A simple measurement exists which you could apply to your white crystals.  This measurement  has bearing on the problem of configuration, i.e. are your white crystals  3 and therefore (S) or 4 (R)..  This measurement is “optical rotation”.  As explained above, “optical isomers” such as 3 and 4, have identical properties such as solubility.  They differ in one property only: 3 and 4 have equal but opposite (in sign) “optical rotation”.  The symbol for "optical rotation" is the Greek letter alpha.

To explain “optical rotation”, let us first consider “polarized light”.  The light beam is composed of two “vectors”, a magnetic vector and an electrical vector.  We will pretend that only the electrical vector is of importance. Shown below is a hypothetical case, where the electric vector of light from a light bulb strikes a sensor.

The sensor would "sense" the electromagnetic vector of the light oscillating in a single plane, i.e. "plane polarized" light.

In actuality, light from the sun, from a candle, or from a light bulb is not "plane-polarized”.  The electrical vectors oscillate in every plane, a few of which are shown below.  The sensor would, in fact, "sense" many vectors of light, each as typified by an "arrow" in the following diagram.

If we pass this "non-polarized" light through an optical device called a “polarizer”, or if we use a beam of light from a laser, “plane-polarized light” results. 

Let us step through the process of determination of "optical rotation".using a sample of 4 (R)-alanine.   .A device called a "polarimeter" will be used  In the "polarimeter", the light source is called a "sodium-D" lamp (A),  This light is rich in orange wavelengths, and it is not polarized.  The non-polarized nature is shown in B.   This light passes through an optical device called a "polarizer" (C)   This light that emanates is now polarized in the vertical plane as shown in D.   If we pass this beam of polarized light through a tube (E) containing a sample alanine (4) dissolved in acetic acid,  the alanine will interact with the light in an interesting way.  The alanine will “bend” or “rotate” the “plane-of-polarization” a certain direction.  As shown in F, the light now oscillates in a new plane of polarization.  The sensor (G), sometimes called an "analyzer", contains a second "polarizer".  The sensor determines  that your alanine has rotated  the “plane-polarized light” “to- the-left” by -29.4^o. 

Arbitrarily, a rotation of the original “plane of polarization” to the right is termed a “positive” or “(+)”  or “d” optical rotation.  A rotation to the left is termed a “negative”, or (-), or “l” optical rotation.  So 4 could also be called "(-)-alanine".

If a sample of 3 had been measured under identical conditions, the plane of polarization would have been rotated in the opposite direction -- to the right.  It is said to have a positive or (+) optical rotation.  Once again, the symbol for ”optical rotation”  is the Greek letter alpha, “a”.   In fact, you would find that 3 and 4 have equal but opposite “optical rotations”.  Thus, for 3, a =+29.4^o, whereas for 4, a=-29.4^o, assuming that the samples of 3 and 4 have the same purity. 

Unfortunately, the sign of optical rotation (+) or (-) is of little help.  Some “S” compounds have a “positive” optical rotation and other “S” compounds have a “negative” optical rotation.  The same occurs for “R” structures such as 4.  in fact another “natural” amino acid “tyrosine” ((S)-5) is found to have a negative (-) optical rotation, even though it has the same configuration as your (S)-3. 

Thus, for 5 a complete and careful designation would be: (S)-(-)-tyrosine, i.e. it belongs to the S family, it happens to have a negative "optical rotation".  Can you see that (S)-(-)-tyrosine has the same arrangement of groups as (S)-(+)-alanine (3)?  It is only an accident that 3 and 5 have opposite "optical rotations".

Thus, your observation of a “negative” optical rotation for your colorless crystals of alanine does not really prove it to be “S” vs. “R” (or 3 or 4).  However, you can search the chemical literature, and you would eventually find a reliable body of research that proved that (-)-alanine (alanine with a negative “optical rotation”) is “R” and (+)-alanine is "S".  For other compounds than alanine, it could be the reverse.

Thus, obtaining colorless crystals of an “optically active” substance (substance with an optical rotation), and the finding if the structure of the substance is in fact 3   or 4 is an arduous task.  This task is called “proof of configuration.”

