Isomerism in Organic Chemistry

isomerism and organic chemistry well I will teach you complete isomerism and add types of isomerism in this special lecture firstly let me teach you about structure formula and molecular formula structure formula of any compound shows the arrangement of atoms or molecules for example consider the structure formula of methane this structure formula of methane shows that the shape of this compound is tetrahedral secondly it shows that one carbon atom is bonded to four hydrogen atoms while molecular formula of any compound shows the total number of atoms for example the molecular formula of methane is CH4 it shows that there is one atom of carbon and there are four atoms of hydrogen present in a single molecule of methane this we learn that structure formula shows the arrangement of atoms in any compound while molecular formula shows the total number of atoms present in a compound hence noted down these fundamental concepts now what are isomers well I always teach this simple analogy consider these gloves we know that this glow has four fingers and it has one left thumb while this glow has also four fingers and it has one right thumb I say both gloves have four fingers and a thumb but they have different Arrangement now listen carefully gloves are isomers of each other because they have same number of fingers and different Arrangement let me repeat it gloves are isomers of each other because they have same number of fingers and different Arrangement so the easy trick to remember the concept of isomer is they must have same number of fingers but different Arrangement now consider these organic compounds let me find the molecular formula of this compound we can see that there are four carbon atoms present and eight while there are 10 hydrogen atoms present in it so the IUPAC name of this compound is nban on the other hand there are also four carbon atoms present in it while there are also 10 hydrogen atoms present in it here I write 1 2 3 the methy CH4 is bonded at the second carbon I write two methy propan now listen carefully this compound and this compound have the same molecular formula C4 h10 but different Arrangement so they are isomers of each other let me repeat it this compound and this compound have the same molecular formula C4 h10 but different Arrangement so they are isomers of each other just remember that these both compounds have same fingers are molecular formula but they have different structure formula or different Arrangement therefore we Define isomers as the chemical compounds which have the same chemical formula but different structure formula are called isomers for example nban and two methy propane are isomers of each other because they have same molecular formula but different structure or different Arrangement just noted down this easy concept of isomers here let me ask you one of my favorite questions what is the difference between isomerism and isomers well believe me many student know its definition but they do not know its concept let consider hands and a nose I say isomerism is possible in hands because we have one left hand and one right hand are one left isomer and one right isomer secondly I say that isomerism is not possible in a nose because we have only one nose it has no other isomer therefore personally I say that isomerism is the tendency ability of any compound to form different compounds having different structures for example our hands have the tendency to form two compounds having different structure while isomers are the outputs of isomerism for example consider CH4 we say that isomerism is not possible in CH4 so it has no isomer th remember that isomerism is the phenomena or tendency of a compound to form different compounds having different structures while isomers are the output of the isomerism here remember this bonus Point isomers have different physical properties for example consider these two isomers this is alcohol and this is ether we know that they have same molecular formula but they have different structures that's why they have different different physical properties for instance this is liquid and this is a gas therefore we say that isomers have different physical properties now we will learn the type of isomerism well there are two types of isomerism structural isomerism and stereo isomerism structural isomerism is further divided into six categories like chain is isomerism position isomerism functional group isomerism metamerism toomis and ring chain isomerism while stereo isomerism is further divided into two categories geometric isomerism and Optical isomerism hence these are the eight different types of isomerism now we will learn chain isomerism and position isomerism well those isomers which have different number of carbons in Man chain are called chain isomers the easy trick is the main chain or the parent chain of carbon must be different while those isomers which have same functional group attached a different position are called position isomers the easy trick is functional group at different position means position isomers for example consider butan we can see that there are four carbon atoms present in the Man chain remember that if the carbon chain is straight we call it n chain so this is nban now I will cut this carbon from the main chain and I will attach it at the second position I get this structure here 1 2 3 this carbon is present at second position remember that if meile group is present at second position we call it ISO so this is isob butan now listen carefully in this nban there are four carbon atoms present in the Mansion our parent chain while in this isob butan there are three car carbon atoms present in the main chain are parent chain according to this trig these isomers have different carbon atoms in the Mansion so they are chain isomers the weite one and two are chain isomers because they have different number of carbons present in the man CH or parent chain now consider pentan I only consider the straight chain of pentan we call this n pentan secondly I cut this carbon and I attach it to this second carbon I get this structure here 1 2 3 4 the methy is present at the second carbon I call it isopentane thirdly I cut this carbon and I attach it to this second carbon I get this structure here 1 2 three remember that if there are two meile groups present at the second carbon we call it Neo hence this is neop pentan now listen carefully we can see that there are five carbon atoms present in the main chain of this compound there are four carbon atoms present in the main chain of this compound and there are three carbon atoms present in the Manion of this compound according to this St these isomers have different carbon atoms and their respective chains so they are chain isomers th one two and three are chain isomers because they have different number of carbons and the main chain are in the parent chain hence noted down now consider these isomers of hegan this is a straight chain of carbon atoms there are six carbon atoms present in it we call it an hexan secondly I write 1 2 3 4 five the methy group is present at the second carbon we call it isohexane thirdly I write 1 2 3 4 5 the methy is present at the third carbon I write three methy pentan fourthly I write 1 2 3 4 One methy is present at the second carbon and another methy is present at the third carbon I write 2 three dimethy butan lastly I write 1 2 3 4 here two methy groups are present at the second carbon we call it new hexen now I write the number of carbon atoms in the Manion the there are six carbon atoms in the Manion there are five carbon atoms in the Mansion there are also five carbon atoms in the Mansion there are four carbon atoms in the mansion and there are also four atoms in the Manion now listen carefully 1 2 and four are chain isomers because first isomer has six carbons in its Manion two R3 has five carbon atoms in its Manion so we call it chain isomers secondly we can see that second and third have five carbon atoms in the main carbon chain they are not chain isomers but wait a minute here the methy group is present at the second carbon while here the methy group is present at the third carbon according to the trick the man CH has five carbon atoms but the functional group ch3 is a different position so they are position isomers the second and third are position isomers similarly in the fourth isomer the M CH our parent chain has four carbon atoms while the fifth isomer has also five carbon atoms in the main chain here the two methy groups are present at the second carbon and at the Third carbon while here the two methy groups are present at second carbon these both isomers have same carbon chains but functional groups are present at different position hence fourth and fifth are also position isomers hence noted down finally consider these isomers pause the video and try to find the isomerism well they both have the same carbon chain having three carbon atoms in the main chain but here the chlorine is present at the first carbon and here the chlorine is present at the second carbon so they are position isomers secondly this compound has four carbon atoms in the M chain and this compound has five carbon atoms in the main chain so they are chain isomers thirdly double bond is present at the second carbon while here the double bond is present at the first carbon so they are position isomers fifthly the main chain has 1 two three carbon atoms while here 1 2 3 4 5 there are five carbon atoms so they are chain isomers therefore remember that chain isomers have different number of carbon at ATS in the M chain while position isomers have the same functional group