Chapter 13.14 - Nomenclature and Isomerism in Alkynes

In the previous section, we completed a discussion on ozonolysis and polymerisation of alkenes. In this section, we will see alkynes.

Let us recall some properties of alkynes that we have seen in earlier chapters. They can be written in 3 steps:
1. Alkynes are unsaturated hydrocarbons. They contain at least one triple bond.
2. If there is one triple bond in an alkyne,
   ♦ it will contain four H atoms less than the corresponding alkane.
   ♦ it will contain two H atoms less than the corresponding alkene.
• For example:
    ♦ Molecular formula of butane is C₄H₁₀
    ♦ Molecular formula of butene is C₄H₈
    ♦ Molecular formula of butyne is C₄H₆
• We know that, the general formula for alkenes is C_nH_2n
• We also know that, the corresponding alkyne has two H atoms less. So the general formula for alkenes is C_nH_2n-2
3. Ethyne is the IUPAC name of the first member. It’s common name is acetylene.
• Acetylene is used for arc welding. In this process, a mixture of acetylene and oxygen is subjected to combustion. As a result, a flame with high temperature and heat is produced. This flame can be used for fusing metal parts together.

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Structure of triple bond

Alkynes have at least one triple bond. We have seen the details about that triple bond in the previous chapters. [see fig.4.156 of section 4.27 ] Let us recall those details. They can be written in 6 steps:
1. The triple bond consists of:
    ♦ One sigma (σ) bond
    ♦ Two pi (𝜋) bonds.
2. The sigma bond is formed by the head-on overlapping of the sp hybridized orbitals.
    ♦ The pi bonds are formed by the sideways overlapping of the 2p orbitals.
3. The sigma bond is a strong bond.
    ♦ Bond enthalpy of a sigma bond is 397 kJ mol^-1
• Pi bond is a weak bond.
    ♦ Bond enthalpy of a pi bond is 284 kJ mol^-1
4. The triple bond is shorter in bond length.
    ♦ The C-C single bond in alkanes has a bond length of 154 pm.
    ♦ The C=C double bond in alkenes has a bond length of 134 pm.
    ♦ The C☰C triple bond in alkynes has a bond length of 120 pm.

5. The triple bond has greater strength.
    ♦ The C-C single bond has a bond enthalpy of 348 kJ mol^-1.
    ♦ The C=C double bond has a bond enthalpy of 681 kJ mol^-1.
    ♦ The C☰C triple bond has a bond enthalpy of 823 kJ mol^-1.
6. We have seen that, ethyne has a linear structure. The electron clouds between the two C atoms are cylindrically symmetrical around the internuclear axis.

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Nomenclature of Alkynes

• We have seen the rules for writing the IUPAC names in an earlier chapter [see section 12.3]
• Let us recall some basic rules which are applicable to alkynes. It can be written in 3 steps:
1. The longest chain should be selected in such a way that, it contains the triple bonds.
2. Numbering must be done in such a way that, the triple bonds get the lowest possible numbers.
3. The suffix ‘yne’ is used instead of ‘ane’ of alkanes.
• Let us see some examples:

Example 1:
Write the IUPAC name of the structure: CH₃-C☰CH
Solution:
1. Applying rule 1, we see that, there are three C atoms in the main chain.
2. Applying rule 2, we see that, ‘prop’ must be used.
3. Applying rule 3, we get propane.
• But since it is an alkyne, we must remove ‘ane’ and replace it with ‘yne’.
• We get: propane – ane + yne = propyne
4. Applying rule 4, we see that, there are no branches.
5. Applying rule 5, we see that the correct way of numbering is from right to left.
    ♦ Numbering from right to left will give number ‘1’ to the triple bond.
    ♦ Numbering from left to right will give number ‘2’ to the triple bond.
6. Rules 6 and 7 are not applicable here because, there are no branches.
• So we can apply rule 8 to assemble the IUPAC name. We get:
Prop-1-yne.
7. Here we note an interesting point:
• In propyne, the triple bond has only two possible positions. Both those positions will give the same name: Prop-1-yne. So we can write the name simply as: Propyne.

Example 2:
Write the IUPAC name of the structure: CH₃-CH₂-C☰CH
Solution:
1. Applying rule 1, we see that, there are four C atoms in the main chain.
2. Applying rule 2, we see that, ‘but’ must be used.
3. Applying rule 3, we get butane.
• But since it is an alkyne, we must remove ‘ane’ and replace it with ‘yne’.
• We get: butane – ane + yne = butyne
4. Applying rule 4, we see that, there are no branches.
5. Applying rule 5, we see that the correct way of numbering is from right to left.
    ♦ Numbering from right to left will give number ‘1’ to the triple bond.
    ♦ Numbering from left to right will give number ‘3’ to the triple bond.
6. Rules 6 and 7 are not applicable here because, there are no branches.
• So we can apply rule 8 to assemble the IUPAC name. We get:
But-1-yne.

