Chemitry

Basic Principles of Organic Chemistry Next: Structural Isomerism >> Classification of Organic Compounds a) Acyclic or open chain compounds: These contain alkanes, alkenes, alkynes and their derivatives. These are also called aliphatic compounds. b) Cyclic or closed chain compounds: • Cyclic compounds whose rings are made up of only one kind of atoms, i.e. carbon atoms are called homocyclic or cabocyclic compounds. Aliphatic cyclic compounds are called alicyclic compounds eg cyclopropane, cyclobutane etc

•

Organic compounds containing one or more fused or isolated benzene rings and their functinalized derivatives are called benzenoids or aromatic compounds, eg benzene, toluene, naphthalene, anthracene etc.

• Cyclic compounds containing one or more heteroatoms (usually O,N, S etc) are called hetrocyclic compounds eg ethylene oxide, tetrahydrofuran (THF), furan, pyrole etc

IUPAC Nomenclature of Organic Compounds:

a) Nomenclature of saturated hydrocarbons: • Select the longest continuous chain of carbon atoms in the molecule. The compound is named as a derivative of this alkane • Number the carbon atoms in the parent chain starting from the end which gives lowest possible sum for the numbers of the carbon atoms carrying the substituents • That set of locants is preferred, which when compared term by term with other set of locants, each in order of increasing magnitude, has the lowest term at the first point of difference. For example the set of locants (2,7,8) is preferred over the set of locants (3,4,9) since 2 comes before 3 even though the sum of locants in the former case is 17 while in the latter case, it is 16

• The correct name is: 2,7,8 - Trimethyldecane and not 3,4,9 Trimethyldecane b) Nomenclature of compounds containing functional group or multiple bonds: • Select the longest continuous chain containing the carbon atoms having the functional group or those involved in the multiple bonds • The numbering of atoms in the parent chain is done in such a way that the carbon atom bearing the functional group or those carrying the multiple bond gets the lowest possible number • While writing the name of alkene (double bond) or alkyne (triple bond), the primary suffix 'ane' of the corresponding alkane is replaced by 'ene' and 'yne' respectively. However, if the multiple bond occurs twice or thrice in the parent chain, the prefix di- or tri- is attached to the primary suffix ene or yne • In naming the organic compounds containing one functional group a suffix known as secondary suffix is added to the primary suffix (giving number of carbon atoms in the chain) to indicate the nature of the functional group. A few important secondary suffixes are: Functional

Secondary

Functional group

Secondary

group

suffix

suffix

Alcohols (-OH)

-ol

Aldehydes (-CHO)

-al

Ketones (>C=O)

-one

Carboxylic acids (-COOH)

-oic acid

Amines (-NH2 )

-amine

Acid amides (-CONH2 )

-amide

Acid chlorides (-COCL)

-oyl chloride

Esters (-COOR)

-oate

Nitrites (-C=N)

-nitrite

Thioalcohols (-SH)

-thiol

Nomenclature of compounds having polyfunctional groups: When an organic compound contains two or more functional groups, one group is called the principal functional group while the others are called the secondary functional groups and are treated as substituents: The order of preference for principal group is: Carboxylic acid > acid anhydrides > esters > acid halides > amides > nitrites > aldehydes > ketone > alcohols > amines > double bond > triple bond When the functional groups act as substituents, they ar named as: Functional group

Prefix

Functional group

Prefix

- COOH

Carboxy

-CHO

Formyl

-COOR

Alkoxy cabonyl or Carbalkoxy

>CO

Oxo or Kelo

-COCL

Chloroformyl

-OH

Hydroxy

-CONH2

Carbamoyl

-SH

Mecaplo

-CN

Cyano

-NH2

Amino

-OR

Akoxy

=NH

Imino

-X

Halo

-NO2

Nitro

Nomenclature of simple aromatic compounds: a) Nuclear substituted: In these the functional group is directly attached to the benzene ring. Most of these compounds are better known by their common and historical names. In the IUPAC system, they are named as derivatives of benzene. b) Side chain substituted: In these the functional group is present in the side chain of the benzene ring. Both in the common and IUPAC systems, these are usually named as phenyl derivatives of the corresponding aliphatic compounds.

Sturctural Isomerism Componds having the same molecular formula but different structures: • Compounds having the same molecular formula but different arrangement of carbon chains are called chain or nuclear isomers. For example (i) pentane, isopentane and neopentane (ii) hexane, 2methulpentane, 3-methylpentane, 2,3-dimethylbutane and 2,2dimethulbutane etc • Compounds having the same structure of the carbon chain but differing in position of multiple bonds, functional groups or substituents are called position isomers. For example, (i) but-1-ene and but-2-ene (ii) prpan-1-ol and propan-2-ol (iii) o-,m- and p- xylenes etc. • Compounds having the same molecular formula but different functional groups are calledfunctional isomers. For example, (i) alcohols and ethers (ii) aldehydes and ketones (iii) alkynes and dienes (iv) various amines (1o , 2o , 3o ) (v) cyanides and isocyanides etc.

Compounds having the same molecular formula but different number of carbon atoms on either side of the functional group (O,S or NH) are caled metamers. For example, (i) diethyl ether and methyl npropyl ether (or isopropyl methyl ether). 2-pentanone and 3pentanone are position isomers as well as metamers. • Isomers obtained by 1,3-migration of a proton from one polyvalent atom to the other within the same molecule are called tautomers. If one tautomer contains a keto group and the other the enol group, such a tautomerism is called keto-enol tautomerism. For example, acetaldehyde and vinyl alcohol. •

• The percentage of enol form is negligible in simple aldehydes and ketones, ie, acetone, acetaldehyde etc. The percentage of enol form, however, increases, if the enol form is stabilized by intramolecular Hbonding. For example, the percentage of enol form in acetoacetic ester is 7% while that in acetylacetone is 76%. •

Only those aldehydes and ketones which contain one or more hydrogens show keto-enol tautomerism.

-

• Primary and secondary nitroalkanes also show tautomerism but tertiary nitroalkanes and nitroarenes do not. • Compounds having the same molecular formula but possessing open chain and cyclic structures are called ring chain isomers. For example, propene and cyclopropane.

Stereoisomerism Isomers which have the same structures but differ in the relative arrangements of atoms or groups in space are called stereoisomers. It is of three types, i.e. conformational isomerism, optical isomerism and geometrical isomerism. Geometrical Isomers: Stereoisomers which have the same structural formula but differ in the spatial arrangement of atoms around the double bond are called geometrical isomers. The isomer in which the two similar atoms/groups lie on the same side of the double bond is called the cisisomer while that isomer in which the two similar atoms/groups lie on the opposite side of the double bond is called the trans-isomer.

