Organic Chemistry

Introduction| Definition of Organic chemistry

Organic chemistry deals with the study of carbon compounds and the reactions involving carbon.  In addition, carbon valency of 4 means different kinds of chemical bonding can occur. For instance, there are single, double and triple bonds and this is also related to the hybridization and the reactivity of carbon. Therefore, this is called functional groups and will be examined further in more detail.

The alkanes (alkane functional group) contain single carbon-carbon and carbon-hydrogen bonds throughout. However, alkenes have one or double carbon-carbon bonds while alkynes have at least one triple carbon-carbon bond.

The presence of atoms of other elements (heteroatoms) in the organic compound introduces other functional groups. For example, common elements such as  O, N, Halogens, S, P are very common in organic molecules. Therefore, there are many other different kinds of functional groups in organic chemistry, and this will be shown later.

The subject of organic chemistry is broad however teaching usually involves the following topics:

Lewis structures and VSEPR theory

Formal charges and Resonance structures

Lewis structure are simple way of writing the structure of organic compounds. This is done by using the sum total of number of outermost electrons in the atoms involved in the bonding.

VSEPR theory represents simple covalent chemical bonds by straight lines. Moreover, the number of outermost electrons around the atoms in a chemical compound is used to find the hybridization and the geometry. Lone pairs are electron pairs not involved in any chemical bonding and they are represented by two dots. Shared pairs (also known as Bonding pairs) are electron pairs involved in a chemical bond.

Moreover, each type of bonding has specific shape in space and this is called the geometry of the compound. The geometry can be either electronic or molecular geometry.

Formal Charge of atom in a compound = [Number of valence electrons in neutral atom (or group number on periodic table)] – [Number of lone pairs on the atom in the compound structure] – [Number of bonding pairs or bonds around the atom.]

Example formal charge of Carbon atom in CO2, O=C=O:

Carbon has 4 outermost electrons =group number is 4., There is no lone pair electron on the Carbon, so the number of lone pair electrons on Carbon in CO2 is zero. The open structure O=C=O shows that the Carbon in CO2 has 4 bonds or 4 bonding electrons around it.

Therefore, Formal Charge of Carbon in CO2 = 4-0-4=0

Resonance involves shift of electron density by moving lone pairs or bonding pairs without breaking any chemical bond. This can change the formal charge on atoms in a compound and also lead to writing of different resonance structures for certain compounds. Most stable resonance structure is also known as the lowest energy; this is also called most preferred structure and is the structure with the least number of formal charges.

Structure and Nomenclature in organic chemistry

The structure and formula of functional groups consist of a fixed number of elements and a specific chemical bonding . This could be single ,double or triple bond. Some common names and categories of functional groups in organic chemistry are listed below;

Functional groups

These are atoms or group of atoms in a molecule or compound that determines the way the molecule reacts with other molecules or atoms. There are several functional groups and their formula. Examples include alkanes RCH-, Alkanols RH2C-OH, Aldehydes RHC=O, Ketones R2C=O, Carboxylic acids RCOOH, Amides RCONH2, Esters RCOOR and more such as nitro R-NO2, amines RCNH2 etc.

The R group is represents an aliphatic group or an alkyl group. Examples includes a  Methyl group -CH3 from parent name ; methane CH4. Therefore, methyl chloride has formula of CH3Cl. Similar examples are Ethyl groups CH3CH2– , Propyl groups CH3CH2CH2 -, Butyl groups CH3CH2CH2CH2 – and Pentyl groups  CH3CH2CH2CH2CH2


The name ‘ane‘ is used to name all alkanes. General formula is CnH2n+2 where n is the number of carbon atoms in the compound. This is also the formula for saturated hydrocarbons. Examples include; Methane CH4, Ethane C2CH6 , Propane C3H8, Butane C4H10, Pentane C5H12, Hexane C6H14, Heptane C7H16, Octane C8H18, Nonane C9H20 , Decane C10H22, Undecane C11H24, Dodecane C12H26

Video of structure and nomenclature of alkanes: click here

Video of structure and nomenclature of alkanes: part 2:

