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How come the given compound in the diagram is optically inactive? The two rings should be in different planes due to steric hindrance between $\ce{COOH}$ and bromine. How am I wrong?

why isn't this compound optically active?

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    $\begingroup$ There's a plane of symmetry $\endgroup$ – Zenix Apr 27 '20 at 17:06
  • $\begingroup$ Can you plz explain further.. What I understand is that if the two rings stay in one plane then sure one can imagine a mirror plane. But aren't the two rings in different planes? How does a plane of symmetry come in the picture then? $\endgroup$ – Loveforphysics Apr 27 '20 at 17:11
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    $\begingroup$ As @Zenix states, planar or not, it's symmetrical. Make a paper cutout, or tinker-toy model, and flip it over. $\endgroup$ – DrMoishe Pippik Apr 27 '20 at 17:43
  • $\begingroup$ Related: chemistry.stackexchange.com/questions/65064/… $\endgroup$ – Tyberius Apr 27 '20 at 18:07
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As users Zenix and DrMoishe Pippik indicate there is symmetry in the compound.

If the two phenyl groups are planar, then there is one plane of symmetry which would be the plane of the image as shown. The bond between the two phenyl groups would be a $C_2$ axis of rotation.

If the two phenyl groups are perpendicular, then there are two planes of symmetry, one with each of the phenyl groups. The two planes would intersect along the bond between the two phenyl groups, and that intersection line would again be a $C_2$ axis of rotation.

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