Isomerism and Ohrganic Chhemistry. 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 ship 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 ChH4. It shows that there is one atom of carbon and there are four atoms of hydrogen present in a single molecule of methane. Thus 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. Chonsider these gloves. We know that this glove has four fingers and it has one left thumb.
While this glove has also four fingers and it has one right thumb. I say both the 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 4 carbon atoms present in it, while there are 10 hydrogen atoms present in it. So, the IUPACh name of this compound is N-butane.
Ohn the other hand, there are also 4 carbon atoms present in it, while there are also 10 hydrogen atoms present in it. Here I write 1, 2, 3. The methyl ChH4 is bonded at the second carbon. I write 2-methylpropane.
Now listen carefully. This compound and this compound have the same molecular formula Ch4H10 but different arrangement. So, they are isomers of each other.
Let me repeat it. This compound and this compound have the same molecular formula Ch4H10 but different arrangement. So, they are isomers of each other. Thus, remember that these both compounds have same fingers or 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, n-butane and 2-methylpropane are isomers of each other because they have same molecular formula but different structure or different arrangement. Thus 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 students know its definition, but they do not know its concept.
Let's consider hands and a nose. I say isomerism is possible in hands because we have one left hand and one right hand, or 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 ChH4. We say that isomerism is not possible in ChH4, so it has no isomer. Thus 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 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 isomerism, position isomerism, functional group isomerism, metamerism, tautomerism, and ring chain isomerism.
While stereo isomerism is further divided into two categories, geometric isomerism and optical isomerism. Hence, these are the at different types of isomerism. Now we will learn chain isomerism and position isomerism.
Well, those isomers which have different number of carbons in main 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 at different position are called position isomers.
The easy trick is functional group at different position means position isomers. For example, consider butane. We can see that there are four carbon atoms present in the main chain. Remember that if the carbon chain is straight, we call it n-chain. So, this is n-butane.
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 methyl group is present at second position, we call it iso. So this is isobutan. Now listen carefully. And this n-butan there are four carbon atoms present in the main chain or parent chain. While in this isobutan, there are three carbon atoms present in the main chain or parent chain.
According to this trick, these isomers have different carbon atoms in the main chain, so they are chain isomers. Thus, we write 1 and 2 are chain isomers because they have different number of carbons. present in the main chain or parent chain. Now consider pentane.
I only consider the straight chain of pentane. We call this n-pentane. Secondly, I cut this carbon and I attach it to this second carbon.
I get this structure. Here, one, two, three, four. The methyl 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 1,2,3. Remember that if there are two methyl groups present at the second carbon, we call it NEOh.
Hence this is neopentane. 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 main chain of this compound.
According to this trick, these isomers have different carbon atoms and their respective chains so they are chain isomers. Thus 1, 2 and 3 are chain isomers because they have different number of carbons and the main chain are in the parent chain. Hence note it down. Now consider these isomers of hexane.
This is a straight chain of carbon atoms. There are 6 carbon atoms present in it. We call it an hexane.
Secondly, I write 1, 2, 3, 4, 5. The methyl group is present at the second carbon. We call it isohexane. Thirdly, I write 1, 2, 3, 4, 5. The methyl is present at the third carbon.
I write 3-methylpentane. Fourthly, I write 1,2,3,4. Ohne methyl is present at the second carbon and another methyl is present at the third carbon.
I write 2,3-dimethylbutane. Lastly, I write 1,2,3,4. Here two methyl groups are present at the second carbon. We call it neohexane. Now I write the number of carbon.
carbon atoms in the main chain. There are 6 carbon atoms in the main chain. There are 5 carbon atoms in the main chain.
There are also 5 carbon atoms in the main chain. There are 4 carbon atoms in the main chain. And there are also 4 atoms in the main chain.
Now listen carefully. 1, 2 and 4 are chain isomers. Because first isomer has 6 carbons in its main chain. 2 are 3 has 5 carbon atoms and its main chain.
So, we call it chain isomers. Secondly, we can see that second and third have 5 carbon atoms and the main carbon chain. They are not chain isomers. But wait a minute. Here, the methyl group is present at the second carbon, while here the methyl group is present at the third carbon.
