Chapter 14.14 - Isomerism

In the previous section, we saw the nomenclature of organic compounds carrying the amino group. In this section, we will see Isomers.

1. Consider the two compounds in fig.14.88 below. We have learned how to name them.

Fig.14.88

• Compound 1 is Propan-1-ol
• Compound 2 is Propan-2-ol
2. In this section we are going to see something more than just nomenclature. So let us write the properties of each:
Compound 1: 
• Molecular formula: C₃H₈O
• Functional group in the compound: Hydroxyl
Compound 2:
• Molecular formula: C₃H₈O
• Functional group in the compound: Hydroxyl
3. We find that:
• The molecular formula is the same. 
• The functional group is also the same. 
• The only difference that we can see is in the position of the functional group.
4. They are two different compounds having different physical and chemical properties. 
• The difference in properties is due to the difference in the 'position of the functional group'.  
■ Compounds having same molecular formula but different arrangement of atoms are called isomers. This phenomenon is called isomerism
5. We can say:
• Propan-1-ol is an isomer of Propan-2-ol
The reverse is also true:
• Propan-2-ol is an isomer of Propan-1-ol

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Now we will see different types of isomerism

Chain Isomerism

1. Consider the two compounds in fig.14.89 below. We have learned how to name them.

Fig.14.89

• Compound 1 is Butane
• Compound 2 is 2-Methylpropane
2. Let us write the details of each:
Compound 1: 
• Molecular formula: C₄H₁₀
• Functional group in the compound: No functional group present
Compound 2:
• Molecular formula: C₄H₁₀
• Functional group in the compound: No functional group present
3. We find that:
• The molecular formula is the same. 
• Both does not have any functional groups. 
■ The only difference that we can see is in the chain structure.
• Compound 1 has an 'open chain structure' 
• Compound 2 has a 'branched chain structure' 
4. They are two different compounds having different physical and chemical properties. 
• The difference in properties is due to the difference in the 'chain structure'.  
■ Compounds having same molecular formula but different chain structures are called chain isomers.
5. We can say:
• Butane is a chain isomer of 2-Methylpropane
The reverse is also true:
• 2-Methylpropane is a chain isomer of Butane

Functional Isomerism

1. Consider the two compounds shown below
• CH₃ㅡCH₂ ㅡOH
• CH₃ㅡOㅡCH₃
We know how to name them.

• Compound 1 is Ethanol
    ♦ Ethanol does not require the position number (of it's functional group ㅡOH) to be written. This is because in ethanol, there is only one possible position that the ㅡOH group can take. We can number it from either left or right. The number should always be the lowest, which is 1. Since there is only one possible position, we need not write it's number.  
• Compound 2 is Methoxymethane
2. Let us write the details of each:
Compound 1: 
• Molecular formula: C₂H₆O
• Functional group in the compound: Hydroxyl
Compound 2:
• Molecular formula: C₂H₆O
• Functional group in the compound: Alkoxy
3. We find that:
• The molecular formula is the same. 
■ The only difference that we can see is in the functional group
• Compound 1 has hydroxyl group
• Compound 2 has alkoxy group
4. They are two different compounds having different physical and chemical properties. 
• The difference in properties is due to the difference in functional groups.  
■ Compounds having same molecular formula but different functional groups are called functional isomers.
5. We can say:
• Ethanol is a functional isomer of Methoxymethane
The reverse is also true:
• Methoxymethane is a functional isomer of Ethanol

Let us see another example for functional isomerism:
1. Consider the two compounds shown below
• CH₃ㅡCOㅡCH₃
• CH₃ㅡCH₂ ㅡCHO
We know how to name them.

• Compound 1 is Propanone
• Compound 2 is Propanal
2. Let us write the details of each:
Compound 1: 
• Molecular formula: C₃H₆O
• Functional group in the compound: Keto
Compound 2:
• Molecular formula: C₂H₆O
• Functional group in the compound: Aldehyde
3. We find that:
• The molecular formula is the same. 
■ The only difference that we can see is in the functional group
• Compound 1 has keto group
• Compound 2 has aldehyde group
4. They are two different compounds having different physical and chemical properties. 
• The difference in properties is due to the difference in functional groups.  
5. We can say:
• Propanone is a functional isomer of Propanal
The reverse is also true:
• Propanal is a functional isomer of Propanone

Position Isomerism

1. Consider the two compounds that we saw in fig.14.88 at the beginning of this section. They are shown again below:
1. Consider the two compounds in fig.14.88 below. We have learned how to name them.