• The necessary and sufficient condition for a molecule to exhibit geometrical isomerism is that each ot the two carbon atoms of the double bond must have different atoms/groups which may be same of different. Gor example , alkenes of the types abC=Cab and abC=Cde show geometrical isomerism • Compounds containing C=N and N=N also show geometrical isomerism • 1,2-1,3 and 1,4-disubstituted cycloalkanes also show geometrical isomerism • The dipole moment of a cis-isomer is higher than that of a transisomer • E and Z configuration of geometrical isomers ○ Give priority to the four groups attached to the double bonded

C-atoms. Higher the atomic number of the atom of the group attached to the double bonded carbon atom higher will be its priority. If priority cannot be decided the atoms directly attached to double bonded C-atoms then next atoms of the group are compared ○ If two higher priority groups are on the same side of the double bond then it is called a Z-isomer otherwise it is called an Eisomer Conformational Isomerism: The infinite number of momentary arrangement of the atoms in space which result through rotation about a single bond are called conformations or rotational isomers • Ethane has infinite number of conformations, of which, only two, i.e. staggered and eclipsed are important. The staggered conformation of ethane is more stable than the eclipsed conformation by about 3.0 kcal or 12.55 kJ mol-1 • Important conformations of n-butane are staggered (or anti), gauche (or skew), partially eclipsed and fully eclipsed. Their relative stability is in the order:staggered>gauche>partially eclipsed>fully eclipsed • Cyclohexane exists in two non-planer conformations, i.e., chair and the boat form. Both these conformations are free from angle strain (i.e., have tetrahedral angles) • The boat form of cyclohexane is less stable than the chair form by about 44 kJ mol-1

• The two conformations of cyclohexane cannot be separated but ar readily interconvertible

BASIC CONCEPTS OF CHEMISTRY Law of Definite Proportion: We know that mass is conserved. If we were to take a sample of a pure compound it will always contain the same elements combined in the same proportion by mass. Law of Multiple Proportion: If one element can combine with another to form more than one compound then the mass ratios of the elements in the compounds are simple whole-number ratios of each other. Law of Definite Proportion by Volume: The volumes of reactants and products, in chemical reactions, are related to each other by simple whole-numbers. Of course the measurements must be taken at the same temperature and pressure. Law of Reciprocal Proportions: The ratio by mass, in which two elements combine with thefixed mass of a third element, is either the same or a simple multiple of the ratio in which they combine with each other. Avogardo's Law: Equal volumes of all gases contain equal number of molecules under similar conditions of temperature and pressure. One Atomic Mass Unit (amu): It is a mass unit equal to exactly one twelfth the mass of a carbon 12 atom. Formula Mass of a Substance: It is the sum of the atomic masses of all atoms in a formula unit of a compound. Molar Mass: The mass of 1 mole of any substance. Molar Volume (Vm): The volume of 1 mole of the substance. Mole (m): • It is a number which is equal to 6.023 x 1023 . It is the amount of substance which contains as many entities (atoms, molecules, ions or other particles) as there are atoms in exactly 12 grams of Carbon 12 isotope. • It is also known as Avogadro's Constant. •

Number of Mole Method: Number of moles of a substance n = w (in gms)/m

also n = volume (in lt) at NTP/22.4lt also n = M x vol in lt (where M is Molarity = number of moles of solute per lt of solution) M= (w/m) x (1000/ Vm ) Number of mole method is used only in the case of balanced chemical equations. The coefficients of balanced equations represent the ratios in terms of number of moles in which the reactants react or products are formed. Number of particles(ions, atoms or molecules) in a given number of moles = number of moles x 6.023 x 1023 . Equivalent Weight of an Element (E): It is defined as the number of parts by weight of the element which combine with or displace from a compund 1 part by weight of Hydrogen, 8 parts by weight of Oxygen or 35.5 parts by weight of Chlorine. E = M ( Molar Mass) / nf (n factor) where: nf = Valency in case of an atom = Total positive or negative oxidation number of an atom in a molecule = Basicity or Acidity. = Change in oxidation number in case of a redox reacton. Number of Gram Equivalents: = weight in gms / E (Equivalent wt) = Normality x vol in lt Normality = (w/E) x (1000/Vm ) Number of Equivalents Method: In this method you need not balance the chemical equation. The basic principle is that the equivalents of each reactant which have dissapeared are equal to the number of equivalents of each of the products formed. The working tool for this is equivalent weight. Number of equivalents = W in gms / Eq Wt = (W x nf ) / M Dulong and Petit's Law: Atomic wt (approx) = 6.4/specific heat(in calories)

Curved Arrows Additional reading recommendation: You may find Chapter 3 of Pushing Electrons by Daniel P. Weeks (Saunders College Publishing; ISBN 0-03-0206936) to be a useful tool for mastering the fundamentals of using curved arrows. Discussion: Chemical reactions are a result of bonding changes. These bonding changes are most easily described by considering the changes in electron sharing between atoms. For example, consider the collision of two water molecules leads to the ionization of water and the formation of hydroxide and hydronium ion.

In this reaction, the oxygen atom of one water molecule collides with a hydrogen atom of the second water molecule. A lone pair of the oxygen atom becomes the new O-H bond in the hydronium ion. Because a hydrogen can only be fully bond to one other atom at a time, the old O-H bond is lost. The electron pair of the old O-H bond becomes a lone pair, sole property of the oxygen atom of hydroxide ion. This very simplistic step-by-step bookkeeping description of all the bonding and electron changes in a reaction is called the reaction mechanism. Study, understanding and prediction of reaction mechanisms is at the very heart of reactions in organic chemistry. Mastering the fundamentals of reaction mechanisms is a fundamental survival skill for students of organic chemistry. You will use them every day that you study organic chemistry. Above we used several lines of text to describe the bonding changes in a single step of a reaction mechanism. A reaction mechanism might have ten steps or more,

making such descriptions very cumbersome. A shorthand notation that summarizes these changes has thus been developed. This notation, called curved arrow formalism, provides a rapid way of drawing bond and electron changes in a given mechanism step. They are also useful to indicate electron changes between a set of contributing resonance structures. Each curved arrow with two barbs on the head represents the shift of one electron pair. (Later we will encounter single-barbed curved arrows that represent the shift of single electrons.) The curved arrows shows the direction of electron flow. The tail shows the electron origin, and always come from an electron source, usually a lone pair or bonding pair from a σ or π bond. The head of the arrow indicates the electron pair destination, either as a new lone pair or a new bond. If the arrowhead points to another atom, that atom must either have an open octet and thus be able to accept the electron pair, or have an electron pair that can be displaced by the incoming electron pair. Electrons never flow from atoms which are electron-poor to atoms which are electron-rich, so a curved arrow will never point from an atom with a positive charge to an atom with a negative charge. New bond formed to X: Bonding pair becomes lone pair; bond broken: The use of curved arrows for the ionization of water are shown below.

