Chapter 14.15 - Isomerism in Alicyclic compounds

In the previous section, we saw the details about Isomerism. In this section, we will see Isomerism in Alicyclic compounds.

• We have seen three types of structures for hydrocarbons:
(i) Straight chain  (ii) Branched chain  (iii) Cyclic
• We have seen some basic details about cyclic compounds in our earlier classes. (Details here)
• Now we will analyse them in greater detail. We will do the analysis by taking Cyclopentane as an example:
1. Consider the chain structure in fig.14.94(a) below:

Fig.14.94

• We know that it is the structure of pentane
    ♦ All the carbon-carbon bonds are single bonds
    ♦ 3 hydrogen atoms are attached to the carbon atom at the left end
    ♦ 3 hydrogen atoms are attached to the carbon atom at the right end also
    ♦ All interior carbon atoms have 2 hydrogen atoms each
2. Let us bend this 'straight chain' into a 'ring' structure. This is shown in fig.14.94(b)
• In fig(b), we have only bend it. The closed ring is not formed yet.
    ♦ To get a closed ring, the end carbons should meet (ie., bond together)
• Note that in this 'bent state', the 'number of hydrogen atoms carried by each of the carbon atoms' has not changed.
3. Now we will bond the end carbons together. For easy comparison, the 'bent state' and the 'ring'state' are shown side by side in fig.14.95 below

Fig.14.95

• Fig.14.95(a) is the same fig.14.94(b) that we saw above.
4. See the completed ring in fig.14.95(b)
• The end carbons have finally met. That is., a bond is formed between them
• What else do we see?
Ans: Each of the end carbon atoms have lost one Hydrogen atom. So, the molecule as a whole, has lost two hydrogen atoms
• Why did the hydrogen atoms leave?
Ans: Because they are no longer required for octet of the end carbon atoms. A pair of electrons is now shared in the newly formed bond between the end carbon atoms 
• When a hydrogen atom leave, it takes it's electron (shown as red dot in the fig.) with it.
    ♦ The green dot which was in bond with that red dot is now free. 
• The same happens in the other end carbon
• The newly freed green dots together form a bond.
• This bond results in the bonding between the end carbon atoms and the ring is completed
• The completed ring in fig.14.95(b) is Cyclopentane
5. Let us write the molecular formulae:
Pentane: C₅H₁₂  
Cyclopentane: C₅H₁₀
• So the cyclopentane has two hydrogen atoms less than the pentane
■ This is true for all alkanes:
The number of hydrogen atoms in any cycloalkane will be 2 less than the corresponding alkane
6. At this time we should recall another result that we saw in previous classes:
■ The number of hydrogen atoms in any alkene will be 2 less than the corresponding alkane
7. Combining (5) and (6) we can write:
• The number of hydrogen atoms in a cycloalkane is same as that in the corresponding alkene
This gives an interesting result:
■ An alkene and it's corresponding cycloalkane are isomers
8. By the word 'corresponding', we mean 'same number of carbon atoms'
Example:
• If the cyclic compound is: Cyclopentane
    ♦ Corresponding alkane is: Pentane
    ♦ Corresponding alkene is: Pentene
    ♦ Cyclopentane and Pentene will be isomers
Another example:
• If the cyclic compound is: Cyclopropane
    ♦ Corresponding alkane is: Propane
    ♦ Corresponding alkene is: Propene
    ♦ Cyclopropane and Propene will be isomers

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• Thus we saw the relation between a cycloalkane and it's corresponding alkene. For this result, we started out  with a cycloalkane.
• Now we will start out another discussion with a cycloalkene. Let us see where we will reach:

1. Consider the chain structure in fig.14.96(a) below:

Fig.14.96

• We know that it is the structure of pentene
    ♦ One of the carbon-carbon bonds is a double bond
    ♦ 3 hydrogen atoms are attached to the carbon atom at the left end
    ♦ 2 hydrogen atoms are attached to the carbon atom at the right end
    ♦ 1 hydrogen atom is attached to the second carbon atom from the right end
    ♦ All other interior carbon atoms have 2 hydrogen atoms each
2. Let us bend this 'straight chain' into a 'ring' structure. This is shown in fig.14.96(b)
• In fig(b), we have only bend it. The closed ring is not formed yet.
    ♦ To get a closed ring, the end carbons should meet (ie., bond together)
• Note that in this 'bent state', the 'number of hydrogen atoms carried by each of the carbon atoms' has not changed.
3. Now we will bond the end carbons together. For easy comparison, the 'bent state' and the 'ring'state' are shown side by side in fig.14.97 below:

Fig.14.97

• Fig.14.97(a) is the same fig.14.96(b) that we saw above.
4. See the completed ring in fig.14.97(b)
• The end carbons have finally met. That is., a bond is formed between them
• What else do we see?
Ans: Each of the end carbon atoms have lost one Hydrogen atom. So, the molecule as a whole, has lost two hydrogen atoms
• Why did the hydrogen atoms leave?
Ans: Because they are no longer required for octet of the end carbon atoms. A pair of electrons is now shared in the newly formed bond between the end carbon atoms 
• When a hydrogen atom leave, it takes it's electron (shown as red dot in the fig.) with it.
    ♦ The green dot which was in bond with that red dot is now free. 
• The same happens in the other end carbon
• The newly freed green dots together form a bond.
• This bond results in the bonding between the end carbon atoms and the ring is completed
• The completed ring in fig.14.97(b) is Cyclopentene
5. Let us write the molecular formulae:
• Pentane: C₅H₁₂  
    ♦ Cyclopentane: C₅H_{10
• Pentene: C5H10  
    ♦ Cyclopentene: C5H8}
• So the cyclopentene has two hydrogen atoms less than the pentene
■ This is true for all alkenes:
The number of hydrogen atoms in any cycloalkene will be 2 less than the corresponding alkene
6. At this time we should recall another result that we saw in previous classes:
■ The number of hydrogen atoms in any alkyne will be 2 less than the corresponding alkene
7. Combining (5) and (6) we can write:
• The number of hydrogen atoms in a cycloalkene is same as that in the corresponding alkyne
This gives an interesting result:
■ An alkyne and it's corresponding cycloalkene are isomers
8. By the word 'corresponding', we mean 'same number of carbon atoms'
Example:
• If the cyclic compound is: Cyclopentene
    ♦ Corresponding alkane is: Pentane
    ♦ Corresponding alkene is: Pentene
    ♦ Corresponding alkyne is: Pentyne
    ♦ Cyclopentene and Pentyne will be isomers
Another example:
• If the cyclic compound is: Cyclopropene
    ♦ Corresponding alkane is: Propane
    ♦ Corresponding alkene is: Propene
    ♦ Corresponding alkyne is: Propyne
    ♦ Cyclopropene and Propyne will be isomers
• Thus we saw the relation between a cycloalkene and it's corresponding alkyne.
■ We can show the above findings in a simple fig:

Fig.14.98

• The green arrow indicates that Alkenes and Cycloalkanes can be isomers
• The yellow arrow indicates that Alkynes and Cycloalkenes can be isomers
■ In this section, we will be seeing problems related to the green arrow only

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But before we see those problems, a more interesting result related to the green arrow is coming up:
1. We know that the pentene itself has isomers. This is based on the position of it's double bond. We saw details here. 
Thus we have:
Pent-1-ene: CH₃ㅡCH₂ ㅡCH₂ ㅡCH＝CH₂
Pent-2-ene: CH₃ㅡCH₂ ㅡCH＝CHㅡCH₃
2. The three compounds: cyclopentane, pent-1-ene and pent-2-ene have the same molecular formula C₅H₁₀.
• We can write:
The three compounds: cyclopentane, pent-1-ene and pent-2-ene are isomers
• pent-1-ene and pent-1-ene are chain isomers
3. For higher alkenes like octene, decene etc., more chain isomers are possible. 
• All those isomers will be isomers of the corresponding cycloalkanes (cyclooctane, cyclodecane etc.,) also.

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Solved example 14.21
Write the IUPAC names and structures of all possible isomers of the compound given below:
CH₃ㅡCH₂ ㅡCH＝CH₂
Solution:
1. The given compound is But-1-ene
It has two possible isomers
(i) The chain isomer: But-2-ene
• It is produced by the change in the position of the double bond
• It's structure is: CH₃ㅡCH＝CHㅡCH₃.
(ii) The corresponding cycloalkane: Cyclobutane
• It's structure is shown in the fig.14.99(a) below:

Fig.14.99

2. So we have three compounds:
But-1-ene, But-2-ene and Cyclobutane
• All three of them have the same molecular formula: C₄H₈.
• All three are isomers.

Solved example 14.22
Write the IUPAC names and structures of all possible isomers of the compound shown in fig.14.99(b) above.
Solution:
1. The given compound is Cyclohexane
It has three possible isomers
(i) Hex-1-ene
• It's structure is: CH₃ㅡCH₂ㅡCH₂ㅡCH₂ ㅡCH＝CH₂
(ii) Hex-2-ene
• It's structure is: CH₃ㅡCH₂ㅡCH₂ㅡCH＝CHㅡCH₃

(iii) Hex-3-ene
• It's structure is:  CH₃ㅡCH₂ㅡCH＝CHㅡCH₂ㅡCH₃
2. So we have four compounds:
Hex-1-ene, Hex-2-ene, Hex-3-ene and Cyclohexane
• All four of them have the same molecular formula: C₆H₁₂
• All four are isomers

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In the next section, we will see Reactions of Organic compounds

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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.

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• 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 . . .

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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'

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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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