To summarize a few take-home lessons:

(1)    Optical isomers can be written as “mirror images” of one another.

    (a)  Mirror image structures are called "enantiomers".

(2)    Optical isomers have identical “physical properties” such as solubility, and identical chemical reactions.

(3)    Optical isomers (enantiomers) have equal but opposite "optical rotations", a.

a.      Such compounds are said to be “optically active”.

b.  Most organic compounds such as acetone, ether, or benzene are NOT "optically active"

(4)    Each optical isomer is a member of either the “S” family of optical isomers, or the “R” family, but telling which-is-which is difficult.

            a.  Telling “which-is-which” is termed “proof of configuration”.

Some new points:

(5)    Optical isomers (generally speaking):

a.      the mirror image is not “superimposable” on the original structure

b.      are “asymmetric” (not symmetric)

        i.  Axis of symmetry is OK, however

c.      possess an “asymmetric center” that is marked by a star, e.g. “C*”

d.      most often have four different groups bonded to C*

e.      are “chiral”

“Superimposability and “Chirality”: 

First, a lesson on "symmetry".   "Symmetry is a difficult concept for students who have not had plane geometry.

Is your nose "symmetric"?   Yes -- the left side is the "mirror image" of the right side of your nose.  An imaginary "plane-of-symmetry" "bisects" the nose.   Each side is identical to the other side, except for being a mirror image.   Are your two ears symmetrical?  Yes -- your right ear is the "mirror image" of your left ear.  Are your two hands "symmetric"?  Yes, for the same reasons outlined above.

There are four main elements of "symmetry".

• a center-of-symmetry

• a plane-of-symmetry

• an axis-of-symmetry

• an alternating axis of symmetry (more complicated -- will not be considered here). 

A "sphere" is an object with a center-of-symmetry.  If we start out at the center-of-symmetry and go in any direction, after we travel "d' the radius of the sphere, we will be at the edge of the sphere.  Going back to the center-of-symmetry, and travel "d" in the OPPOSITE direction, we will come to an identical feature -- the edge of the sphere.

We have already encountered several cases of a "plane-of-symmetry".  Some more examples:  Does the letter "A" have a plane of symmetry?  The answer is "yes".  Does the letter "G" or the numeral "7" have a plane of symmetry?  Well -- yes and no.   The answer is "yes" if we consider the plane of the screen.  If we ignore this plane, the answer would be "no" -- the right hand side of "G" is different than the left-hand side.

  Let's jump to an example involving a chemical structure, i.e. 6.  Does 6 have a "plane-of-symmetry"?  If so, it cannot be "optically active" or "chiral".    The answer is "yes", it does have a "plane-of-symmetry" as shown in 6a.   The "plane-of-symmetry" bisects the C1-C2 bond and also the C3-C4 bond.   The molecule is split into two identical halves.  It is NOT "chiral" and it cannot be "optically active".  Its "optical rotation", alpha, will be zero.  The image 6b shows a "side-view", and it shows the "plane-of-symmetry" from a different angle.

[Is the "plane" shown in 6c a "plane-of-symmetry"?  No -- the right-hand side of the plane (carbons 1 and 2) and the left-side of the plane (carbons 4 and 3) are "different" (not the same).  For example, carbon 1 is attached to a chlorine, but carbon 4 is attached to hydrogens (not shown, but nonetheless present).   Thus the plane shown by dotted lines in 6c is merely a plane, but NOT a "plane-of-symmetry".]

Similarly, molecule 7 has two "planes-of-symmetry", and it cannot be "optically active".  One "plane-of -symmetry" bisects the Cl-C-Cl bond angle, and it passes directly through the carbon and both hydrogens. 

Question:  Draw the second "plane-of-symmetry".  You may have to re-draw the image of 7.

Axis-of-Symmetry

An "axis-of-symmetry" is permissible for "chirality".  A molecule may possess an "axis-of-symmetry" and still be "chiral" and still be "optically active".  An example is 8.   The axis is shown in 8a as a simple line passing through the C1-C2 bond and also the C3-C4 bond.  If we rotate 180 degrees about this axis, we will obtain 8b, which is exactly the same image as 8a.   The numbers 1-4 do not count.  On the molecular level, atoms do not have numbers.  We add them merely for convenience.  This axis is called a "C2" axis since an 180 degree rotation gives the same image as the original structure (360/2 = 180).