Example 3:
Write the IUPAC name of the structure: CH₃-CH☰CH-CH₃
Solution:
1. Applying rule 1, we see that, there are four C atoms in the main chain.
2. Applying rule 2, we see that, ‘but’ must be used.
3. Applying rule 3, we get butane.
• But since it is an alkyne, we must remove ‘ane’ and replace it with ‘yne’.
• We get: butane – ane + yne = butyne
4. Applying rule 4, we see that, there are no branches.
5. Applying rule 5, we see that numbering can be done either from left to right or from right to left.
    ♦ Numbering from left to right will give number ‘2’ to the triple bond.
    ♦ Numbering from right to left will give number ‘2’ to the triple bond.
6. Rules 6 and 7 are not applicable here because, there are no branches.
• So we can apply rule 8 to assemble the IUPAC name. We get:
But-2-yne.

Example 4:
Write the IUPAC name of the structure: CH☰C-C☰CH
Solution:
1. Applying rule 1, we see that, there are four C atoms in the main chain.
2. Applying rule 2, we see that, ‘but’ must be used.
3. Applying rule 3, we get butane.
• But since it is an alkyne, we must remove ‘ane’ and replace it with ‘yne’.
• We get: butane – ane + yne = butyne
4. Applying rule 4, we see that, there are no branches.
5. Applying rule 5, we see that numbering can be done either from left to right or from right to left.
    ♦ Numbering from left to right will give numbers '1,3' to the triple bonds.
    ♦ Numbering from right to left will give number ‘1,3’ to the triple bonds.
6. Rules 6 and 7 are not applicable here because, there are no branches.
• So we can apply rule 8 to assemble the IUPAC name. We get:
Buta-1,3-diyne.
• Note that, since more than one triple bond is present, we write 'buta' instead of 'but'
• Similarly, 'penta' is written instead of 'pent', 'hexa' is written instead of hex', so on . . .

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Isomerism in Alkynes

Some basics can be written in 6 steps:
1. Consider the three structures in fig.13.86 below:

Fig.13.86

   ♦ Structure (a) is Pent-1-yne.
   ♦ Structure (b) is Pent-2-yne.
   ♦ Structure (c) is 3-Methylbut-1-yne.
2. Let us compare the number of atoms:
   ♦ In (a), there are five C atoms and eight H atoms.  
   ♦ In (b) also, there are five C atoms and eight H atoms.
   ♦ In (c) also, there are five C atoms and eight H atoms.
• So all the three structures have the same molecular formula C₅H₈
• It is clear that pent-1-yne, pent-2-yne and 3-Methylbut-1-yne 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 and alkenes.
• Based on those discussions, we can write:
   ♦ Structures (a) and (c) in fig.13.86, are chain isomers.     
   ♦ Structures (b) and (c) in fig.13.86, are chain isomers.
   ♦ Structures (a) and (b) in fig.13.86, 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 alkyne 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 alkynes (ethyne and propyne) do not exhibit isomerism.
• This is because, a minimum of four C atoms are required to obtain different structural arrangements.
5. We have seen that the first two members (ethyne and propyne) of the alkyne series do not exhibit isomerism.
• In examples 2 and 3 above, we have seen the isomers of the third member butyne.
• In fig.13.86 above, we have seen the isomers of the fourth member pentyne. Next let us see the isomers of the fifth member.
• It is explained as as solved example below:

Solved example 13.13
Write all the structures and IUPAC names of the structural isomers of alkynes corresponding to C₆H₁₀.
Solution:
Seven isomers are shown in fig.13.87 below:

Fig.13.87

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In the next section we will see preparation of alkynes.

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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 13.2 - Problems in Nomenclature of Alkanes

In the previous section, we saw the nomenclature and isomerism in alkanes. In this section, we will see some advanced examples.

Example 1:
Write the IUPAC name of the structure shown in fig.13.13 below:

Fig.13.13

Solution:
• We have seen the 8 rules in section 12.3
1. Applying rule 1, we see that, there are six C atoms in the main chain.
2. Applying rule 2, we see that, 'hex' must be used.
3. Applying rule 3, we get hexane.
4. Applying rule 4, we see that, there are two branches: methyl and ethyl
5. Applying rule 5, we see that, the correct way of numbering is indeed from left to right as shown in fig.13.13(a).
• If we number the C atoms from left to right as in fig.b, the branches will get the numbers 2,4. The sum is (2+4) = 6.
• If we number the C atoms from right to left, the branches will get the numbers 3,5. The sum is (3+5) = 8, which is larger and hence not acceptable.
6. Applying rule 6, we get: 2-methyl and 4-ethyl
7. Applying rule 7, we get: 4-ethyl-2-methyl
• This is because, 'e' comes before 'm' in the alphabetical listing.
8. Applying rule 8, we get: 4-Ethyl-2-methylhexane.