Fig.14.88

We have already named them.
• Compound 1 is Propan-1-ol
• Compound 2 is Propan-2-ol

2. Let us write the details of each:
Compound 1:
• Molecular formula: C₃H₈O
• Functional group in the compound: Hydroxyl
Compound 2:
• Molecular formula: C₃H₈O
• Functional group in the compound: Hydroxyl
3. We find that:
• The molecular formula is the same.
• The functional group is also the same. 
■ The only difference that we can see is in the position of the functional group. 
• Compound 1 has hydroxyl group at position 1
• Compound 2 has hydroxyl group at position 2
4. They are two different compounds having different physical and chemical properties. 
• The difference in properties is due to the difference in 'position of same functional group'.  
■ Compounds having same molecular formula and same functional group but different positions for that functional group are called position isomers.
5. We can say:
• Propan-1-ol is a position isomer of Propan-2-ol
The reverse is also true:
• Propan-2-ol is a position isomer of Propan-1-ol

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So we have seen three types of isomerisms: Chain, functional and position. 
We will now see some solved examples:

Solved example 14.16
Write all the possible position isomers of CH₃ㅡCH₂ ㅡCH₂ ㅡCl.
Solution:
1. The given compound is shown in fig.14.89(a) below:

Fig.14.89

• It's IUPAC name is 1-Chloropropane
• It's molecular formula is C₃H₇Cl.
2. A position isomer of the given compound is shown in fig(b) 
• It's IUPAC name is 2-Chloropropane
• It's molecular formula is C₃H₇Cl.
3. The compound in (2) is the only possible position isomer of the given compound
• In fig.(c), another structure is drawn. But it is in fact 1-Chloropropane. 
    ♦ Because in it's case, numbering should be done from left to right
• So it is not a position isomer and hence, a '×' mark is given for it  
4. Note that the position isomers in figs(a) and (b) have the same molecular formula

Solved example 14.17
Examine the compounds given in fig.14.90 below and find out the isometric pairs. To which type do they belong?

Fig.14.90

Solution:
Step 1. Write the molecular formula of each of the compounds:
1. C₅H₁₂      2. C₄H₁₀O       3. C₅H₁₂       4. C₅H₁₂O     5. C₃H₈O     6. C₃H₈O     7. C₄H₈O     8. C₄H₈O
Step 2. Group them into pairs. Pairs are formed in such a way that, both the members of any pair taken, will have the same molecular formula. Thus we get:
1&3 ⟹ [C₅H₁₂, C₅H₁₂]
5&6 ⟹ [C₃H₈O, C₃H₈O]
7&8 ⟹ [C₄H₈O, C₄H₈O]
Step 3. Write the names:
1&3 ⟹ [C₅H₁₂, C₅H₁₂] ⟹ [Pentane, 2,2-Dimethylpropane]
5&6 ⟹ [C₃H₈O, C₃H₈O] ⟹ [Propan-1-ol, Methoxyethane]
7&8 ⟹ [C₄H₈O, C₄H₈O] ⟹ [Propanal, Butanone]
Step 4. Write the type of isomerism:
1&3 ⟹ [Pentane, 2,2-Dimethylpropane] ⟹ [Alkane, Alkane] ⟹Chain isomerism
5&6 ⟹ [Propan-1-ol, Methoxyethane] ⟹ [Alcohol, Ether] ⟹Functional isomerism
7&8 ⟹ [Propanal, Butanone] ⟹ [Aldehyde, Ketone] ⟹Functional isomerism
■ From step 4, we can write a general information:
• An Alcohol may form a functional isomer with an Ether and vice versa 
• An Aldehyde may form a functional isomer with a Ketone and vice versa 

