In CICH2C=CCH2Br, why is bromine given more priority than chlorine?
In CICH2C=CCH2Br, why is bromine given more priority than chlorine?
There are few aspects of answering this.
- While writing, as most of the authors have mentioned - Due to alphabetical preference, Bromo will be mentioned before Chloro, irrespective of their C number (position on longest chain), in IUPAC nomenclature.
- While numbering, however, both Bromo & Chloro being substituents, do not enjoy any numbering priority, in case any major functional group is absent. Here we go by lowest sum rule. For e.g. - CH3CH(Cl)CH2CH(Br)CH2CH3 will be named as 4-Bromo-2-chlorohexane & not as 3-Bromo-5-chlorohexane. (lowest sum rule - I am providing a very brief idea of it below the end of this answer in case you are not knowing it )
- While numbering, if from either side of the longest chain, both the substituents have same position, then lowest sum rule is redundant. In this case, both number wise and naming wise, Bromo is preferred over Chloro. For e.g. - CH3CH(Cl)CH2CH(Br)CH3 will be named as 2-Bromo-4-chloropentane & never as 4-Bromo-2-chloropentane.
Lowest sum rule comes when u need to give names to the compounds. when the functional group are of same priority and is connected to a long chain carbon compound, then this rule comes into play. For example C-CBr- C-C-C.
so if i number from left side the carbon with Br gets a number of two whereas if i start give it number from right side , then it would get 4. Therefore ….. Left side numbering name :- 2 bromo pentane……….. Right side numbering name 4 bromo pentane. Therefore a low number name would be correct i.e., 2 bromo pentane. take another example of C-C-CCl-CBr-C-C-C. here in this case if i number it fro left side then carbon with Cl and C with Br would get a number of 3 and 4 respectively, but if i number it from the right side C with Cl and C with Br would get a number of 5 and 4 respectively.
Then the sum of all locants(number is known as locants)
left side -: 3+4=7
Right Side :- 5+4=9
hence the name 4 bromo 3 chloro heptane is correct.
There are few aspects of answering this.
- While writing, as most of the authors have mentioned - Due to alphabetical preference, Bromo will be mentioned before Chloro, irrespective of their C number (position on longest chain), in IUPAC nomenclature.
- While numbering, however, both Bromo & Chloro being substituents, do not enjoy any numbering priority, in case any major functional group is absent. Here we go by lowest sum rule. For e.g. - CH3CH(Cl)CH2CH(Br)CH2CH3 will be named as 4-Bromo-2-chlorohexane & not as 3-Bromo-5-chlorohexane. (lowest sum rule - I am providing a very brief idea of it below the end of this answer in case you are not knowing it )
- While numbering, if from either side of the longest chain, both the substituents have same position, then lowest sum rule is redundant. In this case, both number wise and naming wise, Bromo is preferred over Chloro. For e.g. - CH3CH(Cl)CH2CH(Br)CH3 will be named as 2-Bromo-4-chloropentane & never as 4-Bromo-2-chloropentane.
Lowest sum rule comes when u need to give names to the compounds. when the functional group are of same priority and is connected to a long chain carbon compound, then this rule comes into play. For example C-CBr- C-C-C.
so if i number from left side the carbon with Br gets a number of two whereas if i start give it number from right side , then it would get 4. Therefore ….. Left side numbering name :- 2 bromo pentane……….. Right side numbering name 4 bromo pentane. Therefore a low number name would be correct i.e., 2 bromo pentane. take another example of C-C-CCl-CBr-C-C-C. here in this case if i number it fro left side then carbon with Cl and C with Br would get a number of 3 and 4 respectively, but if i number it from the right side C with Cl and C with Br would get a number of 5 and 4 respectively.
Then the sum of all locants(number is known as locants)
left side -: 3+4=7
Right Side :- 5+4=9
hence the name 4 bromo 3 chloro heptane is correct.
