Question

What symmetry operations leave trans-1,3-dichlorocyclobutane invariant?

Answer 1

You can carry out four symmetry operations on trans-1,3-dichlorocyclobutane.

Explanation:

The structure of trans-1,3-dichlorocyclobutane is

The ring is puckered, but it undergoes a rapid ring flip between the two bent forms.

We can treat the ring as if it were planar. However, at low temperatures the two conformations would be "frozen out" and the ring would have a different symmetry.

The planar molecule has `"C"_(2h)` symmetry.

This point group contains four symmetry operations:

`bbhat"E" color(white)(ml)"the identity ('do nothing') operation"`
`bbhat"C"_2 color(white)(m)"a rotation about a two-fold axis"`
`bbhat"i" color(white)(mm)"inversion about a single point"`
`bbhatsigma_h color(white)(m)"reflection about the mirror"`
`" "" ""plane perpendicular to the C"_n`
`" "" "color(white)(l)"axis of the highest n"`

Let's apply these operations to the molecule.

`bbhat"E"`, the identity operation (symmetry element: the entire molecule)

This one is easy. It amounts to doing nothing to the molecule.

`bbhat"C"_2`, rotation by `(360^@)/2 = bb(180^@)` about an axis (a `C_n` rotation with `n = 2`; symmetry element: the `C_2` rotation axis)

The `"C"_2` axis passes between carbons 2 and 4, and lines up with the plane of the ring (it is not through the ring).

Rotation by 180° about this axis interchanges the two `"Cl"` atoms, the two `"H"` atoms, and carbons 1 and 3. A front-facing atom is now facing the rear and vice versa.

`bbhat"i"`, the inversion operation (symmetry element: a dot at the center of the molecule)

There is a centre of inversion at the mid-point of the ring.

Inversion through this point interchanges the two `"Cl"` atoms, the two `"H"` atoms, and carbons 1 and 3. In other words, it takes `(x,y,z)` and transforms each coordinate into `(-x,-y,-z)`.

`bbhatσ_"h"`, reflection through a mirror plane perpendicular to the `"C"_2` axis (symmetry element: the mirror plane itself)

A mirror plane passes through the `"H, C"` and `"Cl"` atoms at carbons 1 and 3.

It divides the molecule into two front and back mirror-image halves, and bisects the ring, while being coplanar with the explicit `"H"` and `"Cl"` atoms.

The operation reflects carbons 2 and 4, but the remainder of the atoms stay in place.