Video of structure and nomenclature of alkanes: part 3: click here

You can get more information about structure and nomenclature by watching this video of structural isomerism of butanes and hexane.: click here


These have formula of CnH2n : These are unsaturated hydrocarbons with at one carbon-carbon double bond. The name of these compounds end with suffix ‘ene’. Examples include; Ethene C2H4, Propene C3H6 , Butene C4H8 , Pentene C5H10, Hexene C6H12, Heptene C7H14 , Octene C8H16 , Nonene C9H18


The formula of alkynes is given as CnH2n-2 . They are unsaturated hydrocarbons with at least one carbon-carbon triple bond. The suffix ‘yne’ replaces the ending of the ‘ane’ of the parent alkane. Examples include Ethyne C2H2, Propyne C3H4 , Butyne C4H6 , Pentyne C5H8, Hexyne C6H10, Heptyne C7H12 , Octyne C8H14 , Nonyne C9H16

You can get extra information about alkanes, alkenes and alkynes and other functional groups by watching the video on structural isomerism.

More detailed information on functional groups will be discussed later.

Reaction mechanisms in Organic chemistry

This is referring to details of the stages of the chemical reactions of chemical compounds. Common terms that are used are Nucleophilic Substitution methods (SN1 and SN2) and the Elimination methods (E1 and E2). Most reactions occur due to differences in electron density around atoms and subsequent electrostatic  rearrangements to form new bonds. As a result of this, there are two main common reactive groups and these called Nucleophiles and Electrophiles.

Nucleophiles are atoms or group of atoms in a molecule that are electron rich. These are also Lewis bases and they donate electrons to form new bonds. A base is an atom or group that can pick hydrogens (or protons) attached to other atoms. Some of these can be shown as H2O, OH-, RO-, CH3CH2CH2CH2

Leaving groups are atoms or groups that can hold electrons or negative formal charge by having high polarizability or resonance stabilization can act as good leaving groups. Some of these are ; H2O, Cl-, CO2, Br-

Electrophiles are atoms or group that do not have sufficient electrons and can bond by accepting electrons or associating with electron rich groups. Examples of this are; Ammonium NH4+ , H3O+, NO+, Lewis acids; FeCl3

Substitution and Elimination reactions

There are two common substitution reactions, and they are called SN1 and SN2 reactions. In SN1 occur in two steps; a) the formation of charged intermediate ionic species. b) formation of final neutral molecule. However, SN2 occur in one concerted step to form products. This results in change of stereochemistry at reaction center. E1 is an elimination reaction involving two stages as in SN1 but E2 is a single step reaction as in SN2.

The numbers 1 and 2 in the terminology refers to the rate dependence on the reactants. SN1 reaction rate is determined by how fast the single one intermediate ion  is formed from the reactants. This implies SN1 reaction rate depends on the concentration of only one of the reactants. SN2 reaction rate depends on how fast two reactant molecules react to form products. This implies SN2 reaction rate depends on the concentration of the two reactants. Similarly, the same situation is true of E1 and E2.

Analytical techniques in organic chemistry 

(IR):Infra Red spectroscopy

This is used commonly to identify functional groups and is based on the wavenumbers by which they absorb infrared radiation. For instance, the carbonyl groups show around 1800 to 1650 cm-1 depending on what is attached to it. The hydroxyl group is usually broad ranging at 3300 to 3000cm-1 . A hydrogen attached to an alkane carbon is usually at 3000cm-1 while that in alkene is 3200cm-1 and that for an alkyne is over 3300cm-1. This difference in wavenumbers is due to the stiffness or strength of the C-H bond.

Stronger and shorter bonds require more energy to vibrate and therefore have higher wavenumbers. This also explains slight differences in wavenumbers of O-H, N-H and C-H. The O-H is slightly shorter and stiffer than N-H, which is also stiffer and shorter than C-H due to atomic size and orbital overlap. The other factor involved is the reduced mass of the atoms forming the bond. If the bond is the same between two atoms, then the reduced mass takes effect. Atoms of lower mass or weight vibrate at higher wavenumbers than heavier atoms An example is stretching vibrations of O-H, N-H and C-H occur at higher frequencies than C-O, C-N and C-C.