Solved example 14.18
(i) How many position isomers are possible for the compound
CH₃ㅡCH₂ ㅡCH₂ ㅡCH₂ ㅡCH₂ ㅡOH ?
(ii) Write a functional isomer of this compound
(iii) Which functional group is present in that isomer
(iv) Write the structure and IUPAC name of that isomer
Solution:
Part (i): Two position isomers are possible for the given compound. They are shown in fig.14.91 (a) and (b) below:

Fig.14.91

The given compound is Pentan-1-ol. It's molecular formula is C₅H₁₂O 
The compound in fig.14.91(a) is Pentan-2-ol. It's molecular formula is C₅H₁₂O 
The compound in fig.14.91(b) is Pentan-3-ol. It's molecular formula is C₅H₁₂O
Part (ii): In the previous example 14.17, we saw that:
• An Alcohol may form a functional isomer with an Ether and vice versa 
• An ether which is a functional isomer of the given alcohol is shown in fig(c) 
Part (iii): The functional group in ether is Alkoxy group
Part (iv): The structure is shown in fig(c)
• The IUPAC name is Methoxybutane

Solved example 14.19
How many chain isomers are possible for the compound
CH₃ㅡCH₂ ㅡCH₂ ㅡCH₂ ㅡCH₂ ㅡCH₃ ?
Write them down

Solution:
There are 5 possible chain isomers including the given compound. They are shown in the fig.14.92 below:

Fig.14.92

The IUPAC names are:
1. Hexane
2. 1-Methylpentane
3. 2-methylpentane
4. 2,3-Dimethylbutane
5. 2,2-Dimethylbutane

Solved example 14.20
The structural formulae of some compounds are given in fig.14.93 below. Tabulate them into different pairs of isomers. Write the IUPAC names and the functional groups of each compound.

Fig.14.93

Solution:
Step 1. Write the molecular formula of each of the compounds:
1. C₄H₈O      2. C₆H₁₄       3. _C4H8O       4. C₆H₁₄     5. C₃H₈O     6. C₃H₈O
Step 2. Group them into pairs. Pairs are formed in such a way that, both the members of any pair taken, will have the same molecular formula. Thus we get:
1&3 ⟹ [C₄H₈O, C₄H₈O]
2&4 ⟹ [C₆H₁₄, C₆H₁₄]
5&6 ⟹ [C₃H₈O, C₃H₈O]
Step 3. Write the names:
1&3 ⟹ [C₄H₈O, C₄H₈O] ⟹ [Butanal, Butanone]
2&4 ⟹ [C₆H₁₄, C₆H₁₄] ⟹ [Hexane, 2-Methylpentane]
5&6 ⟹ [C₃H₈O, C₃H₈O] ⟹ [Methoxyethane, Propanol]
Step 4. Write the type of isomerism:
1&3 ⟹ [Butanal, Butanone] ⟹ [Aldehyde, Ketone] ⟹ Functional isomerism
2&4 ⟹ [Hexane, 2-Methylpentane] ⟹ [Alkane, Alkane] ⟹ Chain isomerism
5&6 ⟹ [Methoxyethane, Propanol] ⟹ [Ether, Alcohol] ⟹ Functional isomerism
■ From step 4, we can write a general information:
• An Alcohol may form a functional isomer with an Ether and vice versa 
• An Aldehyde may form a functional isomer with a Ketone and vice versa
We saw this result in solved example 14.17 also

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In the next section, we will see Isomerism in Alicyclic compounds.

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Chapter 14.13 - Amino functional group

In the previous section, we saw the nomenclature of organic compounds carrying the alkoxy group. In this section, we will see the Amino group.

