbon atom. But you already know that not all sp2 carbon atoms needs to be in an aromatic ring. For example, these compounds, these two compounds here, the vinyl helides. So this simply mean that I have a double bond and the carbon that is attached to the double bond involved in the double bond is attached to a hogen atom. So these are examples of vinyl helids. They can be open chain compounds or they can be cyclic compounds. Now we call only those compounds as arh helidates where the

00:16:57 halogen is attached to an sp2 carbon of course but to an aromatic ring. The aromatic ring need not be benzene in all cases. It can be napylene. It can be any other higher order aromatic rings. But as long as they are attached to the hogen atom is attached to an aromatic ring, they are called arelhalides. So this is a very simple way of classifying these compounds. Okay. So now that we have uh talked about how these molecules can be classified, we will look at their nomenclature. or you already know that

00:17:28 once we know about a particular class of compounds we would like to give them names so that whenever we would like to call out a name and immediately a structure associated with the name comes to our mind. So this is the whole function of nomenclature. So the term nomenclature is uh given by UPSC which you know is the international union for pure and applied chemistry. So now what UPSC does what the union does is it sets out a set of rules based on which how chemistry should be seen and discussed

00:17:59 by the class of people who follow chemistry. So one of their main criteria is to set out rules by which compounds can be named in a very systematic manner. So that the name that I am suggesting or I am telling is something that immediately the person who listens to the name is able to associate with a particular chemical structure. So all of the compounds that we see today can have an UPSC name. However, some of the compounds were well established. they were already known even before UPSC started setting out its rules. So such

00:18:33 compounds are sometimes known more commonly by their common names because the chemicals are so common that people would immediately identify them with name common names that were in use for uh decades. So based on this some of the compounds would have an UPSC name and along with them they would have a common name. Now if you do not know the common name that is all right but you should always be able to write an IUPC name for a compound but we would discuss the common names for the uh most common

00:19:01 compounds and then discuss UPSC name for most of the compounds that we would see here. So I have examples here to discuss the points with you. So the first compound I have is a ho alkan that has three carbon atoms and it is attached to a broine atom by one of the sides. So uh just like the nomenclature that you studied in UPSC whenever we see a compound what do we do first is we identify the carbon atom that has this particular substitution then we look at the compound and identify the longest carbon chain. So in this case it is a

00:19:34 straight chain we do not have any issues with that. So there are three carbon atoms in this chain and it is attached to a bromine atom at one end. So the carbon that is attached to broine atom is the one we would call as carbon number one. So this compound has three carbon atoms. So it is a propane. So we would be able to call this compound as a broo substituted propane. And the UPSC name for that compound would be one brooane. Which means that we have a propane where the bromine atom is attached to the first carbon atom. That

00:20:04 means at the end of the chain. Now again this compound is propane. So it is called as normal propel bromide in a common way. So it's it has a trivial name which corresponds to normal propel bromide or nropel bromide. So you would also find this name being used. So what I have done is the red colors that I have drawn here are all the UPSC names and the blue names are the common names. So it is always important that you know the UPSC names and you are able to write the UPSC names. Common names are not

00:20:32 something which any any of us can start giving because those are already established names that were there. So if you do not know them that simply means that you have not heard about them. But UPSC names are something that we are able to give to a molecule even if we we are seeing a molecule for the first time. So those are the names that one should practice and be able to give it to a molecule. So the second compound here so this again is a propane but this time instead of broine I have a chlorine

00:20:58 and the chlorine is attached to the second carbon atom. So it is not at the end of the chain it is in the middle. So then what we do is we start numbering the carbon atoms from one of the side. So we have 1 2 and three and the chlorine atom is attached to the second carbon atom. So we call it as two chloro and then of course propane because there are only three carbon atoms in the chain. So the UPSC name for this is two chloropane. The common name is even more interesting whenever you have a propane chain that

00:21:25 is three carbon chain and if you put a substitution on the middle carbon it is normally called as an isopropile group or this particular compound is called as isopropile chloride. So isopropile is bent. The substitution is on the middle carbon atom of a propile group. Now the third example is isobutile iodide. So you would immediately see that this has some similarities with the isopropile group because I would show you three carbon atoms that actually looks like a propile and the substitution of a CH2i