Example 2:
Write the IUPAC name of the structure shown in fig.13.14 below:

Fig.13.14

Solution:
In this problem, there are branches within a branch. So we must be ready to apply the rules that we saw in section 12.4 also. Fortunately, there is a common name available for that branch. So we can first name the main carbon chain.
1. Applying rule 1, we see that, there are eight C atoms in the main chain.
2. Applying rule 2, we see that, 'oct' must be used.
3. Applying rule 3, we get octane.
4. Applying rule 4, we see that, there are four branches: two ethyl branches, one methyl branch and one isopropyl branch.
5. Applying rule 5, we see that, the correct way of numbering is from right to left as shown in fig.13.14(b).
• If we number the C atoms from right to left, the branches will get the numbers 3,3,4,5. The sum is (3+3+4+5) = 15.
• If we number the C atoms from left to right (fig.a), the branches will get the numbers 4,5,6,6. The sum is (4+5+6+6) = 21, which is larger and hence not acceptable.
6. Applying rule 6, we get: 3-ethyl, 3-ethyl, 4-methyl and 5-isopropyl
7. Applying rule 7, 7A and 7B, we get: 3,3-diethyl-5-isopropyl-4-methyl
• This can be explained in 2 steps:
(i) 'e' comes before 'i' and 'm' in the alphabetical listing.
'di' is not considered as part of the name. So 'd' should not be considered for alphabetical listing.
(ii) 'i' comes before 'm' in the alphabetical listing.
'iso' is considered as part of the name.
8. Applying rule 8, we get: 3,3-diethyl-5-isopropyl-4-methyloctane.

Example 3:
Write the IUPAC name of the structure shown in fig.13.15 below:

Fig.13.15

Solution:
In this problem, there are branches within a branches. So we must be ready to apply the rules that we saw in section 12.4 also. Fortunately, there are common names available for those branches. So we can first name the main carbon chain.
1. Applying rule 1, we see that, there are ten C atoms in the main chain.
2. Applying rule 2, we see that, 'dec' must be used.
3. Applying rule 3, we get decane.
4. Applying rule 4, we see that, there are two branches: one isopropyl branch and and one sec-butyl branch.
5. Applying rule 5, we see that, the correct way of numbering is from left to right as shown in fig.13.15(a).
• If we number the C atoms from left to right, the branches will get the numbers 4,5. The sum is (4+5) = 9.
• If we number the C atoms from right to left (fig.b), the branches will get the numbers 6,7. The sum is (6+7) = 13, which is larger and hence not acceptable.
6. Applying rule 6, we get: 4-isopropyl and 5-sec-butyl
7. Applying rule 7, 7A and 7B, we get: 5-sec-butyl-4-isopropyl
• This can be explained in 3 steps:
(i) 'b' comes before 'i' in the alphabetical listing.
(ii) 'sec' is not considered as part of the name. So 's' should not be considered for alphabetical listing.
(iii) 'iso' is considered as part of the name.
8. Applying rule 8, we get: 5-sec-butyl-4-isopropyl decane.

Example 4:
Write the IUPAC name of the structure shown in fig.13.16 below:

Fig.13.16

Solution:
In this problem, there are branches within a branch. So we must be ready to apply the rules that we saw in section 12.4 also. The branch do not have any common name. So we first name the branch.
1. Applying rule 1, we see that, there are three C atoms.
2. Applying rule 2, we see that, 'prop' must be used.
3. Applying rule 3, we get propane.
4. Applying rule 4, we see that, there are two branches: two methyl groups.
5. Applying rule 5, we see that, the correct way of numbering is from the main branch as shown in fig.13.16(a).
6. Applying rule 6, we get: 2-methyl and 2-methyl
7. Applying rule 7, we get: 2,2-dimethyl
8. Applying rule 8, we get: 2,2-dimethylpropyl.
• This name must be written within parenthesis.

Now we can name the main chain.
1. Applying rule 1, we see that, there are nine C atoms.
2. Applying rule 2, we see that, 'non' must be used.
3. Applying rule 3, we get nonane.
4. Applying rule 4, we see that, there is only one branch.
5. Applying rule 5, we see that, numbering can be done in both ways. Both will give the number 5.
6. Applying rule 6 we get: 5-(2,2-dimethylpropyl)
7. Applying rule 7 is not required because there is only one branch.
8. Applying rule 8, we get: 5-(2,2-dimethylpropyl)nonane.