The tail of the curved arrow on the right starts at the oxygen lone pair, meaning this curved arrow shows a bonding change for this lone pair. The head of the curved arrow points to the space between the oxygen and hydrogen atoms, meaning the electron pair ends in that space as a bond between the oxygen and the hydrogen. The hydrogen that accepts a new bond to oxygen must give up a pair of electrons because it cannot have more than two valence electrons at a time. Thus, the old O-

H bond is displaced by the new electron pair from the other oxygen atom. The curved arrow on the left indicates the electron pair that was the O-H bond becomes a lone pair on the oxygen of the hydroxide ion. If the arrow starts at a bond between two atoms, then that bond is broken. If the arrow ends between two atoms, then a new bond is formed between those atoms. (If the atoms are already bonded, then a double or triple bond results.) The process of drawing a curved arrow mechanism is also commonly called "electron pushing." When drawing curved arrows, the start and stopping points of the arrows are critical. Things that make no difference are where the arrows curve up or down, or whether they start or stop at the top or bottom of an atom. The arrows you draw may therefore look different than the arrows shown in this tutorial. Note also the changes in formal charge that result from the electron changes. If an atom shares a lone pair that it used to have all to itself then its charge decreases by one (i.e., neutral atom become +1). If an atom gains a bonding electron pair all to itself as a lone pair then its charge increases by one unit (i.e., a neutral atom becomes -1). The charge is conserved in this mechanism step. The total charge on the left (zero) is the same as the total charge on the right (-1 +1 = 0). Charge is conserved in all mechanism steps. You should make a habit of checking your work against this point to minimize errors. Lone pairs are often involved in reaction mechanisms, so you should be in the habit of drawing all lone pairs. It is also important that your curved arrows be drawn neatly and precisely, clearly showing the atomic origin of the electron pair at the tail of the curved arrow and the electron pair destination at the head of the arrow. Example 1: Provide the curved arrows for the reaction of hydroxide and hydronium ions to form two molecules of water.

Solution 1: A reasonable approach to an exercise like this is to analyze the bond changes are then draw the corresponding curved arrows. The oxygen of hydroxide ion has shared a lone electron pair with the hydrogen of hydronium ion, forming a new O-H bond. Thus we draw a curved arrow with the tail at the hydroxide ion lone pair and ending at the hydronium ion

hydrogen. This hydrogen atom can only have one covalent bond at a time, so it must sacrifice the bond to the hydronium ion. We draw a curved arrow to show the bonding electron pair between the hydrogen and oxygen of the hydronium ion moving to become the sole property of the oxygen atom. The oxygen of the hydroxide ion is sharing a pair of electrons that it had all to itself before, so its formal charge drops by one unit (-1 to neutral). The oxygen of the hydronium ion gains a lone pair of electrons that is used to share with the hydrogen, so its formal charge decreases by one unit (neutral to +1).

Example 2: Draw the product(s) of the following mechanism step based upon the curved arrows.

Solution 2: The curved arrow that starts at the carboncarbon π bond and ends at the bromine atom indicates the π bond electron pair has shifted to become a carbon-bromine bond. The left-hand carbon of the π bond has lost an electron pair, so its formal charge becomes one unit more positive (neutral to +1). The curved arrow on the right indicates that the electron pair of the bromine-bromine bond is shifting to reside solely on the bromine on the right, resulting in rupture of the bond and formation of bromine with four lone pairs and a negative charge.

Exercises: Provide curved arrows that show how the following mechanism steps might occur.

Provide the product(s) for the following mechanism steps based upon the curved arrows.

Electrophiles and Nucleophiles

Electrophile: A molecule or ion that accepts a pair of electrons to form a new covalent bond (same as Lewis acid).

Leaving group: A molecule or ion that leaves with a pair of electrons that used to be shared in a covalent bond with this fragment. Lewis acid: A molecule or ion that accepts a pair of electrons to form a new covalent bond (same as electrophile). Lewis base: A molecule or ion that shares a pair of electrons to form a new covalent bond (same as nucleophile). Nucleophile: A molecule or ion that shares a pair of electrons to form a new covalent bond (same as Lewis base). Discussion: Consider the reaction of hydroxide ion (a Bronsted base) with hydrogen chloride (a Bronsted acid).

The oxygen of hydroxide ion bears a formal charge of -1. The hydrogen of hydrogen chloride bears a δ+ charge because chlorine is more electronegative than hydrogen and thus the H-Cl bonding electron pair is unequally shared. We can envision the start of the acid-base reaction between hydroxide ion and hydrogen chloride as an electrostatic attraction between the opposite charges. As the reaction proceeds, the oxygen atom of hydroxide ion shares a lone electron pair with the hydrogen atom of hydrogen chloride, as shown with the curved arrow in the reaction above. This simple reaction shares one feature in common with the majority of reactions that you will encounter in your study of elementary organic chemistry. One species in the reaction shares an electron pair (a Lewis base) with another species (a Lewis acid) to make a new covalent bond. Application of this electron sharing idea to any reaction gives you an excellent chance at starting to figure out the mechanism for most common organic reactions. Because of the ubiquity of electron pair sharers and acceptors in organic reactions, we assign special and distinct terms to these species. A molecule or ion that accepts a pair of electrons to make a new covalent bond is called an electrophile (from the Greek for "electron loving"). An electrophile is the same thing as aLewis acid. Any molecule, ion or atom that is electron deficient in some way can behave as an electrophile. Electron deficiency would include a formal positive charge (methyl carbocation), a partial positive charge (δ+), usually in

conjunction with a polar bond (such as H-Cl) or an open octet (borane). "E" or "E+" are common abbreviations for generic electrophiles.

Typical electrophiles: A molecule or ion that donates a pair of electrons to form a new covalent bond is called a nucleophile (from the Greek for "nucleus loving"). A nucleophile is the same thing as a Lewis base. Any molecule, ion or atom that has electrons that can be shared can be a nucleophile. The most common indications that electrons are available to be shared are formal negative charge (iodide ion), a partial negative charge (δ-), usually in conjunction with a polar bond (methyl magnesium bromide), aπ bond (isobutylene) or lone pairs (ammonia). "Nuc" or "Nu" are common abbreviations for generic nucleophiles.

Typical nucleophiles: The study of reaction mechanisms is central to the study of organic chemistry at any level. Therefore identification of electrophiles and nucleophiles is a critical organic chemistry survival skill. Examination of a structure for the features discussed above is one way to identify how a molecule or ion might behave in a reaction. Another way is by considering the curved arrows. Because electrons flow from an electron source to a place of electron deficiency, a curved arrow points away from a nucleophile and to an electrophile. This does not work in every reaction, however. In some reactions, the electron flow could go in either direction, and there are no distinct nucleophiles and electrophiles. Such reactions are uncommon in a course of this level. Electrons always flow from nucleophile to electrophile:

Example 1: Using the curved arrows shown below, label each reactant as a nucleophile or electrophile.

