Stereochemistry part 2

Stereochemistry, a subdiscipline of chemistry, involves the study of the relative spatial arrangement of atoms within molecules. An important branch of stereochemistry is the study of chiral molecules.

Stereochemistry is also known as 3D chemistry because the prefix "stereo-" means "three-dimensionality".
The study of stereochemical problems spans the entire range of organic, inorganic, biological, physical and supramolecular chemistries. Stereochemistry includes methods for determining and describing these relationships; the effect on the physical or biological properties these relationships impart upon the molecules in question, and the manner in which these relationships influence the reactivity of the molecules in question (dynamic stereochemistry).

 

DRAWING STEREOCHEMICAL STRUCTURES FROM A TO Z

 

Structural diagrams which depict stereochemistry must be prepared with extra care to ensure there is no ambiguity.  The ability to proficiently draw and read such structures requires some practice with reference to 3D molecular models. Some simple "do's" and "don'ts" of the art of stereochemical drawing are illustrated below.  In general, the molecules are presented in some kind of perspective drawing, based on the idea that the four substituents of a tetrahedral center can be divided into two pairs, laying in mutually perpendicular planes.  Most often the center and two of such substituents are shown in the plane of the drawing (i.e. the plane of the drawing surface) and their bonds are depicted as plain lines ( ). 
Bonds to the other two substituents are shown with different symbols.  Bonds to atoms above the plane of the drawing (coming out, toward the viewer) are shown with a bold wedge (  ), with the narrow end of the wedge starting at the stereogenic center.  As an alternative bald bar bonds (  ) are used.
Bonds to atoms below the plane (going in, away from the viewer) are shown with hash wedges (  ).  There are two separate conventions in use.  In the American usage the narrow edge points to the central atom, while in the European convention, the wide edge points to the central atom.  As an alternative, a bar of hash lines ( ) is used.  A broken line () or an open wedge (  ) can also be found in some drawings, but their usage is discouraged.
These various presentations are illustrated below on an example of a compound with one stereocenter (in all cases, except where indicated,  the same S absolute stereochemistry is shown). The first structure (A) is the favored presentation, while B shows a rotational variant of A (any rotation within the plane of these drawings is perfectly acceptable). Structure C demonstrates the European convention that has a more consistent perspective view of the wedges.  Structure D shows the use of open wedge (not recommended), while structure E illustrates the use of bar bonds, that are commonly employed to designate relative stereochemistry. In our class we will use wedges (A) to indicate absolute stereochemistry (one enantiomer), and bars (E) to designate racemic mixtures (and relative stereochemistry, see below).

		A B F G H  I J K L/M

In addition to structures with two bonds to the stereocenter in the plane of the drawing (above), some representations may have only one bond in that plane (F and G) and some have none (H and I), as shown below.  In structures like F and G or H and I, three or four wedged lines are used to designate the 3D disposition of substituents,  respectively.
Confusingly, one may often encounter other 3D structures that seem to follow the above conventions, but are in fact different perspective representations. In structures J–M the plain bonds (–––) are not in the plane of the drawing. Their 3D disposition is implied by the wedge lines present.  For example, in structure J the wedge bond is a mast pole on three legs, and in L and M the plain-line bonds are really going in or out of the plane of the drawing, respectively..
All these arrangements may be visualized in 3D by selecting the appropriate radio-button in the Jmol applet. Of course C, D, and E are in the exact same orientation as A and are not repeated. Note that all the manipulations are just simple rotations of the same molecule.
Paying attention to the correct representation of the perspective is crucial.  Some incorrect usage of wedge bonds is illustrated in structures N and O.  Despite appearances, there is no stereochemical information in such structures.   A couple of simple rules can be used to decide adequacy of such drawings:  (1) the bold-wedge and the hash-wedge substituents must be on the same side of an imaginary line connecting the plain-line bonded substituents  (as in structures A or B), and  (2) two bold-wedge subsituents (or two hash-wedge substituents) must be on the opposite sides of an imaginary line connecting the plain-line bonded substituents (as in structures L or M).

Another way to present stereo-centers is with help of Fischer projections which are designed not to employ any wedged lines. In this class we will not cover Fischer projections. The convention used is that all vertical bonds are pointing in, away from the viewer, and all horizontal bonds are pointing out, toward the viewer. Such structures can only have vertical and horizontal bonds on stereocenters, and cannot be rotated by 90° (without a change in stereochemistry).  Often, the most oxidized carbon is placed at the top (but other arrangements are acceptable as well).  Structure P (above) is the Fischer projection, and it could be translated into a "wedge" structure H (or structures L or M, above).  One has to be careful to distinguish Fischer projections, such as P, from structures where no stereochemistry is explicitly shown, i.e. only plain bonds are used (in such cases 90° angles between drawn bonds should be avoided; see also below). 
Hydrogen atoms attached to the stereocenter are often not shown explicitly, as is common for skeletal structures in general. If done carefully, the omission does not lead to complications and the stereo information is perfectly readable, but there are instances where ambiguity or loss of stereo information may occur. In general, at the learning stage, it is advisable to show stereocenter hydrogens explicitly.

These structures, labeled with the same letters as their originals above, illustrate clear and unambiguous examples of hydrogens omitted without any loss of stereochemical information.  Structure P' is fine as long as it is made clear that it is a Fischer projection.  On the other hand, other presentations may be questionable. For example, structure S can be interpreted (correctly) as structure L with hydrogen omitted, but (especially) when drawn by hand it may be interpreted as a poorly drawn structure of R, leading to the opposite 3D information.  Despite appearances, structure T does not carry any stereochemical information (the position of hydrogen with its plain-line bond needs to be explicitly specified). 
If stereochemistry is unspecified, no wedge lines are used, as in structure U. To easily distinguish such situation from Fischer projections, multiple 90° and 180° angles between drawn bonds should be avoided. Alternatively, situations when the stereochemistry is unknown, or a mixture of both enantiomers is present, can be indicated explicitly by a wavy line ()  as shown in structure W.  Similarly,  the unspecified stereochemistry of the double bond (E or Z) may be shown by drawing "extended" formulas as in structure X.  The wavy line in structure Y is an alternative way to indicate unknown stereochemistry or a mixture of isomers.

For molecules with multiple stereocenters other methods of display are also available. The Newman projection in Z(1) unambiguously defines the absolute stereochemistry on both centers (S,S).  Occasionally, a perspective drawing without any wedge or bar bonds is employed as shown in Z(2). This sawhorse representation is a "stretched" version of the Newman projection.  When wedges or bars are used, the plain-line bonds should be used to connect the centers, as illustrated for structure Z(3).  This approach avoids possible ambiguities as to which wedge bond belongs to which center.

	Z(1) Z(2)/Z(3) Z(4)/Z(5)

As illustrated by structures Z(4) 'and Z(5) omitting stereocenter's hydrogens (if done properly) often increases the clarity of the presentation. Structure Z(4) with wedge bonds designate absolute stereochemistry, i.e. enantiomerically pure compound with two S centers. In our class, structure Z(5) with bars designates only the relative stereochemistry, i.e. it corresponds to the racemic mixture of  (S,S) and (R,R) enantiomers.

http://en.wikipedia.org/wiki/Stereochemistry
(http://www.organicchemistryreview.com/STEREOCHEMISTRY.html)

stereochemistry part 1

       Stereochemistry is another name for the optical activity that a molecule expresses. Optical activity is a property of organic molecules       So far several methods have been presented to distinguish between organic molecules. They can be classified as alkanes, alkenes, or alkynes depending on the presence of double and triple bonds, and their functional groups can be identified. A systematic method for naming compounds has also been presented, allowing different compounds to be differentiated.
     Somtimes, two different molecules can possess the same chemical formula. Two distinct molecules with the same chemical formula are called isomers.

Constitutional isomers

Constitutational isomers or structural isomers are molecules with the same chemical formula but different structures of atoms and bonds. For example, both 3-methylpentane and hexane have the same chemical formula, C₆H₁₄, yet they clearly have different structures:

3-methylpentane	hexane

Another example involves functional groups. Methoxy methane, an ether, and ethanol, an alcohol, both have the chemical formula C₂H₆O:

Methoxy methane	Ethanol

Enantiomers

Sometimes, two molecules can have the same chemical formula and even the same structure of atoms bonded to each other but can still be distinct molecules (isomers). How can this be so? Not only the arrangement in which atoms are bonded to each other, but also the actual geometry of those bonds in space, determine the properties of a molecule. Stereoisomers are distinct molecules with the same sequence of bonded atoms but a different orientation of these atoms in space. Consider the following representations of the molecule bromochlorofluromethane:

S enantiomer

R enantiomer

The two representations clearly depict the same structure of atoms bonded to each other. Yet they are clearly different molecules. No rotation will cause the molecule on the right to adopt the same arrangement of bonds in space as the molecule on the left. If bromine is rotated to adopt the same location as the molecule on the left, then the hydrogen and fluorine atoms will have switched places. Clearly the two molecules are different. In fact they are mirror images of each other. What is special about bromochlorofluromethane is that its mirror image can never be manipulated to produce the original molecule; that is, the mirror image can never be superimposed on the original. Two molecules which are non-superimposibile mirror images of each other are called enantiomers. One special property of enantiomers is that despite being distinct molecules, enantiomers have the same physical properties. The only exception is that enantiomers will rotate plane polarized light in different directions. This property is used to distinguish them. The amino acids that compose proteins in living organisms are enantiomers, but only one form exists in most living organisms. Thus, often only one enantiomer of a drug molecule will have an effect in the human body.
The simplest example of enantiomers involve four different groups bonded to a central carbon atom. When the four groups are distinct, the molecule is an enantiomer, and the carbon atom is called a chiral center or stereogenic center. Two enantiomers can be distinguished by assigning a letter to the chiral center, either R or S. First, the three substituents are numbered according to atomic mass. The atom with the highest atomic mass is assigned highest priority (1), the second heaviest is assigned next highest priority (2), and so on. If two atoms are the same, the atoms bonded to them are examined, and so on until the first point of difference, at which point the assignment of priority is made. If a rotation through the numbers 1, 2, and 3 is clockwise, the enantiomer is labeled R; if it is counterclockwise, it is labeled S. Let's assign R or S configurations to the above molecules. In the molecules above, bromine is the heaviest, so it get the highest priority (1). Chlorine comes next (2). Flourine comes third (3). In the molecule on the left, if we follow along passing from 1 to 2 to 3, we find that we have made a rotation in the counterclockwise direction; thus the left molecule is the S enantiomer. If we now follow from 1 to 2 to 3 on the right, we find that we have moved in a clockwise direction; the molecule on the right is the R enantiomer. The system of assigning R or S labels to enantiomers, called the Cahn–Ingold–Prelog system is a simple way of distinguishing between enantiomers.

Diastereomers

Diastereomers are stereoisomers that are not enantiomers; that is, they are distinct molecules with the same structural arrangement of atoms that are non-superimposible, non-mirror images of each other. Consider the following representations of tartaric acid:

(2S,3S)-tartaric acid	(2R,3R)-tartaric acid	(2R,3S)-tartaric acid

In tartaric acid, both of the central carbon atoms are stereogenic centers. Thus, they both must be assigned R or S configurations according to the Cahn–Ingold–Prelog system. The first two molecules shown above, (2S,3S)-tartaric acid and (2R,3R)-tartaric acid, are clearly enantiomers of each other since they are mirror images. But what about the third molecule, (2R,3S)-tartaric acid? It is clearly distinct from the first two, yet the same atoms are bonded to each other. It is also very clearly not a mirror image of either of the first two. Thus, as a non-superimposible, non-mirror image with the same arrangement of atoms, (2R,3S)-tartaric acid is a diasteromer of the first two molecules.
Another type of diastereomer can occur in the presence of a double bond. Consider the following representations of the molecule 2-butene:

cis-2-butene	trans-2-butene

The molecules are clearly distinct; they clearly have the same arrangement of atoms structurally; and they clearly are not mirror images. Thus, these molecules are diastereomers. Diastereomers such as these can be distinguished using the Cahn–Ingold–Prelog system. If there are only two substituents, as above, the molecule is simply inspected to determine whether the higher priority substituents are on the same side of the double bond. If they are, the molecule is designated cis. If not, it is designated trans. If more than two different substituents are present on the double bond, then the highest priority substitunet on each carbon atom composing the double bond is identified. If these highest priority substituents are on the same of the double bond, the molecule is designated Z; if not, it is designated E.

