In the previous section, we saw how the green rectangles are named when there are branches inside those rectangles. In this section, we will see nomenclature of cyclic compounds.
Nomenclature of cyclic compounds
This can be written in 6 steps:
1. We know that, cyclic compounds are formed when the first C atom of a chain bonds to the last C atom of that chain.
2. If the chain is an alkane, we can simply put ‘cyclo’ in front of that alkane. The resulting word will be the IUPAC name of that cyclic compound.
• However, this rule is applicable only if there is only one ring in that cyclic compound.
• An example is given in fig.12.36(a) below. It's name is Cyclopentane.
 |
| Fig.12.36 |
3. If there are side chains, we can use the same eight rules that we saw in the previous sections for alkanes.
• Let us see an example. It can be written in 3 steps:
(i) In fig.12.36(b) above, the numbering is done in such a way that, branches get numbers 1,3.
• Fig.c shows the same compound. The numbering is done in such a way that, branches get numbers 1,5
(ii) Next we calculate the sum of the numbers:
♦ In fig.b, the sum of numbers is: (1+3) = 4
♦ In fig.c, the sum of numbers is: (1+5) = 6
(iii) We must always choose the numbering which gives the smallest sum. So the numbering in fig.b is correct. We must discard fig.c
♦ Based on fig.b, the IUPAC name will be 1,3-Dimethylcyclohexane
♦ Based on fig.c, the name will be 1,5-Dimethylcyclohexane.
✰ This name is not valid.
• Let us see another example. It can be written in 3 steps:
(i) In fig.12.36(d) above, the numbering is done in such a way that, branches get numbers 1,2,4.
• Fig.e shows the same compound. The numbering is done in such a way that, branches get numbers 1,3,4
(ii) Next we calculate the sum of the numbers:
♦ In fig.d, the sum of numbers is: (1+2+4) = 7
♦ In fig.e, the sum of numbers is: (1+3+4) = 8
(iii)
We must always choose the numbering which gives the smallest sum. So
the numbering in fig.d is correct. We must discard fig.c
♦ Based on fig.d, the IUPAC name will be 1,2,4-Trimethylcyclohexane
♦ Based on fig.e, the name will be 1,3,4-Trimethylcyclohexane.
✰ This name is not valid.
4. When two or more different branches are present, the numbering should be done in such a way that, the branch which comes first in the alphabetical listing, gets the lowest number '1'.
• Let us see an example. It can be written in 4 steps:
(i) In fig.12.37(a) below, the numbering is done in such a way that, branches get numbers 1,2.
♦ So methyl group gets the number 1, which gives 1-methyl
♦ Also ethyl group gets the number 2, which gives 2-ethyl
 |
| Fig.12.37 |
(ii) Fig.b shows the same compound. Here also, the numbering is done in such a way that, branches get the same numbers 1,2
♦ But methyl group gets the number 2, which gives 2-methyl
♦ Also ethyl group gets the number 1, which gives 1-ethyl
(iii) In the alphabetical listing, ethyl comes before methyl.
♦ So ethyl must be given the number 1
(iv) Therefore, we must follow the numbering in fig.b. We must discard fig.a
♦ Based on fig.b, the IUPAC name of the compound is: 1-Ethyl-2-methylcyclohexane
♦ Based on fig.a, the name will be 2-Ethyl-1-methylcyclohexane
✰ This name is not valid.
• Let us see another example. It can be written in 4 steps:
(i) In fig.12.37(c) above, the numbering is done in such a way that, branches get numbers 1,3.
♦ So methyl group gets the number 1, which gives 1-methyl
♦ Also propyl group gets the number 3, which gives 3-propyl
(ii) Fig.d shows the same compound. Here also, the numbering is done in such a way that, branches get the same numbers 1,3
♦ But methyl group gets the number 3, which gives 3-methyl
♦ Also propyl group gets the number 1, which gives 1-propyl
(iii) In the alphabetical listing, methyl comes before propyl.
♦ So methyl must be given the number 1
(iv) Therefore, we must follow the numbering in fig.c. We must discard fig.d
♦ Based on fig.c, the IUPAC name of the compound is: 1-Methyl-3-propylcyclohexane
♦ Based on fig.d, the name will be 3-Methyl-1-propylcyclohexane
✰ This name is not valid.
5. In some cases, a C atom in the ring may contain more than one branches. In such cases, the C atom with the greater number of branches should be given the lowest number 1.
• Let us see an example. It can be written in 4 steps:
(i) In fig.12.38(a) below, the numbering is done in such a way that, branches get numbers 1,3.
♦ So methyl groups get the number 1, which gives 1,1-dimethyl
♦ Also ethyl group gets the number 3, which gives 3-ethyl
 |
| Fig.12.38 |
(ii) Fig.d shows the same compound. Here also, the numbering is done in such a way that, branches get the same numbers 1,3
♦ But methyl groups get the number 3, which gives 3,3-dimethyl
♦ Also ethyl group gets the number 1, which gives 1-ethyl
(iii) According to the IUPAC rule, the C atom which has the two methyl groups must be given the number 1
(iv) Therefore, we must follow the numbering in fig.a. We must discard fig.b
♦ Based on fig.a, the IUPAC name of the compound is: 3-Ethyl-1,1-dimethylcyclohexane
(Alphabetically, 'E' of Ethyl, comes before 'm' of methyl. Recall that, 'di' is not considered for alphabetical listing)
♦ Based on fig.b, the name will be 1-Ethyl-3,3-dimethylcyclohexane
✰ This name is not valid.
6. Let us see an example in which there is a branch within the branch. It can be written in 3 steps:
(i) Consider the branch in fig.12.38(c) above. There is a branch within that branch.
(ii) So we must first find the IUPAC name of that branch.
• Applying rule 1, we see that the parent chain in that branch contains five C atoms.
• Applying rule 2, we see that ‘pent’ must be used.
•
Applying rule 3 needs special care. Since the branch is attached to the
ring, there will be one H atom less. So we must use 'pentyl'
instead of 'pentane'
• Applying rule 4, we see that, the branch is methyl group.
•
Applying rule 5 needs special care. We must always start numbering from
that C atom which is attached to the ring.
• Applying rule 6, we get: 4-methyl
• Applying rule 7: There is no need to apply rule 7 because, there is only one branch.
• Applying rule 8, We get: 4-Methylpentyl.
♦ This must be written in parenthesis. We get: (4-Methylpentyl)
(iii) So the final name will be: (4-Methylpentyl)cyclohexane.
In the next section, we
will see nomenclature of compounds having functional groups.
Previous
Contents
Next
Copyright©2021 Higher secondary chemistry.blogspot.com
No comments:
Post a Comment