Chapter 13.17 - Aromatic Hydrocarbons

In the previous section, we completed a discussion on alkynes. In this section, we will see aromatic hydrocarbons.

Three basic features of aromatic hydrocarbons are written below:
1. Aromatic hydrocarbons derived their name from the Greek word ‘aroma’ which means ‘pleasant smelling’.
• Another name for aromatic hydrocarbons is Arenes.
2. Recall that, we have seen some details about the benzene ring in the previous chapter [see fig.12.89 in section 12.14].
• Most of the aromatic compounds contain one or more benzene rings. We know that benzene contains double bonds. That means, benzene is an unsaturated hydrocarbon.
• We also know that, unsaturated hydrocarbons are easily attacked by other reagents. This is because, the loosely held π-electrons are readily available for the attacking reagents. So during the reaction, the double or triple bonds will be converted into single bonds.
• But in the case of aromatic hydrocarbons, the benzene ring is retained even after the reaction. That means, the π-electrons of the benzene ring are not easily available for the attacking reagents. We will see the reason in the next section.
3. Most aromatic hydrocarbons contain one or more benzene rings. But there are a few which do not contain any benzene rings.
• Aromatic compounds which contain benzene rings are known as benzenoids.  
• Aromatic compounds which do not contain benzene rings are known as non-benzenoids.

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Nomenclature and Isomerism

• We have seen the details about the nomenclature and isomerism of aromatic hydrocarbons in the previous chapter [see section 12.8].
• A few more details are written in 2 steps below:
1. All the six H atoms in benzene are equivalent. So if we want a monosubstituted benzene molecule, we will get only one type.
• ‘Monosubstituted’ means:
    ♦ one H atom is removed
    ♦ a group like -CH₃ or -OH takes it’s place.
• We can remove any one of the six H atoms. Then we can put a suitable group in it’s place. We will always get the same product. This is because, all six H atoms are equivalent.
• We have already seen the explanation in the previous chapter. There we wrote:
When there is only one branch, the position of the C atom is not important [see the animation in fig.12.55 in section 12.8].
• A similar example is shown in fig.13.96 below.

Fig.13.96

    ♦ Consider the structure II
    ♦ Imagine an axis such that:
        ✰ It is perpendicular to the plane of the paper.
        ✰ It passes through the center of II.
    ♦ If we rotate II about that axis, we will get I.
    ♦ Both the structures in the fig., represent toluene.
• So we can write:
    ♦ If the benzene ring is monosubstituted, six different arrangements are possible.
    ♦ But all those six arrangements will represent the same molecule.
• We explained the 'equality of the six structures' by using the concept of rotation. However, a more scientific explanation can be given by using the concept of resonance. Due to resonance in benzene, there will not be any difference between single bonds and double bonds. All six bonds are identical. So all the six H atoms are in identical positions. We will see more details about resonance in the next section. 
2. In the case of disubstituted benzene ring, five different arrangements are possible. But there will be only three different molecules. This can be explained in 4 steps:
(i) Fig.13.97(a) below shows two arrangements.

Fig.13.97

• In I, the substituents are in 1, 2 positions.
    ♦ The substituents are at the two ends of a double bond.
• In II, the substituents are in 1, 6 positions.
    ♦ The substituents are at the two ends of a single bond.
• But thanks to resonance, there is no difference between single bonds and double bonds in benzene. All bonds are equivalent. So, the substituents are in identical positions. Consequently, the two structures are the same.
• So if the substituents are in 1, 2 or 1, 6, we give a common name: ortho. It is abbreviated as: o-.
• So we can write:
    ♦ Both structures in fig.13.97(a) represent the same molecule.
    ♦ It’s common name is o-xylene.
• When we use the word ortho, what really matters is: The substituents are attached to adjacent C atoms.
    ♦ It does not matter whether it is 1,2 or 1,6.
• Based on the rules that we saw in the previous chapter, we can write the IUPAC name of this o-xylene. It is 1,2-Dimethylbenzene.
(ii) Fig.13.97(b) above shows two arrangements.
• In III, the substituents are in 1, 3 positions.
• In II, the substituents are in 1, 5 positions.
• Comparing I and II, we see that, the arrangement of single and double bonds between the two substituents is different.
• But thanks to resonance, there is no difference between single bonds and double bonds in benzene. All bonds are equivalent. So, the substituents are in identical positions. Consequently, the two structures are the same.
• So if the substituents are in 1, 3 or 1, 5, we give a common name: meta. It is abbreviated as: m-.
• So we can write:
    ♦ Both structures in fig.13.97(b) represent the same molecule.
    ♦ It’s common name is m-xylene.
• When we use the word meta, what really matters is: The substituents are attached to two C atoms with a C atom in between.
    ♦ It does not matter whether it is 1,3 or 1,5.
• Based on the rules that we saw in the previous chapter, we can write the IUPAC name of this o-xylene. It is 1,3-Dimethylbenzene.
(iii) Fig.13.97(c) above shows the last of the five arrangements.
• In V, the substituents are in 1, 4 positions.
• In this case, we give a common name: para. It is abbreviated as: p-.
• So we can write:
    ♦ The common name of the structure is p-xylene.
• When we use the word para, what really matters is: The substituents are attached to two C atoms with two C atoms in between.
• Based on the rules that we saw in the previous chapter, we can write the IUPAC name of this p-xylene. It is 1,4-Dimethylbenzene.
(iv) Consider the above three molecules again:
    ♦ 1,2-Dimethylbenzene
    ♦ 1,3-Dimethylbenzene
    ♦ 1,4-Dimethylbenzene
• It is clear that, the three are position isomers.