• We have seen that the hydroxyl group is represented as 'ㅡOH'. (see details) 
• We have seen that the aldehyde group can be represented in two ways (see fig.14.63)
• We have seen that the keto group can be represented in two ways (see fig.14.67 (a) & (b))
• We have seen that the carboxyl group can be represented in two ways (see fig.14.72 (a) & (b))
• In the previous sections, we saw that the halo group can be represented as 'ㅡF, ㅡCl, ㅡBr or ㅡI'.
And the alkoxy group can be represented as ㅡOㅡR
■ Now, our present amino group is represented as: 'ㅡNH₂'
■ Note that:
• In the hydroxyl group (ㅡOH), the bond which connects the group to a main chain starts from the oxygen atom  
• In the aldehyde group (ㅡCHO), the bond which connects the group to a main chain starts from the carbon atom
• In the carboxyl group (ㅡCOOH), the bond which connects the group to a main chain starts from the carbon atom
• In the halo group, the bond which connects the group to a main chain starts from halogen atom
• In the alkoxy group (ㅡOㅡR), the bond which connects the group to a main chain starts from the oxygen atom
• In our present amino group (ㅡNH₂), the bond which connects the group to a main chain starts from the nitrogen atom
• We will see all the bonding details at the end of this section. At present, we will see the nomenclature.

---

• The IUPAC names of compounds with keto group end in amine. 
• Such compounds are called amines. 
• In any amine, there will be a ㅡNH₂ group
■ The naming of amines is done by the following 2 steps:
1. Remove the 'e' from the name of the corresponding alkane
2. Put 'amine' in it's place
• So we can write:
Alkane - e + amine → Alkanamine
• Thus we get:
    ♦ Methane - e + amine → Methanamine
    ♦ ethane - e + amine → ethanamine
    ♦ Propane - e + amine → propanamine
so on . . .

---

Consider the fig.14.54 that we saw in a previous section (alcohols). It is shown again below:

Fig.14.54

• In both compounds, there are 3 carbon atoms, 8 hydrogen atoms and 1 oxygen atom. 
    ♦In fig(a), the hydroxyl group is at the end position. 
    ♦ But in fig(b), it is at an interior position. 
• Accordingly we gave two distinct names: Propan-1-ol and propan-2-ol
■ In our present discussion on amines also, we encounter a similar situation.
• This is because, the ㅡNH₂ group can be anywhere in the interior of the structure
    ♦ Obviously, the C that carries the group should get the lowest possible number

• Let us see a solved example:
Solved example 14.14
Write the IUPAC name of the compounds shown in fig.14.84 below:

Fig.14.84

Solution:
We will write the steps:
Case (i):  
1. In the given compound, there are 2 carbon atoms. So the corresponding alkane is ethane
• Remove the 'e' at the end. 
2. Put 'amine' in it's place. We get:
ethane - e + amine= ethanamine
3. The lowest position number is 1 from right. So we get: Ethanan-1-amine
Case (ii):  
1. In the given compound, there are 3 carbon atoms. So the corresponding alkane is propane  
• Remove the 'e' at the end. 
2. Put 'amine' in it's place. We get:
propane - e + amine = propanamine
3. The lowest position number is 1 from right. So we get: Propan-1-amine

Case (iii):  
1. In the given compound, there are 3 carbon atoms. So the corresponding alkane is propane  
• Remove the 'e' at the end. 
2. Put 'amine' in it's place. We get:
propane - e + amine = propanamine
3. The lowest position number is 2 from left as well as right. So we get: propan-2-amine
Case (iv):  
1. In the given compound, there are 6 carbon atoms. So the corresponding alkane is hexane  
• Remove the 'e' at the end. 
2. Put 'amine' in it's place. We get:
hexane - e + amine = hexanamine
3. The lowest position number is 3 from right. So we get: hexan-3-amine

Solved example 14.15
We cannot write 'Propan-3-amine'. Give reason.
Solution:
1. The structure of propan-1-amine is: CH₃ㅡCH₂ ㅡCH₂ㅡNH₂.
• In this case we begin the numbering from the right most carbon atom
2. Then we would be inclined to think about propan-3-amine as: NH₂ ㅡCH₂ ㅡCH₂ㅡCH₃.
• But this is wrong. In this case the numbering should begin from the left most carbon atom.
• So we cannot write  'Propan-3-amine'

---

Now we will see the reverse process. That is., we are given the IUPAC name of an amine. We must write the structural formula. We will learn the method by analysing an example:
1. Given IUPAC name is: Hexan-2-amine
2. Consider the word root. It is hex. So there are 6 carbon atoms
3. Consider the suffix. It is an. So the ketone is derived from an alkane
• So all the carbon-carbon bonds are single bonds
    ♦ alkane: ane minus e gives an
    ♦ alkene: ene minus e gives en
    ♦ alkyne: yne minus e gives yn
4. So we have 6 carbon atoms with single bonds between them. This is shown in fig.14.85(a) below:

5. The functional group is at the position 2.  So we can attach the amino group to the carbon atom at position 2. This is shown in fig.14.85(b)
6. Now fill all the valencies of carbon atoms. 
• We use hydrogen to fill up the valencies. 
• The result is shown in fig.14.85(c)

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■ In the first section in this chapter, we saw how the 'alkyl radical' gets itself attached to an open chain. See details here. 
■ In another previous section we saw how a 'hydroxyl group' that is., 'ㅡOH', gets itself attached to an open chain. See details here.
■ In yet another previous section we saw how an 'aldehyde group' that is., 'ㅡCHO', gets itself attached to an open chain. See details here.
so on . . .  
■ Now we will see the same for the amino group:
• We have seen the covalent bonding between two nitrogen atoms. Two nitrogen atoms share three pairs of electrons between them to achieve octet and thus stability. 
• It can be represented as: N☰N. Details here. The two nitrogen atoms bonded together by a triple bond becomes as Nitrogen molecule (N₂)
• So a nitrogen atom require three electrons to satisfy the valency. In ammonia molecule (NH₃) three hydrogen atoms form bonds with nitrogen.  
• So when a hydrogen atom is removed from the ammonia molecule, the remaining portion becomes an amino group (ㅡNH₂). It will be unstable and will be looking for an electron.
• If a hydrogen atom is removed from a hydrocarbon, the ㅡNH₂ can take it's place. This is shown in fig.14.86 below:

Fig.14.86

• Thus we get a stable molecule

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In the next section, we will see Isomers.

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Chapter 14.12 - Alkoxy functional group

In the previous section, we saw the nomenclature of organic compounds carrying the halo group. In this section, we will see the Alkoxy group.

• We have seen that the hydroxyl group is represented as 'ㅡOH'. (see details) 
• We have seen that the aldehyde group can be represented in two ways (see fig.14.63)
• We have seen that the keto group can be represented in two ways (see fig.14.67 (a) & (b))
• We have seen that the carboxyl group can be represented in two ways (see fig.14.72 (a) & (b))
• In the previous section, we saw that the halo group can be represented as 'ㅡF, ㅡCl, ㅡBr or ㅡI'.
■ Now, our present alkoxy group is represented as: 'ㅡOㅡR'
• We know what 'R' is. It stands for alkyl radicals like methyl radical, ethyl radical etc., We saw it in a previous section of this chapter. Details here.
• So 'ㅡOㅡR' is a combination of oxygen and alkyl radicals. Hence the name alkoxy.
■ Note that:
• In the hydroxyl group (ㅡOH), the bond which connects the group to a main chain starts from the oxygen atom  
• In the aldehyde group (ㅡCHO), the bond which connects the group to a main chain starts from the carbon atom
• In the carboxyl group (ㅡCOOH), the bond which connects the group to a main chain starts from the carbon atom
• In the halo group, the bond which connects the group to a main chain starts from halogen atom
• In our present alkoxy group (ㅡOㅡR), the bond which connects the group to a main chain starts from the oxygen atom
• We will see all the bonding details at the end of this section. At present, we will see the nomenclature.

---

• The IUPAC names of compounds with ㅡOㅡR group end in an ethane. 
    ♦ This end part indicates 'the main alkane to which the group is attached'.
• The beginning part indicates the name of the group
Let us analyse this with an example:
■ Consider the compound: CH₃ㅡOㅡCH₂ㅡCH₃.
• In this, which is the ㅡOㅡR?
    ♦ It seems that 'ㅡOㅡCH₂ㅡCH₃' at the right side is the ㅡOㅡR
    ♦ Where R is the ethyl radical 'ㅡCH₂ㅡCH₃'
• But 'CH₃ㅡOㅡ' on the left side is also an ㅡOㅡR. 
    ♦ Because the methyl radical 'ㅡCH₃' is attached to O
• So we have a situation:
    ♦ There are alkyl groups on either side of O. 
    ♦ which one belongs to the ㅡOㅡR group? and 
    ♦ which one is the main alkane?
■ According to the IUPAC rule:
• The longest alkyl group should be taken as the main alkane    
• The other alkyl group should be the 'R' of the ㅡOㅡR group 
■ Thus in our present case,
• The end part of the name = main alkane = longest alkyl group = ethane
• The beginning part = the other alkyl group = methoxy
• So the IUPAC name is: Methoxyethane

