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I was going through my text book, (Solomon&Frhyle) I came across a couple of of examples which I didn't understand.

1.

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How is the molecule symmetrical? Upon folding the molecule over its plane of symmetry the cooh group will be placed over fluorine.

2.

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Similarly in this in structures 3,4,5,6 a chlorine atom out of the plane and below the plane coincide.

3.enter image description here It's written that for biphenyls to be chiral a can be the same as a'. (Same for b and b') But If a a' and b b' are the same them won't a plane of symmetry lie in the middle perpendicular to the plane of molecule.


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Finding planes of symmetry is simple: try all planes, some (or maybe none) of them will fit.

As for your examples, the first image is just poorly drawn. The plane is supposed to be the one of the ring containing COOH and F, while the other ring is perpendicular to it (which, BTW, I don't believe is the native conformation of the molecule, but whatever).

The second example asserts right things, but for the wrong reasons. (This is not what you would normally expect of a textbook, to put it mildly.) True, each of these structures contains at least one plane of symmetry. Too bad the planes are drawn incorrectly in all structures except for the first two. You may take your time and fiddle around to find out where they really are.

In the third example, you would be right if the molecule were planar, which it is totally not. Steric repulsion between a and a' (also, between b and b') makes that impossible, even if all of the substituents a, a', b, b' are simply hydrogens.

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  • $\begingroup$ What angle do you suggest between the phenyl rings in the two biphenyl systems? $\endgroup$ – Jan Dec 26 '17 at 15:37
  • $\begingroup$ Anything between $40^\circ$ and $60^\circ$ sounds reasonable to me. $\endgroup$ – Ivan Neretin Dec 26 '17 at 15:42
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    $\begingroup$ @IvanNeretin : you're right, example 2 is really bizarre. For a minute I thought they meant axes of symmetry by those dotted lines, but that would only apply to molecule 4, as far as I can see. It will forever remain a mystery what the authors meant, but at face value a really poor didactic example indeed... $\endgroup$ – user6376297 Dec 26 '17 at 17:49
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    $\begingroup$ For the OP. If your point is to check if the molecule is chiral or not, you don't necessarily need to 'find a plane'. Some achiral molecules don't even have (in some conformations) a plane of symmetry, but a rotoreflection axis instead. To keep it simple, the idea is that a molecule, fixed in a rigid conformation that is as symmetrical as possible yet chemically plausible, is chiral if there is no combination of rotation or translation by which it can be superimposed to its mirror image. So just draw it, make its mirror image and check. $\endgroup$ – user6376297 Dec 26 '17 at 17:56

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