NMR: Nuclear Magnetic Resonance

The atoms of some elements have nuclei spins. Therefore this allows them to absorb and emit energy when placed in a magnetic field. This emitted energy is transformed into electrical display on computers to be used to analyze the compound. This is called NMR spectroscopy and it is a very important method of analysis in organic chemistry.

Atoms of different elements display different signals and this depends on the chemical environment in which they are found. Examples  of atoms of elements that show NMR activity are as follows ; H, C, O, N, F and S.

UV : Ultra Violet spectroscopy

Chromophores are chemical bonds or functional groups in a compound can absorb UV radiation. Some examples of this include double and triple bonds, benzene and similar ring systems. Electrons are excited to higher energy levels when the compound absorb UV energy . However , the energy is emitted when electrons fall to lower energy ground state . The ability of a compound to absorb UV energy is used UV spectroscopy .

The absorbance A or transmittance can be calculated to find the molar absorptivity of the compound. A = e c l where e is the molar absorptivity, c is the molarity and l is the pathlength of 1.0 cm.

Moreover, this can similarly be used to analyze inorganic compounds that contain transition metals.

MS  : Mass spectroscopy

The mass of a compound can be obtained from the mass of the total mass of atoms that make the compound. The elements with different isotopes have percent abundance that appear as different peaks in the spectrum. Isotopes of hydrogen are protium H, Deuterium D, and Tritium T. Carbon isotopes are  C-12, C-13, and C-14. Oxygen isotopes are O-16, O-17 and O-18. These can be used to identify the atoms in a compound. In addition to this , it is has been found that a compound will  breaks apart due to some methods used. The fragments are ionized and identified as peaks in a mass spectroscopy and this is a common procedure in organic chemistry. Examples of fragments and their masses include benzene -77, H2O -18 etc.

Special topics in organic chemistry


Besides alkanes, alkenes and alkynes, the presence of heteroatoms will introduce the rest of the functional groups in organic chemistry. Common heteroatoms are oxygen, nitrogen, halogens, sulfur and phosphorus.

IUPAC rank the functional groups with respect to oxidation and several other factors such including;

  1. Oxidation (oxidation number of carbon atom in group or number of oxygens, least number of hydrogens in the functional group.
  2. Acidity, resonance stabilization of the group or conjugate base or carbanion or carbocation, ease of formation of carbanion from alpha carbon; ease of undergoing substitutions and elimination reactions (derivatization).
  3. Number of oxygen atoms (or least number of non-protic hydrogens); presence of electron withdrawing groups; hybridization (sp>sp2>sp3);
  4. Stability of group or carbanion or carbocation formed.

These above factors will rank the functional groups in order of importance in determining the suffix name = [last name] of an organic compound.

Carboxylic acids (-O=C-OH) > Sulfonic acids (CSO3H)> carboxylic acid anhydrides(-H2CO=C-O-C=OCH2-)> esters(-O=COCH3) > acyl halides(O=CCl)

Amides (O=CNH2>, nitriles(-C=N)> aldehydes (HC=O)> ketones(-CH2C=OCH3)

Alcohols(-C-OH) >Mercaptan (-C-SH)>amines(C-NH2) >alkynes(-C=C-)>alkenes(-CH=CH-)

Alkanes CH3-CH2CH3)> alkyl halides (-H2C-Cl)>ethers(-H2C-O-CH2-) >azides or azido alkanes -H2C-N3> nitro groups(-H2C-NO2) [ These will occur as prefixes since they are ranked lower than alkanes]

Ranking order of common functional groups determining last names of organic compounds: Highest determinant to least determinant.

Please watch all the videos on naming of alkanes [1, 2 and 3] and also structural isomerism to familiarize yourself on how to name organic compounds before you proceed to other functional groups.