According to the trick, the main chain has 5 carbon atoms. But the functional group ChH3 is a different position. So, they are position isomers.
The second and third are position isomers. Similarly, in the fourth isomer, the main chain or parent chain has 4 carbon atoms. Well, the fifth isomer has also 5 carbon atoms in the main chain. Here, the two methyl groups are present at the second carbon and at the third carbon, while here, the two methyl 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 note it 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 main chain and this compound has four carbon atoms in compound has 5 carbon atoms in the main chain, so they are chain isomers. Thirdly, the double bond is present in the second carbon, while here the double bond is present in the first carbon, so they are position isomers.
Fifthly, the main chain has 1,2,3 carbon atoms, while here 1,2,3,4,5. Here are 5 carbon atoms. So they are chain isomers. Therefore, remember that chain isomers have different number of carbon atoms in the main chain. While position isomers have the same functional group bonded at different position in 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 aldehyde and ketone functional group. Now I look for the molecular formula of the compounds. The molecular formula of this aldehyde is there are 3 carbon atoms, 6 hydrogen atoms and 1 oxygen atom.
The molecular formula of this ketone is, here are also 3 carbon atoms, 6 hydrogen atoms and 1 oxygen atom. Secondly, the functional group of aldehyde is ChOhH and that of ketone is ChOhChH3. They both have different functional groups.
Thus, this aldehyde and this ketone are functional group isomers of each other. Now consider carboxylic acid group and ester functional group. As usual, I look for the molecular formula of the compounds. In this carboxylic acid, there are 3 carbon atoms, 6 hydrogen atoms and 2 oxygen atoms. While in this ester, There are also 3 carbon atoms, 6 hydrogen atoms and 2 oxygen atoms.
Secondly, the functional group of carboxylic acid is ChOhOhH and that of ester is ChOhOh. They both have different functional groups, so both the conditions are fulfilled. Thus, this carboxylic acid and this ester 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 Oh and that of alcohol is OhH. They both have different functional groups.
Thus, this ether and this alcohol are functional group isomers of each other. Lastly, consider alkene functional group and cycloalkane functional group. I find the molecular formula of these compounds. In the alkene, there are four carbon atoms and and add hydrogen atoms. While in this cycloalkane, there are also four carbon atoms and add hydrogen atoms.
Secondly, the functional group of alkene is double bond and that of cycloalkane is single bond. Both the conditions are fulfilled, thus this alkene and this cycloalkane are functional group isomers of each other. To summarize this whole concept, We have learned that functional group isomerism exists between aldehyde and ketone, carboxylic acid and ester, ether and alcohol, alkene and cycloalkane.
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 alkyl groups are bonded to the same functional group are called metamerism. The condition is same functional groups and different alkyl groups.
Here the question is what are alkyl groups? Well, consider methane ChH4. When we remove one hydrogen from methane, we get ChH3. and we call it methyl. Secondly, consider ethane.
When we remove one hydrogen from it, we get ChH3ChH2 and we call it ethyl. Thus, by this way, we can form alkyl groups. Remember that alkyl group is represented by R'.
Now consider ether functional group. The ether group is R'. Here let this is the left hand side of this ether and this is the right hand side of this ether.
At the left side, there is ethyl while at the right side, there is also ethyl. This ether contains two ethyl groups. We call it diethyl ether. Now listen carefully.
I shift this ChH2 from the left side to the right side. Let me repeat it. I shift this ChH2 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 methyl group while at the right side, there is propyl group.
I write methyl propyl ether. Now listen carefully. These two compounds have the same functional group But at the left side of this molecule, there is ethyl, while at the left side of this molecule, there is methyl. According to the condition, these molecules have same functional group but different alkyl groups are attached to it. Thus we call them two metamers of each other or two isomers of each other.
Hence note it 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 ethyl group and at the right side, there is also ethyl group. So we call it diethylamine. Now I shift this ChH to from the left side to the right side.
I get this new compound. Here at the left side there is methyl group while at the right side there is propyl group. So we call it methylpropylamine. Lastly, I cut these two carbons of the propyl. I attach it to this carbon.
