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 14.13 - 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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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.

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• 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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Chapter 14.10 - Carboxyl Functional group

In the previous section, we saw the nomenclature of organic compounds carrying the keto group. In this section, we will see the carboxylic acid 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))
• Now we will see that the carboxylic acid group can also be represented in two ways:

Fig.14.72

• We can use either (a) or (b) in the above fig.14.72 to represent the carboxylic acid group
■ 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 our present carboxylic acid group (ㅡCOOH), the bond which connects the group to a main chain starts from the carbon atom
• Also there is a double bond within the group. 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 ㅡCOOH group end in oic acid. 
• Such compounds are called carboxylic acids. 
• In any carboxylic acid, there will be a ㅡCOOH group
■ The naming of carboxylic acids is done by the following 2 steps:
1. Remove the 'e' from the name of the corresponding alkane
2. Put 'oic acid' in it's place
• So we can write:
Alkane - e + oic acid → Alkanoic acid
• Thus we get:
    ♦ Methane - e + oic acid → Methanoic acid
    ♦ ethane - e + oic acid → ethanoic acid
    ♦ Propane - e + oic acid → propanoic acid
so on . . .
■ But there are some important points to note:
• The 'newly attaching carboxylic acid group' has one C atom of it's own. 
• So when it is attached, the total number of carbon atoms will increase by one. 
• This 'total number' should be considered while naming the compound. 
• This is a general rule for any functional group which carries a C atom of it's own. We applied it earlier for aldehyde group and keto group 
We can write:
■ If the functional group contains a carbon atom that carbon atom should be treated as part of the main chain.

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■ Then we will be wondering about methanoic acid. 
• Because, One hydrogen atom is removed from Methane and the ㅡCOOH group takes it's place to give the carboxylic acid. 
• But when the group takes it's place, the total number of C atoms will become 2. 
    ♦ So it should be named as Ethanoic acid. 
• That means, the compound methanoic acid cannot exist. 
• However, methanoic acid does exist. It's structure is shown in fig.14.73(a) below:

Fig.14.73

■ So how is methanoic acid formed?
• We know that, the ㅡCOOH group will be looking for an electron.
    ♦ This electron is supplied by a hydrogen atom. 
• Thus we get the methanoic acid shown in fig.14.73(a) above. 
    ♦ The portion shown within the magenta polygon is the carboxylic acid group.
■ How is the compound in fig(b) formed?
• It was originally CH₄. One of it's H atoms was removed and the ㅡCOOH group took it's place. 
    ♦ Now there are a total of two C atoms, and so it is ethanoic acid    
■ How is the compound in fig(c) formed?
• It was originally CH₃ㅡCH₃. One of it's H atoms was removed and the ㅡCOOH group took it's place. 
    ♦ Now there are a total of three C atoms, and so it is propanoic acid
■ The compounds in fig.73 above can be represented in another way also:
• Methanoic acid: HㅡCOOH
• Ethanoic acid: CH₃ㅡCOOH  
• Propanaoic acid: CH₃ㅡCH₂ㅡCOOH

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Consider the fig.14.54 that we saw in the previous section. 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 another position. 
• Accordingly we gave two distinct names: Propan-1-ol and propan-2-ol
■ But in our present discussion on carboxylic acid group, we do not have to learn about such change in positions. 
• That is., In all the carboxylic acids that we see in the present discussion, the ㅡCOOH group will be at the end (or at the beginning).
    ♦ So we will not see names like Propan-1-oic acid, propan-2-oic acid etc., in this discussion.

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Solved example 14.10
Write the IUPAC name of the compound given below:
CH₃ㅡCH₂ㅡCH₂ㅡCH₂ㅡCOOH
Solution:
• As mentioned above, in the present discussion on aldehydes, we will not encounter those carboxylic acids where the ㅡCOOH group takes different positions. 
• We will see those carboxylic acids only where the group is at the end (or at the beginning). 
    ♦ So we do not need to give position numbers.
We will write the steps:  
1. In the given compound, there are 5 carbon atoms. So the corresponding alkane is Pentane  
• Remove the 'e' at the end. 
2. Put 'oic acid' in it's place. We get:
Pentane - e + oic acid = Pentanoic acid  

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Now we will see the reverse process. That is., we are given the IUPAC name of a carboxylic acid. We must write the structural formula. We will learn the method by analysing an example:
1. Given IUPAC name is: Propanoic acid
2. Consider the word root. It is prop. So there are 3 carbon atoms
3. Consider the suffix. It is an. So the aldehyde 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 3 carbon atoms with single bonds between them. 
• Note that this 'total 3 numbers' is including the C atom in the ㅡCOOH group. 
• Also, we do not need to write the numbering. Because the group will be at the end (or the beginning).
• So we can write:
C ㅡC ㅡCOOH
5. Fill all the valencies of carbon atoms. 
• We use hydrogen to fill up the valencies. 
• The result is: CH₃ㅡCH₂ㅡCOOH

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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.   
■ In the just previous section we saw how a 'keto group' that is., 'ㅡCO', gets itself attached to an open chain. See details here.  
■ Now we will see the same for the carboxylic acid group:
• We have seen the structure of ㅡCHO group (keto group) in a previous section. It is shown again in fig.14.74 (a) below:

Fig.14.74

• From fig(a), it is clear that the group is in need for an electron. 
    ♦ This need is indicated by the absence of a dot at the end of the bond line attached to the C atom.
• Now consider fig.14.74(b). It shows the bonding details in the ㅡCOOH group 
    ♦ All the valency needs of top oxygen atom is satisfied (6 cyan dots plus two green dots, giving a total of 8 dots.
    ♦ Those of bottom oxygen atom is satisfied (6 cyan dots plus one green plus one red dot, giving a total of 8 dots. 
    ♦ Those of hydrogen is also satisfied (1 red dot plus 1 cyan dot, giving a total of 2 dots
    ♦ Those of carbon is not fully satisfied. (4 green plus 3 cyan give a total of only 7)
• The need for one electron is indicated by the absence of a dot at the end of the bond line attached to the C atom. 
    ♦ Thus the group as a whole, is in need for one electron. 
• This need will be satisfied when the group gets itself attached to a hydrocarbon chain as shown in fig.14.75 below:

Fig.14.75

• Thus we get a stable molecule

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In the next section, we will see the Halo group.

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