# How can a compound having a plane of symmetry be Chiral?

The compound( Given in the image below) is treated with $$\ce{excess H2 in Pt}$$.

Here, (a) - The given compound. (b) - What the answer should be according to me. (c) - What the actual answer is:

Q1 - How is the chirality center at the top right even a chiral center, as the carbon has 2 identical groups attached to it?

Q2 - How is the compound as a whole a chiral compound? As far as I'm aware, compounds having a plane of symmetry (passing through the two 'chiral' centers in this compound) are supposed to be achiral?

• Who wrote diastereomers under c & d. The book? Because if it was you is surprising that you ask this question. See the detailed answer by Lycoris, anyway. – Alchimista May 9 at 8:21
• If you draw the methyl group attached to the cyclohexane with a dash, then either a) you do not know (e.g., lack of experimental evidence), or b) you do not want to assign the configuration of the carbon atom of the cyclohexane binding to the methyl group (e.g., because it is irrelevant for the discussion, or you know it is racemic mixture about this carbon atom). Wedges as in c) or d) however specify the absolute configuration to be either (R), or (S). – Buttonwood May 9 at 9:36

I think you are getting confused between chirality and E-Z isomerism in cyclic structures. So let's clear that up first.

According to IUPAC:

chirality
The geometric property of a rigid object (or spatial arrangement of points or atoms) of being non-superposable on its mirror image; such an object has no symmetry elements of the second kind (a mirror plane, σ = S1, a centre of inversion, i = S2, a rotation-reflection axis, S2n). If the object is superposable on its mirror image the object is described as being achiral.

Chiral molecules are always dissymmetric (lacking $$S_n$$) but not always asymmetric (lacking all symmetry elements except the trivial identity). However, asymmetric molecules are necessarily always chiral.

For a basic E-Z isomerism definition:

A cycloalkane has two distinct faces, and any substituent on a ring lies toward one of two faces. When two substituents on a ring point to the same face, they are cis. When the two substituents point to opposite faces, they are trans.

Image courtesy I'm sure you must've figured out by now, that the carbon on top isn't a capable of showing optical isomerism. However, due to the dissimilar substituents present ($$\ce{H\text{ and } CH3}$$), the compound can show cis-trans isomerism. Further reading: