Stereochemistry is an important principle in biology, when we consider drugs and receptors. Pharmaceutical drugs often have a family of different isomers, of which only one of these members will successfully carry out the desired role in the body. Other isomers may have no effect or may promote adverse side effects, the most well known example being thalidomide. A basic knowledge of stereochemistry is therefore essential to truly understand some key aspects in pharmacology and biochemistry.
An isomer is a compound with the same molecular formula, but a different structural formula. The same atoms present but they are arranged differently in space.
Stereoisomers are structures which cannot be superimposed onto each other. The structures are arranged differently despite having the same elements in the compound. There are two different types of stereoisomerism. The first is Geometric isomerism, and the second is Optical isomerism.
Geometric isomers are associated with structures which contain double bonds. Groups attached to atoms which form single covalent bonds to other atoms are free to rotate. With double carbon bonds the substituents cannot rotate, they are confined to a single side of the compound. This is due to the double bond having restricted rotation. The figure below shows how this can lead to isomerism.
When it comes to naming geometric isomers, we use phrases to describe the location of a substituent with reference to the double bond. The old phrases used were:
The other notation which is more commonly used is E and Z:
E = Entgegen which means opposite. So the substituents of a double bond are opposite each other- Trans (think of E standing for enemies in a fight which are on opposite sides. So the groups of highest priority are on opposite sides
Z = Zusammen which is together- Cis. So the highest priority groups are on the same side of double bond.
If there are two substituents attached to the carbon double bond, how do we know which one is the highest priority when naming? We choose the substituent with the highest priority. The highest priority is the atom with the highest atomic number. Using Figure 2 as an example, the following points provide a step-by-step guide on how to name it.
1) We start by splitting the compound into two, so that each carbon is considered separately. We'll call them carbon A and carbon B.
2) Start by looking at carbon A; notice how it is bonded to a hydrogen atom and a methyl group.
3) Which one has the higher atomic number - hydrogen or carbon? Using the periodic table we know that carbon has a greater atomic number hence we can say that the highest priority for carbon A is the methyl group. (Carbon = 6 and Hydrogen = 1).
4) Consider carbon B; notice how it is bonded to a bromine atom and a methyl group. Which has higher atomic number bromine or carbon? Using the periodic table we know that bromine is higher so it has greater priority. (Bromine = 35 and carbon = 6).
5) Therefore the priority substituents for carbon A is the methyl group and the priority for carbon B is the bromine atom.
6) One priority group is on the bottom face, whilst the other priority is on the top face, so they are opposite each other.
7) So this compound is trans 2 bromo but-2-ene, or we can say it is (E) 2 bromo but-2-ene, because E is opposite which is the same as trans.
Figure 3 shows an example of how to name isomers using both the cis-trans and E-Z notation.
Before we go into optical isomers you must be familiar with the term chirality.
An atom of carbon is described as being chiral if it is bonded to 4 different groups. Look at Figure 4. The carbon (labelled α) is chiral since it has 4 bonds attached to 4 different substituents. One bond is to a methyl group (CH3), another to an amino group (NH2), one to a carboxyl group (COOH) and finally one to hydrogen.
Another rule that must be followed for carbon to be chiral is that the carbon is sp3 hybridised. This means the carbon contains 4 sigma bonds. If we consider the carbon atom in the COOH group in Figure 4, we notice it has one sigma bond to another carbon, one to the hydroxyl group and a double bond to oxygen. Therefore this carbon contains 3 sigma bonds and one pi bond, hence it is sp2 hybridised so it is not chiral. The important conclusion for all carbon atoms is that if they contain a double bond, they are not chiral.
In order for a compound to be optically active it must have a chiral carbon. Compounds that do not have a chiral centre are optically inactive. The chiral carbon is sometimes referred to as the stereogenic centre. When dealing with chiral carbons we usually reference the carbon as having S or R configuration. We base this characteristic depending on the priorities of each different group attached to the chiral carbon.
The best known examples of chiral molecules in biochemistry include:
Look at Figure 5.
1) Ensure the carbon is sp3 hybridised, in other words check that the carbon has four sigma bonds (No double bonds).
2) Check that this carbon is attached to four different groups to ensure it is chiral.