Rules for naming:

  1. Start the with the name of the number of carbons as in alkanes but replace the ending “ane” with last name of the most important functional group in the compound. [e.g. Write the name of CH3COOH. The line structure is given below,
CH3COOH = Ethanoic acid (aka acetic acid). This name is derived from 2 carbon alkane = ethane and the COOH group carboxylic acid = ‘oic acid
  • CH3CH2OH naming is simple: [2 carbons =parent name is ethane, Functional group present =OH alcohol implies last name is alkanol = ol. Therefore, name of compound is Ethanol. CH3CH2SH name is also derived from 2 carbon parent ethane and functional group present as Sulfur = thiol or mercaptan therefore name is Ethyl mercaptan or Ethanethiol. CH3CH2-NO2 name is derived from 2 carbon parent name ethane, and functional group nitro NO2– is below alkane ranking so can only name it as a prefix “nitro”, therefore the name of the compound is nitroethane.
  • Numbering is similar to alkanes and every other functional group is given a number at the carbon where it begins.
  • Use dashes to separate numbers from names. [e.g. [ e.g BrCH2COOH = 2-Bromoethanoic acid, CH2=CHCOOH = 2-Propenoic acid], BrCH=CHCOOH is called 3-bromo-2-propenoic acid, BrCH=CH-C=C-COOH [this has 5 carbons in longest chain implies pentanoic acid], numbering and naming: Carbon#2 alkyne= yne , Carbon #4 alkene= ene ,number 5 has Bromine=bromo. Therefore, name will be 5-Bromopent-4-en-2-ynoic acid
  • Use comas to separate numbers that represent the same substituent attached multiple times to carbons in the compound. Examples of this is shown in the video for naming alkanes.
  • The carbon of highest ranked functional group in the compound should be included in counting the continuous carbon chain that determines the name of the compound. This applies even if it is not the longest chain. However, check to see if you can get the smallest number as possible for the main functional group and the other functional groups and substituents as well as the overall sum of the numbers used in the numbering. This is very important especially when the functional groups occur on the middle carbons. e.g., Write the name of the compound.
  • CH3C=OCH3 [3 carbons =parent name is propane; functional group present is ketone=alkanone implies last name is “one“. Therefore, name of compound is propanone, however functional group C=O is at number # 2, implies proper name is 2-propanone or Propan-2-one, CH3C=OCH2CH(CH3)CH2COOH This contains a carboxylic acid functional group COOH (most important) so naming should end oic acid. The longest chain containing the COOH group is 6 carbons= hexanoic acid. Carbon number 3 is bearing a methyl substituent (CH3) and carbon number 5 is having a ketone C=O functional group implies first name use for ketone =oxo. Therefore, the name will be 3-methyl-5-oxohexanoic acid. The other example is writing the name of CH3C=OCH2CH=CHC=CH. The line structure is drawn below;
This can be named as in two different ways and then use the rule of SMALLEST integer summation to determine the correct answer. ; First, start the count from left to right 7 carbons = heptane; highest ranked functional group present = ketone=heptanone parent name; Positions =carbon number 2 is ketone implies 2-heptanone; position 4 is alkene = 4-ene and position 6 is alkyne =6-yne. Therefore, name will be 4-hepten-6-yn-2-one. The sum of the number [4+6+2=12]. However, naming from right to left will be; parent name =6-heptanone; Carbons 1 is alkyne=1-yne, and carbon 3 is alkene= 3-ene. Therefore, name will be 3-hepten-1-yn-6-one. The sum of numbers [3+1+6=10]. The name with the smallest sum of numbers is correct which is 3-hepten-1-yn-6-one
  • The other example is finding the name of the compound shown below;
The name of this compound is tricky in that, the main functional group, alcohol OH is not in the longest continuous carbon chain = cyclohexane. Therefore, cyclohexyl group is named as a substituent and the longest continuous chain contains the OH group is 3 carbons= propanol. The OH is on carbon number 2= 2-propanol parent name. The cyclohexyl group is attached to carbon number 1 when counting from left to right. Therefore, the name of the compound is 1-cyclohexyl-2-propanol. Try naming by counting from right to left will give the wrong answer according to the smallest number summation rule. [SMALLEST NUMBER SUMMATION RULE].


