In the previous section, we saw the tetrahedral structure of alkanes. In this section, we will see the nomenclature and isomerism in alkanes.
Some basics can be written in 16 steps:
1. Consider the two structures in fig.13.6 below:
 |
| Fig.13.6 |
• Based on the rules of nomenclature [see section 12.3] that we saw in the previous chapter, we can write:
♦ The structure in fig.13.6(a) is butane.
♦ The structure in fig.13.6(b) is 2-Methylpropane.
2. Let us compare the number of atoms:
♦ In fig.13.6(a), there are four C atoms and ten H atoms.
♦ In fig.13.6(b) also, there are four C atoms and ten H atoms.
• So both structures have the same molecular formula C4H10
• It is clear that butane and 2-Methylpropane are isomers.
• We have discussed about isomers in a previous section. [see section 12.9]
3. Consider the three structures in fig.13.7 below:
 |
| Fig.13.7 |
• Based on the rules of nomenclature that we saw in the previous chapter, we can write:
♦ The structure in fig.13.7(a) is pentane.
♦ The structure in fig.13.7(b) is 2-Methylbutane.
♦ The structure in fig.13.7(c) is 2,2-Dimethylpropane.
4. Let us compare the number of atoms:
♦ In fig.13.7(a), there are five C atoms and twelve H atoms.
♦ In fig.13.7(b) also, there are five C atoms and twelve H atoms.
♦ In fig.13.7(c) also, there are five C atoms and twelve H atoms.
• So all the three structures have the same molecular formula C5H12
• It is clear that pentane, 2-Methylbutane and 2,2-Dimethylpropane are isomers.
5. Consider the two isomers in fig.13.6
♦ The first isomer has a boiling point of 273 K
♦ The second isomer has a boiling point of 261 K
• Though the two isomers have the same numbers of C and H atoms, they differ in their boiling points and other properties.
• The difference in properties is due to the difference in structures. So they are called structural isomers.
6. Consider the three isomers in fig.13.7
♦ The first isomer has a boiling point of 309 K
♦ The second isomer has a boiling point of 301 K
♦ The third isomer has a boiling point of 282.5 K
• Though the three isomers have the same numbers of C and H atoms, they differ in their boiling points and other properties.
• The difference in properties is due to the difference in structures. So they are also structural isomers.
7. Consider the two isomers in fig.13.6
♦ The first isomer has a continuous chain of C atoms.
♦ The second isomer has a branched chain of C atoms.
• Structural isomers which differ in the chain of C atoms are called chain isomers.
• We can write:
Butane and 2-Methylpropane are chain isomers of C4H10
8. Consider the three isomers in fig.13.7
♦ The first isomer has a continuous chain of C atoms.
♦ The second isomer has a branched chain of C atoms.
♦ The third isomer has a different branched chain of C atoms.
• We can write:
Pentane, 2-Methylbutane and 2,2-Dimethylpropane are chain isomers of C5H12
9. In the case of alkanes, only those which have more than three C atoms will exhibit isomerism.
• The first three alkanes (methane, ethane and propane) do not exhibit isomerism.
• This is because, a minimum of four C atoms are required to obtain different structural arrangements.
10. We have seen that the first three members do not exhibit isomerism.
• We have seen the isomers of fourth and fifth members. Next let us see the isomers of the sixth member.
• The sixth member C6H14 have five chain isomers. They are shown in a condensed form in figs.13.8 (a), (b), (c), (d) and (e) below:
 |
| Fig.13.8 |
Here we can note an interesting point. It can be written in steps:
(i) Consider the structure shown on the right side.
• It appears that, an ethyl group is attached to a chain of four C atoms.
• So we are tempted to name it as 2-Ethylbutane.
(ii) But this name is wrong.
• Recall that, the longest chain is to be considered.
• So the correct name is 3-Methylpentane. This structure is already shown on the left side. It is one of the five isomers.
11. Now we will discuss about primary carbon atoms. It can be written in 5 steps:
(i) Consider the C atom numbered as ‘1’ in the n-Hexane in fig.13.8(a) above.
• This C atom is attached to:
♦ three H atoms.
♦ one C atom.
• Since there is only one C atom attached, the C atom numbered as ‘1’ is a primary carbon atom.
(ii) Consider the C atom numbered as ‘6’ in the same n-Hexane.
• This C atom is attached to:
♦ three H atoms.
