2024 Organic Chemistry Lecture Notes PDF

Organic Chemistry: Introduction & Isomerism 2024 Semester 2 CATHOLIC JUNIOR COLLEGE H2 CHEMISTRY (Syllabus 9729) ORGANIC CHEMISTRY: INTRODUCTION & ISOMERISM Content Empirical, mol...

Organic Chemistry: Introduction & Isomerism 2024 Semester 2 CATHOLIC JUNIOR COLLEGE H2 CHEMISTRY (Syllabus 9729) ORGANIC CHEMISTRY: INTRODUCTION & ISOMERISM Content Empirical, molecular and structural formulae Functional groups and the naming of organic compounds Common terms for organic reactions and reactivities Shapes of organic molecules; σ and π bonds Learning Outcomes Students should be able to: Introduction to organic chemistry (a) interpret and use the nomenclature, general formulae and displayed formulae of the following classes of compound: (also will be done in subsequent topics) (i) hydrocarbons (alkanes, alkenes and arenes) (ii) halogen derivatives (halogenoalkanes and halogenoarenes) (iii) hydroxyl compounds (alcohols and phenols) (iv) carbonyl compounds (aldehydes and ketones) (v) carboxylic acids and derivatives (acyl chlorides and esters) (vi) nitrogen compounds (amines, amides, amino acids and nitriles) (b) interpret, and use the following terminology associated with organic reactions: (i) functional group (ii) degree of substitution : primary, secondary, tertiary, quaternary (iii) homolytic and heterolytic fission (iv) carbocation (v) free radical, initiation, propagation, termination (iv) electrophile (Lewis acid), nucleophile (Lewis base) (v) addition, substitution, elimination, condensation, hydrolysis (vi) oxidation and reduction. [in equations for organic redox reactions, the symbols [O] and [H] are acceptable] (c) interpret, and use the following terminology associated with organic reactivities: (i) delocalisation (ii) electronic effect (electron donating and electron withdrawing effect) (iii) steric effect (steric hindrance) P a g e | 8-1 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 (d) describe sp3 hybridisation, as in ethane molecule, sp2 hybridisation, as in ethene and benzene molecules, and sp hybridisation, as in ethyne molecule (e) explain the shapes of, and bond angles in, the ethane, ethene, benzene, and ethyne molecules in relation to σ and π carbon-carbon bonds (f) predict the shapes of, and bond angles in, molecules analogous to those specified in (e) (g) apply (b) and (c) to the understanding of mechanisms in terms of organic structure and bonding (h) recognise that the mechanisms of polar reactions involve the flow of electrons from electron-rich to electron-poor sites Isomerism Content Isomerism: Constitutional (structural); cis-trans; enantiomerism Learning Outcomes Students should be able to: (a) describe constitutional (structural) isomerism (b) describe cis-trans isomerism in alkenes, and explain its origin in terms of restricted rotation due to the presence of π bonds [use of E, Z nomenclature is not required] (c) explain what is meant by a chiral centre (d) deduce whether a given molecule is chiral based on the presence or absence of chiral centres and/or a plane of symmetry (e) recognise that an optically active sample rotates plane-polarised light and contains chiral molecules (f) recognise that enantiomers have identical physical properties except in the direction in which they rotate plane-polarised light [usage of the term diastereomers is not required] (g) recognise that enantiomers have identical chemical properties except in their interactions with another chiral molecule (h) recognise that different stereoisomers exhibit different biological properties, for example in drug action (i) deduce the possible isomers for an organic molecule of known molecular formula (j) identify chiral centres and/or cis-trans isomerism in a molecule of given structural formula P a g e | 8-2 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 Contents 1. INTRODUCTION............................................................................................................................ 4 2. WHY ARE THERE SO MANY CARBON-CONTAINING COMPOUNDS?.................................... 5 3. BONDING IN ORGANIC COMPOUNDS – HYBRIDISATION....................................................... 6 3.1 sp3 hybridisation – forming four sigma bonds........................................................................ 7 3.2 sp2 hybridisation – forming three sigma bonds and one pi bond........................................... 9 3.3 sp hybridisation – forming two sigma bonds and two pi bonds........................................... 