  1. If the number is 1, you can ignore number one in most cases. e.g., 1-ethanol or Ethan-1-ol is usually called Ethanol.
  2. Functional group only takes the last name if it ranks the highest in the compound otherwise only their first names are used. e.g., Ketone is alkanone or last name is “one”. However, if it is not the highest ranked then the first name is “oxo“. Alcohol OH group last name is alkanol =”ol”, but first name is hydroxy if it is not the highest ranked functional group in the compound. e.g., HOCH2CH2CH=O; [3 Carbon = parent name is propane, there are 2 functional groups Aldehyde CH=O alkanal = “al” and also there is alcohol OH group. However, aldehyde is more oxidized than alcohol, so it ranks higher. so, the name of the compound will be 3-hydroxypropanal]. HSCH2CH2OH is called 2-mercaptoethanol because the –OH ranks higher than –SH
  3. Besides the highest ranked functional group, the rest of the names are written in alphabetical order, but numbering can be different. e.g., alkene =ene comes before alkyne =yne but alkyne takes the smallest number when only both are present except in a symmetrical compound when alkene takes the smallest number.
The compound has 7 carbons in longest continuous chain = heptane parent. There are both alkene and alkyne at carbon 1 and 6. Compound is not symmetrical, so alkyne takes smallest number 1 and also the last name = heptyne. There is methyl group on carbon 6. Therefore, name of 6-methyl-6-hepten-1-yne
The compound has 6 carbons in longest continuous chain = hexane parent. There are both alkene and alkyne at carbon 1 and 5. Compound is symmetrical, so alkene takes smallest number but the last name is still alkyne = hexyne. There is methyl group on carbon 6. Therefore, name of 1-hexen-5-yne
The compound is an ester containing. an alkene. The is a methyl group on the right side so the name starts with methyl. The other side has 4 continuous carbon chain = butane parent = butanoate for an ester. The carbon 2 on the left side has alkene or double bond =ene. There is a methyl group on carbon 3 at the left side.
Therefore, the name will be Methyl 3-methyl-2-butenoate.
Compound has a Methyl group on the right side =naming starts with methyl. The left side has 2 longest continuous carbon chain and one methoxy group substituent on carbon 2. Therefore, the name of the compound is Methyl 2-methoxyethanoate.
This is an ether=alkoxy alkane. The longest continuous carbon chain takes the parent name. This is 6 carbons at the right of the oxygen= hexane. The left side of the oxygen can be named as first name. You can choose the longest continuous carbon chain at the left to be 3 carbons= propane parent= propoxy with tert-butyl substituent on carbon 2 counting from the oxygen. This will give the name as Propoxy-(2-tert-butyl)hexane. The other name which is very IUPAC is to ignore the tert-butyl and count the longest continuous carbon on the left to be 4 carbons from the oxygen=butane=Butoxy with methyl substituents on carbon 2 and then two methyl groups on carbon 3. This will give the name as 1(2,3,3-trimethylbutoxy)hexane. Both names are correct, but the latter is the IUPAC name.
This can be named in three different ways. First the longest continuous carbon chain is on the left side = 6 carbons= hexane. The count starts at the oxygen, so Bromine is on carbon 6=6-Bromo. Now the right side of the oxygen gives 3 different ways to name. First, you can take the whole right side as a tert-butyl group so the name will be 6-bromo-1-tert-Butoxyhexane. The other name which is more IUPAC is to count the longest continuous chain on the right side from the oxygen as 3 carbons = propane=propoxy attached at carbon 2 to the oxygen =2-propoxy There is a methyl substituent also on carbon 2= 2-methyl-2-propoxy. Therefore, the name will be 6-bromo-1-[2-propoxy(2-methyl)]hexane or 6-bromo-1-[2-methyl-2-propoxy]hexane. The other best IUPAC name is imagining the compound as a halo alkane with an alkoxy substituent at carbon number 6. Therefore, count the longest chain on the right as 3 carbons=propane=propoxy and the carbon 2 attached to the oxygen = 2-propoxy. The methyl group is also attached at carbon 2. which gives a 2-methyl-2-propoxy. Therefore, the name will be 1-bromo-6-[2-methyl-2-propoxy)]hexane. The last name is the most preferred by IUPAC name because halo alkanes are ranked higher above ethers so the bromine should take carbon number 1 instead of 6. Therefore, the IUPAC name again will be 1-bromo-6-[2-methyl-2-propoxy)]hexane.
Clue: The alkene double bond is higher in rank than the chlorine. Start counting from the right side so that alkene gets a smaller number. The stereochemistry is E at carbon 2 alkene and S at carbon 4 Chlorine. Therefore, the name is (E,4S)-4-chloropent-2-ene.
This is an ester. The acid part has C=O always takes the last name, and, in this case, it has 2 carbons = acetate or ethanoate. Now let’s name the alcohol side. It has 6 carbons in the longest continuous chain in the alcohol side and is attached to the oxygen at carbon3 =3-oxyl. There is an alkyne and an alkene. Let’s count from right to left to give alkyne smallest number carbon1. The alkene C=C will be at carbon 4. There is a methyl substituent at carbon 4. The stereochemistry at the oxygen on carbon 3 is S because the alkene in this case takes precedence over the alkyne because it is bonded to more carbons than the alkyne C-C bonds if you split them to check. The stereochemistry of the alkene is E at carbon 4 to 5. The IUPAC name will be [(E,3S)-4-methyl-4-hexen-1-yn-3-yl]ethanoate or [(E,3S)-4-methyl-4-hexen-1-yn-3-yl]acetate
Counting the carbon at the oxygen as carbon 1 gives the alkene C=C number 1. This is an alkene with alkoxy group oxygen bonded to carbon 1. Longest chain containing C=C is 4 = 1-butene. The longest chain on the right side of the oxygen is 4 carbons = butane and is bonded at carbon 2 to the oxygen = 2-butoxy. There are two methyl substituents on carbon 3 = dimethyl. The stereochemistry of the alkene is E and that at carbon 2 is R. Therefore, the IUPAC name will be [(E,2R)-3,3-dimethylbut-2-oxy]but-1-ene