Solution 1: A nucleophile is a molecule or ion that donates an electron pair to form a new covalent bond. In this example, chloride ion is donating a lone pair to form a new bond with carbon. The chloride ion is at the origin on the curved arrow that indicates this bond change. That the chloride ion bears a formal negative charge further suggests it should function as a nucleophile. Because nucleophile must react with an electrophile that leaves the other molecule (a chlorosulfite ester) to be the electrophile. The ester carbon is accepting an electron pair from the nucleophile to form the new C-Cl bond. The other bond changes within the chlorosulfite ester molecule are inconsequential when defining its role as a nucleophile or electrophile. The chlorine atom that is expelled as chloride ion accepts and electron pair from the S-Cl bond, but it does not make a new covalent bond, so it is neither electrophile or nucleophile. In this reaction, chloride ion is the leaving group.

Example 2: Decide if each molecule or ion shown below will react as a nucleophile or electrophile, or both.

Solution 2: Examine each structure for the charge distribution and electronic features discussed above. a. Bromide ion: This atom has four lone pairs and a formal negative charge, suggesting it is electron-rich and can therefore function as a nucleophile. It has none of the features that would suggest it might behave as an electrophile. b. Ammonium ion: This ion has a formal positive charge, suggesting it is electronpoor and can therefore function as an electrophile. It has no lone pairs or areas of negative charge, suggesting it will not function as a nucleophile.

c. Water: The oxygen atom of water has two lone pairs and a δ- charge (oxygen is more electronegative than hydrogen). This suggests that water can behave an a nucleophile. Each hydrogen atom bears a δ+ charge, so the molecule can behave as an electrophile as well. Many molecules can be both nucleophiles and electrophiles. How they behave depends upon what they react with. For example, if water is reacted with an electrophile, the water will behave as a nucleophile. Exercises: Identify the nucleophiles and electrophiles in each mechanism step shown below.

Decide if each molecule or ion shown below will react as a nucleophile or electrophile, or both.

A Brief Tutorial on Drawing Lewis Dot Structures We will use three molecules (CO2, CO32- and NH4+) as our examples on this guided tour of a simple method for drawing Lewis dot structures. While this algorithm

may not work in all cases, it should be adequate the vast majority of the time. Procedure for Neutral Molecules (CO2) 1. Decide how many valence (outer shell) electrons are posessed by each atom in the molecule.

2. If there is more than one atom type in the molecule, put the most metallic or least electronegative atom in the center. Recall that electronegativity decreases as atom moves further away from fluorine on the periodic chart.

Arrangement of atoms in CO2: 3. Arrange the electrons so that each atom contributes one electron to a single bond between each atom.

4. Count the electrons around each atom: are the octets complete? If so, your Lewis dot structure is complete.

5. If the octets are incomplete, and more electrons remain to be shared, move one electron per bond per atom to make another bond. Note that in some structures

there will be open octets (example: the B of BF3), or atoms which have ten electrons (example: the S of SF5).

6. Repeat steps 4 and 5 as needed until all octets are full. 7. Redraw the dots so that electrons on any given atom are in pairs wherever possible.

Procedure for Negatively Charged Ions (CO32-) Use the same procedure as outlined above, then as a last step add one electron per negative charge to fill octets. Carbonate ion has a 2- charge, so we have two electrons available to fill octets.

Using the procedure above, we arrive a this structure: The two singly-bonded oxygen atoms each have an open octet, so we add one electron to each so as to fill these octets. The added electrons are shown with arrows. Don't forget to assign formal charges as well! The final Lewis structure for carbonate ion is:

Procedure for Positively Charged Ions (NH4+) Use the same procedure as outlined above, then remove one electron per postive charge as needed to avoid expanded octets. When using this procedure for positively charged ions, it may be necessary to have some atoms with expanded octets (nitrogen in this example). For each unit of positive charge on the ion remove on electron from these exapnded octets. If done correctly, your final structure should have no first or second period elements with expanded octets.

Using the basic procedure outlined above, we arrive at a structure in which nitrogen has nine valence electrons. (Electrons supplied by hydrogen are red; electrons supplied by nitrogen are black.) Removal of one of these valence electrons to account for the 1+ charge of ammonium ion solves this octet rule violation.

Resonance: Vocabulary Exercise Solutions

Contributing resonance structures: One or more alternate Lewis structures for a molecule or ion. Double-headed arrow: Drawn between two molecular structures to indicate that they are contributing resonance structures. Resonance: When a molecule can be represented by the weighted hybrid of two or more hypothetical but reasonable Lewis structures that differ only in the distribution of bonding and nonbonding electrons, and in which the positions of the nuclei are constant. Resonance hybrid structure: A structure which is the weighted average of all contributing resonance structures for a molecule or ion. The resonance hybrid structure is the closest representation of reality for that molecule or ion.

Resonance: Drawing Resonance Structures

Additional reading recommendation: You may find Chapter 2 of Pushing Electrons by Daniel P. Weeks (Saunders College Publishing; ISBN 0-03-0206936) to be a useful tool for mastering the fundamentals of resonance structures. Discussion: Consider the Lewis structure of the carbonate ion, CO32-. The Lewis structure for this ion has a carbon-oxygen double bond, and two carbon-oxygen single bonds. Each of the singly bonded oxygen atoms bears a formal charge of 1-. (Review the formal charge tutorial if needed.) But which of the three oxygens forms the double bond? There are three possibilities:

These structures are similar in that the have the same types of bonds and electron positions, but they are not identical. The position of the carbon-oxygen double bond makes them different. In structure A the double bond is with the top oxygen atom, in B with the right hand oxygen atom, and C with the left hand oxygen atom. These oxygen atoms are at different places in space, so these are different structures. Consider this analogy: when the hands on a clock are at a 90o angle, the time could be 3 o'clock, or 6:15. The angle between the hands stays the same, but because they point to different places in space, they indicate a different time. The position of the carbon-oxygen bond is like the fixed angle of the clock hands, but pointing to different places on the clock face. When more than one Lewis structure can be drawn, the molecule or ion is said to have resonance. The individual Lewis structures are termed contributing resonance structures. Resonance is a common feature of many molecules and ions of interest in organic chemistry. Which one of these three structures is the correct one? How could we tell? If structure A was correct, laboratory measurements would show one shorter bond (the carbon-oxygen double bond) and two longer bonds (the carbon-oxygen single bonds). Measurement of structures B and C would give the same results as well. As it turns out, laboratory measurements show that all three bonds are equal and between single and double bond length. This suggests that none of the Lewis structures we have drawn are correct. It further suggests that the actual structure has three equal carbon-oxygen bonds that are intermediate between single and double bonds. Perhaps the three Lewis structures for carbonate ion are in rapid equilibrium. The structures are changing so quickly that all we see can measure is an average blur

(structure D), instead of being able to detect individual structures. By analogy, consider a camera with the shutter left open. The picture would be a blur that looks like A, B and C all at the same time. Structures A, B and C have the same bonds and electron distribution, the only difference is the position of the bond. Thus the three structures have equal stability, and the three structures would occur to the same extent at equilibrium. The blur we see would look like 1/3 A, 1/3 B and 1/3C. The bond lengths would be a blur as well; we would perceive them as being something between single and double bonds. The charge would shift so rapidly that we would see it on all oxygen atoms at once. Since each oxygen atom has a 1- charge in two of the three equilibrium contributors, each oxygen atom would appear to have, on average, a charge of 2/3-.