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Structure of Benzene

Some basics about the structure can be written in 5 steps:
1. Benzene was first isolated by Michael Faraday in 1825. At that time, it’s structural details were not known.
2. We have seen the methods that scientists use to determine molecular formula of unknown compounds [see section 12.22]. Through those methods, the molecular formula of benzene was found to be C₆H₆.
• Six H atoms will not be able to satisfy the valencies of six C atoms through single bonds. So scientists assumed that, double or triple bonds are present in a benzene molecule.
3. To confirm the presence of double or triple bonds, bromine solution was added.
• We have seen test for unsaturation in previous sections [see fig.13.89 in section 13.15].
• We would expect the double or triple bonds to break and add Br atoms. This is shown in fig.13.98(a) below.

Fig.13.98

• But surprisingly, benzene was unreactive to bromine. No such product was obtained.
• Scientists came to this conclusion:
Though double or triple bonds are present, benzene has a stable structure.
4. Later scientists forced benzene to react with bromine. They used a catalyst for this purpose.
• The actual result obtained was unexpected. The double bond did not break. One H atom was removed and a Br atom was substituted for that H atom. This is shown in fig.13.98(b) above.
• We see that, by preferring substitution reaction, the benzene molecule is able to retain the double bonds.
5. After years of research, the German scientist August Kekule proposed the familiar structure (with alternate single and double bonds) that we see today in our basic courses in organic chemistry.
• But this structure was not able to explain why the double bond does not add bromine atoms.
• In other words, the Kekule structure was not able to explain the unusual stability of benzene.

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Scientists were able to give a satisfactory explanation for the stability of benzene by using the concept of resonance. We will see it in the next section.

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Chapter 13.9 - Geometrical Isomerism In Alkenes

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 C_nH_2n 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:

Fig.13.55

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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:

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’

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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:

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:

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.

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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.

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• 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.

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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.

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In the next section we will see preparation of alkenes.

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Chapter 12.9 - Isomerism in Organic Compounds

In the previous section, we saw the nomenclature of substituted Benzene compounds. In this section, we will see isomerism.

We have seen the basics about isomerism in our earlier chemistry classes. Details here. The reader is advised to study those notes thoroughly. In our present discussion, we will see some advanced examples.

Chain isomerism

This can be explained in 3 steps:
1. Consider the three compounds in fig.12.62 below.
• Each of them have five C atoms. Also each of them have twelve H atoms. So they have the same molecular formula: C₅H₁₂

Fig.12.62

2. Even though they have the same molecular formula, they are different compounds.
• This is because, the carbon chains are different.
3. The phenomenon of two or more compounds having the same molecular formula but different carbon chains is known as chain isomerism. Such compounds are known as chain isomers. 

Position isomerism

This can be explained in 3 steps:
1. Consider the two compounds in fig.12.63 below.
• Both of them have three C atoms, eight H atoms and one O atom. So they have the same molecular formula: C₃H₈O

Fig.12.63

• Also note that, they have the same functional group, which is: ㅡOH
2. Even though they have the same molecular formula, they are different compounds.
• This is because, the position of the ㅡOH functional group is different.
3. The phenomenon of two or more compounds having the same molecular formula but different positions of functional groups in the carbon chains is known as position isomerism. Such compounds are known as position isomers. 

Functional group isomerism

This can be explained in steps:
1. Consider the two compounds in fig.12.64 below.
• Both of them have three C atoms, six H atoms and one O atom. So they have the same molecular formula: C₃H₆O

Fig.12.64

• Note that, they have different functional groups:
   ♦ The first compound has the functional group: >C=O
   ♦ The second compound has the functional group: ㅡCHO
2. Even though they have the same molecular formula, they are different compounds.
• This is because, the functional groups are different.
3. The phenomenon of two or more compounds having the same molecular formula but different functional groups is known as functional group isomerism. Such compounds are known as functional isomers. 

Metamerism

This can be explained in 3 steps:
1. Consider the two compounds in fig.12.65 below.
• Both of them have four C atoms, ten H atoms and one O atom. So they have the same molecular formula: C₄H₁₀O

Fig.12.65

• Note that, they have the same functional group: ㅡOㅡ
2. Even though they have the same molecular formula, they are different compounds.
• This is because, the alkyl chains on either sides of ㅡOㅡ are different.
3. The phenomenon of two or more compounds having the same molecular formula but different alkyl chains on either sides of the functional group is known as metamerism. Such compounds are known as metamers.

Stereoisomerism

This can be explained in 3 steps:
1. Consider two hydrocarbons having the same molecular formula.
• They have the same number of double bonds also.
• The positions of the double bonds are also same.
• So they appear to be same in all respects.
2. But in reality, the two compounds are different because, relative positions of atoms are different. We have seen such a case in the animation in fig.4.141 in section 4.25.
• This phenomenon is called stereoisomerism.
3. This type of isomerism can be further classified as geometrical isomerism and optical isomerism. We will see more details in later sections.

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Based on the above discussion, we can prepare a chart showing how the phenomenon of isomerism is classified into various types. It is shown in fig.12.66 below:

Fig.12.66

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• The link below gives the folder containing additional solved examples on this chapter.
• Part 1 is related to nomenclature and isomerism.

Additional solved examples

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In the next section, we will see the fundamental concepts in organic reaction mechanism.


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