11 3.4 Comparing bond length and bond strength formed using sp3, sp2 and sp hybrid orbitals.. 12 4. CLASSIFICATION........................................................................................................................ 15 4.1 Based on the arrangement of the carbon chain.................................................................. 15 4.2 Based on Functional Groups and Homologous series........................................................ 16 5. TYPES OF FORMULAE............................................................................................................... 16 6. NOMENCLATURE - HOW TO NAME ORGANIC MOLECULES (using IUPAC system)............ 18 7. REACTIONS INVOLVING ORGANIC COMPOUNDS................................................................. 22 7.1 Types of bond fission in organic compounds...................................................................... 22 7.2 Types of attacking species (on organic compounds).......................................................... 23 7.3 Types of organic compounds (one classification method).................................................. 24 7.4 Types of reactions (from point of view of organic compounds)........................................... 25 7.5 Basic principles of mechanism drawing using curly arrows................................................ 26 8. ISOMERISM................................................................................................................................. 28 8.1 Constitutional (structural) Isomerism................................................................................... 28 8.1.1 Chain isomerism......................................................................................................... 28 8.1.2 Positional isomerism................................................................................................... 29 8.1.3 Functional group isomerism....................................................................................... 29 8.2 Stereoisomerism.................................................................................................................. 30 8.2.1 Cis-trans isomerism.................................................................................................... 30 8.2.2 Enantiomerism (Optical Isomerism)...................................................................................32 APPENDIX: FULL CLASSIFICATION.......................................................................................... 39 P a g e | 8-3 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 1. INTRODUCTION Organic Chemistry relates solely to the chemistry (structures, reactions, properties) of carbon containing compounds. At the ‘O’ levels, you have encountered alkanes, alkenes, alcohols, carboxylic acids, esters and macromolecules. At the ‘A’ levels, you will encounter a greater range of organic compounds and start to put them into classes, namely: 1) Hydrocarbons (alkanes, alkenes and arenes) 2) Halogen derivatives (halogenoalkanes and halogenoarenes) 3) Hydroxyl compounds (alcohols and phenols) 4) Carbonyl compounds (aldehydes and ketones) 5) Carboxylic acids and derivatives (acyl chlorides and esters) 6) Nitrogen compounds (amines, amides, amino acids and nitriles) It is important to note that organic compounds often also contain other elements: H, O, N, S, P and halogens (F, Cl, Br). The term organic can be misleading; it was derived from the historical fact that most early discoveries of carbon-containing compounds came from living things. Most of the man-made (synthetic) products nowadays do not come from living things anymore but are still called organic compounds. Nevertheless, oxides of carbon (CO, CO2) and carbonates are NOT classified as organic compounds (inorganic – origin is mineral sources). For instance, limestone is the source of calcium carbonate. Importance of Organic Chemistry To date, more than 2.5 million organic compounds are known, and this number continues to grow. All living organisms contain organic compounds, such as proteins, carbohydrates and fats. A sound knowledge of organic chemistry enables chemists to develop and manufacture drugs, agricultural chemicals, dyes and polymers. The manufacturing of drugs is an area that is particularly important in Singapore as the government slowly shifts the country’s dependence on manufacturing toward the pharmaceutical industry in the next few decades. P a g e | 8-4 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 2. WHY ARE THERE SO MANY CARBON-CONTAINING COMPOUNDS? Carbon has the unique