0056- Kiran- Absolute and Relative Configuration PDF

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Absolute and Relative Configuration in Stereochemistry...

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Absolute and Relative Configuration Kiran Shahzadi 0056-BH (E) -CHE-18 Organic Chemistry Sir Shakoor Department Of Chemistry GCU Lahore

Configuration: The molecular configuration of a molecule is the permanent geometry that results from the spatial arrangement of its bonds. The ability of the same set of atoms to form two or more molecules with different configurations is stereoisomerism.

Relative Configuration: Configuration with respect to another stereogenic element Relative configuration compares the arrangement arrangement of atoms in space of one compound compound with those of another cis/trans, E/Z Relative configuration compares the arrangement arrangement of atoms in space of one compound compound with those of another. until the 1950s, all configurations configurations were relative relative As an example, consider 2-methylcyclohexanol. There are two possible relative configurations based on the relative positions of the two substituents and whether they are on the same side or opposite faces of the cyclic structure. However, this system has two chirality centres and therefore there are 4 configurational isomers with the absolute configurations shown. This means that there are two possible configurations of the cis- structure.

The arrangement of atoms in an optically active molecule, based on chemical interconversion from or to a known compound, is a relative configuration. Relative, because there is no way of knowing just by looking at a structure whether the assignment of (+) or (-) is correlated to a particular enantiomer, R- or S-.

What does this mean? Suppose you have two samples X and Y that are pure enantiomers of each other, and you measure the rotation of each sample, and find let's say X = (+) and Y =(-). Assuming you know the structure, you could then draw 3D-structures and assign the absolute configurations (R-) and (S-) to those drawn structures. But how do you know which structure belongs to X and which one to Y ?

Answer... In general terms, based on this information alone you don't because there is no specific link or either R- or S- and (+) or (-). (R) could be (+) or (-). This is because it is not easy to relate the configuration of a drawn structure to the actual configuration of the molecules in the actual sample. However, if we know for a specific case that (R) = (+), then for its enantiomer, we do know that (S) = (-). If the name of a compound includes both the sign of rotation and the designation R and S then the absolute configuration of that compound is known. This means other experiments have allowed the absolute configuration to the matched to a particular sample. Let's see how chemists can determine the relative configurations of optically active compounds by chemically interconverting them.

The reaction of an alcohol with TsCl is known to occur with retention of configuration that is the group priority of the stereogenic centre has not been altered. The reaction of the tosylate with nitrile occurs with inversion, as a result the group priority at the stereogenic centre has been altered. The absolute configuration of the parent is known while only the relative configurations of the tosylate and the nitrile are known.

Absolute Configuration: We can now determine determine the absolute absolute configuration of almost any compound Absolute configuration is the precise arrangement of atoms in space. Or the precise arrangement of substituents at a stereogenic centre is known as the absolute configuration of the molecule.

The absolute configuration at a chiral centre in a molecule is a time-independent and unambiguous symbolic description of the spatial arrangement of ligands (groups) bonded to the chiral centre.

Let’s take a molecule bromochlorofluromethane (CHBrClF) has two stereoisomers that are its R and S enantiomers. If you are given each of these enantiomers in a separate unlabelled bottle with no additional information, you would not know which one is R and which one is S because they have identical properties except for the direction that they rotate plane polarized light. Because of this difference in light rotation, you can determine which is (+) and which is (-). However without knowing the answer to begin with, you cannot say which is R and which is S because there is no connection between R, S configuration and direction of light rotation. This means that you do not know the absolute configurations of these two enantiomers. In order to know the absolute configuration of a stereoisomer with a known specific rotation ([α]) you must be able to specify the R or S configuration at each of its chiral centres.