00:21:53 goes from the middle carbon. So that is why in common name it is called as an isobutile group and it is an iodide. So we call it as isobbutil iodide. Then again you see it is very difficult to give this common name by universal because there are no rules followed in here but whereas the UPSC name is something that we can always give to the molecule. So now let us look at the molecule once again. So this is a molecule that has four carbon atoms and an iodide but you what we need to do is we need to identify the carbon that is

00:22:21 attached to I and then make the longest carbon chain. So no matter which way you go you would see that you can only make a threecarbon chain in this. So one of the methyl groups either this or this has to be seen as a substitution. So therefore the name of this compound would be one iodo 2 methyl propane. This is because between the methyl group substitution. So if I am assuming that this is the chain that I would like to go with. So you see that I start with I CH2 CHCH3 and the second carbon atom has

00:22:52 a CH3 group. And how do I name it is I give the name in such a way that the name that has the first alphabet in the alphabetical order. Uh that has to be that substituent has to be named first. So it has an iodine and carbon number one. So it is one iodo 2 methile propane. Propane because there are only three chains. So it's a very simple rule but once we start using it we would be able to name it better. Now I have another example here. This should be rather straightforward. So this is a four carbon chain with a chlorine at the

00:23:22 end. It is called one chlorobutane. It doesn't matter which side you write chlorine. Chlorine can be written on this side or this side. But as soon as you see a chlorine atom attached to an alkal chain, you number the carbon that has chlorine atom as the first one. If chlorine is in the middle, then that particular carbon should get the lowest possible number um as much as possible. So we will see that example. So this compound here is one chlorobane and or it's trivial name is nbutyl chloride or

00:23:48 normal but chloride. A careful look at these names would suggest that while all the UPSC names look like hollow alkanes the trivial names or the common names are corresponding to alkyhalide. So they name the alkal group first followed by the hallogen that is associated with as a halide. So these names also have more similarity with uh the inorganic compounds that we call sodium chloride for example where sodium is replaced with an nbutile group. So, so this this has a lineage from all the other

00:24:22 chlorides that you would find in nature and different kinds of chemists used to use but whereas organic chemists would like to call them as one chlorobutane so that their names are pretty clear to everyone who hears them. Okay. So this is one chlorobutane. So the second example here is more interesting because here I have four carbon atom but the chlorine is not attached to the first one. So now I can start naming this from this end or this end. So if I start naming it from the left hand side, I

00:24:50 would see that the chlorine atom is attached to the second carbon atom here. But if I start numbering it from the right hand side, I would find that chlorine is attached to the third carbon. So you should always start numbering from the side in such a way that the carbon attached to chlorine gets the lowest number. So this is two chlorobutane and not three chlorobutane. So we would always start naming from here. Butane because there are four chains, two because there is the chlorine is attached to the second

00:25:16 carbon. chloro because it's a chloro. So if I replace chloride it would be with bromine it would be two bromutane and so on and this compound is its common name is secondary but chloride and sometimes repen as sb butil chloride because uh here in butane the chlorine is attached to a secondary carbon atom a carbon atom that is attached to two alkal groups. So it is called a secondary butil chloride in its common name. Now the third one here is uh something that we have already seen. It is similar to this

00:25:46 compound here that I have the isobutil iodide that we have already seen. So this is isobutil chloride. It has to be named as one chloro 2methyl propane. So that is the upc name. So assume that you do not have any issues with this. Now have a look at the last compound I have on this slide. So this particular compound has a rather interesting structure. It's a carbon atom that is attached to three CH3s and a chlorine atom. So definitely this is a tertiary uh hoalk alkan because the carbon here

00:26:16 which is attached to chlorine atom is already attached to three alkal groups. But to name this compound properly in uh following the upsc rules what we need to do is we look at the longest alkal chain that has chlorine atom on it. So you can see that there is a CH3 a C attached to Cl and a CH3. So once you consider this particular alkal chain, you would find that it has a CH3 attached to the carbon that is attached to chlorine. So because the longest chain has only three uh carbon atoms, it is a propane and the