nitriles

What are nitriles? Nitriles contain the -CN group, and used to be known as cyanides. Nitriles Nomenclature Formula 3D structure Functional class = alkyl cyanide Functional group suffix = nitrile or -onitrile Substituent prefix = cyano- Notes : The cyano prefix is used in a very similar manner to haloalkanes The cyano nomenclature is most common when the alkyl group is simple. The nitrile suffix is used in a very similar manner to carboxylic acids. Cyano substituent style: The root name is based on the longest chain with the -C≡N as a substituent. This root give the alkane part of the name. The chain is numbered so as to give the -C≡N group the lowest possible locant number Nitrile style: The root name is based on the longest chain including the carbon of the nitrile group. This root give the alkyl part of the name. Since the nitrile must be at the end of the chain, it must be C1 and no locant needs to be specified. Nitriles can also be named by replacing the -oic acid suffix of the corresponding carboxylic acid with -onitrile. Cyano substituent style: Functional group is an alkane, therefore suffix = -ane The longest continuous chain is C3 therefore root = prop The substituent is a -CN therefore prefix = cyano The first point of difference rule requires numbering from the right as drawn, the substituent locant is 1- 1-cyanopropane Nitrile style: Functional group is a -C≡N, therefore suffix = -nitrile Hydrocarbon structure is an alkane therefore -ane The longest continuous chain is C4 therefore root = but butanenitrile  CH3CH2CH2C≡N Some simple nitriles The smallest organic nitrile is ethanenitrile, CH3CN, (old name: methyl cyanide or acetonitrile - and sometimes now called ethanonitrile). Hydrogen cyanide, HCN, doesn't usually count as organic, even though it contains a carbon atom. Notice the triple bond between the carbon and nitrogen in the -CN group. The three simplest nitriles are: CH3CN ethanenitrile CH3CH2CN propanenitrile CH3CH2CH2CN butanenitrile When you are counting the length of the carbon chain, don't forget the carbon in the -CN group. If the chain is branched, this carbon usually counts as the number 1 carbon.
Physical properties Boiling points The small nitriles are liquids at room temperature. nitrile boiling point (°C) CH3CN 82 CH3CH2CN 97 CH3CH2CH2CN 116 - 118
These boiling points are very high for the size of the molecules - similar to what you would expect if they were capable of forming hydrogen bonds. However, they don't form hydrogen bonds - they don't have a hydrogen atom directly attached to an electronegative element. They are just very polar molecules. The nitrogen is very electronegative and the electrons in the triple bond are very easily pulled towards the nitrogen end of the bond. Nitriles therefore have strong permanent dipole-dipole attractions as well as van der Waals dispersion forces between their molecules.
Solubility in water Ethanenitrile is completely soluble in water, and the solubility then falls as chain length increases. nitrile solubility at 20°C CH3CN miscible CH3CH2CN 10 g per 100 cm3 of water CH3CH2CH2CN 3 g per 100 cm3 of water The reason for the solubility is that although nitriles can't hydrogen bond with themselves, they can hydrogen bond with water molecules. One of the slightly positive hydrogen atoms in a water molecule is attracted to the lone pair on the nitrogen atom in a nitrile and a hydrogen bond is formed. There will also, of course, be dispersion forces and dipole-dipole attractions between the nitrile and water molecules. Forming these attractions releases energy. This helps to supply the energy needed to separate water molecule from water molecule and nitrile molecule from nitrile molecule before they can mix together. As chain lengths increase, the hydrocarbon parts of the nitrile molecules start to get in the way. By forcing themselves between water molecules, they break the relatively strong hydrogen bonds between water molecules without replacing them by anything as good. This makes the process energetically less profitable, and so solubility decreases. Making nitriles  making nitriles from halogenoalkanes The halogenoalkane is heated under reflux with a solution of sodium or potassium cyanide in ethanol. The halogen is replaced by a -CN group and a nitrile is produced. Heating under reflux means heating with a condenser placed vertically in the flask to prevent loss of volatile substances from the mixture. The solvent is important. If water is present you tend to get substitution by -OH instead of -CN. For example, using 1-bromopropane as a typical halogenoalkane: You could write the full equation rather than the ionic one, but it slightly obscures what's going on: The bromine (or other halogen) in the halogenoalkane is simply replaced by a -CN group - hence a substitution reaction. In this example, butanenitrile is formed. Making a nitrile by this method is a useful way of increasing the length of a carbon chain. Having made the nitrile, the -CN group can easily be modified to make other things - as you will find if you explore the nitriles menu. Making nitriles from amides Nitriles can be made by dehydrating amides. Amides are dehydrated by heating a solid mixture of the amide and phosphorus(V) oxide, P4O10. Water is removed from the amide group to leave a nitrile group, -CN. The liquid nitrile is collected by simple