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Phenolphthalein is colorless when pH = 0-8.2 and pink when pH = 8.2-12.

Why does the ring bearing the carboxylate (COO-) group not lying on the same plane as the other 2 rings?

enter image description here

Image source: https://www.learner.org/courses/chemistry/images/lrg_img/Pink.jpg

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    $\begingroup$ In fact, the other two rings aren't exactly in the same plane either. $\endgroup$
    – Ivan Neretin
    Sep 17, 2017 at 12:14
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    $\begingroup$ I suggest drawing it out on paper with constant bond lengths and perfect $120^\circ$ angles. ;) $\endgroup$
    – Jan
    Sep 17, 2017 at 12:36
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    $\begingroup$ @Jan I suggest making this an answer. $\endgroup$
    – Oscar Lanzi
    Sep 17, 2017 at 18:14
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    $\begingroup$ @OscarLanzi Yes, but I didn’t have access to ChemDraw when I commented (and won’t until tomorrow morning). In the meanwhile, DSVA has provided a very nice answer that I could hardly add anything to … $\endgroup$
    – Jan
    Sep 18, 2017 at 15:10

1 Answer 1

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Ok, let's look at this structure. I've drawn/calculated (B3LYP/6-31G) a 3D model of this structure and optimized the geometry.

enter image description here

So what do we see here? Actually no two rings are in one plane, but two are nearly planar (~40° of twist) while one is basically at 90° twist. Let's first look at the one with the carboxy group which is twisted a lot. The carboxy group is in ortho position. If we twist the ring to be in a plane with one of the other rings the oxygen would collide with a CH group. The carbons and oxygens are quite big, so is the repulsion between them and the angle has to be quite big for it to work out. Here's a picture of those groups colliding (I did put it at some degree angle because it's easier to see), keep in mind that this representation of atoms is not the truth. The electron "cloud" around them is actually much larger.

enter image description here

Now why do the other two rings have an angle of about 40° between them? If they are planar they can have conjugation which is favorable. Now What happens if we put them into one plane?

enter image description here

You can see that the two hydrogens in the top get very close. In fact, too close. So there are two ways to avoid this. Twist the two rings or make the angle around the central carbon atom between the rings bigger. In a sp2 carbon like we got here the optimal value would be 120°, we could make it bigger, let's say 145 degree:

enter image description here

Now the hydrogens would not clash any more, but it turns out this costs much more energy than we would lose if we twist the two rings to avoid the hydrogens clashing together. What happens in the end is that this angle gets a little bit bigger (~124° instead of 120) and we get that twist between the aromatic rings which is still low enough for good conjugation between them (see first structure).

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  • $\begingroup$ You only had to explain why the last ring is on 90º and you didn't..there could be an arrange with another ring perpendicular, for example..anyway..the structures look good.. $\endgroup$
    – user43021
    Sep 17, 2017 at 21:32
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    $\begingroup$ @HernanMiraola OP assumed that the other rings are in one plane. They most likely are not so it's almost necessary to explain it, otherwise there will be even more questions. And I don't get the second part. it is twisted almost 90° (perpendicular) to the other rings, I explained that this is because of steric reasons and there is no other way to do this. $\endgroup$
    – DSVA
    Sep 17, 2017 at 21:41

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