Is this compound chiral?

enter image description here

I can't tell where the chiral centres are because this polycyclic backbone is too confusing. I tried to draw a mirror image, but I got really confused and couldn't tell whether it was superimposable. And I don't have a molecular modelling set so I can't build a model.


1 Answer 1


Yes, this compound is chiral.

The polycyclic backbone is called adamantane. It has $T_\mathrm{d}$ symmetry, meaning that as far as chirality goes, it behaves like a perfect tetrahedron, somewhat like methane does.

It also has the interesting property that if you extend the C-F, C-Cl, ... bonds inwards, they will all meet at the same point. Those red dotted lines form a perfect tetrahedron.

enter image description here

Now, since the adamantane backbone behaves exactly like a methane molecule - why don't we just replace it with one? [n.b. this is not 100% rigorous but works here.]

enter image description here

Hopefully you'll agree that this is chiral. It basically behaves like it has a chiral centre right in the middle of the adamantane backbone.

If you really wanted to be sure, you could build a mirror image and verify that they are non-superimposable. (Hint: they are.)

Each of the four substituted carbons is connected to four different groups, and theoretically one would expect $2^4 = 16$ stereoisomers. However, because three of the four groups (the polycyclic backbone) are held in a rigid, fixed geometry, this number is reduced to 2.

There is more discussion of this class of compounds here.

  • 1
    $\begingroup$ I find it fairly obvious that every quaternary carbon (which is tetrahedrally coordinated) has all different substituents and thus breaks all symmetry. However, I find this answer very nicely illustrated, and it makes a good point for more complicated cases. $\endgroup$ Commented Sep 23, 2016 at 5:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.