Chapter 14.6 - Structure from names of Alkenes and Alkynes

In the previous section, we saw the naming procedure for unsaturated hydrocarbons. In this section, we will see a reverse process. That is., we will be given the IUPAC name of an alkene or an alkyne. We must draw it's structure.

We will learn the process with the help of an example:
Given: IUPAC name is hex-3-ene
We will write the procedure in steps:
1. The word root is hex. So there will be 6 carbon atoms. 
• Draw these 6 carbon atoms with out any bonds between them. 
• Number the carbon atoms from left to right
• This is shown in fig.14.50(a) below:

Fig.14.50

2. The suffix is ene. So there will be a double bond.
• The position of the bond is given as 3 
• This indicates that, the bond is in between carbon atoms 3 and 4
    ♦ The bond cannot be in between carbon atoms 2 and 3
    ♦ This is because, the IUPAC name would then be hex-2-ene
• So draw a double bond between carbon atoms 3 and 4
    ♦ All other carbon-carbon bonds can be shown as single bonds
• This is shown in fig.14.50(b)
3. 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.14.50(c) above. So this fig.14.50(c) shows the required structure

Another example:
Given: IUPAC name is but-1-yne
We will write the procedure in steps:
1. The word root is but. So there will be 4 carbon atoms. 
• Draw these 4 carbon atoms with out any bonds between them. 
• Number the carbon atoms from left to right
• This is shown in fig.14.51(a) below:

Fig.14.51

2. The suffix is yne. So there will be a triple bond.
• The position of the bond is given as 1 
• This indicates that, the bond is in between carbon atoms 1 and 2
    ♦ There is no other possibility because, 1 is the lowest possible number
• So draw a triple bond between carbon atoms 1 and 2
    ♦ All other carbon-carbon bonds can be shown as single bonds
• This is shown in fig.14.51(b)
3. 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.14.51(c) above. So this fig.14.51(c) shows the required structure
• Note that, there is only one hydrogen atom for the first carbon and no hydrogen for the second carbon. Reader may write the reason for this in his/her own notebooks.

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Now we will see some solved examples based on what we have discussed in this section and the previous section

Solved example 14.4:
Write the IUPAC name of the compound shown in fig.14.52(a) below:

Fig.14.52

Solution:
1. Two possible methods of 'numbering of carbon atoms' are shown in figs.(b) and (c)
• In fig.(b), the numbering is done from left to right
    ♦ In this, the 'carbon atoms linked to the double bond' gets 4 and 5
• In fig.(c), the numbering is done from right to left
    ♦ In this, the 'carbon atoms linked to the double bond' gets 2 and 3
2. According to the IUPAC rules, 'carbon atoms linked to the double bond' should get the lowest number. In our present problem, 
• The numbering in fig.(b) gives 4 and 5
• The numbering in fig.(c) gives 2 and 3
3. The numbering in fig.(c) gives the lowest number 2. So it is the correct method of numbering.
4. Now we can assemble the name. The rule for assembling is:
Word root+hyphen+position of double bond+hyphen+suffix
• As there are 6 carbon atoms, the word root is hex
• As one carbon-carbon bond is a double bond, the suffix is ene 
So we get: hex-2-ene.

Solved example 14.5
Given: IUPAC name is hex-3-yne. Draw the structure
Solution:
We will write the procedure in steps:
1. The word root is hex. So there will be 6 carbon atoms. 
• Draw these 6 carbon atoms with out any bonds between them. 
• Number the carbon atoms from left to right
• This is shown in fig.14.53(a) below:

Fig.14.53

2. The suffix is yne. So there will be a triple bond.
• The position of the bond is given as 3 
• This indicates that, the bond is in between carbon atoms 3 and 4
    ♦ The bond cannot be in between carbon atoms 2 and 3
    ♦ This is because, the IUPAC name would then be hex-2-yne
• So draw a triple bond between carbon atoms 3 and 4
    ♦ All other carbon-carbon bonds can be shown as single bonds
• This is shown in fig.14.53(b)
3. 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.14.53(c) above. So this fig.14.53(c) shows the required structure
• Note that, there is no hydrogen for the third and fourth carbons. Reader may write the reason for this in his/her own notebooks.

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In the next section, we will see functional groups.

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Chapter 14.5 - IUPAC Nomenclature of Unsaturated hydrocarbons

In the previous section, we saw the naming procedure for saturated hydrocarbons. In this section, we will see unsaturated hydrocarbons.

We have seen the basics of nomenclature of unsaturated hydrocarbons in a previous chapter. Let us refresh our memory:
• Alkenes and Alkynes come under the category of unsaturated hydrocarbons.
    ♦ C₄H₁₀ is an alkane. 
    ♦ C₄H₈ is an alkene
    ♦ C₄H₆ is an alkyne 
• Note that all of them have 4 carbon atoms. But the 'number of hydrogen atoms' is different. 
• We know that the valencies in saturated hydrocarbons (alkanes) are satisfied by single bonds. 
    ♦ We saw an example in fig.14.1 earlier in this chapter.
• We know 'how the valencies are satisfied (with the help of double bonds) in alkenes'. See fig.8.11.
   ♦ A larger molecule is shown in fig.14.43(a) below:

Fig.14.43

• We know 'how the valencies are satisfied (with the help of triple bonds) in alkynes'. See fig.8.14.
    ♦ A larger molecule is shown in fig.14.43(b) above
■ Note that:
• Which ever carbon atom we take from the above fig.14.43(a) or (b), there will be 4 pairs of electrons around it. 
    ♦ So there will be eight electrons available to any carbon atom that we take
• Which ever hydrogen atom we take, there will be 1 pair of electrons around it. 
    ♦ So there will be two electrons available to any hydrogen atom that we take

