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Simun Mishra posted an Question
June 12, 2020 • 12:33 pm 30 points
  • IIT JAM
  • Chemistry (CY)

What is ansa compound ?

3 Answer(s) Answer Now
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  • Suman Kumar

    See the two Pos it has without substitution. Horizontal plane is the paper itself & one vertical plane

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  • Suman Kumar

    These 4 position I am taking about

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    If these 4 are substituted then they will be Optically active

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    these 4 are substituted then they will be optically active means

  • Suman Kumar

    If any of those four carbon is substituted then ansa compound will become optically active

  • Suman Kumar best-answer

    Ansa compound simply is both para position of benzene is substituted with hetero atom & they are connected through polymethylene chain

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    substituted means

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    replaced

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    Shivaji khadake Best Answer

    Ansa Compounds are derivatives of Hydroquinone If two Para Position of Aromatic Ring is attached to hetero atom and they are connected through the polymethylene chain. cyclophane in which the para positions of a benzene ring are connected by a bridge of at least 10-12 atoms. By extension, any arene bridged by a chain constrained to lie over one of the two faces of the arene.13-Bromo-1,10-dioxa[8]paracyclophane is an example of a chiral and optically active ansa compound. The chirality plane lies inside the plane of the aromatic ring. The rotation of the aromatic ring is restricted by steric interactions between the aliphatic bridge and the bromine at the aromatic ring. An ansa-metallocene is a type of organometallic compound containing two cyclopentadienyl ligands that are linked by a bridging group such that both cyclopentadienyl groups are bound to the same metal. The link prevents rotation of the cyclopentadienyl ligand and often modifies the structure and reactivity of the metal center. Some ansa-metallocenes are active in Ziegler-Natta catalysis, although none are used commercially. I feel now you may get the idea regarding Ansa compounds which looks like exactly given in the question very much for asking this question if you have any doubt you can ask

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