Optical isomerism

Left hand, right hand

Hold up your hands. Your left hand is the mirror image of your right hand. But you cannot place one on top of the other so that every finger matches. A left glove does not fit a right hand. An object that cannot be placed on top of its mirror image is chiral, from the Greek word for hand.

In Class 11 you met chirality in carbon compounds. A carbon atom with four different groups has a mirror image that does not match it. A complex needs no such atom. The ligands arranged round the metal can be chiral all by themselves.

The two mirror-image forms are called enantiomers, or optical isomers.

The test: a plane of symmetry

Is a complex chiral? Look for a plane of symmetry, a plane that cuts the complex into two halves that are mirror images of each other.

• If there is a plane of symmetry, the complex matches its own mirror image. It is achiral.
• If there is none, the complex is chiral and has an enantiomer.

Apply this to octahedral complexes. Start with $[\text{Co}(\text{en})_{3}]^{3+}$. Each en ring spans two cis corners. Seen along the three-fold axis, the three rings spiral like the blades of a propeller. A propeller can spiral to the right or to the left, and neither matches the other.

The top row shows the two forms of $[\text{Co}(\text{en})_{3}]^{3+}$. The bottom row shows $[\text{Co}(\text{en})_{2}\text{Cl}_{2}]^{+}$ with its two chloride ligands cis. This form has no plane of symmetry, so it also exists as a mirror-image pair.

Now put the two chloride ligands trans. The four nitrogen atoms lie in one plane, and that plane cuts the complex into two equal halves. It is a plane of symmetry. So the trans form is achiral.

Which complexes are chiral

An octahedral complex is chiral when the ligands are arranged without any plane of symmetry. These complexes often are.

• $[\text{M}(\text{AA})_{3}]$, where AA is a bidentate ligand such as en or oxalate. Example: $[\text{Fe}(\text{ox})_{3}]^{3-}$.
• The cis form of $[\text{M}(\text{AA})_{2}\text{B}_{2}]$ and of $[\text{M}(\text{AA})_{2}\text{BC}]$.
• $[\text{M}(\text{AB})_{3}]$ with an unsymmetrical ligand. Both the fac and the mer forms are chiral.
• $\text{MA}_{2}\text{B}_{2}\text{C}_{2}$ in the form where all three pairs are cis.

These are not chiral.

• All octahedral complexes of the types $\text{MA}_{6}$, $\text{MA}_{5}\text{B}$, $\text{MA}_{4}\text{B}_{2}$, $\text{MA}_{3}\text{B}_{3}$, $\text{MA}_{4}\text{BC}$ and $\text{MA}_{3}\text{B}_{2}\text{C}$.
• Every trans form of $[\text{M}(\text{AA})_{2}\text{B}_{2}]$.
• Square planar complexes. All the atoms lie in one plane, and that plane is a plane of symmetry.

A tetrahedral complex with four different ligands is chiral in principle. It is rare, because such complexes tend to turn into their mirror images quickly.

The mirror-image forms are named by the handedness of the propeller. A right-handed propeller is $\Delta$ (delta) and a left-handed one is $\Lambda$ (lambda).

Optical activity

Ordinary light vibrates in every plane. A polariser lets through only one plane. That light is plane-polarised. A solution of one enantiomer turns the plane of the light. This is optical activity.

• The form that turns the plane clockwise, seen by the observer looking at the source, is dextrorotatory. It is marked (+).
• The form that turns it anticlockwise is laevorotatory. It is marked (-).
• The two enantiomers turn the plane by the same angle in opposite directions.

The sign cannot be read from the drawing. It has to be measured.

A mixture of equal amounts of the (+) form and the (-) form is a racemic mixture. The two rotations cancel, so it shows no rotation. The rotation of any mixture follows from the shares of the two forms.

Here $p_+$ and $p_-$ are the shares of the (+) form and the (-) form in percent.

The difference of the two shares is the excess of one form. A sample with a 60 percent excess gives 60 percent of the rotation of the pure form.

Put the slider at 50 percent. Then try 100 percent and 0 percent.

Practise with new numbers

Splitting a racemic mixture

A chemist who makes a chiral complex usually gets the racemic mixture, since both forms are equally likely. To separate them, add a chiral partner such as a tartrate salt. This gives two products that are no longer mirror images of each other. They dissolve differently, so one crystallises first. The method is resolution. Werner resolved a cobalt complex that contained no carbon, which showed that chirality does not need a carbon atom.

Counting all the stereoisomers

The number of stereoisomers of a complex is the number of geometrical isomers plus the number of mirror-image pairs. Here are some results.

• $[\text{M}(\text{AA})_{3}]$: 1 geometrical isomer and 2 stereoisomers.
• $[\text{M}(\text{AA})_{2}\text{B}_{2}]$: 2 geometrical isomers (cis and trans) and 3 stereoisomers.
• $[\text{M}(\text{AB})_{3}]$: 2 geometrical isomers (fac and mer) and 4 stereoisomers.
• $\text{MA}_{2}\text{B}_{2}\text{C}_{2}$: 5 geometrical isomers and 6 stereoisomers.
• $\text{MABCDEF}$: 15 geometrical isomers and 30 stereoisomers.

Try it in Python

The program places rings as well as single ligands on the octahedron. Each ring takes two cis corners. It counts the geometrical isomers (a mirror pair counts once) and the stereoisomers (a mirror pair counts twice). The difference is the number of mirror-image pairs.

Common mistakes

• Saying every chelate complex is chiral. The trans form of $[\text{M}(\text{en})_{2}\text{Cl}_{2}]^{+}$ has a plane of symmetry.
• Calling a square planar complex chiral. Its own plane is a plane of symmetry.
• Counting a mirror-image pair as one stereoisomer. Two enantiomers are two stereoisomers. They are one geometrical isomer.
• Expecting a racemic mixture to be a pure compound that is inactive. It is a mixture of two active forms, and the rotations cancel.
• Reading the sign from the drawing. The sign, (+) or (-), is measured. It does not follow from $\Delta$ or $\Lambda$.

Remember these

• A chiral complex has no plane of symmetry. Its mirror image is a different compound, an enantiomer.
• Chiral octahedral types: $[\text{M}(\text{AA})_{3}]$, cis-$[\text{M}(\text{AA})_{2}\text{B}_{2}]$, $[\text{M}(\text{AB})_{3}]$, and the all-cis form of $\text{MA}_{2}\text{B}_{2}\text{C}_{2}$.
• Square planar and trans-$[\text{M}(\text{AA})_{2}\text{B}_{2}]$ complexes are achiral.
• The enantiomers rotate plane-polarised light by equal angles in opposite directions. A racemic mixture does not rotate it.
• Stereoisomers: geometrical isomers plus the mirror-image pairs.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Is trans-$[\text{Co}(\text{en})_{2}\text{Cl}_{2}]^{+}$ optically active? Give a reason.

2. How many stereoisomers does $[\text{Co}(\text{en})_{2}\text{BrCl}]^{+}$ have?

3. A sample is a racemic mixture. What rotation does it give, and why?

4. Why is a square planar complex such as $[\text{Pt}(\text{en})\text{Cl}_{2}]$ not chiral?

5. A pure (+) form rotates light by $+6^\circ$. A mixture gives $+3^\circ$ under the same conditions. What is the share of the (+) form?