bonded a different position and the same carbon chain now we will learn the third type of isomerism which is known as functional group isomerism well those isomers which have same molecular formula but different functional groups are called functional group isomerism in order to spot functional group isomerism always look at these two conditions firstly the two compounds must have the same molecular formula secondly the two compounds must have different functional groups for example consider alahh and Ketone functional group now I look for the molecular formula of the compounds the molecular formula of this alide is there are three carbon atoms six hydrogen atoms and one oxygen atom the molecular formula of this Ketone is here are also three carbon atoms six hydrogen atoms and one oxygen atom secondly the functional group of alahh is co and that of Ketone is C3 they both have different functional groups thus this alahh and this Ketone are functional group isomers of each other now consider carboxilic acid group and Easter functional group as usual I look for the molecular formula of the compounds and this carboxilic acid there are three carbon atoms six hydrogen atoms and two oxygen atoms while in this Easter there are also three carbon atoms six hydrogen atoms and two oxygen atoms secondly the functional group of carboxilic acid is coh and that of Easter is coo they both have different functional groups so both the conditions are fulfilled thus this carboxilic acid and this Easter are functional group isomers of each other now consider ether functional group and alcohol functional group as usual I find the molecular formula of compounds in this ether compound there are two carbon atoms six hydrogen atoms and one oxygen atom while in this alcohol compound there are also two carbon atoms six hydrogen atoms and one oxygen atom secondly the functional group of ether is O and that of alcohol is o Edge they both have different functional groups th this ether and this alcohol are functional group isomers of each other lastly consider alen functional group and cyc alkan functional group I find the molecular formula of these compounds in the alen there are four carbon atoms and eight hydrogen atoms well in this cyc alkan there are are also four carbon atoms and eight hydrogen atoms secondly the functional group of alen is double bond and that of cyc alkan is single Bond both the conditions are fulfilled thus this alen and this cyc alkan are functional group isomers of each other to summarize this whole concept we have learned that functional group isomerism exists between alide and Ketone carboxilic acid and Easter ether and alcohol alen and cyclo alkan thus remember that functional group isomers are those isomers which have same molecular formula but different functional groups now we will learn the four type of isomerism which is metamerism well when different alkal groups are bonded to the same functional group are called metamerism the condition is same functional groups and different alkal groups here the question is what are alkal groups well consider methane CH4 when we remove one hydrogen from methane we get ch3 and we call it methy secondly consider Ethan when we remove one hydrogen from it we get ch3 ch2 and we call it eile thus by this way we can form alkal groups remember that alkal group is represented by R Das now consider ether functional group The Ether group is r d o d r here that this is the left hand side of this ether and this is the right hand side side of this ether at the left side there is eile while at the right side there is also eile this ether contains two ethy groups we call it dthy ether now listen carefully I shift this ch2 from the left side to the right side let me repeat it I shift this ch2 from the left side to the right side I get this new compound here in this compound this is the left side and this is the right side at the left side there is meile group while at the right side there is propile group I write methy propile ether now listen carefully these two compounds have the same functional group which is ether but at the left side of this molecule there is ethy while at the left side of this molecule there is methy according to the condition these molecules have same functional group but different alkal groups are attached to it thus we call them two metamers of each other are two isomers of each other hence noted down now consider amine functional group we know that the functional group of amine is r- nh- r that this is the left side of this compound and this is the right side of this compound at the left side there is eile group and at the right side there is also eile group so we call it dthy amine now I shift this ch2 from the left side to the right side I get this new compound here at the left side there is methy group while at the right side there is propile group so we call it methy propile amine lastly I cut these two carbons of the propile I attach it to this carbon I get this new compound now at the left side there is methal group while at the right side there are two methy groups bonded to Second carbon we call it isopropyl isopropyl amine now listen carefully these three compounds have the same functional groups which is amine NH at the right side of this compound there is eile and at the right side of this compound there is condition is fulfilled th these three are metamers are three isomers of each other so note it down finally consider Ketone functional group we know that its functional group is r d c d r now consider these two compounds pause the video and try to find its isomerism here I write 1 2 3 4 5 the functional group of Ketone is present at the second carbon so we call it two Penton now listen carefully at the left side of this compound there is eile group while at the left side of this compound there is methy group hence they both are metamers or isomers of each other th remember that if compounds have same functional group but different alkal groups are Bond to it it is known as metamoris to summarize this whole concept we have learned that metamerism is possible in ether functional group amine functional group and Ketone functional group now we will learn the fifth type of isomerism which is toomis the phenomena in which single compound exist and two are more interconvertible structures are called Toom merism remember that Toom merism is possible in carbonal compounds having Co group like Ketone and alide personally I teach this condition for Toom merism it is CH single Bond a double bond B if any compound has this pattern there is 90% chances that totom merism will occur here I write 1 2 3 I shift hydrogen from first position to the third position and I shift double bond from second position to the first position th using this conversion we can easily do totom merism of any compound for example consider this Ketone here this carbon double bonded to oxygen is like a double bonded to B I write AB this ch3 is like CH so the condition CH single Bond a double bonded to be is fulfilled hence too merism is possible in this organic compound I write 1 2 3 now according to the condition I shift hydrogen from the first position to the third position and I shift double bond from the second position position to the first position I write ch2 because I shift one hydrogen double bond carbon single Bond o and ch3 this compound contains double bond of alen I write en n and for this o I write o so I get an hence these are the two toomer are two isomers of each other we also call toomis as keto enol isomerism because we get Ketone and enol this double arrow means that Ketone and anol are in dynamic equilibrium they are interconvertible that's why we also call it dynamic isomerism because they are moving from one form or from one structure to another structure remember that we will get 99% of Keto and 1% of enol because ketones are very stable than enol let me repeat it ketones are very stable than inol now consider this alide here this carbon double bonded to oxygen is like a double bonded to B I write a and b this ch3 is like CH the condition C H single Bond a double bonded to be is fulfilled hence Toom merism is possible in this compound I write 1 2 3 now according to the condition I shift hydrogen from the first position to the third position and double bond from the second position to the first position as a result I get ch2 double bond C single Bond o and H this is enol as usual we get 99% of alahh and 1% of enol hence these are the two toomer are two isomers of each other now consider this organic compound pause the video and try to find its doomers well this C double bonded to oxygen is like a double bonded to oxygen I write A and B this carbon has no hydrogen here the CH is not present I mean this carbon has no hydrogen the condition CH single Bond a double bonded to be is not fulfilled theism is not possible in this compound now consider this organic compound this N double bonded too is like a double bonded to B I write a and b this ch2 is like CH the condition CH single Bond a double bonded to be is fulfilled I write 1 2 three I shift hydrogen from the first position to the second position and I shift double bond from the second position to the first position as a result I get this compound th by this way we can easily do Toom merism of organic compounds now we will learn the exception cases of Toom merism usually these exception cases exist in cyclic organic compounds for these cases we have to learn two important Concepts aromatic anti-aromatic and conjugation for aromatic and anti-aromatic compounds I use this Drake I write 2 and four now I increase both the numbers by 4 2 + 4 is = to 