00:26:47 chlorine atom is attached to the middle carbon. So it is two chloro and the same carbon is also attached to a methile group. So it is two chloro two methyl propane. So this is the way we would name it. So we would name this compound with the longest chain being propane. We will call it as a propane with a chloro and a methile substituent. Now which one to name first? We would always name chloro before methyl because chloro starts with a C which comes first in the alphabetical sequence and methyl comes

00:27:11 second because it has an M and the names the numbers have to be such that these substituents have the lowest possible numbers. And now I have other class of compounds. So we have left the simple hoalk alkanes there and we are going further and looking at these compounds that have double bonds. So see what you see here is an ethine molecule. So you know all of you know ethine. Ethine is CH2 double bond CH2. Now one of the hydrogen is replaced with a chlorine atom. So it is called chloroeththine and its common name is vinyl chloride.

00:27:41 So there is no particular reason why it is called vinyl chloride. The compounds that are attached to a double bond a simple double bond are called vinyl compounds. So if it is a chloride we call it as a vinyl chloride. So this is a common name. All of you might have heard PVC polyvinyl chloride which is a polymer that is actually obtained by polymerizing this particular compound. That is a common name for the compound. The UPSC name is chloroethine. So it is ethine attached to a chlorin atom. Now I

00:28:07 have a second example. This is something we have seen. This is an ali compound is an alil bromide. But uh the UPSC name does not allow you to call it as an alil bromide. Instead what we would call it as is a propene derivative because there are three carbon atoms and a double bond. And you would know that propene is the parent hydrocarbon here. There's no more propane because there is a double bond. So propane is the parent hydrocarbon and the bromine atom is attached not on the double bond but on

00:28:35 the carbon that comes after the double bond. So we would start numbering this uh molecule in such a way that the double bond gets the lowest possible number. So the double bond start from carbon number one. It goes to carbon number two and in the third carbon we have the bromine atom attached. So it is called three bromopropene or and we put a one in the middle just to tell you that the double bond is actually on not on the carbon that has bromine. But once you mention three broine, it is pretty evident that the

00:29:05 double bond is not exactly on the carbon that has bromine atom. However, a more appropriate way of numbering this would be three broodine. Three brooen is also fine. Now this compound, this is something again was something we have seen. This is a benzile compound. This is benzile fluoride because benzile group is attached to florine but the upsc name is quite interesting here because this has to be seen as a methile group that is attached to a phenile and a florine. So we call this as fluo phenile methane.

00:29:34 It's not phenile fluoromthane. This can also be given a different name. It can also be called called as a methile fluoride attached to a benzene. So it is fluoromthane. Benzene is methile benzene is another name for this. But you can always see it as a methile group attached to two florine and a phenile ring. So it is called fluorophenyl methane in UPSC. Okay. So now we'll go ahead and we look at polyhalogenated or dihalogenated compounds. So here I have ethane molecule that is attached to two

00:30:07 chlorine atoms. So I'm talking about the example here in the middle sequence the first one. So it is an ethane molecule that is attached to two chlorine atoms. We call it as one one dchloroeththane. This is just because the chlorine atoms are on the same carbon atom. So we give and when two substituents are attached to the same carbon atom we number it we give the number twice. So we say 1 comma 1. So it is 1 one dchloroeththane and its trivial name is ethyidine chloride. So it is ethylidine actually comes from

00:30:37 an a concept that it is something that is similar to having a double bond because we have replaced two hydrogen atoms. So these compounds are called as ethyidine chloride in the common name. So something you would know only if you know the names by heart but whereas the UPSC names are always something that we can give. So the second compound here is one two dchloroeththane. So this is an isomer of this. The second compound is an isomer of the first one except that the chlorine has moved to the second

00:31:01 one. So we say that it has two substituents on one two carbon atom and then it is an ethane one two dchloro ethane and it is normally called as ethylene dicchloride. You know what ethylene is? Ethylene is ethane. So it is as if chlorine has added on to an ethine or an ethylene and two chlorine atoms are present on each of the carbon atoms of an ethylene to start with. So therefore we call them as ethylene dchloride in the common name. These two compounds also have a very interesting name. Sometimes people refer to them as