Another example:
Consider the compound: CH₃ㅡCH₂ㅡCH₂ㅡCH₂ㅡOㅡCH₂ㅡCH₃.
1. The end part of the name = main alkane 
= longest alkyl group = butane (4 carbon atoms on the left side of O)
2. The beginning part = the other alkyl group = ethoxy
3. So the IUPAC name is: Ethoxybutane.

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■ Compounds with the alkoxy group (ㅡOㅡR) are called ethers.

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Now we will see the reverse process. That is., we are given the IUPAC name of an ether. We must write the structural formula. We will learn the method by analysing an example:
1. Given IUPAC name is: Methoxyethane 
2. The end part of the name = main alkane = ethane
• So the 'ethyl radical' is the main part. It should be written on one side (either left or right) of 'O'
    ♦ We will write it on the right side of O
3. The beginning part = the other alkyl group = methoxy  
• So methyl radical is the R of the ㅡOㅡR.
    ♦ We will write it on the left side of O
4. Thus we get: CH₃ㅡOㅡCH₂ㅡCH₃.

Another example:
1. Given IUPAC name is: Ethoxyethane 
2. The end part of the name = main alkane = ethane
• So the 'ethyl radical' is the main part. It should be written on one side (either left or right) of 'O'
    ♦ We will write it on the right side of O
3. The beginning part = the other alkyl group = ethoxy  
• So ethyl radical is the R of the ㅡOㅡR.
    ♦ We will write it on the left side of O
4. Thus we get: CH₃ㅡCH₂ㅡOㅡCH₂ㅡCH₃.
• Note that on both sides, we have the ethyl radicals

One more example:
1. Given IUPAC name is: Methoxypropane 
2. The end part of the name = main alkane = propane
• So the 'propyl radical' is the main part. It should be written on one side (either left or right) of 'O'
    ♦ We will write it on the right side of O
3. The beginning part = the other alkyl group = methoxy  
• So methyl radical is the R of the ㅡOㅡR.
    ♦ We will write it on the left side of O
4. Thus we get: CH₃ㅡOㅡCH₂ㅡCH₂ㅡCH₃.

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■ In the first section in this chapter, we saw how the 'alkyl radical' gets itself attached to an open chain. See details here. 
■ In a previous section we saw how a 'hydroxyl group' that is., 'ㅡOH', gets itself attached to an open chain. See details here.
■ In another previous section we saw how an 'aldehyde group' that is., 'ㅡCHO', gets itself attached to an open chain. See details here.
so on . . .
■ Now we will see the same for the alkoxy group:
In the ㅡOㅡR group, a bond is starting out from the oxygen atom. This indicates that the O is in deed of an electron. This will be clear when we draw the electron dot diagram of the group. It is shown in the fig.14.82 below:

Fig.14.82

• In the above fig.14.82, a methyl radical (CH₃) is attached to an oxygen atom.
• The valency needs of all H atoms are satisfied. Since they have one red dot and one green dot
• The valency needs of C atom is satisfied. Since it has 3 red plus 4 green plus 1 cyan, giving a total of 8 dots
• Oxygen has it's own 6 cyan dots plus the green dot of C. This gives a total of only 7. So one more electron is required. Thus the group as a whole requires one more electron
• Note: In the fig.14.82, The oxygen atom is attached to a methyl radical. It is the 'R' in 'ㅡOㅡR.' Instead of methyl, we can show ethyl, propyl etc., also 
• This need will be satisfied when the group gets itself attached to another alkyl radical as shown in fig.14.83 below:

Fig.14.83

• Thus we get a stable molecule

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In the next section, we will see Amino group.

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