Example 5:
Write the IUPAC name of the structure shown in fig.13.17 below:

Fig.13.17

Solution:
• We have seen the 8 rules in section 12.3
1. Applying rule 1, we see that, there are seven C atoms in the main chain.
2. Applying rule 2, we see that, 'hept' must be used.
3. Applying rule 3, we get heptane.
4. Applying rule 4, we see that, there are two branches: methyl and ethyl
5. Applying rule 5:
   ♦ We see that, the numbering can be done in both ways.
   ♦ In both fig. a and b, the numbers are 3 and 5.
• So we apply rule 5A:
If two branches are present in equivalent positions, then the lower number should be given to that branch which comes first in the alphabetical order.
   ♦ Here, ethyl comes first in the alphabetical order.
   ♦ Numbering in fig.a is correct.
6. Applying rule 6, we get: 3-ethyl and 5-methyl
7. Applying rule 7, we get: 3-ethyl-5-methyl
• This is because, 'e' comes before 'm' in the alphabetical listing.
8. Applying rule 8, we get: 3-Ethyl-5-methylheptane.

Solved example 13.3
Write the IUPAC names of the following compounds:
(i) (CH₃)₃CCH₂C(CH₃)₃
(ii) (CH₃)₂C(C₂H₅)₂
(iii) tetra-tert-butylmethane
Solution:
Part (i):
• We are given the condensed formula. Based on the condensed formula, we can draw the structural formula. It is shown in fig.13.18 below:

Fig.13.18

• We have seen the 8 rules in section 12.3
1. Applying rule 1, we see that, there are five C atoms in the main chain.
2. Applying rule 2, we see that, 'pent' must be used.
3. Applying rule 3, we get pentane.
4. Applying rule 4, we see that, there are four branches: four methyl branches.
5. Applying rule 5:
   ♦ We see that, the numbering can be done in both ways.
   ♦ In both fig. a and b, the numbers are 2 and 4.
6. Applying rule 6, we get: 2-methy, 2-methyl, 4-methy, 4-methyl
7. Applying rule 7, we get: 2,2,4,4-tetramethyl
8. Applying rule 8, we get: 2,2,4,4-Tetramethylpentane.

Part (ii):
• We are given the condensed formula. Based on the condensed formula, we can draw the structural formula. It is shown in fig.13.19 below:

Fig.13.19

• We have seen the 8 rules in section 12.3
1. Applying rule 1, we see that, there are five C atoms in the main chain.
2. Applying rule 2, we see that, 'pent' must be used.
3. Applying rule 3, we get pentane.
4. Applying rule 4, we see that, there are two branches: two methyl branches.
5. Applying rule 5:
   ♦ We see that, the numbering can be done in both ways.
   ♦ In both fig. a and b, the number is 3.
6. Applying rule 6, we get: 3-methy, 3-methyl
7. Applying rule 7, we get: 2,2-dimethyl
8. Applying rule 8, we get: 2,2-Dimethylpentane.

Part (iii):
• We have seen the 8 rules in section 12.3
• In this problem, there are branches within a branch. So we must be ready to apply the rules that we saw in section 12.4 also.
• We are given the common name tetra-tert-butylmethane.
   ♦ This is similar to the common name: tetra-chloromethane.
• In tetrachloromethane, the four H atoms of methane are replaced by four Cl atoms.
• In the same way, in tetra-tert-butylmethane, the four H atoms of methane are replaced by four tert-butyl groups.
• We saw the structure of tert-butyl in an earlier section [see fig.12.33 in section section 12.4]. It is shown again in fig.13.20(a) below:

Fig.13.20

• Based on the structure of tert-butyl, the structure of tetra-tert-butylmethane will be as shown in fig.b. Now we can write the IUPAC name.
1. Applying rule 1, we see that, there are five C atoms in the main chain.
2. Applying rule 2, we see that, 'pent' must be used.
3. Applying rule 3, we get pentane.
4. Applying rule 4, we see that, there are six branches: four methyl branches and two tert-butyl branches.
5. Applying rule 5:
   ♦ We see that, the numbering can be done in two ways.
   ♦ In both fig. b and c, the numbers are 2, 3 and 4.
6. Applying rule 6, we get: 2-methy, 2-methyl, 4-methy, 4-methyl, 3-tert-butyl, 3-tert-butyl
7. Applying rule 7, we get: 3,3-di-tert-butyl-2,2,4,4-tetramethyl
• Remember that, 'tetra' is not considered as part of name. So 't' cannot be considered in the alphabetical listing.
8. Applying rule 8, we get: 3,3-Di-tert-butyl-2,2,4,4-tetramethylpentane.