00:31:30 this one as a geminal dihalide. So by the time geminal what we really mean is that both the halogen atoms are attached to the same carbon atom. So then we call it as geminal dihalide. If the halogen atoms are attached to adjacent carbon atoms close to each other then we say both these car hogen atoms are in the vicinity of each other then they can be called as visinal dihalates. So these are two names. So this is a geminal dihalate this is a visinal dihalate. So you might come across these names being

00:31:59 used in organic chemistry that only means that whenever you hear the term geminal which is the short form is gem gem m. So that means that both the halogen atoms are on the same carbon atom and vinyl means they are on adjacent carbon atoms not too far away from each other but on adjacent carbon atoms. The third example is a methane derivative. Two halogens are attached to methane. It is called dchloromthane. See dupin name is very simple. We say chloromthane and if there are more than one chlorine then we add a prefix to

00:32:28 that. So in this case d because there are two chlorine atoms. So it is dchloromthane. And uh the last one is tri broomeomthane because there are three bromine. Uh the trivial names for these two compounds dicchloromthane is methylene chloride. So it is similar to ethylene because two hydrogen atoms are replaced with chlorine. And the last one has a more interesting name broomem. And the corresponding chlorine analog is chloroform which you already know. uh bropform and chloroform are also accepted by UPSC although they are not

00:33:00 according to the strict rules of UPSC naming they are terms similar to benzene toine and so on which can be accepted okay now I have put the final example here so that we can summarize all the points that we were discussing so the compounds that I have each of these compounds are uh pentine so they all have five carbon atoms present in them but their UPS names vary If you follow the rules properly, we'll be able to name them. And they do not necessarily have common names. So these compounds are complex enough that they

00:33:31 are not commonly named. So a person who wants to talk about these compound will not have a name that he can immediately remember and say. So they only have UPSC names that we use. But upsc. So that is the advantage. So let us look at this molecule. So this molecule has a double bond and it has a bromine atom. We said because the double bond changes the name of the compound. The parent hydrocarbon is now known as a pentene. It ends with an ene which corresponds to alkanes. So it's an alken. So the double bond should get the

00:34:04 smallest number the lowest number possible. So we start naming from this end from the CH3 so that this particular compound gets the number two. Then the double bond continues to the number three and finally the bromine is attached to the fourth carbon followed by CH3. So you can see there is a longest aliphatic chain and that chain has a double bond between carbon number two and three and bromine is attached to the fourth carbon. So we call this compound as four bromo pentene because it has five carbon atoms. pent 2in

00:34:34 because the double bond is on the second carbon. So this is called four broo pentene. Now let us look at the second compound. Here again there are five carbon atoms. But if you look at this we would only be able to draw four carbon chains. And this CH3 here has to be always seen as a substituent. And we would name it in such a way that the double bond as I mentioned before gets the lowest number. So the double bond starts from one of the N. So we would start numbering one on the CH2 that has the double bond. CH2 double bond and the

00:35:04 broine is on the third carbon and that is a CH3. So the CH3 which is here this has to be seen as a substituent. So this CH3 is a substituent here. So therefore when we try to name it we say that bromine is on the third carbon atom 1 2 3 third carbon atom. Then the methyl group is on the second one. So it is three bromo 2 meil and it is a butine because there are only four carbons in the longest chain. So three bromo two methyl but one in. So this one in would immediately tell you that the double

00:35:35 bond is on the first carbon. So what is good thing with these upc names is if if this compound is given to anybody and if anybody follows the UPSC rule strictly they will all come up with the same name. There's no even if the compounds are being seen for the first time they would be able to write a unique name following these rules. Similarly if this name is given to you and if you are asked to write a structure if you know rules you will only draw one structure that corresponds to these molecules. So

00:36:00 that is the interesting fact about this that is this is the exact reason why uh the international union for pure and applied chemistry that comes up with this set of rules so that every compound every organic compound that we come across can be named uniquely and can be distinguished based on the name. So let us look at the last compound to complete this. So this again has a methyl group as a substituent in a four carbon chain. the bromine is attached to the last one. But we would always have to name this

00:36:28 compound in such a way that the double bond gets the smallest possible number. So it doesn't matter if I number from here or from here. Uh the double bond is always starting from the second carbon. Then we would start naming it from the carbon that is attached to bromine because then the bromine substitution also gets a lower number. So this compound has to be named as one bromemo and a medial group on the second carbon. one bromemo 2 methile but 2 in. So, so this is the important thing. If you