ability to form an enormous number of compounds due to its two important properties: 1. Carbon can form single, double and triple bonds not only with itself (bond orders 1, 2 and 3 respectively), but also with other elements. The stability of the C−C bond can be seen by comparing bond energies: RECALL: Chemical bonding Bond Bond energy / kJ mol−1 C−C 350 N−N 160 O−O 150 Si−Si 222 P-P 200 S-S 264 Cl−Cl 244 C−O 360 C−Cl 340 o Strength of C−C bond is comparable to C−O and C−Cl bonds. This indicates that the C−C bond is actually quite strong. Compare this with silicon, which is also in Group 14. The Si−Si bond is significantly weaker. o Hence, the ability of carbon to form strong bonds to itself means that it can catenate (able to bond with atoms of same element to form long chains and structures). 2. Carbon can form 4 covalent bonds; a chain of carbon atoms can have many different groups attached. This leads to a wide diversity of compounds. Why Carbon forms Covalent Bonds rather than Ionic Bonds? Carbon belongs to Group 14 and has 4 valence electrons. It involves a very high energy change in order to gain or lose 4 valence electrons. Hence, simple C4− or C4+ ions do not exist. It is energetically more favourable for C to share electrons to form covalent bonds. RECALL: Carbon is in the second period of the Periodic Table and so, can hold a maximum of 8 electrons only (cannot expand its octet). Hence, each carbon atom can only form 4 bonds. P a g e | 8-5 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 3. BONDING IN ORGANIC COMPOUNDS – HYBRIDISATION Learning Outcome (d) Describe sp3 hybridisation, as in ethane molecule, sp2 hybridisation, as in ethene and benzene molecules, and sp hybridisation, as in ethyne molecule. (e) Explain the shapes of, and bond angles in, the ethane, ethene, benzene, and ethyne molecules in relation to σ and π carbon-carbon bonds. (f) Predict the shapes of, and bond angles in, molecules analogous to those specified in (e). Carbon is able to form single, double and triple bonds. Each type of bonding pattern corresponds to a different shape around the carbon atom. For example: Methane Ethene Ethyne Tetrahedral Trigonal Planar Linear However, the ground state electronic configuration of a carbon atom (1s22s22p2) does not agree with the observation above. Take methane for example. The carbon atom is covalently bonded to 4 other hydrogen atoms. However, there are only two 2p orbitals, each containing one unpaired electron that can be used to form covalent bonds with other atoms. This may be explained by promoting an electron from the 2s orbital to the 2p orbitals to give four unpaired electrons. The four unpaired electrons reside in different orbitals with different orientation in space. If each electron was used to form bonds, we will get bonds with two different lengths and strengths due to the differences in relative energy of the bonding electrons. Furthermore, since P a g e | 8-6 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 the p orbitals are perpendicular to one another, we will get 3 bonds that are 90o to each other, something that is definitely not the case! Therefore, something else must be happening. In actual fact, carbon needs to form equivalent orbitals of same shapes and energies by a process known as hybridisation before bonding can occur. This process involves combining atomic orbitals of different energies to produce new hybrid orbitals (sp3, sp2 and sp). [You may access the following YouTube video to learn more about hybridisation: http://youtu.be/g1fGXDRxS6k (1min:36s)] 3.1 sp3 hybridisation – forming four sigma bonds The “combination” of one 2s orbital with the three 2p orbitals to form four identical sp3 hybrid orbitals is known as sp3 hybridisation. bond angle:109.5o four tetrahedral sp3 hybrid orbitals The orbitals are of a shape different from that of the s or p orbitals they originate from, and they are directed towards the corners of a regular tetrahedron. The bond angle is 109.5o. P a g e | 8-7 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 Each hybrid orbital has one large lobe and one small lobe. RECALL: Sigma & pi Example: ethane (C2H6) bonds (Chemical Bonding) Each C atom is sp3 hybridised (Bond angle 109.5o) All the bonds are  bonds (C-C and C-H single bonds). C Ball and stick model of ethane H P a g e | 8-8 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 3.2 sp2 hybridisation – forming three sigma bonds and one pi bond sp2 hybridisation is the “combination” of the 2s orbital with two 2p orbitals to form three sp2 hybrid orbitals