All laboratory experiments have failed to detect structures A, B and C. No matter what experiments are performed, analysis has always concluded that D is the best description for the structure of the carbonate ion. This suggests that A, B and C do not exist, and are not adequate descriptions for carbonate ion at any time. The actual structure is D, and not equilibrium between A, B and C. That D appears to be a combination of A, B and C still appears to be a useful way to determine the actual structure of carbonate ion. The only problem is that A, B,C and D are not in equilibrium. Can they be interacting in some other way? The answer is that the true structure of carbonate ion appears to be a simultaneous hybrid of the three resonance contributors A, B and C. Structure D has features derived from A, B, and C, but is never just A or just B or just C. For example, Dhas some double bond character, and each oxygen has some negative charge. Hybrid structure D is termed the resonance hybrid structure. It is a much better representation of reality for carbonate ion than any of the individual contributing resonance structure. We use a double headed arrow to show that individual structures are related by resonance.

It is important to understand that structure D is a hybrid of A, B and C. It has features common to A, B and C, but carbonate ion does not rapidly interconvert between A, B and C. By analogy, consider the case of a mule. A mule is the offspring of a horse and donkey. It has features similar to both, such as long tail and mane, but it does not rapidly interconvert between the two. A mule is not a horse one second, then a donkey the next. (An animal that changed would certainly be the highlight of any zoo!) Drawing contributing resonance structures. Resonance is an important feature of many organic molecules. It can have a profound influence on their structure, chemical reactions, and physical properties. Key to understanding resonance is the ability to draw contributing resonance structures and the resonance hybrid structure. Resonance structures are simply alternate Lewis structures for a given ion or molecule. Thus we can draw all resonance structures by drawing all of the possible Lewis structures. However, it is not always easy to see what all these Lewis structures might be. A set of Lewis structures for a give ion or molecule must have the same number of electrons as the Lewis structures are constructed from the same atoms. The only difference between the Lewis structures is the placement of the electrons. The position of the atoms in space is held constant. We can use this facts to assist us in drawing resonance structures. Because the number of electrons is conserved, electrons taken away from one atom must appear somewhere else in the structure. When drawing resonance structures, it is most convenient to shift these electrons between adjacent atoms. Electrons that can be moved between adjacent atoms in resonance structures are lone pairs or π electrons. (Unpaired electrons present in radicals will be considered in section 15.5 of the text.) These electrons reside on a "resonance donor atom." The "resonance acceptor atom" must be adjacent to the donor atom. The acceptor atoms must also have an open octet, be able to accommodate an expanded octet (text section 1.1C; most commonly this will be chlorine, bromine, sulfur and phosphorus) or have another electron pair (lone pair or π) that can be displaced. An atoms with a formal positive charge can also be an resonance acceptor atom, as long as the atom does not accept more electrons that it can normally accommodate. We use curved arrows as a bookkeeping tool to indicate the electron changes that differentiate resonance structures. Although these are the same curved arrows we use to indicate that bond changes are occurring in a reaction mechanism, they do not mean exactly the same thing. A reaction step requires a finite amount of time to occur, whereas a shift between resonance structures never actually occurs. Recall that individual resonance structures do not exist, they are simply alternate Lewis structures for the same molecule or ion. In the example below, the curved

arrow indicates that the lone pair of electrons on the left-hand carbon is moving to a position where it can be shared between the two carbons, thus becoming the carbon-carbon π bond. This use of curved arrows is often termed "electron pushing."

Because electrons are shifting around, the formal charge distribution will vary between resonance structures. The formal charge of an atom that gains a pair of electrons through resonance becomes one unit more negative. Conversely, the formal charge of an atom that shares a pair of electrons that it did not share previously becomes one unit more positive. Rules for drawing contributing resonance structures. Some rules must be considered when drawing resonance structures. Rule 1: All resonance structures must have the same number of valence electrons. Electrons are not created or destroyed, nor are they lost or gained to other molecules or ions during resonance.

The rule is violated because structure E has 12 valence electrons (four bonding pairs and two lone pairs), whereas structure F has 14 valence electrons (five bonding pairs and two lone pairs). Therefore these cannot be resonance structures of the same ion. (Structure F also violates rule 2.) Rule 2: The octet rule must be obeyed. Hydrogen may never have more than two valence electrons. Lithium through fluorine may never have more than eight valence electrons. (A carbon with five attachments, often called a "pentavalent carbon" has ten valence electrons. This is forbidden because carbon does not have the space in its orbitals to accommodate ten electrons. You should take care to avoid this common mistake made by inexperienced organic chemistry students.) Certain elements commonly encountered in organic chemistry may have ten valance electrons. These elements are in periods three and higher in the periodic

table, and include chlorine, bromine, iodine, phosphorus and silicon. Of all the atoms commonly encountered in organic chemistry, only sulfur can routinely expand its octet to include twelve valence electrons. These atoms expand their octets so as to improve the importance of the resonance structure.

Resonance structure G is acceptable. Structure H is not acceptable because the carbon has ten valance electrons.

Structures I and J are both acceptable resonance contributors for bisulfate ion, the conjugate base of sulfuric acid. The sulfur atom of structure J has 12 valence electrons, an expanded octet. Sulfur is a third row element, so an expanded octet is allowed. Structure J is the more important resonance structure because it maximizes the number of covalent bonds and minimizes the number atoms with a nonzero formal charge. Rule 3: Nuclei do not change positions in space between resonance structures. Resonance structures differ only in the arrangement of valence electrons.

Structures K and L are both acceptable Lewis structures, but they are not related by resonance because the circled hydrogen atom has changed position in space. Exercises: Draw all reasonable resonance forms for the structures shown below. Use curved arrows to indicate electron pair changes.

Resonance: Most Important Resonance Contributor Discussion: Many molecules or ions that participate in an organic reaction have resonance. When writing the mechanism for the reaction, the best representation of reality would be achieved by using the resonance hybrid structure. However, because the resonance hybrid does not show explicitly show electron pairs that are shared by resonance, its use in mechanisms can be unclear. Thus we often use a single resonance contributor instead of the hybrid. When deciding which resonance contributor to use, it makes sense to use the one that makes the greatest contribution to the resonance hybrid. If we cannot use the hybrid, then we should use the next closest structure. In addition, many (but not all) reactions of molecules or ions with resonance proceed as if this most important resonance contributor was the actual reactant. Thus we need a set of rules to determine the most important resonance contributor. These rules are based on the idea that if individual resonance contributors did indeed exist, the most thermodynamically stable structures would make more significant contributions to the resonance hybrid. Factors that enhance thermodynamic stability are maximization of covalent bonding and minimization of charge. Resonance increases stability by increasing the bonding between adjacent atoms and by distributing charge over a greater number of atoms. Preference 1: The most important contributor has the maximum number of atoms with full octets. This preference gets priority over the other three rules for determining the most important resonance contributor.