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• In the above discussion, we were given the structures of hydrocarbons. We wrote the corresponding IUPAC names.
• We must be able to do the reverse also. That is., we will be given the IUPAC names. We must draw the corresponding structures.
• Let us see an example:
Draw the structure of 3-Ethyl-2,2-dimethylpentane
Solution:
1. In the given name, we have ‘pent’ as the root. Then there will be five C atoms in the longest chain. So we first draw a chain of five C atoms. This is shown in fig.13.21(a) below:

Fig.13.21

2. Next we number the C atoms from 1 to 5. This is shown in fig.b
3. Attaching the branches:
‘3-Ethyl’ indicates that, there is an ethyl group at the C atom number 3
So we draw an ethyl group at that C atom.
‘2,2-dimethyl’ indicates that, there are two methyl groups at the C atom number 2
So we draw two methyl groups at that C atom
This is shown in fig.c
4. Now we put the required number of H atoms to satisfy the valencies of the C atoms.
Thus we get the final structure shown in fig.d

Solved example 13.4
Write the structural formulas for the following compounds:
(i) 3,4,4,5-Tetramethylheptane
(ii) 2,5-dimethylhexane
Solution:
Part (i):
1. In the given name, we have ‘hept’ as the root. Then there will be seven C atoms in the longest chain. So we first draw a chain of seven C atoms. This is shown in fig.13.22(a) below:

Fig.13.22

2. Next we number the C atoms from 1 to 7. This is shown in fig.b
3. Attaching the branches:
‘3,4,4,5-tetramethyl’ indicates that, there are four methyl groups at the three C atoms as written below:
   ♦ One methyl group at C atom number 3
   ♦ Two methyl groups at C atom number 4
   ♦ One methyl group at C atom number 5
• This is shown in fig.c
4. Now we put the required number of H atoms to satisfy the valencies of the C atoms.
• Thus we get the final structure shown in fig.d

Part (ii):
1. In the given name, we have ‘hex’ as the root. Then there will be six C atoms in the longest chain. So we first draw a chain of six C atoms. This is shown in fig.13.23(a) below.

Fig.13.23

2. Next we number the C atoms from 1 to 6. This is shown in fig.b
3. Attaching the branches:
‘2,5-dimethyl’ indicates that, there are two methyl groups at the two C atoms as written below:
   ♦ One methyl group at C atom number 2
   ♦ One methyl group at C atom number 5
• This is shown in fig.c
4. Now we put the required number of H atoms to satisfy the valencies of the C atoms.
• Thus we get the final structure shown in fig.d

Solved example 13.5
Write structures for each of the following compounds. Why are the given names incorrect. Write the correct IUPAC names.
(i) 2-Ethylpentane
(ii) 5-Ethyl-3-methylheptane
Solution:
Part (i):
1. In the given name, we have ‘pent’ as the root. Then there will be five C atoms in the longest chain. So we first draw a chain of five C atoms. This is shown in fig.13.24(a) below:

Fig.13.24

2. Next we number the C atoms from 1 to 5. This is shown in fig.b
3. Attaching the branches:
‘2-ethyl indicates that, there is an ethyl group at the C atom number 2.
• This is shown in fig.c
4. Now we put the required number of H atoms to satisfy the valencies of the C atoms.
• Thus we get the final structure shown in fig.d
5. For this structure, the name 3-Ethylpentane is wrong. The correct name can be obtained in 4 steps:
(i) For this structure, the C atoms should be numbered as shown in fig.e
(ii) Now we see that, there are six C atoms in the longest chain.
(iii) We also see that, the branch is methyl, not ethyl. The branch is at the third C atom.
(iv) So the correct IUPAC name is: 3-Methylhexane.

Part (ii):
1. In the given name, we have ‘hept’ as the root. Then there will be seven C atoms in the longest chain. So we first draw a chain of seven C atoms. This is shown in fig.13.25(a) below:

Fig.13.25

2. Next we number the C atoms from 1 to 7. This is shown in fig.b
3. Attaching the branches:
   ♦ ‘5-ethyl' indicates that, there is an ethyl group at the C atom number 5.
   ♦ ‘3-methyl' indicates that, there is a methyl group at the C atom number 3.
• This is shown in fig.c
4. Now we put the required number of H atoms to satisfy the valencies of the C atoms.
• Thus we get the final structure shown in fig.d
5. For this structure, the name 5-Ethyl-3-methylheptane is wrong. The correct name can be obtained in 4 steps:
(i) For this structure, the C atoms can be numbered as shown in fig.e also
(ii) Both methods of numbering will give the same numbers 3 and 5.
• That means, the branches are at equivalent positions.
(iii) So we must apply rule 5A. [see section 12.3]
Then 'ethyl' gets the lower number because, it comes first in the alphabetical order.
(iv) So the numbering in fig.e is the correct method.
• Based on this numbering, the correct IUPAC name is: 3-Ethyl-5-methylheptane.

