In the previous section, we completed a discussion on the nomenclature of alkenes. In this section, we will see isomerism in alkenes.
Some basics can be written in 6 steps:
1. Consider the three structures in fig.13.54 below:
 |
| Fig.13.54 |
♦ Structure I is But-1-ene.
♦ Structure II is But-2-ene.
♦ Structure III is 2-Methylprop-1-ene.
2. Let us compare the number of atoms:
♦ In I, there are four C atoms and eight H atoms.
♦ In II also, there are four C atoms and eight H atoms.
♦ In III also, there are four C atoms and eight H atoms.
• So all the three structures have the same molecular formula C4H8
• It is clear that but-1-ene, but-2-ene and 2-Methylprop-1-ene are isomers.
3. We have discussed about isomers in a previous section [see section 12.9]. Also we have discussed about the various isomers among alkanes.
• Based on those discussions, we can write:
♦ Structures I and III in fig.13.54, are chain isomers.
♦ Structures II and III in fig.13.54, are chain isomers.
♦ Structures I and II in fig.13.54, are position isomers.
(Recall that position isomers have the same carbon chain. But the positions of the functional groups will be different. In our present case, the functional group is the alkene group)
4. In the case of alkanes, we have seen that, only those which have more than three C atoms will exhibit isomerism.
• In the same way, the first two alkenes (ethene and propene) do not exhibit isomerism.
• This is because, a minimum of four C atoms are required to obtain different structural arrangements.
5. Based on the general formula CnH2n for alkenes, we expect CH2 to be the first member of the alkene series. But it can be seen that, the valencies of C atom are not satisfied in CH2. So CH2 has a very short life. That is the reason for considering ethene (common name ethylene) as the first member.
6. We have seen that the first two members of the alkene series do not exhibit isomerism.
• In fig.13.54 above, we have seen the isomers of the third member. Next let us see the isomers of the fourth member.
It is explained as as solved example below:
Solved example 13.9
Write all the structures and IUPAC names of the structural isomers of alkenes corresponding to C5H10.
Solution:
Five isomers are shown in fig.13.55 below:
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| Fig.13.55 |
Geometrical Isomerism in Alkenes
• This can be written in steps:
1. Consider the alkene but-2-ene (CH3-CH=CH-CH3) shown in fig.13.56(a) below:
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| Fig.13.56 |
• C atom on the left is marked as C1
♦ Two valencies of this C atom is satisfied by the double bond.
♦ One valency of this C atom is satisfied by the H atom.
♦ The remaining valency of this C atom is satisfied by the CH3 group.
• C atom on the right is marked as C2
♦ Two valencies of this C atom is satisfied by the double bond.
♦ One valency of this C atom is satisfied by the H atom.
♦ The remaining valency of this C atom is satisfied by the CH3 group.
2. In the given alkene in fig.13.56, the atoms participating in the double bond are C1 and C2
• For any C atom participating in a double bond, two of it’s valencies will be satisfied by the double bond. We need to consider the manner in which the remaining two valencies are satisfied.
• In our present case,
♦ the remaining two valencies of C1 are satisfied by H and CH3
♦ the remaining two valencies of C2 are also satisfied by H and CH3
3. Here we note three interesting points:
(i) The remaining two valencies of C1 are satisfied by two different species: H and CH3
♦ This H and CH3 can be represented as X and Y
(ii) The remaining two valencies of C2 are also satisfied by the same X and Y.
(iii) Thus we can write:
X and Y are two different species, and also X and Y are available for both C1 and C2
When the above three points are satisfied, we can represent the alkene as shown in fig.13.56(b) above.
4. Now we see an interesting fact. It can be written in 4 steps:
(i) The structure in fig.13.56(b) is shown again in fig.13.57(a) below:
 |
| Fig.13.57 |
• Alkene in fig.13.57(a) above is exactly the same as that in the previous fig.13.56(b). The repetition is done for a better comparison.
• The alkene in fig.13.57(a) is shown again in fig.13.57(b) with some modifications.
• In fig.a,
♦ both species are Y on the upper side of the double bond.
♦ both species are X on the lower side of the double bond.
• In fig.b,
♦ one species is X and the other is Y on the upper side of the double bond.
♦ one species is Y and the other is X on the lower side of the double bond.
(ii) Recall that, rotation about a double bond is not possible. We saw the details of restricted rotation in an earlier section [see fig.4.151 in section 4.26]
• So the rotation indicated by the yellow curved arrow in fig.13.57(b) is not possible.