♦ one C atom.
• Since there is only one C atom attached, the C atom numbered as ‘6’ is a primary carbon atom.
(iii) Consider the C atom in the methyl group in the 2-Methylpentane in fig.13.8(b) above.
• This C atom is attached to:
♦ three H atoms.
♦ one C atom.
• Since there is only one C atom attached, the C atom in the methyl group is a primary carbon atom.
◼ In fact, terminal carbon atoms are always primary carbon atoms.
• Both C atoms in ethane are primary carbon atoms.
(iv) Consider the C atom in methane.
• This C atom is attached to:
♦ three H atoms.
♦ zero C atoms.
• Carbon atoms attached to no other carbon atoms are also called primary carbon atoms.
• So the single C atom in methane is a primary carbon atom.
(v) Primary carbon atom is abbreviated as 1o carbon atom.
12. Now we will discuss about secondary carbon atoms. It can be written in 4 steps:
(i) Consider the C atom numbered as ‘2’ in the n-Hexane in fig.13.8(a) above.
• This C atom is attached to:
♦ two H atoms.
♦ two C atoms.
• Since there are two C atoms attached, the C atom numbered as ‘2’ is a secondary carbon atom.
(ii) Consider the C atom numbered as ‘5’ in the same n-Hexane.
• This C atom is attached to:
♦ two H atoms.
♦ two C atoms.
• Since there are two C atoms attached, the C atom numbered as ‘5’ is a secondary carbon atom.
(iii) Based on this information, we can write:
• The C atoms numbered as '3' and '4' in n-Hexane are also secondary carbon atoms.
• The C atoms numbered as '3' and '4' in 2-Methylpentane (fig.13.8.b) are also secondary carbon atoms.
• The reader may try to identify the secondary carbon atoms in the other structures in fig.13.8
(iv) Secondary carbon atom is abbreviated as 2o carbon atom.
13. Now we will discuss about tertiary carbon atoms. It can be written in 4 steps:
(i) Consider the C atom numbered as ‘2’ in the 2-Methylpentane in fig.13.8(b) above.
• This C atom is attached to:
♦ one H atom.
♦ three C atoms.
• Since there are three C atoms attached, the C atom numbered as ‘3’ is a tertiary carbon atom.
(ii) Consider the C atom numbered as ‘3’ in the 3-Methylpentane in fig.13.8(c) above.
• This C atom is attached to:
♦ one H atom.
♦ three C atoms.
• Since there are three C atoms attached, the C atom numbered as ‘3’ is a tertiary carbon atom.
(iii) Based on this information, we can write:
• The C atoms numbered as '2' and '3' in 2,3-Dimethylbutane (fig.13.8.d) are also tertiary carbon atoms.
(iv) Tertiary carbon atom is abbreviated as 3o carbon atom.
14. Now we will discuss about quaternary carbon atoms. It can be written in 4 steps:
(i) Consider the C atom numbered as ‘2’ in the 2,2-Dimethylbutane in fig.13.8(e) above.
• This C atom is attached to:
♦ zero H atoms.
♦ four C atoms.
• Since there are four C atoms attached, the C atom numbered as ‘2’ is a quaternary carbon atom.
(ii) Based on this information, we can write:
• The C atom numbered as '2' in the earlier fig.13.7.c is also quaternary carbon atom.
(iii) Quaternary carbon atom is abbreviated as 4o carbon atom.
15. We have seen that the first three members of the alkane series do not exhibit isomerism.
• Isomers exist for members with more than 3 carbon atoms.
♦ We have seen the two isomers of the fourth member.
♦ We have seen the three isomers of the fifth member.
♦ We have seen the five isomers of the sixth member.
• It is clear that, number of isomers increase with the number of C atoms.
♦ The seventh member C7H16 has nine isomers.
♦ The tenth member C10H22 has seventy five isomers.
16. We have seen the importance of alkyl groups while naming an organic compound. [See fig.12.24 in section 12.3]
• Alkyl groups are obtained by removing one H atom from the corresponding alkanes. So they will have a general formula CnH2n+1.
• In fig.13.8 (b) above, a methyl group is attached to the carbon numbered '2'
♦ -CH3 is methyl group.
♦ -CH2-CH3 is ethyl group. It can be written in condensed form as -C2H5.
♦ -CH2-CH2-CH3 is propyl group. It can be written in condensed form as -C3H7.
