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How many numbers of isomers are possible for disubstituted benzene?

Last updated date: 23rd Jul 2024
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Hint :We know that the isomer can be defined as the isomer which differs in position of the substituted atom in the chain or ring. The number of position isomers possible for naphthalene is equal to the number of hydrogen atoms present on the ring or we can say the number of replaceable hydrogen atoms present.

Complete Step By Step Answer:
As we know, structural isomers are those molecules that have the same chemical formula but different chemical structures. The three structural isomers of substituted benzene rings are named as ortho, meta and para. The prefixes ortho (o), meta (m) and para (p) can be used to indicate the relative position of substituents in double substituted benzenes. The relative position of substituents in ortho-substituted benzenes let us understand the structure of the compound given to us. We have dibromo i.e. two bromine atoms are there and a naphthalene ring is present. Further, it is said that each ring has one halogen i.e. one bromine atom. We already know that naphthalene consists of two benzene rings that are fused. These rings are resonance stabilized and have many properties similar to benzene. Each ring in naphthalene consists of one bromine atom substituted at any place.
The structures of ortho and Meta isomers are not so symmetrical like para isomers and they don’t form a close lattice structure. Thus, the para form will have a higher melting point than ortho and meta forms. The solid that will be in the form of residue as given in the question, would be para form since both the other forms would be liquid at lower temperatures than that of para isomer. For disubstituted benzene, the possible isomers are ortho, Meta and para.

Note :
Remember that there might be a possible chance of confusion between boiling and melting point of these isomers. The melting point order is p-dihalo benzene $ > $ o-dihalo benzene $ > $ m-dihalo benzene. The boiling point order is o-dihalo benzene $ > $ p-dihalo benzene $ > $ m-dihalo.