6 6 + 4 is = to 10 and so on here 4 + 4 is = to 8 8 + 4 is = to 12 and so on remember that this number represents the total number of electrons also noted down that if there is negative charge it represents two electrons if there is a lone pair it represents two electrons and if there is double bond it also represents two electrons on the other hand if there is alternate single and double bond it shows High conjugation so on the basis of these two concepts we will easily learn the exception cases of totom ISM the first exception case is CH single Bond a double bonded to B here this condition exist but totom merism is not exhibited by the compound for example consider this organic compound firstly I write CH C and ch2 this C double bonded to O is like a double bonded to B and this CH or this ch2 is like CH so the condition CH single Bond a double bonded to be is fulfilled I can take both this CH or this ch2 but I consider this ch2 here this h of ch2 is shifted to third position and the double bond is shifted to the second position I get this compound but wait a minute we know that these both compounds are cyclic so I have to check either these compounds are aromatic or anti-aromatic this cyclic compound has one double bond I write to electrons we know that if a compound has two electrons it is aromatic hence this compound is aromatic on the other hand this compound has two double bonds I write 2 + 2 I get four electrons we know that if a compound has four electrons it is anti-aromatic here this compound is anti-aromatic now listen carefully aromatic compounds are highly stable while anti-aromatic compounds are less stable so this compound does not exist because it is anti-aromatic and it is less stable as a result we get 100% of this compound because it is aromatic and it is very stable Compound on the other hand we get 0% of this compound because it is anti-aromatic and it is less stable as usual this is Ketone part and this is enol part we get greater Ketone part and lesser inol part therefore we say that this compound does not exhibit Toto merism even though it fulfill the condition CH single Bond a double bond it to B hence noted down this exception case secondly anol part is greater than Ketone part we have learned that we always get greater Ketone part part and lesser enol part but in this exception case we will get greater enol part and lesser Ketone part let consider this example I write C and ch2 this c dou bond o is like a double bonded to B while this ch2 is like CH the condition is satisfied I shift hydrogen to the third position and the double bond to the first position as a result I get this compound these both compounds are cyclic I check aromatic and anti-aromatic compounds this compound has two double bond I write 2 + 2 is equal to four electrons we know that if there are four electrons it is an anti- aromatic compound thus this compound is anti-aromatic and it is less stable while this compound has three double bonds I write 2 + 2 + 2 is equal to six electrons we know that if there are six electrons it is aromatic compound th this compound is aromatic compound and it is highly stable so we get 100% of aromatic compound because it is highly stable while we get 0% of anti-aromatic compound because it is less stable we know that this is Ketone part and this is enol part hence we get greater percentage of enol part than Ketone percentage let noted down this exception case thirdly consider this exception case when there are two possible isomers firstly I write ch3 c ch2 c and ch3 now this C double bonded o is like a double bonded to be this ch2 and this ch3 are like CH let this ch3 gives it hydrogen to this oxygen the double bond will shift here I get this compound secondly let this ch2 gives its hydrogen to this oxygen this double bond shifts here I get this compound now listen carefully this compound has single Bond double bond single Bond single Bond double bond while this compound has single Bond double bond single Bond double bond Etc we know that if a compound has alternate single and double bond it is more conjugated and more stable this compound has alternate single and double bond it is more conjugated and more stable while this compound has not alternate single and double bond it is less conjugated and less stable this we will get greater content of this compound we know that this is Ketone part and this is inol part I get 85% of Ketone part and 15% of enol part part hence noted down this exception case therefore using aromatic anti-aromatic and conjugated concept we can easily learn the exception cases of totemism now we will learn the last STP of isomerism which is ring chain isomerism well those isomers having same molecular formula but different structures like cyclic and open chain are called ring chain isomerism here the word ring chain isomerism teachs the whole concept the ring stands for cyclic isomer and the chain stands for open chain so ring chain isomerisms are those which has one open chain compound and one cyclic compound for example consider propine it is an open chain of propine no I convert it to cyclic structure I get this structure we know that this is a cyclopropane the molecular formula of the both compounds are the same which is c3h6 but they have different structure open and cyclic so they both are known as ring chain isomers secondly consider hexine I convert it to cyclic compound I I get this cyclic structure it has no double bond I call it cyclohex here they both have the same molecular formula which is C6 h12 but they have different structures like open chain and cyclic so they both are ring chain isomers of each other thirdly consider buttine pause the video and try to make its ring chain as Summers well it is super easy just convert it to cyclic structure having single bonds I get this structure which is known as cycl butan the molecular formula of the both compounds are the same which is C4 h10 thus remember that ring chain isomers are those isomers which has one open chain and one cyclic ring so we have learned the six different types of structural isomerism now we will learn stereo isomerism remember that in sterio isomerism isomers have same molecular formula same structural formula but different arrangement of atoms and the space hence noted down this important definition of stereo isomers now we will learn the first type of stereo isomerism which is geometrical is isomerism remember that geometrical isomerism is also known as cyron isomerism those isomers which have different position of the identical group in the space are called geometrical isomers remember that CIS means when two identical groups are present on the same side of double bond while trans means when two identical groups are present on the opposite side of the double bond for example consider this two buttine here I use the three idiots mov it track up down up down this methy ch3 is up and this hydrogen is down here this ch3 is also up and this hydrogen is down so both the ch3 are up and both the hydrogens are down hence we call it cis2 buttin now listen carefully I just interchange the up and down position of ch3 and hydrogen let me repeat it I just interchange the up and down position of ch3 and hydrogen I get this compound here this ch3 is up this ch3 is down this hydrogen is up and this hydrogen is down we call it trans to buttine these both compounds have same molecular formula and same structures but they have different orientation of atoms in the space hence these both compounds are geometrical isomers of each other remember that trans isomers are more stable than CIS isomer because there is less steric repulsion in trans isomers secondly consider this organic compound is geometrical Isom ISM possible in8 the answer is no because even if we change the position of this chlorine from up to down position we still get the chlorine which is an upward position therefore remember that the two groups on each carbon must be different let me repeat it the two groups on each carbon must be different for example consider this compound here two different groups chlorine and hydrogen are bonded to the same carbon hence geometrical isomerism is possible I write up and down up and down now I just interchange the chlorine and hydrogen I get this compound this is CIS compound and this is trans compound so these are the two geometrical isomers of each other now consider this compound which is known as mic acid we know that different groups coh and hydrogen are bonded to each carbon hence geometrical isomerism is possible in it as usual I write up and down up and down now I just interchange the up and down position of coh and hydrogen I get this compound which is known as fumic acid in this compound The Identical groups coh are present on the same side of a double bond so it is cis2 buttin dioic acid while in this compound The Identical groups coh are present on the opposite side of double bond so it is trans2 buttin dioic acid hence melic acid is a CIS compound and fumic acid is a trans compound therefore remember that melic acid and fumic acid are geometrical isomers of each other th noted down these important geometrical isomers which are sometime asked in mcqs now consider open chain alkan and alkine remember that geometrical isomerism is not possible in open chain alkan and alkine it is because alkanes have carbon carbon single Bond or Sigma bond in Sigma Bond atoms constantly shift back and forth I mean atom constantly move up and down so geometrical isomerism is not possible while in alkine there are already triple bonds between carbon and carbon we need to Branch are two more Bonds on each carbon which is not possible because