Now we will see a solved example
Solved example 13.2
Write structures of different isomeric alkyl groups corresponding to the molecular formula C5H11. Write IUPAC names of alcohols obtained by attachment of –OH groups at different carbons of the chain.
Solution:
1. The molecular formula C5H11 satisfies the general formula CnH2n+1
• So it indicates an alkyl group.
• The simplest alkyl group corresponding to C5H11 is the pentyl group. It is shown in fig.13.9(1.a) below:
 |
| Fig.13.9 |
• Any alkyl group will be unstable. It requires an electron to attain stability.
• If we attach an -OH group, it will become a stable molecule. We already know how the valencies are satisfied when the -OH group is attached. [See fig.14.62 in section 14.7 of our earlier chemistry classes]
• In our present case, when the -OH group is attached, the structure will be as shown in fig.13.9(1.b).
• But when the -OH group is attached, the whole structure will be called an alcohol. We must write the IUPAC name of that alcohol. [See section 12.6]
• In fig.13.9(1.b), the C atoms are numbered in such a way that the -OH group gets the smallest number. So the root name is pentan-1-ol
• Since there are no branches, the final name is also pentan-1-ol.
2. Another possible structure of the C5H11 is shown in fig.13.9(2.a) above.
• When the -OH group is attached, the structure will be as shown in fig.13.9(2.b).
• We must write the IUPAC name of this alcohol.
• In fig.13.9(2.b), the C atoms are numbered in such a way that the -OH group gets the smallest number. So the root name is pentan-2-ol
• Since there are no branches, the final name is also pentan-2-ol.
3. Another possible structure of the C5H11 is shown in fig.13.10(3.a) below:
 |
| Fig.13.10 |
• When the -OH group is attached, the structure will be as shown in fig.13.10(3.b).
•
We must write the IUPAC name of this alcohol.
•
In fig.13.10(3.b), the C atoms can be numbered from left to right or from right to left. In both methods, the -OH
group gets the smallest number 3. So the root name is pentan-3-ol
• Since there are no branches, the final name is also pentan-3-ol.
4. Another possible structure of the C5H11 is shown in fig.13.10(4.a) above.
• When the -OH group is attached, the structure will be as shown in fig.13.10(4.b).
•
We must write the IUPAC name of this alcohol.
•
In fig.13.10(4.b), the C atoms are numbered in such a way that the -OH
group gets the smallest number. So the root name is butan-1-ol
• Since there is a methyl group attached, the final name is 3-Methyl-butan-1-ol.
5. Another possible structure of the C5H11 is shown in fig.13.11(5.a) below:
 |
| Fig.13.11 |
• When the -OH group is attached, the structure will be as shown in fig.13.11(5.b).
•
We must write the IUPAC name of this alcohol.
•
In fig.13.11(5.b), the C atoms are numbered in such a way that the -OH
group gets the smallest number. So the root name is butan-1-ol
• Since there is a methyl group attached, the final name is 2-Methyl-butan-1-ol.
6. Another possible structure of the C5H11 is shown in fig.13.11(6.a) above.
• When the -OH group is attached, the structure will be as shown in fig.13.11(6.b).
•
We must write the IUPAC name of this alcohol.
•
In fig.13.11(6.b), the C atoms are numbered in such a way that the -OH
group gets the smallest number. So the root name is butan-2-ol
• Since there is a methyl group attached, the final name is 2-Methyl-butan-2-ol.
7. Another possible structure of the C5H11 is shown in fig.13.12(7.a) below:
 |
| Fig.13.12 |
• We must write the IUPAC name of this alcohol.
• In fig.13.12(7.b), the C atoms are numbered in such a way that the -OH group gets the smallest number. So the root name is propan-1-ol
• Since there are two methyl groups attached, the final name is 2,2-Dimethyl-propan-1-ol.
8. The last possible structure of the C5H11 is shown in fig.13.13(8.a) above.
• When the -OH group is attached, the structure will be as shown in fig.13.13(8.b).
• We must write the IUPAC name of this alcohol.
• In fig.13.11(8.b), the C atoms are numbered in such a way that the -OH group gets the smallest number. So the root name is butan-2-ol
• Since there is a methyl group attached, the final name is 3-Methyl-butan-2-ol.
In the next section we will see a few more solved examples.
Copyright©2022 Higher secondary chemistry.blogspot.com
No comments:
Post a Comment