of equivalent energy leaving a singly-occupied 2p orbital unhybridised. unhybridised p orbital three sp2 hybrid orbitals at 120o The three sp2 hybrid orbitals are planar and are at angles of 120o to each other. The unhybridised p orbital is perpendicular to this plane. Hybridise + + Each hybrid orbital has one large lobe and one small lobe. P a g e | 8-9 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 Example: ethene (C2H4) sigma bond H H C=C 120o H H When these two p-orbitals overlap, a  bond is formed. Ethene is a planar molecule with each C atom being sp2 hybridised (trigonal planar, bond angle 120o). The carbon-carbon double bond consists of a  bond and a  bond and thus, is stronger than the C−C single bond in ethane. Ethene is an unsaturated compound as C is bonded to only 3 other atoms. Example: benzene (C6H6) [will be covered in detail under Arenes] six unhybridised p 6  electrons are delocalised orbitals overlap (side- - they do not belong to any particular C atom but are free to way) to give  bonds move throughout the entire ring. Benzene is a planar molecule with each C atom being sp2 hybridised (trigonal planar, bond angle 120o). Each of the remaining unhybridised p-orbitals in the 6 C atoms forms an extensive overlap and thus, 6  electrons are delocalised around the ring. Thinking Time 1 A molecule with two C=C bonds adjacent to each other is non-planar. WHY? e.g. propadiene, CH2=C=CH2 P a g e | 8-10 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 3.3 sp hybridisation – forming two sigma bonds and two pi bonds sp hybridisation is the “combination” of the 2s orbital with one 2p orbital to form two sp hybrid orbitals of equivalent energy leaving two singly-occupied 2p orbitals unhybridised. 2 unhybridised p orbitals two sp hybrid orbitals at 180 o The two sp hybrid orbitals are collinear to each other (bond angle of 180o). The two unhybridised p orbitals are perpendicular to the sp hybrid orbitals. Hybridise + Each hybrid orbital has one large lobe Example: ethyne (C2H2) and one small lobe.  bonds H C C H Ethyne is a linear molecule. The carbon-carbon triple bond (CC) bond consists of 1 σ and two π bonds, making it even stronger than a C=C bond in ethene. It is also of a higher bond order than the C=C bond. Ethyne is an unsaturated compound as C is bonded to only two other atoms. P a g e | 8-11 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 3.4 Comparing bond length and bond strength formed using sp3, sp2 and sp hybrid orbitals An electron in an s orbital is more strongly attracted to the nucleus compared to an electron in a p orbital. sp hybrid orbital resembles s orbital the most / resembles p orbital the least (50% s- character). sp3 hybrid orbital resembles s orbital the least / resembles p orbital the most (25% s-character). Hence the bond length due to the overlap of sp3 orbitals is greater than the overlap of sp orbitals. The bond length arising from the overlap of hybrid orbitals decreases in this order: sp3–sp3 > sp2–sp2 > sp–sp. Hence the strength of σ bond arising from the overlap of hybrid orbitals increases in this order: sp3–sp3 < sp2–sp2 < sp–sp. Note: Bonds to carbon are shortened by increasing the s-character of the carbon hybrid orbital involved. This is most often explained by the fact that, as the percentage of the s- character in a hybrid orbital increases, the orbital becomes more like an s orbital and hence the electrons are held more tightly by the nucleus than those in an orbital with less s-character. Self-Practice 1: 1. The bond lengths in propyne differ from those which might be expected. The carbon-carbon bond length in ethane is 0.154 nm and in ethyne, 0.120 nm. The single C2–C3 bond in propyne, however, is shorter than the single bond in ethane: it is 0.146 nm. (clue: identify the hybridisation of the carbons) What helps to explain this C2–C3 bond length in propyne? A It has a partial sp2–sp2 character. B It is an sp–sp3 overlap. C The sp3–sp3 bonding is pulled shorter by a p–p ( bond) overlap. D The electrons in the C3 carbon atom are attracted towards the  bonds between C1 and C2. [A-Level 2010/P1/Q19 (modified)] 2. Which substances consist of planar molecules? 