carbon can form only four bonds hence geometrical isomerism is not possible in alkin now consider the cyc alkan the question is is geometrical isomerism possible in cyc Alcan well the answer is yes for instance in this compound I write up and down up and down so this is a CIS cyc Alan now I just enter inter change the up and down position of the ch3 group and hydrogen group I get this compound here the alkal group ch3 is up and here the alkal group is down hence it is a trans cyc Alan th remember that geometrical isomerism is possible in cyc Alan let me teach you one another important question why tutin shows geometrical isomerism but not one buttin well consider two buttine and one buttine now I write 1 2 3 4 the double bond is present at the second position we call it two buttine on the second and third carbon two different groups are attached to the same carbon atom hence geometrical isomerism is possible in it while in this compound I write 1 2 3 the double bond is present at the first position I write one buttine here on the second carbon there is alkal group and hydrogen group so it is all right but here at first carbon the two same groups hydrogen are attached to the first carbon we have already learned that if identical groups are attached to the same carbon geometrical isomerism is not possible therefore geometrical isomerism is not possible in one buttin but it is possible and totin hence noted down this important question the final concept of geometrical isomerism is z and E form consider this organic compound here the four groups bonded to two carbons are different in such cases we decide on the basis of atomic size Florine has the smaller size than chlorine while bromine has the smaller size than iodine now the smaller size groups are at the same side of double bond hence it is AIS form are we call it Z form now as usual I interchange the smaller size and larger size groups of bromine and iodine I get this compound here Florine has smaller size than chlorine while iodine has larger size than bromine the smaller size groups like Florine and bromine are at the opposite side of the double bond so we call it transform or we call it eform thus these are the two geometrical isomers of each other and we call it Z form and E form hence noted down this important concept finally let me teach you the most difficult isomerism which is Optical isomerism Optical isomers have these four properties they have same chemical formula they have same molecular formula they have same chemical and physical properties the only difference between them is that they have different behaviors towards light now I will teach you only two conditions for optical isomerism and it will help you to master this concept the first one is optically active substance and the second one is asymmetrical carbon for example consider this compound if this compound is optically active and it has asymmetrical carbon then Optical isomerism is possible in it if it is not optically active or it has no asymmetrical carbon then it will not show Optical isomet ISM now what is optically active compounds well to learn it we have to learn polarized light polarimeter optically active compounds and optically inactive compounds now the light in One Direction is called polarized light for example this light is going in One Direction it is known as polarized light while the light produced by the bulb is called UNP polarized light because it spreads in all Direction secondly polarimeter is an instrument which is used to measure the polarization of light and Optical activity of a compound thirdly the substances which can easily rotate the polarized light are called optically active substances for example consider glucose it is optically active compound because it can EAS rotate polarized light fourthly those substances which cannot rotate the polarized light are called optically inactive substances for example consider glycine it is optically inactive compound because it cannot rotate polarized light let's remember these important Concepts now how can we find the optical activity of a compound well we use Nicole and polarimeter to find the optical activity of a compound for example we take a light source it produces light in all directions here it is unpolarized light I place nicool prism in front of it now the Nick prism simply convert the unpolarized light to polarized light secondly I make a solution of a compound and place it in the poeter now there are three possible outcomes if the light is rotated in the left direction or anticlockwise Direction the substance is optically active and we call it leur rotatory compound we represent it by negative sign if the light is rotated in right direction or clockwise Direction the substance is optically active and we call it dextr rotatory compound we represent it by positive sign lastly if the light is not rotated in the left or in the right direction it means that compound is optically inactive th remember that if a compound rotate light in left Direction it is Le rotatory if a compound rotate lights in right direction it is De rotatory now we will learn the second condition for optical isomerism which is asymmetrical carbon firstly we will learn asymmetrical compounds for example consider these compounds here we can easily divide this compound into two equal parts so it is symmetrical compound while we cannot divide this substance into two equal parts because this part is different from this part so it is asymmetrical compound now consider this compound here four different groups are attached to the carbon we call this carbon as asymmetrical carbon because if we divide this compound at this carbon we get two different parts so it is asymmetrical carbon to summarize this whole concept we have learned that for optical isomerism we need optically active compounds secondly the compound must have one asymmetrical carbon now we will learn the optical isomers of lectic acid well consider the lectic acid I write 1 2 3 its IUPAC name is two hydroxy propenoic acid I look for the conditions it has asymmetrical carbon so it is optically active compound now I place a mirror in front of it we know that in the mirror the right side becomes left and the left side becomes right I get this compound now listen carefully it rotates polarized light towards right direction we call it dextr rotatory and we represent it by positive sign secondly it rotates polarized light towards left we call it leur rotatory and we represent it by negative sign this first and second isomers are called eners eners are those isomers which are mirror images of each other let me repeat it iners are those isomers which are mirror images of each other just remember that the first and the second isomers are called inimers because they both are mirror images of each other so these are the two Optical isomers of lectic acid now consider secondary butle chloride and form its Optical isomers well we can see that this carbon is asymmetric because all the four carbon attached to this carbon are different so it is optically active compound now I place mirror here as a result I get this comp bound this ch3 is here and this C2 H5 is here now it depends on you to call this compound negative or positive let this is negative we call it leur rotatory isomers this is positive we call it dextr rotatory we know that these both are iners of each other because they are mirror images of each other so these are the two Optical isomers of this compound now consider this organic compound is Optical isomerism possible in it the answer is no this compound is symmetrical if I cut this compound from here I get equal parts on both the sides hence it does not show Optical isomerism finally consider Optical isomerism and tartaric acid well consider its structure remember that hydroxy groups and O group are present at opposite side this compound is asymmetric hence Optical isomerism is possible in it I write 1 2 3 4 at second and third position hydroxy group are present I write 23 dihydroxy butan while at first and fourth position carboxilic acid groups are present I write one for dioic acid so its IUPAC name is 23 dihydroxy butane 14 dioic acid now as usual I place mirror in front of this compound I get this compound this o shifts here and this o shifts here let we call this textr rotatory tartaric acid and we represent it by positive sign we call this leur rotatory tartaric acid and we represent it by negative sign these both isomers are iners of each other because they are both mirror images of each other the third isomers of tartaric acid is this one here the O groups are present at the same site we can see that this carbon is as symmetric because different groups are attached to it and similarly this carbon is also asymmetric because different groups are attached to this carbon but wait a minute if I draw a symmetrical line here I get two equal parts hence this compound is symmetrical compound and Optical isomerism is not possible in it so we say that it is also optically inactive compound we call it mesot tartaric acid this Optical isomerism is not possible in it here you must learn the concept of dumers those isomers which are not mirror images of each other are called dumers for example first isomer and third isomers are not mirror images of each other we say that first and third isomers are dud rumors second and third are not mirror images of each other we say that second and third are dumers while we can see that first and second isomers are mirror images of each other we call it enumer th these are the three different isomers of tartaric acid this was all about isomerism