(iii) Since the rotation is restricted,
♦ Structure in (a) cannot become the structure in (b)
♦ Structure in (b) cannot become the structure in (a)
(iv) So they are two entirely different structures.
• They differ in their properties like melting point, boiling point, dipole moment, solubility etc.,
5. It is clear that, the two structures in fig.13.57 are isomers.
• Let us write how the concept of isomerism can be applied to the two structures. It can be written in 4 steps:
(i) First we will see stereoisomerism.
• We know that, in chain isomerism, the C atoms of various isomers are not joined up in the same order. For example:
♦ In one chain isomer, the methyl group may branch off from the third C atom
♦ In another chain isomer, the methyl group may branch off from the fourth C atom.
• We also know that, in position isomerism, the various C atoms in the various isomers are joined up in the same order (that is., position isomers have the same chain). But the positions of the functional group will be different. For example,
♦ In pent-1-ene, the double bond comes after the first C atom.
♦ In pent-2-ene, the double bond comes after the second C atom.
• Stereoisomerism is different from both chain isomerism and position isomerism.
◼ In stereoisomerism, the C atoms of the various stereoisomers are joined up in the same order. But the atoms and groups occupy different positions in space.
(ii) Geometrical isomerism is a special type of stereoisomerism.
• Geometrical isomerism arises due to the restricted rotation about the C=C double bond.
• Due to the restricted rotation, various atoms or groups are locked in position.
(iii) There are two types of geometrical isomers.
• In the first type, identical atoms or groups lie on the same side of the double bond.
♦ This type of isomer is called cis isomer.
♦ So the isomer in fig.13.57(a) is a cis isomer.
• In the second type, identical atoms or groups lie on the opposite sides of the double bond.
♦ This type of isomer is called trans isomer.
♦ So the isomer in fig.13.57(b) is a trans isomer.
(iv) Geometrical isomerism is also known as cis/trans isomerism.
♦ cis is a Latin word which means ‘on this side’
♦ trans is also a Latin word which means ‘across’
Next we will discuss about the polarity in cis/trans isomers. It can be explained using two examples.
Example 1
This can be written in 4 steps:
1. Consider the molecule 1,2-dichloroethene (ClCH=CHCl). It’s structure is drawn in fig.13.58(a) below:
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| Fig.13.58 |
2. We know that, Cl is more electronegative than C. So the C-Cl bond will be a polar bond.
• The dipole moment in this bond is indicated by the yellow crossed arrow.
• We have discussed about polarity and dipole moment in an earlier section [see fig.4.65 of section 4.10]
3. Note that, the isomer in fig.13.58(a) is the cis isomer. Both the Cl atoms are on the same side.
♦ The dipole moment in the left C-Cl bond acts towards top left.
♦ The dipole moment in the right C-Cl bond acts towards top right.
• So the two moments add up to give a net dipole moment for the whole molecule.
4. Note that, the isomer in fig.13.58(b) is the trans isomer. The Cl atoms are on the opposite sides.
♦ The dipole moment in the bottom C-Cl bond acts towards bottom right.
♦ The dipole moment in the top C-Cl bond acts towards top left.
• So the two moments are acting in the opposite directions. They cancel each other to give zero net dipole moment for the whole molecule.
Example 2
This can be written in 4 steps:
1. Consider the molecule but-2-ene (CH3-CH=CH-CH3). It’s structure is drawn in fig.13.59(a) below:
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| Fig.13.59 |
2. The CH3 group has a tendency to push electrons. Thus the group will acquire a partial +ve charge. So the C-CH3 bond will be a polar bond.
• The dipole moment in this bond is indicated by the yellow crossed arrow.
• We have discussed about polarity and dipole moment in an earlier section [see fig.4.65 of section 4.10]
3. Note that, the isomer in fig.13.59(a) is the cis isomer. Both the CH3 groups are on the same side. Also, both H atoms are on the same side.
♦ The dipole moment in the left C-CH3 bond acts towards bottom right.
♦ The dipole moment in the right C-CH3 bond acts towards bottom left.
• So the two moments add up to give a net dipole moment for the whole molecule.
4. Note that, the isomer in fig.13.59(b) is the trans isomer. The CH3 groups are on the opposite sides.
♦ The dipole moment in the bottom C-CH3 bond acts towards top left.
♦ The dipole moment in the top C-Cl bond acts towards bottom right.
•
So the two moments are acting in the opposite directions. They cancel
each other to give zero net dipole moment for the whole molecule.
