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 . . .

---

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.

---

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

---

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 14.4 - Structure from IUPAC name

In the previous section, we saw the naming procedure for hydrocarbons with repeating branches. In this section, we will see a summary of the various cases that we have seen so far in this chapter.
• Later in this section, we will see how to do a reverse process. That is., when we are given an IUPAC name, we will see how to draw it's structure

The various cases that we saw so far, can be presented in the form of a flow chart as shown in fig.14.35 below:

Fig.14.35

Now we will see some solved examples related to the 'Branched chain' case as a whole:
Solved example 14.4
Write the IUPAC names of the four compounds shown in fig.14.36 below:

Fig.14.36

Solution:
The fig.14.37 below shows the required markings. Based on that, the naming can be done easily.

Fig.14.37

The compounds are:
1. 3-Methylhexane
2. 2,4-Dimethylpentane
3. 3,3-Diethylpentane
4. 3,4-Diethylhexane
• The reader may try different numbering options and write the steps in his/her own notebooks.

Solved example 14.5
Write the IUPAC names of the four compounds shown in fig.14.38 below:

Fig.14.38

Solution:
The fig.14.39 below shows the required markings. Based on that, the naming can be done easily.

Fig.14.39

The compounds are:
1. 3-Ethyl-2-methylhexane
2. 2,3,5-Trimethylhexane
3. 3,4-Dimethylhexane
4. 4-Ethyl-2,3-Dimethylhexane

• The reader may try different numbering options and write the steps in his/her own notebooks.

Structure from IUPAC name

• So far, we have been discussing how to give appropriate names when the structure of various hydrocarbons are given. 
■ Now we will see a reverse process. That is:
We will be given the IUPAC name of a hydrocarbon. We must draw the structure of that hydrocarbon.
• We will analyse the process using an example:
Consider 2,3-Dimethylbutane. We will try to draw it's structure using the following steps:
Step 1: Note the 'word root+suffix'. It is 'butane'
• From this, we get two information:
    ♦ There are 4 carbon atoms in the main chain
    ♦ All the carbon-carbon bonds are 'single bonds'
• So we can draw the main chain in this step 1. 
• The 'numbering of carbon atoms in the main chain' can also be done at this stage. 
• This is shown in fig.14.40(a) below:

Fig.14.40

Step 2: Note the branches and their positions. We have: '2,3-Dimethyl'
• From this we get the following information:
    ♦ There are two branches. 
    ♦ Both are methyl radicals
    ♦ Their positions are at 2 and 3
• So the two branches can be attached to the main chain that we obtained in step 1
• The result is shown in fig.14.40(b) above
Step 3: This is the final step. 
• Fill all the valencies of carbon atoms. 
• Since it is a hydrocarbon, only carbon and hydrogen will be present. 
    ♦ So we use hydrogen to fill up the valencies. 
• The result is shown in fig.14.40(c) above. So this fig.14.40(c) shows the required structure

Another example: 4-Ethyl-3,3-dimethylheptane
Step 1: Note the 'word root+suffix'. It is 'heptane'
• From this, we get two information:
    ♦ There are 7 carbon atoms in the main chain
    ♦ All the carbon-carbon bonds are 'single bonds'
• So we can draw the main chain in this step 1. 
• The 'numbering of carbon atoms in the main chain' can also be done at this stage. 
• This is shown in fig.14.41(a) below:

Fig.14.41

Step 2: Note the branches and their positions. We have: '4-Ethyl and 3,3-Dimethyl'
• From this we get the following information:
    ♦ There are three branches. 
    ♦ One ethyl radical and two methyl radicals
    ♦ The ethyl radical is at 4
    ♦ Both the methyl radicals are at position 3
• So the three branches can be attached to the main chain that we obtained in step 1
• The result is shown in fig.14.41(b) above
Step 3: This is the final step. 
• Fill all the valencies of carbon atoms. 
• Since it is a hydrocarbon, only carbon and hydrogen will be present. 
    ♦ So we use hydrogen to fill up the valencies. 
• The result is shown in fig.14.41(c) above. So this fig.14.41(c) shows the required structure
• Note that, there are no hydrogen atoms attached to the carbon atom at position 3 (there are two branches at position 3)
• Also, there is only one hydrogen atom attached to the carbon atom at position 4 (there is only one branch at position 4)

One more example: 3,3-Diethylpentane
Step 1: Note the 'word root+suffix'. It is 'pentane'
• From this, we get two information:
    ♦ There are 5 carbon atoms in the main chain
    ♦ All the carbon-carbon bonds are 'single bonds'
• So we can draw the main chain in this step 1. 
• The 'numbering of carbon atoms in the main chain' can also be done at this stage. 
• This is shown in fig.14.42(a) below:

Fig.14.42

Step 2: Note the branches and their positions. We have: '3,3-Diethyl'
• From this we get the following information:
    ♦ There are two branches. 
    ♦ Both are ethyl radicals
    ♦ Both the ethyl radicals are at position 3
• So the two branches can be attached to the main chain that we obtained in step 1
• The result is shown in fig.14.42(b) above
Step 3: This is the final step. 
• Fill all the valencies of carbon atoms. 
• Since it is a hydrocarbon, only carbon and hydrogen will be present. 
    ♦ So we use hydrogen to fill up the valencies. 
• The result is shown in fig.14.41(c) above. So this fig.14.41(c) shows the required structure
• Note that, there are no hydrogen atoms attached to the carbon atom at position 3 (there are two branches at position 3)

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In the next section, we will see Nomenclature of Unsaturated hydrocarbons.