I get this new there is methyl group while at the right side there are two methyl groups bonded to second carbon. We call it isopropyl thus we get methyl isopropyl amine. Now listen carefully, these three compounds have the same functional groups which is amine and H.
At the right side of this compound there is ethyl and at the right side of this compound compound, there is isopropyl, hence the condition is fulfilled. Thus, these three are metamers or three isomers of each other. So, note it down.
Finally, consider ketone functional group. We know that its functional group is R'ChOh'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 2-pentanone. Now, listen carefully. At the left side of this compound, there is ethyl group.
While at the left side of this compound, there is methyl group. Hence, they both are metamers or isomers of each other. Just remember that If compounds have same functional group but different alkyl groups are bonded to it, it is known as metamerism. 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 tautomerism.
The phenomena in which single compound is exist and two are more interconvertible structures are called tautomerism. Remember that tautomerism is possible in carbonyl compounds having ChOh group like ketone and aldehyde. Personally, I teach this condition for tautomerism.
It is ChH, single bond A, double bond B. If any compound has this pattern, there is 90% chances that tautomerism 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. Thus using this conversion, we can easily do tautomerism of any compound.
For example, consider this ketone. Here, this carbon double bonded to oxygen is like A double bonded to B. I write A.
This ChH3 is like ChH. So the condition ChH, single bond, A double bonded to B is fulfilled. Hence, tautomerism 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 to the first position.
I write ChH2 because I shift one hydrogen. Double bond carbon, single bond OhH and ChH3. This compound contains double bond of alkene. I write EN.
And for this OhH, I write OhL. So I get a NOhL. Hence these are the two tautomers are two isomers of each other.
We also call tautomerism as ketoenol isomerism because we get ketone and enol. This double arrow means that ketone and enol 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 ketone and 1% of enol because ketones are very stable than enol. Let me repeat it.
Ketones are very stable than enol. Now consider this aldehyde. Here, this carbon double bonded to oxygen is like A double bonded to B. I write A and B. This ChH3 is like ChH.
The condition ChH single bond A double bonded to B is fulfilled. Hence, tautomerism 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 ChH2 double bond Ch, single bond OhH and H. This is enol. As usual, we get 99% of aldehyde and 1% of enol. Hence, these are the two tautomers or two isomers of each other.
Now consider this organic compound. Pause the video and try to find its tautomers. Well, this Ch double bonded to oxygen is like A double bonded to oxygen.
I write A and B. This carbon has no hydrogen. Here, the ChH is not present. I mean, this carbon has no hydrogen.
The condition ChH, single bond, A double bonded to B is not fulfilled. The stratum erism is not possible in this compound. Now consider this organic compound. This N double bonded to Oh is like A double bonded to B. I write A and B.
This ChH2 is like ChH. The condition ChH single bond A double bonded to B is fulfilled. I write 1, 2, 3. 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. Thus by this way, we can easily do tautomerism of organic compounds.
Now we will learn the exception cases of tautomerism. 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 trick. I write 2 and 4. Now I increase both the numbers by 4. 2 plus 4 is equal to 6. 6 plus 4 is equal to 10 and so on. Here 4 plus 4 is equal to 8. 8 plus 4 is equal 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. Ohn 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 tautomerism.
The first exception case is ChH single bond A double bonded to B. Here this condition exists but tautomerism is not exhibited by the compound. For example, consider this organic compound. Firstly, I write ChH, Ch and ChH2.
This This Ch double bonded to Oh is like A double bonded to B. And this ChH or this ChH2 is like ChH. So, the condition ChH single bond A double bonded to B is fulfilled.
I can take both this ChH or this ChH2. But I consider this ChH2. Here, this H of ChH2 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 two electrons. We know that if a compound has two electrons, it is aromatic.
Hence, this compound is aromatic. Ohn the other hand, this compound has two double bonds. I write 2 plus 2. I get 4 electrons.
We know that if a compound has 4 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. Ohn 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 enol part. Therefore, we say that this compound does not exhibit tautomerism even though it fulfilled the condition ChH single bond A double bonded to B.