00:36:56 start naming from the other end, what we would name this as a four bromo 3methyl but toine. So, although the double bond is still on carbon number two, the substituents methyl and bromine goes on to carbon number three and four. So, we should not allow that to happen. We should always name it in such a way that these two substituents also get the lowest possible numbers as much as possible. So this has to be named as one bromo two methyl butin. So with this I will stop naming the alifhatic the ho

00:37:26 alkanes and I would go forward and start talking about hoarines. So all halo aren normally do have the same common name and UPC name. So so for some reason they from even before the upcup had set out the rules they always were being called as hello arraigns. So the simplest one here I have a bromeobenine here. So it is called bromemoenzene in UPSC as well as in common. So it is never called as benzil bromide because you know benzil bromide is something else. It is not even called as phenile bromide or anything. It is

00:37:58 normally called as bromaben in the UPSC nomencle as well as in the common way of calling these molecules. Now if I have two broine atoms again it is very simple in UPSC. You just number it in such a way that these two broine atoms get the lowest possible number. So it is one two dybrenzene. Uh so the only difference in common naming is so you have learned these terms ortho meta and para. So if the substituent is on an adjacent carbon atom you call it as orthodibrine. If it is one carbon away from each other

00:38:31 then we call it as meta bromobenzene. And if they are on one four position then we call it as parody bromoben. So orthome and par are associated with the common associated with the common naming pattern. Whereas 1 2 1 3 and 1/4 are the ways of naming these compounds in UPSC. Now here I so I have put an example quite intentionally here. So here you can see that there is a bromine atom and there are two chlorine atoms. So now how do we name them? Where do we start to name them? So if I call the carbon that

00:39:00 has this chlorine as one then I have the second chlorine atom on two and the broine atom on carbon number four. But whereas if I start numbering from bromine then I would see that the carbon number one gets bromine then three and four gets chlorine. So although the first carbon in both cases are substituted the this the second substituent here. So I I would try to write these numbers here. So I hope you would be able to uh see them better when I do that. So what I would is I'll call this as 1 2

00:39:32 3 and four. So when I do that you are able to see that the substituents are on one two and fourth carbon atoms. Now on the other hand if I go ahead and start numbering broine as one. So I can also give this number here. If I call this as one, this as two, this as three and this as four. So this is another way of naming it. So then broine atom is on the first carbon. Then my second two substitutions are on three and third and fourth carbon. So this is something that I don't want because then uh my numbers

00:40:01 are going higher. So you should always name it in such a way that when you look at the number as a whole when you sum up these numbers you get the lowest possible number. So therefore we would name it as 1 2 and four substituted compound. But now what do we write first? Do we write chloro first or a bromo first? Of course we write bromo first. Even if the bromine is a substitution on carbon number four, we do not we do number that first. We do not associate the number associated with with that particular substituent. We

00:40:30 only look for the alphabet. So in this case although bromemo is on the fourth carbon atom it is B comes first in alphabetical order. So therefore we call it as four bromemo one two dicchlorobenzine. So this has to be the name. Any other name that you give to this compound according to the UPC rules will be wrong. Okay. So now look at the last compound that I have. So this is a compound that has a chlorine atom and a CH3. So we know a benzene ring attached to a CH3 group attached to a methile group is toine. So toluine is also

00:41:00 accepted is an accepted UPSC name. So benzene toluine both these are accepted. Otherwise you can also call it as methylbenzene because upscu also accepts this name. So in this compound you see that you have a one chloro two methylbenzene. So you would name chlorine as one because chlorine in this case has the is the first alphabet that comes. C always precedes them. So that is why we call it as one chloro 2 methylbenzene or it can also be called as two chloroine. So it is toluine with the substituent on carbon number two.