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In the next section we will see preparation of alkanes.

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Chapter 13.1 - Nomenclature And Isomerism In Alkanes

In the previous section, we saw the tetrahedral structure of alkanes. In this section, we will see the nomenclature and isomerism in alkanes.

Some basics can be written in 16 steps:
1. Consider the two structures in fig.13.6 below:

Fig.13.6

• Based on the rules of nomenclature [see section 12.3] that we saw in the previous chapter, we can write:
   ♦ The structure in fig.13.6(a) is butane.  
   ♦ The structure in fig.13.6(b) is 2-Methylpropane.
2. Let us compare the number of atoms:
   ♦ In fig.13.6(a), there are four C atoms and ten H atoms.  
   ♦ In fig.13.6(b) also, there are four C atoms and ten H atoms.
• So both structures have the same molecular formula C₄H₁₀
• It is clear that butane and 2-Methylpropane are isomers.
• We have discussed about isomers in a previous section. [see section 12.9]
3. Consider the three structures in fig.13.7 below:

Fig.13.7

• Based on the rules of nomenclature that we saw in the previous chapter, we can write:
   ♦ The structure in fig.13.7(a) is pentane.  
   ♦ The structure in fig.13.7(b) is 2-Methylbutane.
   ♦ The structure in fig.13.7(c) is 2,2-Dimethylpropane.
4. Let us compare the number of atoms:
   ♦ In fig.13.7(a), there are five C atoms and twelve H atoms.  
   ♦ In fig.13.7(b) also, there are five C atoms and twelve H atoms.
   ♦ In fig.13.7(c) also, there are five C atoms and twelve H atoms.
• So all the three structures have the same molecular formula C₅H₁₂
• It is clear that pentane, 2-Methylbutane and 2,2-Dimethylpropane are isomers.
5. Consider the two isomers in fig.13.6
   ♦ The first isomer has a boiling point of 273 K
   ♦ The second isomer has a boiling point of 261 K
• Though the two isomers have the same numbers of C and H atoms, they differ in their boiling points and other properties.
• The difference in properties is due to the difference in structures. So they are called structural isomers.
6. Consider the three isomers in fig.13.7
   ♦ The first isomer has a boiling point of 309 K
   ♦ The second isomer has a boiling point of 301 K
   ♦ The third isomer has a boiling point of 282.5 K
• Though the three isomers have the same numbers of C and H atoms, they differ in their boiling points and other properties.
• The difference in properties is due to the difference in structures. So they are also structural isomers.
7. Consider the two isomers in fig.13.6
   ♦ The first isomer has a continuous chain of C atoms.
   ♦ The second isomer has a branched chain of C atoms.
• Structural isomers which differ in the chain of C atoms are called chain isomers.
• We can write:
Butane and 2-Methylpropane are chain isomers of C₄H₁₀
8. Consider the three isomers in fig.13.7
   ♦ The first isomer has a continuous chain of C atoms.
   ♦ The second isomer has a branched chain of C atoms.
   ♦ The third isomer has a different branched chain of C atoms.
• We can write:
Pentane, 2-Methylbutane and 2,2-Dimethylpropane are chain isomers of C₅H1₂
9. In the case of alkanes, only those which have more than three C atoms will exhibit isomerism.
• The first three alkanes (methane, ethane and propane) do not exhibit isomerism.
• This is because, a minimum of four C atoms are required to obtain different structural arrangements.
10. We have seen that the first three members do not exhibit isomerism.
• We have seen the isomers of fourth and fifth members. Next let us see the isomers of the sixth member.
• The sixth member C₆H₁₄ have five chain isomers. They are shown in a condensed form in figs.13.8 (a), (b), (c), (d) and (e) below:

Fig.13.8

Here we can note an interesting point. It can be written in steps:
(i) Consider the structure shown on the right side.
• It appears that, an ethyl group is attached to a chain of four C atoms.
• So we are tempted to name it as 2-Ethylbutane.
(ii) But this name is wrong.
• Recall that, the longest chain is to be considered.
• So the correct name is 3-Methylpentane. This structure is already shown on the left side. It is one of the five isomers.
11. Now we will discuss about primary carbon atoms. It can be written in 5 steps:
(i) Consider the C atom numbered as ‘1’ in the n-Hexane in fig.13.8(a) above.
• This C atom is attached to:
   ♦ three H atoms.
   ♦ one C atom.
• Since there is only one C atom attached, the C atom numbered as ‘1’ is a primary carbon atom.    
(ii) Consider the C atom numbered as ‘6’ in the same n-Hexane.
• This C atom is attached to:
   ♦ three H atoms.
   ♦ one C atom.
• Since there is only one C atom attached, the C atom numbered as ‘6’ is a primary carbon atom.
(iii) Consider the C atom in the methyl group in the 2-Methylpentane in fig.13.8(b) above.
• This C atom is attached to:
   ♦ three H atoms.
   ♦ one C atom.
• Since there is only one C atom attached, the C atom in the methyl group is a primary carbon atom.
◼ In fact, terminal carbon atoms are always primary carbon atoms.
• Both C atoms in ethane are primary carbon atoms.
(iv) Consider the C atom in methane.
• This C atom is attached to:
   ♦ three H atoms.
   ♦ zero C atoms.
• Carbon atoms attached to no other carbon atoms are also called primary carbon atoms.
• So the single C atom in methane is a primary carbon atom.
(v) Primary carbon atom is abbreviated as 1^o carbon atom.
12. Now we will discuss about secondary carbon atoms. It can be written in 4 steps:
(i) Consider the C atom numbered as ‘2’ in the n-Hexane in fig.13.8(a) above.
• This C atom is attached to:
   ♦ two H atoms.
   ♦ two C atoms.
• Since there are two C atoms attached, the C atom numbered as ‘2’ is a secondary carbon atom.    
(ii) Consider the C atom numbered as ‘5’ in the same n-Hexane.
• This C atom is attached to:
   ♦ two H atoms.
   ♦ two C atoms.
• Since there are two C atoms attached, the C atom numbered as ‘5’ is a secondary carbon atom.   
(iii) Based on this information, we can write:
• The C atoms numbered as '3' and '4' in n-Hexane are also secondary carbon atoms. 
• The C atoms numbered as '3' and '4' in 2-Methylpentane (fig.13.8.b) are also secondary carbon atoms.
• The reader may try to identify the secondary carbon atoms in the other structures in fig.13.8
(iv) Secondary carbon atom is abbreviated as 2^o carbon atom.
13. Now we will discuss about tertiary carbon atoms. It can be written in 4 steps:
(i) Consider the C atom numbered as ‘2’ in the 2-Methylpentane in fig.13.8(b) above.
• This C atom is attached to:
   ♦ one H atom.
   ♦ three C atoms.
• Since there are three C atoms attached, the C atom numbered as ‘3’ is a tertiary carbon atom.    
(ii) Consider the C atom numbered as ‘3’ in the 3-Methylpentane in fig.13.8(c) above.
• This C atom is attached to:
   ♦ one H atom.
   ♦ three C atoms.
• Since there are three C atoms attached, the C atom numbered as ‘3’ is a tertiary carbon atom.  
(iii) Based on this information, we can write:
• The C atoms numbered as '2' and '3' in 2,3-Dimethylbutane (fig.13.8.d) are also tertiary carbon atoms.
(iv) Tertiary carbon atom is abbreviated as 3^o carbon atom.
14. Now we will discuss about quaternary carbon atoms. It can be written in 4 steps:
(i) Consider the C atom numbered as ‘2’ in the 2,2-Dimethylbutane in fig.13.8(e) above.
• This C atom is attached to:
   ♦ zero H atoms.
   ♦ four C atoms.
• Since there are four C atoms attached, the C atom numbered as ‘2’ is a quaternary carbon atom.    
(ii) Based on this information, we can write:
• The C atom numbered as '2' in the earlier fig.13.7.c is also quaternary carbon atom.
(iii) Quaternary carbon atom is abbreviated as 4^o carbon atom.
15. We have seen that the first three members of the alkane series do not exhibit isomerism.
• Isomers exist for members with more than 3 carbon atoms.
   ♦ We have seen the two isomers of the fourth member.
   ♦ We have seen the three isomers of the fifth member.
   ♦ We have seen the five isomers of the sixth member.
• It is clear that, number of isomers increase with the number of C atoms.
   ♦ The seventh member C₇H₁₆ has nine isomers.
   ♦ The tenth member C₁₀H₂₂ has seventy five isomers.
16. We have seen the importance of alkyl groups while naming an organic compound. [See fig.12.24 in section 12.3]
• Alkyl groups are obtained by removing one H atom from the corresponding alkanes. So they will have a general formula C_nH_2n+1.
• In fig.13.8 (b) above, a methyl group is attached to the carbon numbered '2'
   ♦ -CH₃ is methyl group.
   ♦ -CH₂-CH₃ is ethyl group. It can be written in condensed form as -C₂H₅.
   ♦ -CH₂-CH₂-CH₃ is propyl group. It can be written in condensed form as -C₃H₇.