Background of Absolute Configuration: Absolute configurations for chiral centers in compounds were unknown until 1951. In that year a Dutch chemist J. M. Bijvoet (1892) reported his use of X-ray diffraction to determine that (+)-tartaric acid and (-)-isoleucine were the (R,R) stereoisomers that we show in Figure [graphic 4.60]. [graphic 4.60] These X-ray diffraction experiments on (+)-tartaric acid and (-)isoleucine also provided absolute configurations for chiral centers in other molecules whose relative configurations were known with respect to the chiral centers in (+)- tartaric acid or (-)-isoleucine such as those related to (+)-tartaric acid that we show here:

(+)-tartaric acid ↓ (+)-malic acid ↓ (+)-isoserine ↓ (-)-glyceric acid ↑ (+)-glyceraldehyde

HO2C-C*H(OH)-C*H(OH)-CO2H ↓ HO2C-C*H(OH)-CH2-CO2H ↓ HO2C-C*H(OH)-CH2-NH2 ↓ HO2C-C*H(OH)-CH2-OH ↑ H(O=)C-C*H(OH)-CH2-OH

The configuration at C* is the same in each of these compounds and is not changed by the chemical reactions (indicated by the arrows) that interconvert these compounds. As a result, if C* in (+)tartaric acid has the configuration shown in Figure [graphic 4.61], then the configuration at C* in (+)glyceraldehyde is that shown in this same figure. [graphic 4.61]

Since Bijovet showed that this Fischer projection for (+)-tartaric acid has the correct absolute configuration at each chiral C, the Fischer projection for (+)-glyceraldehyde also has the correct configuration. Two Chiral Centers in (+)-Tartaric Acid. Note that: (+)-tartaric acid has two chiral C's while there is only one in (+)-glyceraldehyde. Which chiral C in (+)-tartaric acid becomes the chiral C in (+)- glyceraldehyde? It turns out that it makes no difference since the two chiral C's in (+)-tartaric acid are chemically and configurationally identical (they are both R). No matter which C*H(OH)-CO2H group of tartaric acid becomes CH2-OH in glyceraldehyde, the stereochemical result for glyceraldehyde is the same. Long before Bijvoet carried out his experiments, Emil Fischer (1852-1919) arbitrarily assigned the structure in Figure [graphic 4.61] to (+)-glyceraldehyde. [graphic 4.61] He knew that the odds were 50/50 that it was correct, so chemists were pleased when Bijvoet's structure determination of (+)tartaric acid showed that Fischer's guess was correct.

Absolute Configuration or the R and S system: How to Determine the R and S configuration?

What is the R and S Configuration and why we need it? If we name these two alkyl halides based on the IUPAC nomenclature rules, we get the name as 2chlorobutanbe for both:

However, they are not the same compound – they are enantiomers:

So, we need an extra piece of information to distinguish enantiomers (and other stereoisomers) by their names. Cahn, Ingold, and Prelog developed a system which, regardless of the direction we are looking at the molecule, will always give the same name (unlike the wedge and dash notation).

And that is why this is also known as the Absolute Configuration or most commonly referred to as the R and S system. Let’s see how it works by looking first at the following molecule and we will get back to the 2chlorobutane after that:

Assigning R and S Configuration: Steps and Rules To assign the absolute configuration, we need to first locate the carbon(s) with four different groups (atoms) connected to it. These are called chirality centers (chiral center, stereogenic center). In our molecule, we only have one carbon with four different groups and that is the one with the bromine and we are going to assign the absolute configuration of this chiral center.

For this, you need to follow the steps and rules of the Cahn-Ingold-Prelog system. Step 1: Give each atom connected to the chiral center a priority based on its atomic number. The higher the atomic number, the higher the priority. So, based on this, bromine gets priority one, the oxygen gets priority two, the methyl carbon is the third and the hydrogen is the lowest priority-four:

Step 2: Draw an arrow starting from priority one and going to priority two and then to priority 3:

If the arrow goes clockwise, like in this case, the absolute configuration is R. As opposed to this, if the arrow goes counter clockwise then the absolute configuration is S. As an example, in the following molecule, the priorities go Cl > N > C > H and the counter clockwise direction of the arrow indicates an S absolute configuration:

So, remember: Clockwise – R, Counter clockwise – S. Now, let’s see what would be the absolute configuration of the enantiomer:

The priorities are still the same since all the groups around the carbon are the same. Starting from the Bromine and going to the oxygen and then the carbon, we can see that this time the arrow goes counter clockwise. If the arrow goes counter clockwise, the absolute configuration is S.