The highest ranked functional group present in the compound is C=O ketone= alkanone =one located on carbon 1(shortest route count from left to right). The benzene ring aka phenyl group is also attached to carbon 1.

1-phenyl-1-propanone or 1-phenylpropan-1-one or 1-phenylpropanone

The highest ranked functional group present in the compound is C=O ketone= alkanone =one located on carbon 2 (shortest route count from right to left). The benzene ring aka phenyl group is attached to carbon 1.

1-phenyl-2-propanone or 1-phenylpropan-2-one

The first compound is benzaldehyde and the second compound is called cyclohexanone.

Practice naming these compounds below


Isomerism refers to the situation involving compounds of the same formula but differ in structure or spatial orientation. Therefore, we can have two main forms of isomerism, namely, structural isomerism and stereoisomerism.

Structural isomerism:

This refers to compounds with the same formula but different structural formula. Knowledge and practice of how to write open chain structures, branching and functional groups of organic compounds is very important. In this case it is very useful to know the Degree of Unsaturation when given the formula of an organic compound.

For any formula containing C, H, O, N or halogen, you can calculate Degree of Unsaturation (DU) by ignoring the presence of oxygen and sulfur group elements and use the formula below.

Degree of Unsaturation = ([Hydrogens in parent straight chain alkane-Hydrogens in given formula] + [Nitrogen in given formula]- [Halogens in given formula]) / 2

Example: Find DU for a compound of formula C5H4NCl;

The given formula has five carbons, therefore n=5, and the parent straight chain alkane CnH2n+2 = C5H12 will have 12 Hydrogens

The given formula has 4H ,1N and 1 halogen (Cl) therefore DU= (12-4+1-1)/2 = 8/2 = 4

4 degrees of unsaturation means several possibilities for the structure of the compound and its structural isomers:

A) presence of 4 double bonds, B). 3 double bonds and a ring. C). two double bonds and 2 rings. or D). 1 double bond and 3 rings E) 2 triple bonds., F)1 triple bond and 2 double bonds, G)1 triple bond and 2 rings or H)1 triple bond and a ring and a double bond.

This should be very helpful in finding the structure of the compound and its isomers.