Hence, noted down this exception case. Secondly, Enol part is greater than ketone part. We have learned that we always get greater ketone part and lesser enol part.
But in this exception case, we will get greater enol part and lesser ketone part. Let's consider this example. I write Ch and ChH2.
This Ch double bond Oh is like A double bonded to B, while this ChH2 is like ChH. The condition is satisfied. I shift hydrogen to this 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 plus 2 is equal to 4 electrons.
We know that if there are 4 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 plus 2 plus 2 is equal to 6 electrons. We know that if there are 6 electrons, it is aromatic compound. Thus 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. Thus, note it down this exception case. Thirdly, consider this exception case when there are two possible isomers.
Firstly, I write ChH3Ch, ChH2Ch and ChH3. Now this Ch double bonded Oh is like A double bonded to B. This ChH2 and this ChH3 are like ChH.
Let this ChH3 gives its hydrogen to this oxygen. The double bond will shift here. I get this compound. Secondly, let this ChH2 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. Thus we will get greater content of this compound.
We know that this is ketone part and this is enol part. I get 85% of ketone part. and 15% of enol part, hence noted down this exception case. Therefore, using aromatic, anti-aromatic and conjugated concept, we can easily learn the exception cases of tautomerism. Now we will learn the last step 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 teaches 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 propene.
It is an open chain of propene. Now I convert it to cyclic structure. I get this structure.
We know that this is acyclopropane. The molecular formula of the both compounds are the same which is Ch3H6. But they have different structure, open and cyclic.
So they both are known as range chain isomers. Secondly, consider hexene. I convert it to cyclic compound. I get this cyclic structure. It has no double bond.
I call it cyclohexane. Here they both have the same molecular formula which is Ch6H12. But they have different structures like open chain and cyclic. So they both are ring chain isomers of each other.
Thirdly, consider butene. Pause the video. and try to make its ring chain isomers. Well, it is super easy.
Just convert it to cyclic structure having single bonds. I get this structure, which is known as cyclobutane. The molecular formula of the both compounds are the same, which is Ch4H10. Just 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 stereo isomerism, isomers have same molecular formula, same structural formula but different arrangement of atoms in the space. Hence, note it down this important definition of stereo isomers. Now, we will learn the first type of stereo isomerism. which is geometrical isomerism.
Remember that geometrical isomerism is also known as cis-trans 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 these two butene. Here I use the three idiots movie trick.
Up down, up down. This methyl ChH3 is up and this hydrogen is down. Here this ChH3 is also up and this hydrogen is down. So both the ChH3 are up and both the Hydrogens are down. Hence we call it cis-2-butene.
Now listen carefully. I just interchange the up and down position of ChH3 and Hydrogen. Let me repeat it.
I just interchange the up and down position of ChH3 and Hydrogen. I get this compound. Here this ChH3 is up, this ChH3 is down, this Hydrogen is up, and this hydrogen is down.
We call it trans-2-butene. These both compounds have same molecular formula and same structures but they have different orientation of atoms and 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 isomerism possible in it? 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 malic acid. We know that different groups ChOhOhH 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 ChOhOhH.
and hydrogen. I get this compound which is known as fumaric acid. In this compound, the identical groups ChOhOhH are present on the same side of a double bond.
So, it is cis-2-butyndioic acid. While in this compound, the identical groups ChOhOhH are present on the opposite side of double bond. So, it is trans- 2-butenedioic acid. Hence, malic acid is a cis compound and fumaric acid is a trans compound. Therefore, remember that malic acid and fumaric acid are geometrical isomers of each other.
Thus, note it down these important geometrical isomers which are sometimes asked in MChQs. Now, consider open chain alkane and alkyne. Remember that geometrical isomerism is not possible in open chain alkane and alkyne. 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 alkyne, there are already triple bonds between carbon and carbon.
We need two 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 all kinds. Now consider this cycloalkane. The question is, is geometrical isomerism possible in cycloalkane? Well, the answer is yes.
For instance, in this compound, I write up and down, up and down. So this is a cis-cycloalkane. Now I just interchange the up and down position of the ChH3 group and hydrogen group. I get this compound.