And this is another important thing to remember: All the chirality centres in enantiomers are inverted (every R is S, every S is R in the enantiomer). So, we discussed the roles of priorities 1, 2, and 3 but what about the lowest priority? We did not mention anything about the arrow going to it. Is it part of the game and how do you use it? The lowest priority does not affect the direction of the arrow. However, this is very important, and it is a requirement when assigning the R and S configuration, that;

The lowest priority must point away from the viewer.

In other words, the lowest priority must be a dashed line to assign the R and S based on the direction of the arrow as we just did:

With that in mind, how can we assign the absolute configuration of this molecule where the hydrogen is a wedge line pointing towards us? R and S When the lowest priority is a wedge

You have two options here: Option one. Turn the molecule 180o such that the hydroxyl is now pointing towards you and the hydrogen is pointing away. This allows to have the molecule is drawn as needed, the lowest priority pointing backward as it is supposed to be for determining the R and S configuration:

Next, assign the priorities; chlorine-number one, oxygen-two, carbon-three and the H as number four.

The arrow goes clockwise, therefore the absolute configuration is R. The problem with this approach is that sometimes you will work with larger molecules and it is impractical to redraw the entire molecule and swap every single chirality centre. For example, look at biotin with all these hydrogens pointing forward. Not the best option to redraw this molecule changing all the hydrogens and keeping the rest of the molecule as it should be.

This is why we have the second approach which is what everyone normally follows. Here, you leave the molecule as it is with the hydrogen pointing towards you. Continue as you would normally do by assigning the priorities and drawing the arrow.

The only thing you have to do at the end is change the result from R to S or from S to R.

In this case, the arrow goes counter clockwise but because the hydrogen is pointing towards us, we change the result from S to R. Of course, either approach should give the same result as this is the same molecule drawn differently.

R and S When Group #4 is not a Wedge or a Dash There is a third possibility for the position of group 4 and that is when it is neither pointing away or towards you. This means we cannot determine the configuration as easily as if the lowest priority was pointing towards or away from us, and then switch it at the end as we did when group 4 was a wedge line. As an example, what would be the configuration of this molecule?

For this, there is this simple yet such a useful trick making life a lot easier. Remember it: Swapping any two groups on a chiral centre inverts its absolute configuration (R to S, S to R):

Notice that these are different molecules. We are not talking about rotating about an axis or a single bond, in which case the absolute configuration(s) must stay the same. We are actually converting to a different molecule by swapping the groups to make it easier determining the R and S configuration. Let’s do this on the molecule mentioned above:

The lowest priority group is in the drawing plane, so what we can do is swap it with the one that is pointing away from us (Br). After determining the R and S we switch the result since swapping means changing the absolute configuration and we need to switch back again.

The arrow goes counter clockwise indicating S configuration and this means in the original molecule it is R. Alternatively, which is more time-consuming, you can draw the Newman projection of the molecule looking from the angle that places group 4 in the back (pointing away from the viewer):

The lowest priority group is pointing and therefore, the clockwise direction of the arrow indicates an R configuration.

R and S when Atoms (groups) are the same Sometimes it happens that two or more atoms connected to the chiral centre are the same and it is not possible to assign the priorities right away. For example, let’s go back to the 2-chlorobutane starting with the wedge chlorine:

Chlorine is the first priority, then we have two carbons and a hydrogen which gets the lowest priority. We need to determine the second priority comparing two carbon atoms and there is a tie since they both (obviously) have the same atomic number. What do you do? You need to look at the atoms connected to the ones you compare:

The carbon on the left (CH3) is connected to three hydrogens, while the one on the right is connected to two hydrogens and one carbon. This extra carbon gives the second priority to the CH2 and the CH3 gets priority three. The arrow goes clockwise, so this is the (R)-2-chlorobutane. And if these atoms were identical as well, we’d have to move farther away from the chiral centre and repeat the process until we get to the first point of difference. It is like layers: the first layer is the atoms connected to the chiral centre and you are comparing those and only move to the second layer if there is a tie.