Example : Find degree of unsaturation for the formula of ethanol C2H6O

The Formula C2H6O means, n=2, therefore the parent straight chain alkane CnH2n+2, = C2H6, will have 6 hydrogens. Ignore oxygen atoms (oxygen and its group are not counted). Nitrogens in given formula =0, Halogens in given formula=0. Therefore DU= (6-6+0-0)/2 = 0. This means the structure of the compound and its isomers will have no unsaturation (double bond or triple bond) or a ring. The compound and its isomers can have only single saturated bonds.

NOTE: Degree of Unsaturation (DU) only indicate the presence of unsaturation (double or triple bonds) or a ring (cyclic) in the structure of the compound or its isomers of the given formula. It does not tell you how many structural isomers that can exist for a given formula of a compound. Therefore DU=0 means the compound has no unsaturation and cannot have any isomer that is unsaturated or a ring (cyclic).

Example: Use DU to explain the if there will be unsaturation or ring isomers for a compound with formula C5H12.

This has DU= (12-12)/2 = 0. This means the compound is a saturated organic compound and can only have saturated structural isomers. Therefore, no unsaturated or CYCLIC (ring) isomer is possible in this case.

The other lesson on structural isomers is shown in this video below. This includes other functional groups in addition to alkanes, alkenes and alkynes.


This refers the spatial arrangement of the structure of compounds. In organic compounds, this includes, Newman projections, E/Z isomerism (alkenes), Cis /Trans isomers (alkenes and alkanes), Enantiomers (alkanes) and Diastereomers (alkenes and alkanes) and more.

Newman projections: This involves isomers of carbon compounds drawn such that one of the main carbons is in front and another one is behind. These carbons are represented as a dot (front carbon) in a circle (back carbon). The substituents on these carbons are drawn as seen in the actual structural formula. Different Newman projections can be drawn for the same compound because they have different energies or potential energies.

Cis/Trans isomers: These are stereoisomers that originates from the positions of the same two substituents around a bond or in a ring. Examples of this is shown in cyclohexane chair isomers and alkene isomers.

Example: Draw the most stable configuration


  1. Determine the most stable Newman projection for a) Hexane and b) 3-methyl-2-butanol
  2. Draw the most stable chair conformation for a) 1,2-cyclohexanediol and b) 1,2-difluoro-2-isopropyl-1-methylcyclohexane.

Alkene stereo-isomers:

Alkene stereoisomers are diastereomers. This occur when the substituents at one phase of the double bond have the same spatial orientation or arrangement, but the other phase has different spatial orientation of the substituents. This includes.

Cis/trans; is used when the double bond has only two different substituents.

E/Z isomers; is used when double bond has two or more different substituents.

Alkane stereoisomers:

Carbon in alkanes have 4 single bonds to substituents (atoms or groups). This gives them a 3-dimensional structure in space. Part of the structure directly pointing out towards the observer, the other part points away from the observer while the rest remain in the middle plane. This is like x,y,z spatial dimensions.

Chiral compounds and R/S Configuration:

Chiral compounds have 4 different substituents attached to the central carbon. The nature of the four substituents can cause the compound to rotate plane polarized light to the left or right. The four substituents allow characterization of chiral organic compounds in terms of R / S configuration.

Substituent priorities in chiral organic compounds:

Substituent distribution in R and S configuration depends on the molar mass of substituent directly attached to the chiral center. They are ranked 1= highest ,2,3,4 etc. Direction of molar mass distribution around chiral center determines the R or S configuration. Clockwise is R and counterclockwise is S.

However, there are special substituents that are ranked by the oxidation state of the atom present in the substituent group. A higher oxidation state implies a higher priority. For example, carboxylic acid carbon is more oxidized and has a higher oxidation sate than the carbon in an ester. Also, the carbon in a triple bond is more oxidized than carbon in an alkene. In other words, a higher number of hydrogens lowers oxidation state, but a higher number of oxygen atoms increases the oxidation state of a substituent and hence its priority.

3-Dimensional arrangement of a molecule can be determined from spatial orientation of each chiral center and the whole molecule. Model kits are very useful in this topic.

Video illustration: How to understand 3-D arrangement of chiral center of enantiomers: Click here

Assigning R and S configuration:

Do not assign if your lowest priority is in the plane of the page unless you are using a model kit. You can always re-sketch the structure so that the lowest priority is pointing away from you. You can do this simply by interchanging any two substituents followed by the remaining two.