1 benzene 2 ethene 3 cyclohexene [A-Level 2002/P1/Q38] P a g e | 8-12 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 In summary: Structures Hybridisation No. of hybrid orbitals No. of unhybridised orbitals % of Bond angle Shape of C (= no. of σ bonds) (= no. of π bonds) s-character around C around C sp3 4 0 25% 109o tetrahedral Alkane Carbonyl Alkene carbon trigonal sp2 3 1 33% 120o planar Carbocation Benzene sp 2 2 50% 180o linear Alkyne Diene P a g e | 8-13  Organic Chemistry: Introduction & Isomerism 2024 Semester 2 Additional Reading: Other than carbon, hybridisation can also occur in other elements such as nitrogen and oxygen. When there is a presence of lone pair of electrons around the atom, the lone pair can be considered equivalent to a sigma bond. Structures Hybridisation No. of hybrid orbitals No. of unhybridised % of Bond Shape of N (≈ no. of σ bonds + lone orbitals s-character angle around N pair of electrons) (= no. of π bonds) around N ammonia trigonal sp3 4 0 25% 107o pyramidal amine trigonal sp2 3 1 33% 120o planar amide pyridine (Presence of conjugation where there is overlap of p orbitals on C and N atoms) sp 2 2 50% 180o linear nitrile P a g e | 8-14 Organic Chemistry: Introduction & Isomerism 2023 Semester 2 4. CLASSIFICATION 4.1 Based on the arrangement of the carbon chain i. Aliphatic molecules contain open chains of carbon. The carbon chain could be a straight chain or branched. ii. Alicyclic molecules consist of closed rings of carbon atoms. The rings may contain single or multiple bonds. iii. Aromatic molecules contain at least one benzene ring (ring of 6 carbon atoms where electrons in p-orbitals are delocalised, forming a π electron cloud). [will learn this more under the topic Arenes] P a g e | 8-15 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 4.2 Based on Functional Groups and Homologous series Learning Outcome (b) Interpret, and use the following terminology associated with organic reactions: (i) functional group A functional group consists of an atom or group of atoms or bonds common to a series or family of compounds and governs the principal chemical properties of the series. A series of compounds containing the same functional group is called a homologous series. In a homologous series, successive members i. can be represented by a general molecular formula and have the same functional group; ii. differ in formula by a regular –CH2– unit; iii. have physical properties that vary in a regular manner down the series; iv. can be prepared by similar methods; and v. have similar structures and similar chemical properties. Refer to the Appendix attached at the back for the full table of classification of the homologous series by their functional groups. They will be encountered in future topics. 5. TYPES OF FORMULAE The formula of an organic compound can be expressed in several ways: Empirical formula – gives the simplest ratio of the number of atoms of each element present in one molecule. Molecular formula – gives the actual number of atoms of each element present in one molecule. Structural formula – shows how the constituent atoms of a molecule are joined together; i.e. relative placing of atoms. Displayed formula – shows detailed structure of a molecule with both the relative placing of atoms and the number of bonds between them. (also known as Full structural formula) Skeletal formula – simplified representation of the structure of a molecule showing only the carbon-carbon bonds in the carbon skeleton and the associated functional groups. (H atoms bonded to C atoms of the carbon skeleton are not shown) Stereochemical show spatial arrangement of bonds, atoms and groups in a formula – molecule in 3-D, especially for showing enantiomers. P a g e | 8-16 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 Empirical Molecular Structural Skeletal Examples Displayed Formula Formula Formula Formula Formula propane C3H8 C3H8 CH3CH2CH3 propene CH2 C3H6 CH3CH=CH2 benzene CH C6H6 C6H6 cyclohexane CH2 C6H12 propanoic CH2O C3H6O2 CH3CH2CO2H acid P a g e | 8-17 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 6. NOMENCLATURE - HOW TO NAME ORGANIC MOLECULES (using IUPAC system) Learning Outcome (a) Interpret and use the nomenclature, general formulae and displayed formulae of the following classes of compound: (i) hydrocarbons (ii) halogen derivatives (iii) hydroxyl compounds (iv) carbonyl compounds (v) carboxylic acids and derivatives (vi) nitrogen compounds In general, the name (nomenclature) of an organic compound is made up of two parts: parent chain & substituent groups Step 1: Naming the parent chain The parent chain is the longest continuous unbranched carbon chain. The prefix in the parent chain name denotes the number of carbon atoms in it. Prefix meth- eth- prop- but- pent- hex- hept- oct- non- dec- No. of C 1 2 3 4 5 6 7 8 9 10 atoms The suffix denotes the name of the functional group. The position of the functional group is given the smaller