distillation. For example, you will get ethanenitrile by dehydrating ethanamide. Making nitriles from aldehydes and ketones Aldehydes and ketones undergo an addition reaction with hydrogen cyanide. The hydrogen cyanide adds across the carbon-oxygen double bond in the aldehyde or ketone to produce a hydroxynitrile. Hydroxynitriles used to be known as cyanohydrins. For example, with ethanal (an aldehyde) you get 2-hydroxypropanenitrile: With propanone (a ketone) you get 2-hydroxy-2-methylpropanenitrile: In every example of this kind, the -OH group will be on the number 2 carbon atom - the one next to the -CN group. The reaction isn't normally done using hydrogen cyanide itself, because this is an extremely poisonous gas. Instead, the aldehyde or ketone is mixed with a solution of sodium or potassium cyanide in water to which a little sulphuric acid has been added. The pH of the solution is adjusted to about 4 - 5, because this gives the fastest reaction. The reaction happens at room temperature. The solution will contain hydrogen cyanide (from the reaction between the sodium or potassium cyanide and the sulphuric acid), but still contains some free cyanide ions. This is important for the mechanism. These are useful reactions because they not only increase the number of carbon atoms in a chain, but also introduce another reactive group as well as the -CN group. The -OH group behaves just like the -OH group in any alcohol with a similar structure. For example, starting from a hydroxynitrile made from an aldehyde, you can quite easily produce relatively complicated molecules like 2-amino acids - the amino acids which are used to construct proteins. The hydrolysis of nitriles Introduction When nitriles are hydrolysed you can think of them reacting with water in two stages - first to produce an amide, and then the ammonium salt of a carboxylic acid. For example, ethanenitrile would end up as ammonium ethanoate going via ethanamide. In practice, the reaction between nitriles and water would be so slow as to be completely negligible. The nitrile is instead heated with either a dilute acid such as dilute hydrochloric acid, or with an alkali such as sodium hydroxide solution. The end result is similar in all the cases, but the exact nature of the final product varies depending on the conditions you use for the reaction. Acidic hydrolysis of nitriles The nitrile is heated under reflux with dilute hydrochloric acid. Instead of getting an ammonium salt as you would do if the reaction only involved water, you produce the free carboxylic acid. For example, with ethanenitrile and hydrochloric acid you would get ethanoic acid and ammonium chloride. Why is the free acid formed rather than the ammonium salt? The ethanoate ions in the ammonium ethanoate react with hydrogen ions from the hydrochloric acid to produce ethanoic acid. Ethanoic acid is only a weak acid and so once it has got the hydrogen ion, it tends to hang on to it. Alkaline hydrolysis of nitriles The nitrile is heated under reflux with sodium hydroxide solution. This time, instead of getting an ammonium salt as you would do if the reaction only involved water, you get the sodium salt. Ammonia gas is given off as well. For example, with ethanenitrile and sodium hydroxide solution you would get sodium ethanoate and ammonia. The ammonia is formed from reaction between ammonium ions and hydroxide ions. If you wanted the free carboxylic acid in this case, you would have to acidify the final solution with a strong acid such as dilute hydrochloric acid or dilute sulphuric acid. The ethanoate ion in the sodium ethanoate will react with hydrogen ions as mentioned above. The reduction of nitriles using LiAlH4 The reducing agent Despite its name, the structure of the reducing agent is very simple. There are four hydrogens ("tetrahydido") around the aluminium in a negative ion (shown by the "ate" ending). The "(III)" shows the oxidation state of the aluminium, and is often left out because aluminium only ever shows the +3 oxidation state in its compounds. To make the name shorter, that's what I shall do for the rest of this page. The structure of LiAlH4 is: In the negative ion, one of the bonds is a co-ordinate covalent (dative covalent) bond using the lone pair on a hydride ion (H-) to form a bond with an empty orbital on the aluminium. The overall reaction The nitrile reacts with the lithium tetrahydridoaluminate in solution in ethoxyethane (diethyl ether, or just "ether") followed by treatment of the product of that reaction with a dilute acid. Overall, the carbon-nitrogen triple bond is reduced to give a primary amine. Primary amines contain the -NH2 group. For example, with ethanenitrile you get ethylamine: Notice that this is a simplified equation - perfectly acceptable to UK A level examiners. [H] means "hydrogen from a reducing agent". The reduction of nitriles using hydrogen and a metal catalyst The carbon-nitrogen triple bond in a nitrile can also be reduced by reaction with hydrogen gas in the presence of a variety of metal catalysts. Commonly quoted catalysts are palladium, platinum or nickel. The reaction will take place at a raised temperature and pressure. It is impossible to give exact details because it will vary from catalyst to catalyst. For example, ethanenitrile can be reduced to ethylamine by reaction with hydrogen in the presence of a palladium catalyst.