Thus, in 8a, a 180 degree rotation (a "pancake-flip") places C1 where C2 formerly was.   Similarly, C4 takes the place of C3 and vice-versa.  The two chlorines likewise exchange positions.

Question:  Does the letter "A" have an "axis-of-symmetry"?  How about "X"?  How about "G".  The answers are "yes", "yes -- two axes-of-symmetry", and "no"

Superimposibility:

Identical objects are “superimposable”   Thus, if you have two identical gloves for the left-hand, you could pick one glove up, lay it down on the second glove, and find that the thumbs, the palms of the gloves and each finger matches perfectly. 

Now, lets repeat the “superimposability” test with a left- and a right-handed glove.  The two gloves are mirror images of one another, just like your two hands, if your hands are placed palm to palm.  If you take the right-handed glove and attempt to “superimpose” it on the left-handed glove, there always will be a mismatch no matter how many times you turn one glove and attempt to “superimpose” it on  the other.  Either the thumbs won’t match, or the fingers, or the palms, etc.  The two gloves are "non-superimposable".

Identical "mirror image" structures are NOT “enantiomers" nor "chiral" nor "optically active" For example, methylene chloride (7) can be written as a mirror image.  However, the "mirror images" ARE “superimposable”.  You could pick one molecule up, and set it down on the second molecule, and you would find that like groups match (see below).   Remember, that numbers do not count.  Thus, 7 and its mirror image are "identical" ("superimposable"), and 7 is "optically inactive".

 Now let's make two changes in 7.   One Cl becomes a bromine Br, and one hydrogen becomes a fluorine, F, as shown in 9.   Now, 9 and its mirror image 10 are NOT SUPERIMPOSABLE.

        If we "pick up" 10 and set it down on top of 9 (attempted superposition), we would find that the F in 10 matches the F in 9 and the H in 10 matches the H in 9,  However, the Br of 10 would be "on top of" the Cl of 9 and vice versa.  No matter how we turn 9 or 10, we could NEVER get all groups to match one another.  Make molecular models of 9 and of 10 to prove this to yourself.   Molecules 9 and 10 are "enantiomers". and they are "chiral".   Each, by itself, is "optically active".  One has a positive and the other an equal negative "optical rotation".  One is a member of the "R" family and the other is an "S" isomer.  The carbon is called an "asymmetric carbon" or a "stereogenic" carbon.  This carbon is the "chiral center", usually marked by a star.

    As a general rule, "optical isomers" have four different groups attached to C*, such as hydrogen, fluorine, chlorine and bromine.

            .

The "non-superimposibility" of 9 upon 10 or vice versa, is the one true test of "optical isomerism".  Let's try it with anther case, i.e. 11 and 12.  Note that C* has four different groups present.   The question is:  "Are 11 and 12 "enantiomers" or are they identical? .  Image 12a has merely been written in such a way to confuse the viewer, and to make the question less obvious. 

  Let's rotate about the horizontal axis shown in 12a.  One image that would result is 12b,   What has happened is that methyl in 12a takes the place of hydrogen in 12a, hydrogen tqkes the place of OH and OH takes the place of methyl.  The resulting image is 12b.  Images 12a and 12b are the same basic organic compound, just written differently.   Once again, the mirror image of 11, i.e. 12b, is NOT superimposible upon 11.  Once again, you must make molecular models to prove this fact to yourself.  The failure of the "superimposibility test" proves that 11 and 12 are "enantiomers", they are "chiral", and each, by itself is "optically active".   One molecule is a member of the "R" family and the other is a member of the "S".

An interesting fact is that living systems themselves utilize and are sensitive to just one of two "enantiomers"..  Molecules 11 and 12 are "lactic acid".  It turns out that 12 is a "pheromone" for the mosquito.  A "pheromone" causes a metabolic response in the life form in question; in this case, it signals "food".  Thus, if you are working in your garden, lactic acid, 12, (S)-(+) lactic acid,  plus carbon dioxide, exudes from the pores of your skin.  The insect senses the trail of 12 in the atmosphere, and follows this trail to your skin.  The mosquito could not care less about 11, (R)-(-)-lactic acid.

