What are the minimum number of 'C' atoms for alkanes that are required to show positional isomerism.
Note that in question alkanes are mentioned not alkenes.
Please expalin the answer.
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first of all you spelled explain wrong, its not expalin its explain. But the answer is 64.
Yuvrajpaul:
Answer is 6 bro not 64..Do you study in 6th standard??
? or 5th
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Position isomerism requires the presence of a functional group, substituent, a double bond or a triple bond.
So unless there is a functional group or substituent in the alkane, it can't exhibit position isomerism.
If we take three basic functional methyl (CH3-) group, then the number of carbons on the main chain would be 5, as the group can be placed on either the 3rd carbon or the 2nd carbon.
Hope you find my answer helpful:)
So unless there is a functional group or substituent in the alkane, it can't exhibit position isomerism.
If we take three basic functional methyl (CH3-) group, then the number of carbons on the main chain would be 5, as the group can be placed on either the 3rd carbon or the 2nd carbon.
Hope you find my answer helpful:)
It can exhibit without any functional group.Remember it can have carbon substituents as well...
And hi
....
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