The carbon of structure A has an open octet. All the atoms of structure B have full octets. Therefore contributor B is more important than contributor A, despite the

fact that the positive charge is on the more electronegative oxygen atom instead of the less electronegative carbon atom. Preference 2: If a resonance contributor must have formal charge, the most importnat contributor has these charge(s) on the atoms most willing to accommodate them. Negative charges are best accommodated on more electronegative atoms, whereas positive charges are best accommodated on the least electronegative atoms.

All atoms of resonance contributors C and D have a complete octet, so we turn to other preferences to determine the most important resonance contributor. A negative charge is best accommodated by a more electronegative atom. Because oxygen is more electronegative than carbon, contributor D is more important than contributor C. (If the ion shown above was a cation, then the resonance contributor with the positive charge on carbon would be more important than the contributor with the positive charge on oxygen.) Preference 3: The most significant contributor has the maximum number of covalent bonds. Contributor B (above) is more important than contributor Abecause B has the carbon-oxygen π bond absent in A. Preference 4: The most significant contributor will have the least number of formal charges.

Resonance contributor F is more significant than contributor E because F has no atoms with formal charges, whereas E has two atoms with formal charges. (Contributor F is also favored by Preference 3 as well.) Preference 5: The most significant contributor has the least number of unpaired electrons.

For example, contributors G and H each have one unpaired electron, and thus are preferred over contributor I which has three unpaired electrons. Resonance contributors that include avoidable unpaired electrons are rarely of any consequence and thus should not be considered. There is one common exception: molecular oxygen. Due to molecular orbital considerations, molecular oxygen is best described as having two unpaired electrons and an oxygen-oxygen single bond (contributor J) and not as lacking unpaired electrons with an oxygen-oxygen double bond bond (contributor K).

Exercises: Determine the most significant resonance contributor for each set of contributing resonance structures drawn previously.

Stereochemistry: Vocabulary Exercise Solutions

Achiral: An object that is not chiral. Chiral: An object that is not superposable with its mirror image.

Constitutional isomer: One of a set of isomers that differs in the sequence of attachment or bonding of the atoms. Dextrorotatory: Rotation of plane polarized light in a clockwise direction. Diastereomer: One of a set of stereoisomers that are not enantiomers. Enantiomer: One of a pair of stereoisomers that are nonsuperposable mirror images. Isomer: One of a set of compounds with identical molecular formulas. Levorotatory: Rotation of plane polarized light in a counterclockwise direction. Meso compound: A compound that has at least two stereocenters but is achiral. Optically active: Rotation of plane polarized light. Plane polarized light: Light in which the electric field component oscillates in a single plane. Polarimeter: A device to measure rotation of plane polarized light. Racemic mixture: An equimolar mixture of a pair of mutual enantiomers. Resolution: Separation of mutual enantiomers. Specific rotation: The magnitude of rotation of plane polarized light of compound in a 10.0 cm path length, and at concentration of 1.00g/mL or neat if the compound is a liquid. Stereocenter: An atom that has four different groups attached to it. (Not limited to just carbon.)

Stereoisomer: One of a set of isomers that have the same connectivity of atoms, but differ in the position of the atoms in space. Stereoisomers cannot be interconverted by rotation around a single bond. (Conformational isomers are not stereoisomers.) Stereochemistry: Identifying Stereocenters Discussion: A stereocenter is a carbon atom that has four different attachments. The difference can be anything, except a conformational change. Thus, we consider n-propyl and isopropyl groups to be different attachments. Propyl groups that differ only in rotation about a single bond (conformational differences) are identical because they can rotate to adopt identical conformations.

Figure 1. Left: This molecule has a stereocenter. The carbon indicated by the arrow has four different attachments: CH3, CH3CH2CH2, (CH3)2CH and Cl. Right: This molecule does not have a stereocenter. The carbon indicated by the arrow has three different attachments: CH3, CH3CH2CH2(two different conformations but still the same group) and Cl. Most instances of chiral molecules that you will encounter at this level of organic chemistry owe their chirality to the presence of one or more stereocenters. It is not necessary for a molecule to have a stereocenter to be chiral. Make models of the enantiomers of 2,3-pentadiene (1,3-dimethylallene) shown below. Having a stereocenter does not mean a molecule will always be chiral. Explore this point with a molecular model of meso-tartaric acid, a compound critical to Pasteur's studies of optical activity in organic compounds.

In the next section we explore the importance of recognizing chirality in molecular structure. Because stereocenters are the origin of chirality of most chiral organic molecules, it is useful to be able to recognize stereocenters within a molecule. Example 1: In each molecule shown below, identify the stereocenter(s).

Solution 1: a. When looking for stereocenters, pay careful attention to the "four different attachments" criterion. Recall that hydrogens of "stick structures" are usually not drawn. Also note attachments that differ only in conformation do not make a stereocenter. The only carbon of 2-butanol that bears 4 different attachments is the alcohol carbon.

b. 2-Methyl-3-pentanol has only one stereocenter: the alcohol carbon. Don't neglect the hydrogen on this carbon just because it isn't shown. The other carbon bearing wedge and broken line bonds is not a stereocenter, because it has two of the same thing (methyl groups) attached. We often use the wedge and broken line notation to imply a three dimensional arrangement of atoms, but its does not always occur on a stereocenter.

Exercises: Locate all stereocenters in each structure.

Stereochemistry: Determining Molecular Chirality Discussion: Chiral objects are not superposable with their mirror images. An excellent example of this is your hands. Hold your hands out in front of you, with the palms facing together. Neglecting unnatural additions such as jewelry, note that your hands are mirror images. Now turn your hands so that both palms face the same direction. Note that the thumbs now point in opposite directions. When the thumbs point in the same direction, the palms are opposite. Your hands are mirror images, but not superposable. Each hand is therefore chiral. Achiral objects may be superposed on their mirror image. Examine two sheets of blank paper in the same way as you experimented with your hands. Notice the sheets of paper are mirror images, but superposable. The sheets of paper are therefore achiral. All objects can be classified as chiral or achiral, including molecules. If our hands were molecules, they would be a pair of enantiomers. We know from basic biology that interaction of molecules, such as the docking of a substrate to an enzyme, is vital to living organisms. Because enzymes and their substrates may be chiral, it is useful to understand how achiral and chiral molecules can interact. (enzymes are constructed from a group of about 20 small molecules called amino acids. All amino acids except one are chiral, so the enzymes they make are chiral as well.) The way a hand slips into a glove provides

