you don't literally draw the mirror images of all the letters and
numbers! It is, however, quite useful to reverse large groups -
look, for example, at the ethyl group at the top of the diagram.
It doesn't matter in the least in what order you draw the four
groups around the central carbon. As long as your mirror image is
drawn accurately, you will automatically have drawn the two isomers.
So which of these two isomers is (+)butan-2-ol and which is
(-)butan-2-ol? There is no simple way of telling that. For A'level
purposes, you can just ignore that problem - all you need to be able
to do is to draw the two isomers correctly.
2-hydroxypropanoic acid (lactic acid)
Once again the chiral centre is shown by a star.
The two enantiomers are:
It is important this time to draw the COOH group backwards in the
mirror image. If you don't there is a good chance of you joining it
on to the central carbon wrongly.
If you draw it like this in an exam, you won't get the mark for
that isomer even if you have drawn everything else perfectly.
2-aminopropanoic acid (alanine)
This is typical of naturally-occurring amino acids. Structurally,
it is just like the last example, except that the -OH group is
replaced by -NH2
The two enantiomers are:
Only one of these isomers occurs naturally: the (+) form. You
can't tell just by looking at the structures which this is.
It has, however, been possible to work out which of these
structures is which. Naturally occurring alanine is the right-hand
structure, and the way the groups are arranged around the central
carbon atom is known as an L- configuration. Notice the use
of the capital L. The other configuration is known as D-.
So you may well find alanine described as L-(+)alanine.
That means that it has this particular structure and rotates the
plane of polarisation clockwise.
Even if you know that a different compound has an arrangement of
groups similar to alanine, you still can't say which way it will
rotate the plane of polarisation.
The other amino acids, for example, have the same arrangement of
groups as alanine does (all that changes is the CH3
group), but some are (+) forms and others are (-) forms.
It's quite common for natural systems to only work with one of
the enantiomers of an optically active substance. It isn't too
difficult to see why that might be. Because the molecules have
different spatial arrangements of their various groups, only one of
them is likely to fit properly into the active sites on the enzymes
they work with.
In the lab, it is quite common to produce equal amounts of both
forms of a compound when it is synthesised. This happens just by
chance, and you tend to get racemic mixtures.