00:41:30 However in the common system this molecule is a toluine molecule with a chlorine atom on orthoposition. So it has to be called as orthocchloroine. So this is generally about the UPC names and common names. Try to remember all the common names that you can mean as much as possible if something that you see very often and if people refer to them by common names whenever you come across them it is okay to know them. It is okay to remember some of those. But even if you cannot, it is not a big mistake because you are not supposed to

00:41:58 remember all the names that you see. Uh you may have to remember a few of them like secondary butil, isopropyl, isobutail and so on. But other than that all the other common names are something that you may even skip but make sure that you are able to give a number proper UPSC number to any molecule that comes your way. So you should be able to follow these rules and name them correctly. Your textbook has a lot of examples. It textbook has even some problems that are associated with this nomenclature of this compound. So I

00:42:26 would suggest that you write them down and make sure that you are able to write all the UPSC names possible. Okay. So now we have already discussed about how we can name these compounds. So now you are able to give names to these compounds. So once we are able to do that the next thing is now we start really looking at the chemistry of these compounds. So you see a organo halogen compound you are now able to identify this even if you're not able to identify you are now able to give a proper name

00:42:54 to it so that another person can listen from you about the compound that you are talking about but the next important thing is now we start talking about reactions of these compounds and also about how these compounds can be prepared. So once you do not have a hoalk alkan you want to make them how do we make them? So to know that it is always important that we know what exactly a h hallogen atom does to a carbon atom when it when it is linked. So of course you would have guessed that halo alkanes have different properties

00:43:22 in comparison with uh simple hydrocarbons. Why are they different? So simply speaking a halogen atom is more electrogative than a carbon atom. So whenever you have a carbon atom attached to a halogen atom it is no more like two carbon atoms attached to each other. Now we have a carbon that is attached to a more electronegative atom. All the halogens group 17 elements are more electrogative than carbon. So what do they do is they try to pull electrons more towards them. So when a carbon and halogen atom are involved in bonding the

00:43:56 electrons that form the bond are pulled more towards the halogen and away from carbon atom. As a result, as I have shown in this slide, you will see that the carbon always feels slight positive charge because it has already contributed an electron for the bond. But that electron is now has moved a little more towards hogen than carbon would have liked. So therefore now carbon has a slight positive charge and the hogen is happy to retain a negative charge. So if you want to draw this bond, it may be even ideal to draw a

00:44:25 slight positive charge which is normally represented by a small delta delta sign delta positive and a delta negative. So the delta positive resides with carbon atom and delta negative resides with halogen atom. So this is to say that this particular compound now is polarized. This particular bond now is polarized. So all organo halogen compounds are generally polarized. So they are these bonds are polarized. So these molecules have a dipole moment and the dipole moment as you know is normally measured in device in the unit

00:44:54 dubai. Okay. Now uh I have listed the methile highlights here. Uh so the halomethane molecules. So starting with florine, chlorine, bromine and iodine. So I have fluomethane, chloromthane, brooome and iodomethane on my table here and the left hand side uh column. So in this column you would find them and what you have in the immediately following column is the bond length. So you can see uh when carbon and florine are attached the carbon florine bond length. So these are given in pometers. So the

00:45:25 carbon florine bond length is 139 pometers. So uh or 1.39 anstrong. So you would find these bond lengths are increasing as you go down. So when you go from florine to iodine there is a large difference in the bond length. This is expected right? Because bond length is exactly the distance between the center of the atoms that actually form the bond. So when you have a larger atom like iodine. So because iodine is very much lower down the group. So you would find that iodine is very big in comparison with florine. So therefore

00:45:58 when iodine and carbon bonds together the bond is longer and so so it goes sequentially. So the carbon chlorine bond is longer than the carbon florine bond. carbon bromine bond is longer than the carbon chlorine bond and carbon iodine bond is longer than carbon bromine bond. So, so that is expected. So this is the trend you would find. Now what about the bond strength? Which bonds are more stable? So you can also see that the methyl group when it is attached to florine the carbon has a uh two sp3 hybridized orbital right carbon

00:46:31 falls in the uh second period. So it has its valence electrons in the second orbital. uh so you would find that it is it has a 2 sp3 orbital that forms the bond and so is florine. So therefore these two are of similar size their orbitals are of similar size when so therefore according to the very classical concepts of bond being formed by overlap of orbitals you would find that we actually have two orbitals that are almost the same size. So therefore their overlap would satisfy both the atoms that are involved. So you get a