isomerism and organic chemistry well I will teach you complete isomerism and add types of isomerism in this special lecture firstly let me teach you about structure formula and molecular formula structure formula of any compound shows the arrangement of atoms or molecules for example consider the structure formula of methane this structure formula of methane shows that the shape of this compound is tetrahedral secondly it shows that one carbon atom is bonded to four hydrogen atoms while molecular formula of any compound shows the total number of atoms for example the molecular formula of methane is CH4 it shows that there is one atom of carbon and there are four atoms of hydrogen present in a single molecule of methane this we learn that structure formula shows the arrangement of atoms in any compound while molecular formula shows the total number of atoms present in a compound hence noted down these fundamental concepts now what are isomers well I always teach this simple analogy consider these gloves we know that this glow has four fingers and it has one left thumb while this glow has also four fingers and it has one right thumb I say both gloves have four fingers and a thumb but they have different Arrangement now listen carefully gloves are isomers of each other because they have same number of fingers and different Arrangement let me repeat it gloves are isomers of each other because they have same number of fingers and different Arrangement so the easy trick to remember the concept of isomer is they must have same number of fingers but different Arrangement now consider these organic compounds let me find the molecular formula of this compound we can see that there are four carbon atoms present and eight while there are 10 hydrogen atoms present in it so the IUPAC name of this compound is nban on the other hand there are also four carbon atoms present in it while there are also 10 hydrogen atoms present in it here I write 1 2 3 the methy CH4 is bonded at the second carbon I write two methy propan now listen carefully this compound and this compound have the same molecular formula C4 h10 but different Arrangement so they are isomers of each other let me repeat it this compound and this compound have the same molecular formula C4 h10 but different Arrangement so they are isomers of each other just remember that these both compounds have same fingers are molecular formula but they have different structure formula or different Arrangement therefore we Define isomers as the chemical compounds which have the same chemical formula but different structure formula are called isomers for example nban and two methy propane are isomers of each other because they have same molecular formula but different structure or different Arrangement just noted down this easy concept of isomers here let me ask you one of my favorite questions what is the difference between isomerism and isomers well believe me many student know its definition but they do not know its concept let consider hands and a nose I say isomerism is possible in hands because we have one left hand and one right hand are one left isomer and one right isomer secondly I say that isomerism is not possible in a nose because we have only one nose it has no other isomer therefore personally I say that isomerism is the tendency ability of any compound to form different compounds having different structures for example our hands have the tendency to form two compounds having different structure while isomers are the outputs of isomerism for example consider CH4 we say that isomerism is not possible in CH4 so it has no isomer th remember that isomerism is the phenomena or tendency of a compound to form different compounds having different structures while isomers are the output of the isomerism here remember this bonus Point isomers have different physical properties for example consider these two isomers this is alcohol and this is ether we know that they have same molecular formula but they have different structures that&#39;s why they have different different physical properties for instance this is liquid and this is a gas therefore we say that isomers have different physical properties now we will learn the type of isomerism well there are two types of isomerism structural isomerism and stereo isomerism structural isomerism is further divided into six categories like chain is isomerism position isomerism functional group isomerism metamerism toomis and ring chain isomerism while stereo isomerism is further divided into two categories geometric isomerism and Optical isomerism hence these are the eight different types of isomerism now we will learn chain isomerism and position isomerism well those isomers which have different number of carbons in Man chain are called chain isomers the easy trick is the main chain or the parent chain of carbon must be different while those isomers which have same functional group attached a different position are called position isomers the easy trick is functional group at different position means position isomers for example consider butan we can see that there are four carbon atoms present in the Man chain remember that if the carbon chain is straight we call it n chain so this is nban now I will cut this carbon from the main chain and I will attach it at the second position I get this structure here 1 2 3 this carbon is present at second position remember that if meile group is present at second position we call it ISO so this is isob butan now listen carefully in this nban there are four carbon atoms present in the Mansion our parent chain while in this isob butan there are three car carbon atoms present in the main chain are parent chain according to this trig these isomers have different carbon atoms in the Mansion so they are chain isomers the weite one and two are chain isomers because they have different number of carbons present in the man CH or parent chain now consider pentan I only consider the straight chain of pentan we call this n pentan secondly I cut this carbon and I attach it to this second carbon I get this structure here 1 2 3 4 the methy is present at the second carbon I call it isopentane thirdly I cut this carbon and I attach it to this second carbon I get this structure here 1 2 three remember that if there are two meile groups present at the second carbon we call it Neo hence this is neop pentan now listen carefully we can see that there are five carbon atoms present in the main chain of this compound there are four carbon atoms present in the main chain of this compound and there are three carbon atoms present in the Manion of this compound according to this St these isomers have different carbon atoms and their respective chains so they are chain isomers th one two and three are chain isomers because they have different number of carbons and the main chain are in the parent chain hence noted down now consider these isomers of hegan this is a straight chain of carbon atoms there are six carbon atoms present in it we call it an hexan secondly I write 1 2 3 4 five the methy group is present at the second carbon we call it isohexane thirdly I write 1 2 3 4 5 the methy is present at the third carbon I write three methy pentan fourthly I write 1 2 3 4 One methy is present at the second carbon and another methy is present at the third carbon I write 2 three dimethy butan lastly I write 1 2 3 4 here two methy groups are present at the second carbon we call it new hexen now I write the number of carbon atoms in the Manion the there are six carbon atoms in the Manion there are five carbon atoms in the Mansion there are also five carbon atoms in the Mansion there are four carbon atoms in the mansion and there are also four atoms in the Manion now listen carefully 1 2 and four are chain isomers because first isomer has six carbons in its Manion two R3 has five carbon atoms in its Manion so we call it chain isomers secondly we can see that second and third have five carbon atoms in the main carbon chain they are not chain isomers but wait a minute here the methy group is present at the second carbon while here the methy group is present at the third carbon according to the trick the man CH has five carbon atoms but the functional group ch3 is a different position so they are position isomers the second and third are position isomers similarly in the fourth isomer the M CH our parent chain has four carbon atoms while the fifth isomer has also five carbon atoms in the main chain here the two methy groups are present at the second carbon and at the Third carbon while here the two methy groups are present at second carbon these both isomers have same carbon chains but functional groups are present at different position hence fourth and fifth are also position isomers hence noted down finally consider these isomers pause the video and try to find the isomerism well they both have the same carbon chain having three carbon atoms in the main chain but here the chlorine is present at the first carbon and here the chlorine is present at the second carbon so they are position isomers secondly this compound has four carbon atoms in the M chain and this compound has five carbon atoms in the main chain so they are chain isomers thirdly double bond is present at the second carbon while here the double bond is present at the first carbon so they are position isomers fifthly the main chain has 1 two three carbon atoms while here 1 2 3 4 5 there are five carbon atoms so they are chain isomers therefore remember that chain isomers have different number of carbon at ATS in the M chain while position isomers have the same functional group bonded a different position and the same carbon chain now we