In the following cyclopentane deratives, why can't I say that a,b,e have a center of symmetry so they are optically inactive and c as it has a plane of symmetry as shown is also optically inactive? The book tells that only c is optically inactive.
$\begingroup$
$\endgroup$
1
-
3$\begingroup$ Your "centre of symmetry" works only in one direction, that is to say, the left-right direction. Try inverting the front two carbons through your centre of symmetry; you'll find that it actually isn't a centre of symmetry. You can't pick and choose which atoms are inverted and which aren't. Also, it's helpful if you can rotate your image 90° before posting it, and I don't mean that you should post it upside down. $\endgroup$– orthocresolCommented Jan 5, 2020 at 2:04
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Alright, from what I know, I am certain that any cycloalkane with an odd number of carbon atoms will never have a center of symmetry. So your argument of $a$, $b$, $e$ having centre of symmetry is outright wrong, since upon inverting the front most carbon in cyclopentane derivatives you would see that cyclopentanes do not have a center of symmetry. Cycloalkanes with odd number of carbons can however have plane of symmetry or axis of symmetry.
