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Could anyone explain why compound L is not optically active?

(1r,3R,5S)-3,5-dimethyl-4-methylidenecyclohexan-1-ol

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    $\begingroup$ Draw the mirror image of the compound. Can you prove the mirrored version is different from the original? $\endgroup$ – Nicolau Saker Neto Jun 9 '19 at 4:42
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    $\begingroup$ For the same reason why meso-tartaric acid does not manifest chirality. Its mirror image can be identified with the original, even if it has 2 chiral centers. $\endgroup$ – Poutnik Jun 9 '19 at 13:49
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    $\begingroup$ Because it has a quite obvious plane of symmetry, period. That's enough of a justification, no matter what the chiral centers are trying to say. $\endgroup$ – Ivan Neretin Jun 9 '19 at 20:15
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The given molecule,(1r,3R,5S)-3,5-dimethyl-4-methylenecyclohexanol , is optically inactive .This molecule has a plane of symmetry (as shown below) and, therefore, is achiral. enter image description here

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Imagine yourself cutting the molecule in half along the axis determined by the OH group and the double bond, perpendicular to the plane of the cyclohexane ring. Observe that the 2 halves match each other perfectly, in other words you found a plane of symmetry. For any molecule to be chiral, and thus be optically active, it must not have any plane of symmetry.

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