Boiling points and melting points of cis/trans isomers
This can be written in 2 steps:
1. Based on the above two examples, we can write:
• cis isomers will be polar, while trans isomers will be non-polar.
• Due to the polar nature, the inter molecular attractions will be greater in the case of cis isomers.
• So cis isomers will have higher boiling points than trans isomers.
2 . Since the inter molecular attractions are greater, we would expect the melting points also to be higher for cis isomers. But the truth is that, solid cis isomers have lower boiling points than trans isomers. The reason can be written in 3 steps:
(i) cis isomers do not have symmetry. Similar atoms or groups occupy the same sides. So they do not pack well in the solid structure.
(ii) Trans isomers have greater symmetry. The atoms and groups are distributed uniformly. So they pack well in the solid structure.
(iii) Due to the better packing, inter molecular attractions will be greater in trans isomer solids. So they will have a higher melting point than cis isomer solids.
• In the above discussions, we were dealing with alkenes having two different species.
♦ Those two species were denoted by the letters X and Y.
♦ So the alkenes showing cis/trans isomerism were denoted as XYC=CYX
• This type of isomerism can be shown by alkenes having three different species also.
♦ We can denote those three species by the letters X, Y and Z.
♦ So such alkenes can be denoted as: XYC=CXZ.
♦ We will learn about their cis/trans isomers in higher classes.
• This type of isomerism can be shown by alkenes having four different species also.
♦ We can denote those four species by the letters X, Y, Z and W.
♦ So such alkenes can be denoted as: XYC=CZW.
♦ We will learn about their cis/trans isomers in higher classes.
Now we will see some solved examples
Solved example 13.10
Draw the cis and trans isomers of the following compounds. Also write their IUPAC names.
(i) CHCl=CHCl
(ii) C2H5CCH3=CCH3C2H5
Solution:
Part (i):
Fig.13.60 below shows the isomers and their IUPAC names:
 |
| Fig.13.60 |
Part (ii):
• We are given the condensed formula. Fig.13.61(a) below shows the expanded formula. This expanded formula helps us to give proper numbers to the C atoms.
The two isomers and their IUPAC names are shown in figs. (b) and (c).
 |
| Fig.13.61 |
Solved example 13.11
Which of the following compounds will show cis/trans isomerism?
(i) (CH3)2C=CH-C2H5
(ii) CH2=CBr2
(iii) C6H5CH=CH-CH3
(iv) CH3CH=CClCH3
Solution:
Part (i):
1. Figs.13.62(a) and (b) below shows the two possible arrangements.
 |
| Fig.13.62 |
2. They are not cis/trans isomers. This is because, if we rotate the molecule in (b) as a whole as indicated by the two yellow curved arrows, we will get the molecule in (a).
3. So this compound will not show cis/trans isomerism.
4. We can write:
Consider the left C atom involved in the double bond. Two identical groups (CH3 groups) are attached to this C atom. So the compound cannot show cis/trans isomerism.
Part (ii):
1. In this case, it is not even possible to draw different arrangements.
2. This is because,
♦ The left C atom has two identical groups.
♦ The right C atom also has two identical groups.
3. We can write:
• Consider the left C atom involved in the double bond.
♦ Two identical
atoms (H atoms) are attached to this C atom.
• Consider the right C atom involved in the double bond.
♦ Two identical
atoms (Br atoms) are attached to this C atom.
• A compound cannot
show cis/trans isomerism if even one of the C atoms involved in the double bond, has identical atoms or groups attached to it. So this compound cannot show cis/trans isomerism.
Part (iii):
◼ We can write:
• Consider the left C atom involved in the double bond.
♦ Two different
atoms/groups are attached to this C atom.
♦ They are: C6H5 group and H atom.
• Consider the right C atom involved in the double bond.
♦ Two different
atoms/groups are attached to this C atom.
♦ They are: CH3 group and H atom
• Different atoms/groups are attached to both the C atoms involved in the
double bond. So this
compound will show cis/trans isomerism.
Part (iv):
◼ We can write:
• Consider the left C atom involved in the double bond.
♦ Two different
atoms/groups are attached to this C atom.
♦ They are: CH3 group and H atom.
• Consider the right C atom involved in the double bond.
♦ Two different
atoms/groups are attached to this C atom.
♦ They are: CH3 group and Cl atom
• Different atoms/groups are attached to both the C atoms involved in the
double bond. So this
compound will show cis/trans isomerism.
In the next section we will see preparation of alkenes.
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