will learn the third type of isomerism which is known as functional group isomerism well those isomers which have same molecular formula but different functional groups are called functional group isomerism in order to spot functional group isomerism always look at these two conditions firstly the two compounds must have the same molecular formula secondly the two compounds must have different functional groups for example consider alahh and Ketone functional group now I look for the molecular formula of the compounds the molecular formula of this alide is there are three carbon atoms six hydrogen atoms and one oxygen atom the molecular formula of this Ketone is here are also three carbon atoms six hydrogen atoms and one oxygen atom secondly the functional group of alahh is co and that of Ketone is C3 they both have different functional groups thus this alahh and this Ketone are functional group isomers of each other now consider carboxilic acid group and Easter functional group as usual I look for the molecular formula of the compounds and this carboxilic acid there are three carbon atoms six hydrogen atoms and two oxygen atoms while in this Easter there are also three carbon atoms six hydrogen atoms and two oxygen atoms secondly the functional group of carboxilic acid is coh and that of Easter is coo they both have different functional groups so both the conditions are fulfilled thus this carboxilic acid and this Easter are functional group isomers of each other now consider ether functional group and alcohol functional group as usual I find the molecular formula of compounds in this ether compound there are two carbon atoms six hydrogen atoms and one oxygen atom while in this alcohol compound there are also two carbon atoms six hydrogen atoms and one oxygen atom secondly the functional group of ether is O and that of alcohol is o Edge they both have different functional groups th this ether and this alcohol are functional group isomers of each other lastly consider alen functional group and cyc alkan functional group I find the molecular formula of these compounds in the alen there are four carbon atoms and eight hydrogen atoms well in this cyc alkan there are are also four carbon atoms and eight hydrogen atoms secondly the functional group of alen is double bond and that of cyc alkan is single Bond both the conditions are fulfilled thus this alen and this cyc alkan are functional group isomers of each other to summarize this whole concept we have learned that functional group isomerism exists between alide and Ketone carboxilic acid and Easter ether and alcohol alen and cyclo alkan thus remember that functional group isomers are those isomers which have same molecular formula but different functional groups now we will learn the four type of isomerism which is metamerism well when different alkal groups are bonded to the same functional group are called metamerism the condition is same functional groups and different alkal groups here the question is what are alkal groups well consider methane CH4 when we remove one hydrogen from methane we get ch3 and we call it methy secondly consider Ethan when we remove one hydrogen from it we get ch3 ch2 and we call it eile thus by this way we can form alkal groups remember that alkal group is represented by R Das now consider ether functional group The Ether group is r d o d r here that this is the left hand side of this ether and this is the right hand side side of this ether at the left side there is eile while at the right side there is also eile this ether contains two ethy groups we call it dthy ether now listen carefully I shift this ch2 from the left side to the right side let me repeat it I shift this ch2 from the left side to the right side I get this new compound here in this compound this is the left side and this is the right side at the left side there is meile group while at the right side there is propile group I write methy propile ether now listen carefully these two compounds have the same functional group which is ether but at the left side of this molecule there is ethy while at the left side of this molecule there is methy according to the condition these molecules have same functional group but different alkal groups are attached to it thus we call them two metamers of each other are two isomers of each other hence noted down now consider amine functional group we know that the functional group of amine is r- nh- r that this is the left side of this compound and this is the right side of this compound at the left side there is eile group and at the right side there is also eile group so we call it dthy amine now I shift this ch2 from the left side to the right side I get this new compound here at the left side there is methy group while at the right side there is propile group so we call it methy propile amine lastly I cut these two carbons of the propile I attach it to this carbon I get this new compound now at the left side there is methal group while at the right side there are two methy groups bonded to Second carbon we call it isopropyl isopropyl amine now listen carefully these three compounds have the same functional groups which is amine NH at the right side of this compound there is eile and at the right side of this compound there is condition is fulfilled th these three are metamers are three isomers of each other so note it down finally consider Ketone functional group we know that its functional group is r d c d r now consider these two compounds pause the video and try to find its isomerism here I write 1 2 3 4 5 the functional group of Ketone is present at the second carbon so we call it two Penton now listen carefully at the left side of this compound there is eile group while at the left side of this compound there is methy group hence they both are metamers or isomers of each other th remember that if compounds have same functional group but different alkal groups are Bond to it it is known as metamoris to summarize this whole concept we have learned that metamerism is possible in ether functional group amine functional group and Ketone functional group now we will learn the fifth type of isomerism which is toomis the phenomena in which single compound exist and two are more interconvertible structures are called Toom merism remember that Toom merism is possible in carbonal compounds having Co group like Ketone and alide personally I teach this condition for Toom merism it is CH single Bond a double bond B if any compound has this pattern there is 90% chances that totom merism will occur here I write 1 2 3 I shift hydrogen from first position to the third position and I shift double bond from second position to the first position th using this conversion we can easily do totom merism of any compound for example consider this Ketone here this carbon double bonded to oxygen is like a double bonded to B I write AB this ch3 is like CH so the condition CH single Bond a double bonded to be is fulfilled hence too merism is possible in this organic compound I write 1 2 3 now according to the condition I shift hydrogen from the first position to the third position and I shift double bond from the second position position to the first position I write ch2 because I shift one hydrogen double bond carbon single Bond o and ch3 this compound contains double bond of alen I write en n and for this o I write o so I get an hence these are the two toomer are two isomers of each other we also call toomis as keto enol isomerism because we get Ketone and enol this double arrow means that Ketone and anol are in dynamic equilibrium they are interconvertible that&#39;s why we also call it dynamic isomerism because they are moving from one form or from one structure to another structure remember that we will get 99% of Keto and 1% of enol because ketones are very stable than enol let me repeat it ketones are very stable than inol now consider this alide here this carbon double bonded to oxygen is like a double bonded to B I write a and b this ch3 is like CH the condition C H single Bond a double bonded to be is fulfilled hence Toom merism is possible in this compound I write 1 2 3 now according to the condition I shift hydrogen from the first position to the third position and double bond from the second position to the first position as a result I get ch2 double bond C single Bond o and H this is enol as usual we get 99% of alahh and 1% of enol hence these are the two toomer are two isomers of each other now consider this organic compound pause the video and try to find its doomers well this C double bonded to oxygen is like a double bonded to oxygen I write A and B this carbon has no hydrogen here the CH is not present I mean this carbon has no hydrogen the condition CH single Bond a double bonded to be is not fulfilled theism is not possible in this compound now consider this organic compound this N double bonded too is like a double bonded to B I write a and b this ch2 is like CH the condition CH single Bond a double bonded to be is fulfilled I write 1 2 three I shift hydrogen from the first position to the second position and I shift double bond from the second position to the first position as a result I get this compound th by this way we can easily do Toom merism of organic compounds now we will learn the exception cases of Toom merism usually these exception cases exist in cyclic organic compounds for these cases we have to learn two important Concepts aromatic anti-aromatic and conjugation for aromatic and anti-aromatic compounds I use this Drake I write 2 and four now I increase both the numbers by 4 2 + 4 is = to 6 6 + 4 is = to 10 and so on here 4 + 4 is = to 8 8 + 4 is = to 12 and so on remember that this number represents the total number of electrons also noted down that if there is negative charge it represents two electrons if there is a lone pair it represents two electrons and if there is double bond it also represents two electrons on the other hand if there is alternate single and double bond it shows High conjugation so on the basis of these two concepts we will easily learn the exception cases of totom ISM the first exception case is CH single Bond a double bonded to B here this condition exist but totom merism is not exhibited by the compound for example consider this organic compound firstly I write CH C and ch2 this C double bonded to O is like a double bonded to B and this CH or this ch2 is like CH so the condition CH single Bond a double bonded to be is fulfilled I can take both this CH or this ch2 but I consider this ch2 here this h of ch2 is shifted to third position and the double bond is shifted to the second position I get this compound but wait a minute we know that these both compounds are cyclic so I have to check either these compounds are aromatic or anti-aromatic this cyclic compound has one double bond I write to electrons we know that if a compound has two electrons it is aromatic hence this compound is aromatic on the other hand this compound has two double bonds I write 2 + 2 I get four electrons we know that if a compound has four electrons it is anti-aromatic here this compound is anti-aromatic now listen carefully aromatic compounds are highly stable while anti-aromatic compounds are less stable so this compound does not exist because it is anti-aromatic and it is less stable as a result we get 100% of this compound because it is aromatic and it is very stable Compound on the other hand we get 0% of this compound because it is anti-aromatic and it is less stable as usual this is Ketone part and this is enol part we get greater Ketone part and lesser inol part therefore we say that this compound does not exhibit Toto merism even though it fulfill the condition CH single Bond a double bond it to B hence noted down this exception case secondly anol part is greater than Ketone part we have learned that we always get greater Ketone part part and lesser enol part but in this exception case we will get greater enol part and lesser Ketone part let consider this example I write C and ch2 this c dou bond o is like a double bonded to B while this ch2 is like CH the condition is satisfied I shift hydrogen to the third position and the double bond to the first position as a result I get this compound these both compounds are cyclic I check aromatic and anti-aromatic compounds this compound has two double bond I write 2 + 2 is equal to four electrons we know that if there are four electrons it is an anti- aromatic compound thus this compound is anti-aromatic and it is less stable while this compound has three double bonds I write 2 + 2 + 2 is equal to six electrons we know that if there are six electrons it is aromatic compound th this compound is aromatic compound and it is highly stable so we get 100% of aromatic compound because it is highly stable while we get 0% of anti-aromatic compound because it is less stable we know that this is Ketone part and this is enol part hence we get greater percentage of enol part than Ketone percentage let noted down this exception case thirdly consider this exception case when there are two possible isomers firstly I write ch3 c ch2 c and ch3 now this C double bonded o is like a double bonded to be this ch2 and this ch3 are like CH let this ch3 gives it hydrogen to this oxygen the double bond will shift here I get this compound secondly let this ch2 gives its hydrogen to this oxygen this double bond shifts here I get this compound now listen carefully this compound has single Bond double bond single Bond single Bond double bond while this compound has single Bond double bond single Bond double bond Etc we know that if a compound has alternate single and double bond it is more conjugated and more stable this compound has alternate single and double bond it is more conjugated and more stable while this compound has not alternate single and double bond it is less conjugated and less stable this we will get greater content of this compound we know that this is Ketone part and this is inol part I get 85% of Ketone part and 15% of enol part part hence noted down this exception case therefore using aromatic anti-aromatic and conjugated concept we can easily learn the exception cases of totemism now we will learn the last STP of isomerism which is ring chain isomerism well those isomers having same molecular formula but different structures like cyclic and open chain are called ring chain isomerism here the word ring chain isomerism teachs the whole concept the ring stands for cyclic isomer and the chain stands for open chain so ring chain isomerisms are those which has one open chain compound and one cyclic compound for example consider propine it is an open chain of propine no I convert it to cyclic structure I get this structure we know that this is a cyclopropane the molecular formula of the both compounds are the same which is c3h6 but they have different structure open and cyclic so they both are known as ring chain isomers secondly consider hexine I convert it to cyclic compound I I get this cyclic structure it has no double bond I call it cyclohex here they both have the same molecular formula which is C6 h12 but they have different structures like open chain and cyclic so they both are ring chain isomers of each other thirdly consider buttine pause the video and try to make its ring chain as Summers well it is super easy just convert it to cyclic structure having single bonds I get this structure which is known as cycl butan the molecular formula of the both compounds are the same which is C4 h10 thus remember that ring chain isomers are those isomers which has one open chain and one cyclic ring so we have learned the six different types of structural isomerism now we will learn stereo isomerism remember that in sterio isomerism isomers have same molecular formula same structural formula but different arrangement of atoms and the space hence noted down this important definition of stereo isomers now we will learn the first type of stereo isomerism which is geometrical is isomerism remember that geometrical isomerism is also known as cyron isomerism those isomers which have different position of the identical group in the space are called geometrical isomers remember that CIS means when two identical groups are present on the same side of double bond while trans means when two identical groups are present on the opposite side of the double bond for example consider this two buttine here I use the three idiots mov it track up down up down this methy ch3 is up and this hydrogen is down here this ch3 is also up and this hydrogen is down so both the ch3 are up and both the hydrogens are down hence we call it cis2 buttin now listen carefully I just interchange the up and down position of ch3 and hydrogen let me repeat it I just interchange the up and down position of ch3 and hydrogen I get this compound here this ch3 is up this ch3 is down this hydrogen is up and this hydrogen is down we call it trans to buttine these both compounds have same molecular formula and same structures but they have different orientation of atoms in the space hence these both compounds are geometrical isomers of each other remember that trans isomers are more stable than CIS isomer because there is less steric repulsion in trans isomers secondly consider this organic compound is geometrical Isom ISM possible in8 the answer is no because even if we change the position of this chlorine from up to down position we still get the chlorine which is an upward position therefore remember that the two groups on each carbon must be different let me repeat it the two groups on each carbon must be different for example consider this compound here two different groups chlorine and hydrogen are bonded to the same carbon hence geometrical isomerism is possible I write up and down up and down now I just interchange the chlorine and hydrogen I get this compound this is CIS compound and this is trans compound so these are the two geometrical isomers of each other now consider this compound which is known as mic acid we know that different groups coh and hydrogen are bonded to each carbon hence geometrical isomerism is possible in it as usual I write up and down up and down now I just interchange the up and down position of coh and hydrogen I get this compound which is known as fumic acid in this compound The Identical groups coh are present on the same side of a double bond so it is cis2 buttin dioic acid while in this compound The Identical groups coh are present on the opposite side of double bond so it is trans2 buttin dioic acid hence melic acid is a CIS compound and fumic acid is a trans compound therefore remember that melic acid and fumic acid are geometrical isomers of each other th noted down these important geometrical isomers which are sometime asked in mcqs now consider open chain alkan and alkine remember that geometrical isomerism is not possible in open chain alkan and alkine it is because alkanes have carbon carbon single Bond or Sigma bond in Sigma Bond atoms constantly shift back and forth I mean atom constantly move up and down so geometrical isomerism is not possible while in alkine there are already triple bonds between carbon and carbon we need to Branch are two more Bonds on each carbon which is not possible because carbon can form only four bonds hence geometrical isomerism