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Now we will see a solved example

Solved example 13.2
Write structures of different isomeric alkyl groups corresponding to the molecular formula C₅H₁₁. Write IUPAC names of alcohols obtained by attachment of –OH groups at different carbons of the chain.
Solution:
1. The molecular formula C₅H₁₁ satisfies the general formula C_nH_2n+1
• So it indicates an alkyl group.
• The simplest alkyl group corresponding to C₅H₁₁ is the pentyl group. It is shown in fig.13.9(1.a) below:

Fig.13.9

• Any alkyl group will be unstable. It requires an electron to attain stability.
• If we attach an -OH group, it will become a stable molecule. We already know how the valencies are satisfied when the -OH group is attached. [See fig.14.62 in section 14.7 of our earlier chemistry classes]
• In our present case, when the -OH group is attached, the structure will be as shown in fig.13.9(1.b).
• But when the -OH group is attached, the whole structure will be called an alcohol. We must write the IUPAC name of that alcohol. [See section 12.6]
• In fig.13.9(1.b), the C atoms are numbered in such a way that the -OH group gets the smallest number. So the root name is pentan-1-ol
• Since there are no branches, the final name is also pentan-1-ol.
2. Another possible structure of the C₅H₁₁ is shown in fig.13.9(2.a) above.
• When the -OH group is attached, the structure will be as shown in fig.13.9(2.b).
• We must write the IUPAC name of this alcohol.
• In fig.13.9(2.b), the C atoms are numbered in such a way that the -OH group gets the smallest number. So the root name is pentan-2-ol
• Since there are no branches, the final name is also pentan-2-ol.
3. Another possible structure of the C₅H₁₁ is shown in fig.13.10(3.a) below:

Fig.13.10

• When the -OH group is attached, the structure will be as shown in fig.13.10(3.b).
• We must write the IUPAC name of this alcohol.
• In fig.13.10(3.b), the C atoms can be numbered from left to right or from right to left. In both methods, the -OH group gets the smallest number 3. So the root name is pentan-3-ol
• Since there are no branches, the final name is also pentan-3-ol.
4. Another possible structure of the C₅H₁₁ is shown in fig.13.10(4.a) above.
• When the -OH group is attached, the structure will be as shown in fig.13.10(4.b).
• We must write the IUPAC name of this alcohol.
• In fig.13.10(4.b), the C atoms are numbered in such a way that the -OH group gets the smallest number. So the root name is butan-1-ol
• Since there is a methyl group attached, the final name is 3-Methyl-butan-1-ol.
5. Another possible structure of the C₅H₁₁ is shown in fig.13.11(5.a) below:

Fig.13.11

• When the -OH group is attached, the structure will be as shown in fig.13.11(5.b).
• We must write the IUPAC name of this alcohol.
• In fig.13.11(5.b), the C atoms are numbered in such a way that the -OH group gets the smallest number. So the root name is butan-1-ol
• Since there is a methyl group attached, the final name is 2-Methyl-butan-1-ol.
6. Another possible structure of the C₅H₁₁ is shown in fig.13.11(6.a) above.
• When the -OH group is attached, the structure will be as shown in fig.13.11(6.b).
• We must write the IUPAC name of this alcohol.
• In fig.13.11(6.b), the C atoms are numbered in such a way that the -OH group gets the smallest number. So the root name is butan-2-ol
• Since there is a methyl group attached, the final name is 2-Methyl-butan-2-ol.
7. Another possible structure of the C₅H₁₁ is shown in fig.13.12(7.a) below:

Fig.13.12

• When the -OH group is attached, the structure will be as shown in fig.13.12(7.b).
• We must write the IUPAC name of this alcohol.
• In fig.13.12(7.b), the C atoms are numbered in such a way that the -OH group gets the smallest number. So the root name is propan-1-ol
• Since there are two methyl groups attached, the final name is 2,2-Dimethyl-propan-1-ol.
8. The last possible structure of the C₅H₁₁ is shown in fig.13.13(8.a) above.
• When the -OH group is attached, the structure will be as shown in fig.13.13(8.b).
• We must write the IUPAC name of this alcohol.
• In fig.13.11(8.b), the C atoms are numbered in such a way that the -OH group gets the smallest number. So the root name is butan-2-ol
• Since there is a methyl group attached, the final name is 3-Methyl-butan-2-ol.

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In the next section we will see a few more solved examples.

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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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