3) Identify the group which has the highest atomic number- highest priority. In this scenario we notice that the chlorine is the highest priority so we label it 1 (chlorine =17)
4) The second highest priority group is slightly tricky, since we have two bonds going to two carbon atoms, so we look at both carbons and see which atoms are bonded to these carbon atoms. For the methyl carbon it is bonded to three hydrogen atoms only, whereas the propyl carbon is bonded to two hydrogen atoms and a carbon atom. Since this carbon is attached to another carbon which is higher priority compared to the carbon attached to the hydrogen we can conclude that the propyl group is second highest priority. We label it 2.
5) The third highest priority would be the methyl group, labelled 3.
6) The fourth highest priority is the hydrogen atom, labelled 4.
7) So now we have prioritised each group we can move onto the final step, which is drawing a curved arrow from the first priority group (1) to the second priority group (2) shown in Figure 5.
8) If the arrow is going clockwise then the chiral carbon is R configuration
If the arrow is going anticlockwise then it is S configuration
Hint: If you find it difficult to remember S and R directions, use this trick. When you turn the steering wheel of a car clockwise the car moves to the right R.
In this example the arrow is going anticlockwise hence we know that this carbon is S configuration.
Enantiomers: are compounds with one stereogenic centre which are exactly mirror images of each other. Enantiomers are non-superimposable images. Notice how one enantiomer is R configuration whilst the other is S configuration.
If a compound has more than one stereogenic centre (chiral carbon) then it is described as being a diastereoisomer. Diastereoisomers have different configurations around one or more chiral carbons but are not mirror images of each other. If only one stereogenic centre has a different conformation between isomers then they are called epimers. The most important difference between diastereoisomers and enantiomers is that the former have differing physical properties but the latter do not.
Mesoforms are achiral (not chiral) compounds which contain two stereogenic centres. Each stereogenic centre has an opposite configuration e.g. one carbon atom is S whilst the other carbon atom is R configuration. The two opposite configurations cancel each other out, so the molecule is not optically active.
What happens if we have a compound with more than one chiral carbon? Usually in this case we have to start at one chiral carbon and determine whether it is R or S configuration. We then repeat the same process for the second chiral centre.
In the previous section we mentioned briefly how single carbon bonds are able to freely rotate, however we will find that not every conformation is approved in free space. Conformation of a compound may be described as:
Staggered when the bonds of adjacent carbons are at a maximum distance away opposite from each other.
Eclipsed when the bonds are in line with each other.
When the bonds are in line with each other (eclipsed conformation) the electrons in the bonds repel each other. This means that rotation around the carbon bond is not completely free because the staggered conformation is more energetically favourable. The dihedral angle is a measure between the hydrogen atoms of adjacent carbons, which has a value of 0° in eclipsed conformation and 60° in staggered conformation. Figure 8 shows the example ethane. Newman projections (an alternative way of representing this) are shown next to the ball and stick diagrams as it is likely that your would need to know how to read or draw one in an exam.
Using this graph (Figure 9) we can explain why compounds such as hexane etc. exist as staggered shapes. The staggered conformations have a lower energy value than the eclipsed conformations. The eclipsed forms of a compound correspond to the peaks of the graphs, with the highest potential energy. The troughs of the graph are reflective of the staggered conformation of the compound with the lowest potential energy.
This results in the activation barrier of the formation of this compound being low. Since reactions proceed via the lowest energy pathway most compounds are formed via the staggered barrier. The eclipsed structure requires more input energy to form, since there is repulsion between the bonds leading to additional energy- high activation energy.
However the trend of most compounds existing as staggered configurations is not strictly true, some compounds favour eclipsed conformations. An example is ethane 1, 2 diol. Eclipsed formation is favoured due to hydrogen bonding between the hydrogen and oxygen bonded to adjacent carbon atoms. The force of hydrogen bonding overpowers the steric repulsion.
Ethane is a useful example to start with because all the groups are hydrogen, so the molecule can only rotate between the staggered and eclipsed conformations. But what happens when we have different groups around the carbon? Taking the example of butane, the larger methyl groups are going to repel each other more than the smaller hydrogens. The conformations of butane in order of increasing stability (shown in Figure 10) are;
This changes the shape of the graph in Figure 9. The fully eclipsed peak of the graph would be higher than the two eclipsed peaks as fully eclipsed is the state with the highest potential energy. The anti conformation trough would be lower than the two gauche troughs because anti conformation is the most energetically favourable state.
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