00:47:04 bond that is pretty strong and the orbital overlaps are rather complete. So this results in an extremely strong bond and that is reflected in the enalpy of formation of this bond. So you will find that 452 kJ per mole for a carbon florine bond whereas it comes down to 234 for a carbon iodine bond. So imagine iodine coming it with its big orbitals and whereas the carbon now only has a smaller orbital to offer. So whenever they overlap the iodine the overlap is not completely satisfactory. So you

00:47:36 would find that the bonding itself is slightly weakened and as a result you would always find that the bond energy of a carbon iodine bond is weaker than a carbon bromine bond which is weaker than carbon chlorine bond and the carbon florine bond being the strongest. Now what about the dipole moment? There isn't much difference. you would see that uh all the hogen atoms as I mentioned are more electrogative and then there is this difference in patterns of bonding difference in patterns of bond energies and so on. So

00:48:05 ultimately the dipole moment is not much different they all fall in 1.8 near about except for carbon iodine bond because that is the weakest of all. So in this particular case so iodine starts to be uh starts to get less and less electro negative. So only in this case it is slightly lower. Then again it is not too lower. It's 1.64. So all these carbon halogen bonds are polarized. They all have dipole moment and comparable dipole moments except for carbon iodine bond which is reasonably lower than the

00:48:35 others. So, so this gives you a fairly good idea about how a CX bond is. So, whenever you see that there is a hallogen attached to a carbon immediately remember that the carbon is positively charged or in other words the carbon now is looking for electrons whereas the halogen is ready to leave with a negative charge. It has already pulled the electrons from carbon towards itself a little bit. Now it is ready to leave carbon and leave carbon with a positive charge. So once carbon gets electrons from somewhere it would allow

00:49:03 the halogen to leave. So, so this this has a lot to do with the way organo halogen compounds or halo alkanes react. So, we would come to their reactions later. So, this is something that is uh that is worth remembering as we go along. Okay. Now, um since we have talked about all these things, the next thing that I would discuss is about the preparation of hall alkanes. So, how can halo alkanes be prepared? So the most simple preparation of halo alkanes should of course be from the most available set of

00:49:34 molecules. So the most easily available molecules are hydrocarbons right because they come from petrochemicals. After that those are alcohols. Now alcohols are a good starting point for the synthesis of more many organic molecules just because the hydrocarbons are now already functionalized. That is already a carbon oxygen bond. So if you want to functionalize an alcohol all we need to do is break the carbon oxygen bond and put the new bond there. So therefore the preparation of hall alkanes also do

00:50:04 start from alcohols the simplest of them I mean. So now I have a very simple reaction written here. The first one if you see I have an alcohol which is represented as an R O where R stands for the alkal group. O for the hydroxal group that results in the alcohol structure. So when an alcohol is treated with a hydrohilic acid, hydrochloric, hydrofluoric, hydro bromic, hydroic and so on. So when it is treated with a hydrohilic acid, what do we get is a halo alkan plus a water molecule. So now if you would like to know how this may

00:50:35 be happening, why did this break? What exactly happens is uh the hydrohalic acid dissociates, right? It dissociates in solution to H+ and X minus. Now this H+ that is present in any acid for example would like to interact with the oxygen atom of the alcohol and it would protonate the oxygen atom of the alcohol. So if you look at the alcohol here the O is already bonded to an R group and a hydrogen. Now if another H+ comes and binds to oxygen. So it is an H+ it doesn't have an electron. It comes

00:51:06 and binds with oxygen. So the oxygen is protonated and oxygen gets a positive charge. And because of the positive charge now this O2 group is actually like a water molecule. So what I mean to say here is that um now your RO when you treat it with an H+ becomes R O2 positive. So this this particular group has retained a positive charge. So now this is something that would like to go out as water. So that leaves your alkal group wanting a new group. So we already mentioned that hydrohilic acids are

00:51:45 polarized. So I have H+ and X minus. Now when water leaves what does this R needs is something with a negative charge. So this would then react with the X minus that is present there and would give you RX. So this is how alkalhalides are made from alcohols. So if you take an alcohol and put it with an HX hydrohilic acid, we would get the halo alkan which I have shown here along with the molecule of water. So you would see that water comes out. So that is the simplest way of representing this reaction.