is not possible in alkin now consider the cyc alkan the question is is geometrical isomerism possible in cyc Alcan well the answer is yes for instance in this compound I write up and down up and down so this is a CIS cyc Alan now I just enter inter change the up and down position of the ch3 group and hydrogen group I get this compound here the alkal group ch3 is up and here the alkal group is down hence it is a trans cyc Alan th remember that geometrical isomerism is possible in cyc Alan let me teach you one another important question why tutin shows geometrical isomerism but not one buttin well consider two buttine and one buttine now I write 1 2 3 4 the double bond is present at the second position we call it two buttine on the second and third carbon two different groups are attached to the same carbon atom hence geometrical isomerism is possible in it while in this compound I write 1 2 3 the double bond is present at the first position I write one buttine here on the second carbon there is alkal group and hydrogen group so it is all right but here at first carbon the two same groups hydrogen are attached to the first carbon we have already learned that if identical groups are attached to the same carbon geometrical isomerism is not possible therefore geometrical isomerism is not possible in one buttin but it is possible and totin hence noted down this important question the final concept of geometrical isomerism is z and E form consider this organic compound here the four groups bonded to two carbons are different in such cases we decide on the basis of atomic size Florine has the smaller size than chlorine while bromine has the smaller size than iodine now the smaller size groups are at the same side of double bond hence it is AIS form are we call it Z form now as usual I interchange the smaller size and larger size groups of bromine and iodine I get this compound here Florine has smaller size than chlorine while iodine has larger size than bromine the smaller size groups like Florine and bromine are at the opposite side of the double bond so we call it transform or we call it eform thus these are the two geometrical isomers of each other and we call it Z form and E form hence noted down this important concept finally let me teach you the most difficult isomerism which is Optical isomerism Optical isomers have these four properties they have same chemical formula they have same molecular formula they have same chemical and physical properties the only difference between them is that they have different behaviors towards light now I will teach you only two conditions for optical isomerism and it will help you to master this concept the first one is optically active substance and the second one is asymmetrical carbon for example consider this compound if this compound is optically active and it has asymmetrical carbon then Optical isomerism is possible in it if it is not optically active or it has no asymmetrical carbon then it will not show Optical isomet ISM now what is optically active compounds well to learn it we have to learn polarized light polarimeter optically active compounds and optically inactive compounds now the light in One Direction is called polarized light for example this light is going in One Direction it is known as polarized light while the light produced by the bulb is called UNP polarized light because it spreads in all Direction secondly polarimeter is an instrument which is used to measure the polarization of light and Optical activity of a compound thirdly the substances which can easily rotate the polarized light are called optically active substances for example consider glucose it is optically active compound because it can EAS rotate polarized light fourthly those substances which cannot rotate the polarized light are called optically inactive substances for example consider glycine it is optically inactive compound because it cannot rotate polarized light let&#39;s remember these important Concepts now how can we find the optical activity of a compound well we use Nicole and polarimeter to find the optical activity of a compound for example we take a light source it produces light in all directions here it is unpolarized light I place nicool prism in front of it now the Nick prism simply convert the unpolarized light to polarized light secondly I make a solution of a compound and place it in the poeter now there are three possible outcomes if the light is rotated in the left direction or anticlockwise Direction the substance is optically active and we call it leur rotatory compound we represent it by negative sign if the light is rotated in right direction or clockwise Direction the substance is optically active and we call it dextr rotatory compound we represent it by positive sign lastly if the light is not rotated in the left or in the right direction it means that compound is optically inactive th remember that if a compound rotate light in left Direction it is Le rotatory if a compound rotate lights in right direction it is De rotatory now we will learn the second condition for optical isomerism which is asymmetrical carbon firstly we will learn asymmetrical compounds for example consider these compounds here we can easily divide this compound into two equal parts so it is symmetrical compound while we cannot divide this substance into two equal parts because this part is different from this part so it is asymmetrical compound now consider this compound here four different groups are attached to the carbon we call this carbon as asymmetrical carbon because if we divide this compound at this carbon we get two different parts so it is asymmetrical carbon to summarize this whole concept we have learned that for optical isomerism we need optically active compounds secondly the compound must have one asymmetrical carbon now we will learn the optical isomers of lectic acid well consider the lectic acid I write 1 2 3 its IUPAC name is two hydroxy propenoic acid I look for the conditions it has asymmetrical carbon so it is optically active compound now I place a mirror in front of it we know that in the mirror the right side becomes left and the left side becomes right I get this compound now listen carefully it rotates polarized light towards right direction we call it dextr rotatory and we represent it by positive sign secondly it rotates polarized light towards left we call it leur rotatory and we represent it by negative sign this first and second isomers are called eners eners are those isomers which are mirror images of each other let me repeat it iners are those isomers which are mirror images of each other just remember that the first and the second isomers are called inimers because they both are mirror images of each other so these are the two Optical isomers of lectic acid now consider secondary butle chloride and form its Optical isomers well we can see that this carbon is asymmetric because all the four carbon attached to this carbon are different so it is optically active compound now I place mirror here as a result I get this comp bound this ch3 is here and this C2 H5 is here now it depends on you to call this compound negative or positive let this is negative we call it leur rotatory isomers this is positive we call it dextr rotatory we know that these both are iners of each other because they are mirror images of each other so these are the two Optical isomers of this compound now consider this organic compound is Optical isomerism possible in it the answer is no this compound is symmetrical if I cut this compound from here I get equal parts on both the sides hence it does not show Optical isomerism finally consider Optical isomerism and tartaric acid well consider its structure remember that hydroxy groups and O group are present at opposite side this compound is asymmetric hence Optical isomerism is possible in it I write 1 2 3 4 at second and third position hydroxy group are present I write 23 dihydroxy butan while at first and fourth position carboxilic acid groups are present I write one for dioic acid so its IUPAC name is 23 dihydroxy butane 14 dioic acid now as usual I place mirror in front of this compound I get this compound this o shifts here and this o shifts here let we call this textr rotatory tartaric acid and we represent it by positive sign we call this leur rotatory tartaric acid and we represent it by negative sign these both isomers are iners of each other because they are both mirror images of each other the third isomers of tartaric acid is this one here the O groups are present at the same site we can see that this carbon is as symmetric because different groups are attached to it and similarly this carbon is also asymmetric because different groups are attached to this carbon but wait a minute if I draw a symmetrical line here I get two equal parts hence this compound is symmetrical compound and Optical isomerism is not possible in it so we say that it is also optically inactive compound we call it mesot tartaric acid this Optical isomerism is not possible in it here you must learn the concept of dumers those isomers which are not mirror images of each other are called dumers for example first isomer and third isomers are not mirror images of each other we say that first and third isomers are dud rumors second and third are not mirror images of each other we say that second and third are dumers while we can see that first and second isomers are mirror images of each other we call it enumer th these are the three different isomers of tartaric acid this was all about isomerism

Transcript for:Isomerism in Organic Chemistry

Transcript for:
Isomerism in Organic Chemistry