Here the alkyl group ChH3 is up and here the alkyl group is down. Hence it is a trans cycloalkane. Just remember that geometrical isomerism is possible in cycloalkane.
Let me teach you one another important question. Why 2-butene shows geometrical isomerism but not 1-butene? Well, consider 2-butene and 1-butene.
Now I write 1, 2, 3, 4. The double bond is present at the second position. We call it 2-butene. Ohn 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 1 butene. Here on the second carbon, there is alkyl group and hydrogen group. So it is alright. But here at first carbon, the two same groups hydrogen are attached to the first carbon.
We have already learnt that If identical groups are attached to the same carbon, geometrical isomerism is not possible. Therefore, geometrical isomerism is not possible in 1-butene, but it is possible in 2-butene. Hence, note it down this important question. The final concept of geometrical isomerism is Z and E form. Chonsider this organic compound.
Here, the four groups bonded to two carbons are different. In such cases, we decide on the basis of atomic size. Fluorine 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 a cis form or we call it a Z form. Now as usual, I interchange the smaller size and larger size groups of bromine and iodine, I get this compound. Here, fluorine has smaller size than chlorine, while iodine has larger size than bromine.
The smaller size groups like fluorine and bromine are at the opposite side of the double bond, so we call it transform or we call it e-form. 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. Ohptical 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 isomerism. 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 unpolarized 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 easily 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.
Thus remember these important concepts. Now how can we find the optical activity of a compound? Well, we use Nicole Prism 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 Nicole Prism in front of it. Now the Nicole Prism simply convert the unpolarized light to polarized light.
Secondly, I make a light source. a solution of a compound and place it in the polarimeter. 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 Leo-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 dextrorotatory 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. Thus, remember that if a compound rotates light in left direction, it is leu-rotatory.
If a compound rotates light in right direction, it is dextrorotatory. 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 lactic acid. Well, consider the lactic acid. I write 1, 2, 3. Its IUPACh name is 2-hydroxypropanoic acid.
I look for the conditions. It has asymmetrical carbon. So, it is optically active. 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 texturotatory and we represent it by positive sign. Secondly, it rotates polarized light towards left. We call it leurotetry and we represent it by negative sign.
This first and second isomers are called enantiomers. Enantiomers are those isomers which are mirror images of each other. Let me repeat it.
Enantiomers are those isomers which are mirror images of each other. Just remember that the first and the second isomers are called enantiomers because they both are mirror images of each other. So, these are the two optical isomers of lactic acid.
Now, consider secondary butyl 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 compound. This ChH3 is here and this Ch2H5 is here. Now it depends on you to call this compound negative or positive. Let this is negative.
We call it leu-rotatory isomers. This is positive. We call it dextro-rotatory.
We know that these both are enantiomers 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 sides. Hence it does not show optical isomerism. Finally, consider optical isomerism and tartaric acid.
Chonsider its structure. Remember that hydroxyl groups and OhH 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, hydroxyl group are present. I write 2, 3-dihydroxybutane. While at first and fourth position, carboxylic acid groups are present. I write 1,4-dioic acid.
So its IUPACh name is 2,3-dihydroxy butane 1,4-dioic acid. Now as usual, I place mirror in front of this compound. I get this compound. This OhH shifts here and this OhH shifts here.
Let we call this dextrorotatory tartaric acid. and we represent it by positive sign. We call this leu-rotatory tartaric acid and we represent it by negative sign.
These both isomers are enantiomers of each other because they are both mirror images of each other. The third isomer of tartaric acid is this one. Here, the OhH groups are present at the same site. We can see that this carbon is asymmetric 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 mesotartaric acid.
Thus, optical isomerism is not possible in it. Here, you must learn the concept of disterumers. Those isomers which are not mirror images of each other are called disterumers.
For example, first isomer and third isomers are not mirror images of each other. We say that First and third isomers are diastereomers. Second and third are not mirror images of each other. We say that second and third are diastereomers. While we can see that first and second isomers are mirror images of each other, we call it enantiomers.
Thus, these are the three different isomers of tartaric acid. This was all about isomerism.