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Now we will see their nomenclature:

■ For our present discussion on 'nomenclature of alkenes and alkynes', we will be considering only 'straight chains'. 
• That means, there will not be any 'branches'. 
• So for our present discussion on alkenes and alkynes, we do not need to worry about 'highlighting the main chain' as we did in the case of alkanes in the previous sections.
■ Consider the two alkenes in fig.14.44 below:

Fig.14.44

• The first compound has 4 carbon atoms and 8 hydrogen atoms. 
    ♦ So it's molecular formula is C₄H₈
• The second compound has 4 carbon atoms and 8 hydrogen atoms. 
    ♦ So it's molecular formula is also C₄H₈
■ But the structure of each is different from the other. This is due to the difference in the position of the double bond. 
• So it is clear that, each of them should be given distinct names. 
• Let us see how this is done. We will analyse it in steps:
1. The first compound is shown in fig.14.45(a) below:

Fig.14.45

2. Two possible methods of 'numbering of carbon atoms' are shown in figs.(b) and (c)
• In fig.(b), the numbering is done from left to right
    ♦ In this, the 'carbon atoms linked to the double bond' gets 1 and 2
• In fig.(c), the numbering is done from right to left
    ♦ In this, the 'carbon atoms linked to the double bond' gets 3 and 4
3. According to the IUPAC rules, 'carbon atoms linked to the double bond' should get the lowest number. In our present problem, 
• The numbering in fig.(b) gives 1 and 2
• The numbering in fig.(c) gives 3 and 4
4. The numbering in fig.(b) gives the lowest number 1. So it is the correct method of numbering.
5. Now we can assemble the name. The rule for assembling is:
Word root+hyphen+position of double bond+hyphen+suffix
• As there are 4 carbon atoms, the word root is but
• As one carbon-carbon bond is a double bond, the suffix is ene 
So we get: But-1-ene

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1. The second compound is shown in fig.14.46(a) below:

Fig.14.46

2. Two possible methods of 'numbering of carbon atoms' are shown in figs.(b) and (c)
• In fig.(b), the numbering is done from left to right
    ♦ In this, the 'carbon atoms linked to the double bond' gets 2 and 3
• In fig.(c), the numbering is done from right to left
    ♦ In this, the 'carbon atoms linked to the double bond' gets 2 and 3
3. According to the IUPAC rules, 'carbon atoms linked to the double bond' should get the lowest number. In our present problem, 
• The numbering in fig.(b) gives 2 and 3
• The numbering in fig.(c) gives 2 and 3
4. The numbering by both the methods gives the same numbers 2 and 3. So the lowest number is 2.
5. Now we can assemble the name. The rule for assembling is:
Word root+hyphen+position of double bond+hyphen+suffix
• As there are 4 carbon atoms, the word root is but
• As one carbon-carbon bond is a double bond, the suffix is ene 
So we get: But-2-ene

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The same steps can be adopted for alkynes also. The fig.14.47 below shows two alkynes. 

Fig.14.47

• The first compound has 4 carbon atoms and 6 hydrogen atoms. 
    ♦ So it's molecular formula is C₄H₆
• The second compound has 4 carbon atoms and 6 hydrogen atoms. 
    ♦ So it's molecular formula is also C₄H₆
■ But the structure of each is different from the other. This is due to the difference in the position of the triple bond. 
• So it is clear that, each of them should be given distinct names.
■ By following the same steps that we saw for alkenes, we will get:
• Name of the first compound: But-1-yne
• Name of the first compound: But-2-yne.
The reader may write the steps in his/her own notebooks

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Another example:
1. An alkene is shown in fig.14.48(a) below:

Fig.14.48

2. Two possible methods of 'numbering of carbon atoms' are shown in figs.(b) and (c)
• In fig.(b), the numbering is done from left to right
    ♦ In this, the 'carbon atoms linked to the double bond' gets 3 and 4
• In fig.(c), the numbering is done from right to left
    ♦ In this, the 'carbon atoms linked to the double bond' gets 2 and 3
3. According to the IUPAC rules, 'carbon atoms linked to the double bond' should get the lowest number. In our present problem, 
• The numbering in fig.(b) gives 3 and 4
• The numbering in fig.(c) gives 2 and 3
4. The numbering in fig.(c) gives the lowest number 2. So it is the correct method of numbering.
5. Now we can assemble the name. The rule for assembling is:
Word root+hyphen+position of double bond+hyphen+suffix
• As there are 5 carbon atoms, the word root is pent
• As one carbon-carbon bond is a double bond, the suffix is ene 
So we get: Pent-2-ene.

One more example:
1. An alkyne is shown in fig.14.49(a) below:

Fig.14.49

2. Two possible methods of 'numbering of carbon atoms' are shown in figs.(b) and (c)
• In fig.(b), the numbering is done from left to right
    ♦ In this, the 'carbon atoms linked to the triple bond' gets 4 and 5
• In fig.(c), the numbering is done from right to left
    ♦ In this, the 'carbon atoms linked to the triple bond' gets 1 and 2
3. According to the IUPAC rules, 'carbon atoms linked to the triple bond' should get the lowest number. In our present problem, 
• The numbering in fig.(b) gives 4 and 5
• The numbering in fig.(c) gives 1 and 2
4. The numbering in fig.(c) gives the lowest number 1. So it is the correct method of numbering.
5. Now we can assemble the name. The rule for assembling is:
Word root+hyphen+position of double bond+hyphen+suffix
• As there are 5 carbon atoms, the word root is pent
• As one carbon-carbon bond is a triple bond, the suffix is yne 
So we get: Pent-1-yne

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In the next section, we will see the reverse process. That is., we will be given the IUPAC name of an alkene or an alkyne. We must draw it's structure.

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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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