a useful way to model this effect. The glove is the enzyme, and the hand is the substrate that must fit properly into the enzyme pocket for the enzyme to be able to act upon the substrate. (Verify that a pair of gloves are chiral in the same way you explored the chirality of your hands.) Your right hand fits nicely into the right handed glove, but does not fit well into the left-handed glove. Likewise, your left hand fits well into the left-handed glove, but the right hand does not. Imagine the glove represents an enzyme and your hand the substrate. The left-handed enzyme/glove would accept the left-handed substrate/hand readily, and would be able to act upon the substrate. The left-handed enzyme/glove cannot readily accept the right hand/substrate, as so this enzyme cannot readily act upon this substrate. This simple model implies that an enzyme will act on one enantiomer more readily than another. Thus, enantiomers of drugs can have different effects in the body, because they are acted upon differently by enzymes, despite the fact that they have the same set of functional groups. That enantiomers of drugs can have different biological effects has been demonstrated in many instances, but perhaps none so dramatically as the in the case of the drug thalidomide. In the late 1950s, the racemic form of this drug was prescribed as a sedative or hypnotic for pregnant women. Some women who took the drug delivered children with severe birth defects. A substance that causes fetal abnormalities is called a teratogen. Further research revealed that one enantiomer of thalidomide has the desired sedative effects, while the other enantiomer was teratogenic. The enantiomers of thalidomide were acting differently in the body, because they interacted differently with chiral biomolecules such as enzymes. The drug was quickly removed from the market.

Chiral molecules are nonsuperposable with their mirror images. This can be tested on paper or with molecular models using the two methods described below. Internal mirror plane. We can look for a plane of symmetry in the molecule. Imagine this plane as a mirror through the middle of the molecule. If one half of the molecule is reflected into the other half, then the molecule is achiral. If no such mirror plane exist, the molecule is usually chiral. (There are symmetry elements other than a mirror plane that may render a molecule achiral, but these are rarely encountered and thus beyond the scope of an introductory organic chemistry course.) Molecular models can be used in the same way.

Example 1: Using the method of an internal mirror plane, determine if cyclohexanol is chiral or achiral. Solution 1: To determine if cyclohexanol is chiral using the internal mirror method, draw a mirror plane through the middle of molecule. If there are any unique functional groups or atoms within the molecule, these must lie within the mirror plane, so that one half of the atom or functional group is reflected into the other half. In this case, there is only one alcohol functional group, so it must be contained in the mirror plane. Figure 1 shows that the mirror plane bisects the molecule into two equivalent halves, so cyclohexanol is achiral.

Figure 1. Two views of cyclohexanol showing the internal mirror plane. The mirror plane is indicated by the dashed line.

Figure 2. Cyclohexanol molecular model. A vertical mirror plane bisects the molecule through the middle of the picture. Superposable models. To determine if a molecule is chiral using the superposability requirement, build a molecular model of the molecule in question, then a build a mirror image of this model. Now try to superpose the models by aligning them so that all the atoms match up. The models may be manipulated in any way, such as rotation around single bonds (changing molecular

conformation) or changing perspective, but bonds cannot be broken. If all the atoms can be made to line up, the molecule is achiral. If they cannot be aligned, the molecule is chiral. Example 2: Using the method of superposable molecular models, determine if cyclopentanol and 2-chlorobutane are chiral or achiral. Solution 2:

Figure 3. Left: Molecular model of cyclopentanol and its mirror image. Right: A top view showing that these cyclopentanol models can be aligned (superposed), so cyclopentanol is achiral.

Figure 4. Left: Molecular model of 2-chlorobutane and its mirror image. Right: The same models stacked. The Cl-C-H portions of the models can be made to superpose, but at the same time the methyl and ethyl groups do not. The 2chlorobutane models cannot be superposed, so the molecule is chiral.

Exercises: Using either method discussed above, determine if the molecules shown below are chiral or achiral.

Stereochemistry: Determining Molecular Chirality Exercise Solutions

a. 3-Methylhexane: This molecule is chiral because the mirror images are not superposable. The case is very much like 2-chlorobutane, except the molecule has a propyl group instead of a chlorine atom. Each molecule has a single stereocenter. Examination of other molecules with single stereocenters should rapidly convince you that molecules with a single stereocenter are always chiral.

b. Benzene: Benzene is achiral because it has many internal mirror planes. Perhaps the most obvious of these is the molecular plane that contains all twelve atoms.

c. Cyclohexane: Cyclohexane can exist in many different conformations (chair, boat, etc.). In cases of conformationally fluxional molecules, if there is one conformation that is

achiral, then the molecule as a whole is achiral. The chair conformation has a mirror plane through the middle of the ring, containing C1 and C4, so this conformation is achiral. Because the chair conformation is achiral, we consider cyclohexane to be an achiral molecule.

This molecule contains no mirror planes, and mirror image molecular models are not superposable. Therefore this molecule is chiral.

This molecule has two stereocenters, so at first guess, it might appear to be chiral. However, it has an internal mirror plane:

This is a meso compound, and is achiral, as are all meso compounds. Note that for a compound to be meso, the pairs of stereocenters must have the same groups attached to them.

At first glance, this compound appears to be chiral because it contains no apparent internal mirror plane. As with the cyclohexane case, we need to examine all conformations to see if any are achiral. A good way to do this in a molecule with multiple stereocenters is to hold one stereocenter still (the left hand one in this case), and rotate around the single bonds to try and make the stereocenters align. Such a rotation in this molecule reveals an internal mirror plane:

This molecule is therefore a conformational isomer of the molecule in (e), and is thus achiral.

At first glance, we might suspect this to be a meso compound because the two stereocenters have the same attachments. However, rotation around the bond that attaches the stereocenters reveals that they are not reflected by an internal mirror plane:

Careful examination will reveal that this molecule has no conformations with an internal mirror plane and so it is chiral.

In the quest for an internal mirror plane, rotation around the bond connecting the two stereocenters reveals the molecule to be meso and therefore achiral:

Stereochemistry: Drawing Enantiomers and Diastereomers Discussion: For the same reasons as it is important to recognize and classify stereoisomers it is valuable to be able to draw enantiomers and diastereomers as well. Drawing enantiomers: Recall that enantiomers are molecules which are nonsuperposable mirror images. Because these are mirror image molecules, and because stereoisomers can only have two absolute configurations, all the stereocenters of one enantiomer will be the mirror image of the other enantiomer. A mirror image stereocenter can be draw by switching the place of two groups attached to that stereocenter. We can also draw the mirror image of the enantiomer, using an imaginary mirror. Example 1: Draw the enantiomer of (R)-2-chlorobutane shown below.

Solution 1: Let's switch the places of the methyl and ethyl groups. This gives us (S)-2-chlorobutane. (You should label the stereocenter as R or S to verify this.)

Example 2: Draw the enantiomer of (2R,3R)-tartaric acid, shown below.