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Chapter 14.3 - Hydrocarbons with repeating branches

In the previous section, we saw the naming procedure for hydrocarbons with more than one branch. In this section, we will see a special case coming under it.

This special case can be written as:

• The hydrocarbon has more than one branch     ♦ Some of those branches are repeated

An example is given below in fig.14.23:

Fig.14.23

We will first name it using the usual procedure:
Step 1: The longest chain is highlighted in red colour in fig.14.24 below. It is the main chain:

Fig.14.24

Step 2:
• There are two branches. Both are Methyl 
• In fig.14.24(a), the numbering is done from left to right 
    ♦ In this numbering, the first branch is the left methyl radical and it gets number 2
• In fig.14.24(b), the numbering is done from right to left 
    ♦ In this numbering, the first branch is the methyl radical on the right and it gets number 4
So the numbering in fig.14.24(a) is correct
    ♦ Position of the first branch is 2
    ♦ Position of the second branch is 4
Step 3:
• The 'finalized number of carbon atoms in the main chain' is 7
    ♦ So the word root is hept
• All the carbon-carbon bonds in the molecule are single bonds
   ♦ So the suffix is ane
Step 4:
• The names of the alkyl radicals which come as branches are: 
    ♦ methyl at position 2
    ♦ methyl at position 4
Step 5:
Assembling the name:
[2-methyl]
+hyphen
+[4-methyl]
+heptane
So the methyl radical is repeating. Based on what we have seen so far, the name would be: 
2-Methyl-4-methylheptane.
But . . .
• If a branch comes two times, it's name should not be written two times
    ♦ Instead, the prefix 'di' should be used 
• If a branch comes three times, it's name should not be written three times
    ♦ Instead, the prefix 'tri' should be used
• If a branch comes four times, it's name should not be written four times
    ♦ Instead, the prefix 'tetra' should be used
so on . . .
• The positions should be separated by commas
So the name of our compound is:
2,4-Dimethylheptane

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Another example:

Fig.14.25

We will first name it using the usual procedure:
Step 1: The longest chain is highlighted in red colour in fig.14.26 below. It is the main chain:

Fig.14.26

Step 2:
• There are two branches. Both are Methyl 
• In fig.14.26(a), the numbering is done from left to right 
    ♦ In this numbering, the first branch is the left methyl radical and it gets number 2
• In fig.14.26(b), the numbering is done from right to left 
    ♦ In this numbering, the first branch is the methyl radical on the right and it gets number 2
So the numbering in both fig.14.24(a)&(b) are correct
    ♦ Position of the first branch is 2
    ♦ Position of the second branch is 4
Step 3:
• The 'finalized number of carbon atoms in the main chain' is 5
    ♦ So the word root is pent
• All the carbon-carbon bonds in the molecule are single bonds
   ♦ So the suffix is ane
Step 4:
• The names of the alkyl radicals which come as branches are: 
    ♦ methyl at position 2
    ♦ methyl at position 4
Step 5:
Assembling the name:
[2-methyl]
+hyphen
+[4-methyl]
+pentane
So the methyl radical is repeating. Based on what we have seen in previous sections, the name would be: 
2-Methyl-4-methylpentane
But as we have seen in the first example above in this section,
• If a branch comes two times, it's name should not be written two times
    ♦ Instead, the prefix 'di' should be used 
• If a branch comes three times, it's name should not be written three times
    ♦ Instead, the prefix 'tri' should be used
• If a branch comes four times, it's name should not be written four times
    ♦ Instead, the prefix 'tetra' should be used
so on . . .
• The positions should be separated by commas
So the name of our compound is:
2,4-Dimethylpentane

---

We will now see an interesting case:
Consider the compound shown in fig.14.27 below:

Fig.14.27

Let us name it by the usual procedure:
Step 1: The longest chain is highlighted in red colour in fig.14.28 below. It is the main chain:

Fig.14.28

Step 2:
• There are three branches. All are Methyl 
• In fig.14.28(a), the numbering is done from left to right 
    ♦ In this numbering, the first branch is the left methyl radical and it gets number 2
• In fig.14.28(b), the numbering is done from right to left 
    ♦ In this numbering, the first branch is the methyl radical on the right and it gets number 2
• The numbering in both fig.14.28(a)&(b) seems to be correct
From fig.14.28(a), we get:
    ♦ Position of the first branch is 2
    ♦ Position of the second branch is 4
    ♦ Position of the second branch is 5
From fig.14.28(b), we get:
    ♦ Position of the first branch is 2
    ♦ Position of the second branch is 3
    ♦ Position of the second branch is 5
■ So we have a situation:
• Two methods of numbering gives the same number for the first branch
• In such a situation, we must consider the second branch. That second branch must get the lowest number.
• In our case, the numbering in fig.14.28(b) gives a lower number for the second branch. So it is correct
• The method in fig.14.28(a) is wrong. So an '×' mark is given above it.
Step 3:
• The 'finalized number of carbon atoms in the main chain' is 6
    ♦ So the word root is hex
• All the carbon-carbon bonds in the molecule are single bonds
   ♦ So the suffix is ane
Step 4:
• The names of the alkyl radicals which come as branches are:
    ♦ methyl at position 2
    ♦ methyl at position 3
    ♦ methyl at position 5
Step 5:
Assembling the name:
[2-methyl]
+hyphen
+[3-methyl]
+hyphen
+[5-methyl]
+hexane
• So the methyl radical is repeating. Based on what we have seen in previous sections, the name would be: 
2-Methyl-3-methyl-5-methylhexane
• Base on IUPAC rules that we saw in this section, we get:
2,3,5-Trimethylhexane.

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Another example in this category:
Consider the compound shown in fig.14.29 below:

Fig.14.29

Let us name it by the usual procedure:
Step 1: The longest chain is highlighted in red colour in fig.14.30 below. It is the main chain:

Fig.14.30

Step 2:
• There are three branches. All are Methyl 
• In fig.14.30(a), the numbering is done from left to right 
    ♦ In this numbering, the first branch is the left methyl radical and it gets number 2
• In fig.14.30(b), the numbering is done from right to left 
    ♦ In this numbering, the first branch is the methyl radical on the right and it gets number 2
• The numbering in both fig.14.30(a)&(b) seems to be correct
From fig.14.30(a), we get:
    ♦ Position of the first branch is 2
    ♦ Position of the second branch is 3
    ♦ Position of the second branch is 6
From fig.14.28(b), we get:
    ♦ Position of the first branch is 2
    ♦ Position of the second branch is 5
    ♦ Position of the second branch is 6
■ So we have a situation:
• Two methods of numbering gives the same number for the first branch
• In such a situation, we must consider the second branch. That second branch must get the lowest number.
• In our case, the numbering in fig.14.30(a) gives a lower number for the second branch. So it is correct
• The method in fig.14.30(b) is wrong. So an '×' mark is given above it.
Step 3:
• The 'finalized number of carbon atoms in the main chain' is 7
    ♦ So the word root is hept
• All the carbon-carbon bonds in the molecule are single bonds
   ♦ So the suffix is ane
Step 4:
• The names of the alkyl radicals which come as branches are:
    ♦ methyl at position 2
    ♦ methyl at position 3
    ♦ methyl at position 6
Step 5:
Assembling the name:
[2-methyl]
+hyphen
+[3-methyl]
+hyphen
+[6-methyl]
+heptane
• So the methyl radical is repeating. Based on what we have seen in previous sections, the name would be: 
2-Methyl-3-methyl-6-methylheptane
• Based on IUPAC rules that we saw in this section, we get:
2,3,6-Trimethylheptane.

---

Next we will see another interesting case in this category
Consider the compound shown in fig.14.31 below:

Fig.14.31

Let us name it by the usual procedure:
Step 1: The longest chain is highlighted in red colour in fig.14.32 below. It is the main chain:

Fig.14.32

Step 2:
• There are two branches. Both are Methyl 
• In fig.14.32(a), the numbering is done from left to right 
    ♦ In this numbering, both the branches get the number 2
• In fig.14.32(b), the numbering is done from right to left 
    ♦ In this numbering, both the branches get the number 2
• The numbering in both fig.14.32(a)&(b) are correct
Step 3:
• The 'finalized number of carbon atoms in the main chain' is 3
    ♦ So the word root is prop
• All the carbon-carbon bonds in the molecule are single bonds
   ♦ So the suffix is ane
Step 4:
• The names of the alkyl radicals which come as branches are:
    ♦ methyl at position 2
    ♦ methyl at position 2
Step 5:
Assembling the name:
[2-methyl]
+hyphen
+[2-methyl]
+hyphen

+propane
• So the methyl radical is repeating. Based on what we have seen in previous sections, the name would be: 
2-Methyl-2-methylpropane
• Base on IUPAC rules that we saw in this section, we get:
2,2-Dimethylhexane.
That is., if two identical branches have the same position number, then that number should be repeated.

---

Let us see another example:

Fig.14.33

• The numbering options are shown in fig.14.34 below:

Fig.14.34

• The reader may write the steps in his/her own notebooks. The IUPAC name of this compound is:
2,2,3-Trimethylpentane

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In the next section, we will see a summary of what we have seen so far in this chapter. We will also see some solved examples.

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