You should never compare any atom of the second layer to a first layer atom regardless of its atomic number.

Double and triple bonds in the R and S configurations Let’s do the R and S for this molecule:

Bromine is the priority and the hydrogen is number four. Carbon “a” is connected to one oxygen and two hydrogens. Carbon “b” is connected to one oxygen and one hydrogen. However, because of the double bond, carbon “b” is treated as if it is connected to two oxygens. The same rule is applied for any other double or triple bond. So, when you see a double bond count it as two single bonds when you see a triple bond cut it as three single bonds.

The arrow goes clockwise, however, the absolute configuration is S, because the hydrogen is pointing towards us.

More Tricks in the R and S configurations 

What if you are comparing two carbons; one connected to three high-atomic number elements, and the other one with two hydrogens and a heteroatom. Which one gets a higher priority?

Let’s see this with this molecule:

Even if only one atom has a higher atomic number than the highest one on the other carbon, the group gets higher priority. So, one S beats N, O, F because it has a higher atomic number than the others individually.

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Carbon is not the only atom designated by R and S. In theory, any atom with four different groups is chiral and can be described by the R and S system. For example, phosphorous and sulphur chiral centres are often assigned as R or S.

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Hydrogen is not always the lowest priority. A lone pair of electrons is lower.

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Carbanions are achiral because the lone pair rapidly flips from one side to another unless at very low temperatures:

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R and S do not apply to the nitrogen in amines for the same reason as for carbanions. Quaternary ammonium groups, however, can be chiral.

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The same element can get different priorities based on its isotopes. For example, tritium atom has a higher priority than deuterium: T > D > H

Fischer Projections What are the Fischer projections? Fischer projections are just another way of drawing compounds contacting chirality centres. They were initially proposed by Emil Fischer for making it easier to draw the structures of compounds containing multiple chirality centres with the main idea of not having to draw the wedge and dash lines for every single chiral centre. This is especially applicable and used mostly for drawing sugars.

How to draw Fischer projections? Suppose you have this compound with one chirality centre:

Before getting to draw its Fischer projection, let’s number the carbons in any order (no IUPAC rules needed). Remember, numbering carbons will always be helpful no matter what you need to do with an organic structure.

Here is what Fischer suggested: If you look at the molecule from the top, you will see the following representation where the two groups on the side are pointing towards and the ones on the top and on the bottom are pointing away from you. We will show the ones on the sides with wedge lines and the others with dashed lines:

There are two wedge and two dash lines which may look strange to you since we always have one of each and then the two solid lines but it is okay-it all depends on the direction we are a looking at the molecule.

This, however, is not the Fischer projection yet, since, remember, we said the main idea was to avoid showing wedge and dash lines yet being able to convey absolute configuration of the chirality centres (R, S). For this, we are going to draw the molecule and simply show all the bonds with plane solid lines, keeping in mind that the horizontal groups are pointing towards you and the ones on the vertical line are pointing away from you:

So, how do you remember which ones are pointing towards you? Well, you can remember that Fischer projections like you and they are coming to give you a hug with open arms:

Or, you look at the Fischer projection like you are in the gym and need to grab the molecule. In this case, as well, the horizontal groups have to be pointing towards you.

Determining the absolute configuration (R, S) of Fischer projections To determine the absolute configuration of chirality centres in a Fischer projection, we need to follow the same steps as we do for any other representation such as Bond-line or Newman, according to the Cahn-Ingold-Prelog rules. In this molecule, for example, we need to assign the priorities of the groups on the chiral centre based on the atomic numbers:

Next, draw the arrow going from priority 1-2-3:

The arrow goes clockwise which indicates R configuration. However, this is where you need to remember that the horizontal groups (Cl and H) are pointing towards you, therefore, the configuration must be switched from R to S. This is because one of the rules of Cahn-Ingold-Prelog system is that the lowest priority must point away from th...