Enantiomers have opposite R and S configurations, and they have non-superimposable mirror images, or they are non-superimposable mirror image of each other. They have the same formula and mass. Molecules with single chiral centers can only have enantiomers, however if there is more than one center, we can have Diastereomers in addition to enantiomers.  A diastereomers are two molecules that have one or more chiral centers of the same configuration (R/S) but opposite configuration at one chiral center.  We can example of molecules with one chiral center, or two or more centers and cyclic chiral molecules.

Fisher projections reveal another way to draw 3-dimensional structure of organic molecules with two more chiral centers. This combines simple lines representations with bold wedges. This is very useful in writing open structures of bio-organic molecules such as glucose, sucrose, fructose and some aldose molecules.

The above information in YouTube videos listed below.

How to assign substituent priorities and R/S configuration: Click video

Enantiomers of one chiral center molecules: click for video

There is also a way to get the R/S configuration of chiral centers in cyclic compounds. Example is shown in this video. Click here

Question 1:

a) Identify the chiral centers in each molecule and determine the R and S configuration for each chiral center.

b) Determine which pair of molecules are enantiomers, diastereomers, same or different.

c) Which molecule is optically active and why?

Question 2:

a) Identify the chiral centers in this molecule and determine the R and S configuration for each chiral center.

b) How many stereoisomers can be obtained from this molecule

c) Draw the enantiomer and two diastereomers of the molecule.

d) Draw the Fischer projection of the molecule


Identify the chiral centers and determine the R/S configuration for the structures below.


Compounds are the SAME: If the number of atoms is the same, the atoms are the same, the bonds are the same kind and number, and the stereochemistry is the same.

Axial Chirality

Axial chirality is the chirality that a compound possesses due to the shape of its structure but not due to the presence of any chiral center.    Compounds that show axial chirality include allene, spirocyclic compounds and Atropsisomers.

A) Allene and odd numbered cumulenes have terminal carbons that are twist out of phase making them non-planar. If the terminal carbons have different substituents, the allene will be chiral.   R and S configuration is determined by the two substituents on each terminal carbon i.e. [ high(h) priority or low(L) priority]. Therefore, these compounds have chiral axis and are optically active. The priorities of substituents are assigned in the same way as in any other chiral compound. These compounds can be achiral if the substituents are the same on one carbon; [that is A=B or A’=B’] since this will create a plane of symmetry in the molecule. Video lecture

B) In atropsiomers, there is restriction of free rotation around a single bond caused by certain substituents. This creates a chiral axis in the structure. The chiral axis that is used to determine the R and S configuration is located on four carbons bearing the substituents that restrict free rotation around single bonds. Chiral atropisomers are optically active. The substituents priorities are determined as in any other chiral compound. The substituents should be large enough to restrict free rotation of the rings at room temperature or below. In this case A and B can be the same substituent but both must not be hydrogen at the same time to maintain chirality.  The substituents priorities [high(h) or low(L)], are determined as in any other chiral compound. Chirality is maintained even if two substituents are the same. i.e. [(A and A’), (B and B’), (A and B), (A’ and B’), (A and B’) or (A’ and B).] Video lecture

C) Spirocyclic compound in which a spiral twisted part of a ring is out of phase with the other part. The spiral twist in the structure of compounds can give them chirality. In this example, the R and S configuration is determined by the substituent priorities (High(h) or Low(L)) on the four carbon atoms next to the spiral twist center. However, unlike in ALLENES and in ATROPISOMERS, the Spirocyclic compounds, have A and A’ (as well as B and B’) assigned diagonally from each other to get the chiral axis.

The chiral axis originates from the 3-DIMENSIONAL representation of the structure of the compound.

a) parts of the structure in the flat plane (simple lines).

b) part of the structure going into the plane of the page pointing away from observer(dashes).

c) part of the structure that is coming out of the plane of the page and pointing towards observer (solid bold).

Video lecture

Try and use the above procedure to identify the R/S configuration for BINAP and its derivatives.


Symmetry and Point group for Molecules. Click here

Molecular Orbital Theory

Free radical reactions

Electrocyclic reactions:

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