“R” and “S” Assignment:

In 9, let us write the ATOMIC NUMBER of each atom above each atom (except C*).  We will have ¹H, ⁹F, ¹⁷Cl, and ³⁵Br. 

a.                 Assign priorities to each atom based on DECREASING atomic number.  Thus Br has priority I, chlorine, II, fluorine III and hydrogen last, IV.

b.                 Arrange the molecule 9 so that you are sighting down the C* -H (IV) bond (lowest priority bond.  The lowest priority atom IV is behind the C* and thus is not visible.

c.                  Draw an arrow from group I to group II to III.    This will be from Br to Cl to F, in this case.

(R)-Bromochlorofluoromethane

d.                 If the arrow is “clockwise”, by definition, the molecule is a member of the “R” family. . If "counterclockwise", the molecule is "S".

              So 9 is   “R”.  How about 10?  If we repeat the process for 10, we will find that the arrow is “counterclockwise”.  The molecule is a member of the “S” family.

(S)-Bromochlorofluoromethane

Unfortunately, in actual practice, assignment of “R” or “S” is not quite so easy, because the image of the molecule also is not known.   Let's return to the idea of the colorless crystals of "alanine"  that you "isolated" from soybeans.  You have no idea if the image of these colorless crystals is 3 or 4.  Since a "picture" of the molecule is not available, you cannot immediately assign “R” or “S" configuration.

Let’s return to the question of “R” or “S“ assignment.  Consider molecule 11. 

Two atoms attached to C* have the same atomic number, namely 6.   To decide whether methyl (CH3) or carboxyl (COOH) has the higher priority, "go-out" until you find a "point-of-difference".   The "next atoms" "out" from the carbon of CH3 are three hydrogens, which have atomic number of 1.  For the carboxyl group, the next atoms "out" from carbon are oxygens, which have the atomic number 8.  On this basis, carboxyl gains higher priority than methyl.

After arrows are drawn from I to II and from II to III, a counterclockwise pattern is found.  Thus, 11 is (S)-lactic acid.

 Racemates

"Racemates" or "racemic mixtures" are mixtures of enantiomers.  The mixture is "optically inactive".  Thus, if you "synthesize" (i.e. "prepare" or "make") lactic acid in the laboratory, you will get a 50:50 mixture of 11 and 12.  Since the two isomers have equal but opposite "optical rotations", the "racemate" will be optically inactive.    Generally, single enantiomers only are found in living systems, or are derived from living systems.  Compounds with chiral centers made in the laboratory will be "racemates".  The two enantiomers, may, in theory be separated into equal amounts of two enantiomers.   Thus, a "racemic mixture" of 50% 11 and 50% 12 may be separated into pure components.  The process is called a "resolution".   "Resolutions" are extremely difficult.  The process almost always involves the use of another substance that is already "optically active" (often derived from nature).

Diastereomers

Diastereomers have more than one chiral center.  Two cases in point are 13 and 14.  Diastereomers have the same configuration,  “R“ or “S“, at some chiral centers, and opposite configuration at other chiral centers.  Diastereomers have different boiling points, meltings points, solubilities, and generally different physical properties.  Diastereomers do NOT have equal but opposite optical rotations.  Thus, it may occur that one diastereomer has a = +10^o and the other a = +5^o.  Diastereomers are NOT mirror images of one another.

Compound 13 is “R” at C2 and “S” at C3.  Compound 14 is “S” are both centers. 

[Note that an easy way (on paper) to change from "R" to "S" at a given C* is to switch any two groups.  In the laboratory, it is much more difficult.]

What about compound 15?  Compound 15 is “R “ at both chiral centers.  Compounds 14 and 15 are “enantiomers”.  They "may be written" as mirror images.   This means that it is possible to find a geometry of the molecules (a conformation) in which 14 and 15 may be seen as mirror image structures.  They are opposite configuration at each and every C*. 

    Compounds 13 and 15 are again “diastereomers”, since they are the same configuration at C2 and opposite configuration at C3.

    Some exercises for students will follow later.

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