00:52:16 Uh now but not all alcohols give you the reaction with the same ease. The general order of reactivity is tertiary reacting faster than secondary reacting faster than primary and methile reacting the slowest. Now primary and secondary alcohols react with HX that is hydrohilic acids effectively only in the presence of zinc chloride as a catalyst. So otherwise the reactions do happen but they are extremely slow. You may have to heat it. You may have to leave the reaction for a longer time whereas tertiary alcohols would react

00:52:47 immediately with the x. So so why is this different? So it it has some relation with the equation that I have written here. I have shown that the O gets protonated and this R starts to feel the positive charge too. So generally the one that can hold this positive charge better. So the alkal group that can hold a positive charge better would be the one that reacts better and you would later also see that tertiary alkal groups are better in handling a positive charge better in handling an O2 group leaving away from

00:53:17 it. So that is the reason why it reacts faster. But once you go into secondary and primary the reactivity therefore reduces. Now what exactly does zinc chloride does is zinc also has an affinity for oxygen. So therefore when you take zinc chloride which is a Louis acid it binds to O even before hydrogen binds to it. So therefore it allows the breaking of the carbon oxygen bond and removal of the hydroxy groups faster. So that is why we use a catalyst in this reaction. So zinc chloride can help if

00:53:45 the reaction is slow. Okay. Now sometimes you would have heard about hydrochloric acid. You would have you would have gone to your labs and then seen that there are hydrochloric acids present in your lab uh in most of the labs in your schools. But now some of the other hydrohilic acids are not commonly available. We may have to make them in the reaction mixture. So one simple way of doing it is to take an al alcohol like before and treat it with sodium iodide or potassium iodide, sodium bromide or potassium bromide and

00:54:15 so on along with an acid. So if you are using sodium iodide to make an alkal iodide what you need to do is you do not have to use hydroiic acid. You may be able to use a sodium salt of iodine and then put an acid along with this. So in this case you can use phosphoric acid. So what it would give is the alkal iodide along with the sodium or potassium salt of phosphoric acid. So therefore you can also use an acid with a salt that might inside in C2 in the reaction mixture generate the hydrohalic acid that you need. A preparation of

00:54:46 alkal bromides is given here. So you take alcohol treat it with sodium bromide and H2SO4 then we will get the alkal bromide and a sodium salt of it plus water. So so this is a very simple way of making this molecule. Okay. So I continuing with what can be done with alcohols. I will show you another example here. So if you take alcohol and treat it with phosphosphorus trihalide. Phosphorus has an affinity for oxygen. So it would be able to pull oxygen out of any molecule. So there are three halogens. So three molecules of alcohol

00:55:16 can react with this and then give you a halo alkan along with phosphorus acid H3 P3. So instead of using a hydrohalic acid you can also use PX3. And sometimes when the X is broine and you do not even need the phosphorus trihalide you can always make it in C2 from red phosphorus and the corresponding hogen. So in that case you can take alcohol treat it with red phosphorus and the hogen and you would generate this PX3 species in C2 and get this molecule. You can also do the reaction with PCL5 if you want a

00:55:45 chloride. Then the product that you get is PL3 as the side product along with HCl and alkalhalide. So here again you can see that phosphorus pulls out this oxygen and HCl goes out and then one of the chlorine atoms goes and attaches with this. The final reaction is the one with the chloride and the most interesting because the chloride when it reacts with an alcohol gives you RCL then sulfur dioxide and HCl. These two byproducts that are formed in the reaction are gases. you will always end up getting the alkalhalide that you

00:56:14 want. So in this case chloride because you're using chloride. So the along with the alkal chloride or the uh chloroalkkan you get sulfur dioxide and HCl these two are gases. So they escape from the reaction mixture and you end up having just the product you want. So um to summarize what I wanted to say about the preparation so far. This is the simplest simplest synthesis possible. You make them from alcohols. You can treat them with HX. You can treat them with phosphorus chlorides or you can

00:56:40 treat them with chloride. The one using thyel chloride to make alkal chlorides is the most easy one to do because the byproducts are gious. So I would stop here for uh this class and then we'll continue to discuss the preparation of halo alkanes in the next class. Thank you.