Solution 2: Let's use the mirror image technique for this example. Once again, you may wish to verify the answer by labeling the stereocenters of the mirror image molecule as R or S.

Drawing diastereomers: Recall that stereoisomers differ in the position of the atoms in space, and that diastereomers are stereoisomers that are not enantiomers. Some thought on this will suggest that we can draw the diastereomer of a given structure by inverting one or more, but not all of the stereocenters. Example 3: Draw all the diastereomers of (2R,3R)-tartaric acid (structure shown above). Solution 3: How many diastereomers can there be? Recall that for a molecule with n stereocenters, the molecule can have 2n stereoisomers (section 4.4 of the text). Thus, (2R,3R)-tartaric acid, which has two stereocenters, can have at most 22 = 4 stereoisomers. The (2S,3S) stereoisomer cannot be a diastereomer of (2R,3R) because both stereocenters have been inverted. This leaves two possibilities, each of which has one stereocenter the same as the given molecule: (2R,3S) and (2S,3R). The structures can be drawn by inverting the stereocenter that is altered.

(In this particular case, the new diastereomers are meso compounds, and are identical. This happens when the two stereocenters have the same attachments.) Exercises: Draw the enantiomer for each compound.

Draw a diastereomer for each structure.

Stereochemistry: Specific Rotation and Related Calculations Discussion: With one exception, the chemical and physical properties of enantiomers are identical in the absence of other chiral materials. The exception is the rotation of plane polarized light. Because plane polarized light is effected differently by enantiomers, it can be a useful laboratory tool to quantify the amount of an enantiomer or enantiomers present in a sample. Recall that equal amounts of mutual enantiomers rotate plane polarized light to an equal extent but in opposite directions. Rotation of plane polarized light occurs at the molecular level, so the extent of the rotation is proportional to the amount of optically active material that is present as expressed in the following equation. [α]λT = α/lc Where:

[α]λT = specific rotation in degrees (The correct units are deg cm2 g-1, but are usually just given as degrees). λ is the wavelength of light used for the observation (usually 589 nm, the D line of a sodium lamp unless otherwise specified. This wavelength is responsible for the orange-yellow color of the common sodium vapor street light.), and T is the temperature in oC. This value is characteristic for a given compound, just like the melting point. Concentration and solvent data is included if relevant. Example: (c = 10, CH3OH) after the rotation means the specific rotation was determined at a concentration of 10 g ml-1 in methanol. α = observed rotation in degrees. l = cell path length in decimeters (1 decimeter = 1 dm = 10 cm. A standard polarimeter tube is 1.00 dm in length.) c = concentration in g ml-1 (the density if the sample is a pure liquid). Armed with these facts, and some basic concepts from general chemistry, we can perform just about any sort of simple calculation involving optical activity and enantiomeric composition of a sample. Example 1: A sample of pure (S)-2-butanol was placed in a 10.0 cm polarimeter tube. Using the D line of a sodium lamp, the observed rotation at 20oC was +104o. The density of this compound is 0.805 g ml-1. What is the specific rotation of (S)2-butanol? Solution: Plugging the numbers into [α]λT = α/lc we get: [α]λT = (+104o) / (1.00 dm) (0.805 g ml-1) = +129o. Thus we would write [α]D20 = +129o (neat). ("Neat" refers to a liquid that has not been diluted.) Example 2: Calculate the observed rotation of a solution of 0.5245g of (S)-1amino-1-phenylethane diluted to a volume of 10.0 mL with methanol at 20oC, using the D line of a sodium lamp and a 1.00 dm tube. Specific rotation of this material: [α]D23 = -30.0o.

Solution: Solving the specific rotation equation for observed rotation, we get α = [α]λTlc. The sample size is 0.5245 g, but this has been diluted to 10.0 mL, so the sample concentration is 0.5245 g/10.0 mL = 0.05245 g ml-1. Plugging in the numbers, we get α = (-30.0o) (1.00 dm) (0.05245 g ml-1). Solving, we find α = -1.57o. Exercises: a. Calculate the specific rotation of (2R, 3R)-tartaric acid based on the following observation: A 0.856 g sample of the pure acid was diluted to 10.0 mL with water and placed in a 1.00 dm polarimeter tube. The observed rotation using the 589 nm line of a sodium lamp at 20.0 oC was + 1.06o. b. What is the expected observed rotation of a 1.0 x 10-4 M methanol solution of the potent anticancer drug paclitaxel (also called taxol)? [α]D20 = -49o (c = 1, CH3OH). Paclitaxel has a molecular weight of 853.93 g mole-1. c. A certain compound has a specific rotation of -43.2o (c = 5, toluene). What is the observed rotation of a sample of 1.24 g of the enantiomer of this compound when diluted to a concentration of 1.00 g ml-1 in the same solvent? d. In his classic studies of stereochemistry and optical activity in organic compounds, Pasteur measured the optical activity of many solutions. For the naturally occurring enantiomer of tartaric acid, [α]D20 = +12.4o (c = 20, H2O). What can be concluded about the ratio of tartaric acid enantiomers present in the solution if the observed rotation is (i) -6.0o, or (ii) 0o?

Stereochemistry: Specific Rotation and Related Calculations Exercise Solutions a. 0.856 g in 10.0 mL of solution = 0.0856 g ml-1. [α]λT = α/lc So: [α]D20.0 = +1.06o / (1.00 dm) (0.0856 g ml-1) = +12.38o b. [α]λT = α/lc c = 0.0001 M = 0.0001 moles L-1 = (0.0001) (853.93 g mole-1) (0.001 L) = 0.00085 g ml-1 Assume l = 1.0 dm (standard polarimeter tube) Solving for α: α = [α]λTlc = (-49o) (1.0 dm) (0.00085 g ml-1) = -0.042o.

c. Solutions of enantiomers of equal concentrations rotate plane polarized light to an equal extent but in opposite directions. If a certain compound has a specific rotation of -43.2o (c = 5, toluene) then the enantiomer has a specific rotation of +43.2o (c = 5, toluene). From the previous question, α = [α]λTlc = (+43.2o) (1.00 dm) (1.00 g ml-1) = +43.2o. We could have anticipated this result without doing a calculation, as the rotation is observed under the "standard conditions" of 1.00 dm tube and 1.00 g ml-1 concentration. d. i) If the sample is levorotatory, then the concentration of the (-) enantiomer is greater than the concentration of the (+) enantiomer. We cannot calculate the exact concentration, as we do not know how much (-) enantiomer is being canceled by the (+) enantiomer. For example, the sample might contain 1.1 g ml-1 of the (-) enantiomer and 1.0 g ml-1 of the (+) enantiomer. This solution would have the same rotation as a 0.1 g ml-1 solution of the (-) enantiomer alone. ii) Enantiomers have equal but opposite rotations. An observed rotation of zero implies the enantiomers are present in equal concentrations (a racemic mixture).