number. Example: hexan-2-ol prefix suffix o prefix: “hexan” indicates that the parent chain has six carbon atoms. o suffix: “-2-ol” indicates presence of an alcohol functional group at carbon-2. P a g e | 8-18 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 Step 2: Naming the side chain / substituent and indicate its position Side chain / substituent refers to branches from the parent chain. o side chain that contains only carbon and hydrogen atoms are called alkyl groups (often represented by symbol R). –CH3 methyl (from methane) –R –CH2CH3 ethyl (from ethane) –CH2CH2CH3 propyl (from propane) o Other examples: –Cl is called chloro, –Br is called bromo. The position of a side chain is indicated by the smaller number. A hyphen, “–“, is used to separate the numerals from alphabets in the name. Example: The following compound is named 2-methylpentane, not 4-methylpentane. 1 2 3 4 5 5 4 3 2 1 CH3CHCH2CH2CH3 CH3CHCH2CH2CH3 CH3 CH3 smaller numbers  “1” cannot 2-methylpentane 4-methylpentane start here o “pentane” indicates that parent chain has 5 carbon atoms. o “2-methyl” indicates that the methyl substituent is on carbon-2. If there are more than one side chains, they are arranged in alphabetical order. Example: smaller numbers alphabetical order  “1” starts here  “e” before “m” NOT 2-methyl-5-ethylheptane If the same side chain occurs more than once, it is indicated by prefixes like di (for two), tri (for three), tetra (for four), penta (for five). If more than one same side chain joined to same carbon atom, the same position number is repeated and is separated by a comma, “,”. Example: “di”  2 CH3 side chains; “2,2”  both at carbon-2 parent chain with 3 C atoms P a g e | 8-19 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 Summary: (position — prefix substituent) Parent ❖ Indicates position of ❖ Indicates ❖ Arrange in ❖ Longest, substituent on the the number alphabetical continuous carbon carbon chain. of same order chain (functional ❖ Number should be as substituent. group at smallest small as possible. carbon number) (position — prefix substituent) Parent 1,2- di No. of C atoms 1,2,2- tri meth- 1C eth- 2C 1,1,2,3- tetra prop- 3C bromo −Br Bonding in chain chloro −Cl -an- C-C single bonds only iodo −I -en- one C=C double bond -yn- one C≡C triple bond fluoro −F Functional group phenyl −C6H5 -e (hydrocarbon) −H only (benzene ring) -ol (alcohol) −OH group methyl −CH3 -al (aldehyde) −CHO group -one (ketone) >C=O -oic acid (carboxylic acid) −CO2H group -amine (amine) −NH2 group P a g e | 8-20 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 Worked Example 1: 1. Name the following compounds. (a) 2-methylhexane (b) pent-2-ene (c) 4-methylpent-1-ene (d) 2-chlorobutane Lecture Practice 1 1. Write the structural formula for (a) heptane (b) 2-chloro-3-methylhexane (c) 3-bromo-2-chloroheptane (d) pentan-2-ol heptane 2-chloro-3-methylhexane 3-bromo-2-chloroheptane pentan-2-ol 2. Draw the skeletal formula for the compounds in question 1. P a g e | 8-21 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 Self-Practice 2: 1. Name the following compounds. ❖ ❖ (CH3)2C=CH2 2-methylpropene 2-bromo-3-chloropentane 2. Write the structural formula and skeletal formula for (a) hex-2-ene (b) 3-bromo-2-chloropentane CH3CHClCHBrCH2CH3 7. REACTIONS INVOLVING ORGANIC COMPOUNDS Learning Outcome (b) Interpret, and use the following terminology associated with organic reactions: (iii) homolytic and heterolytic fission (v) free radical, initiation, propagation, termination (iv) electrophile (Lewis acid), nucleophile (Lewis base) (v) addition, substitution, elimination, hydrolysis (vi) oxidation and reduction. [in equations for organic redox reactions, the symbols [O] and [H] are acceptable] 7.1 Types of bond fission in organic compounds In a reaction, bonds must be broken (fission) in order for new compounds to be formed. Covalent bonds may be broken in two main ways: (i) Homolytic fission Breaking of a single covalent bond such that an electron from the bond is transferred to each of the two bonded atoms. This leads to formation of free radicals. Example: Note: → 2Cl Half arrow is used to indicate the transfer of 1 electron. Arrow head points to 2 electrons per atom that receives the 1 electron. covalent bond P a g e | 8-22 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 (ii) Heterolytic fission Breaking of a single covalent bond such that the pair of electrons is transferred onto the more electronegative atom, resulting in formation of ions. A positively charged carbon ion formed is referred to as a carbocation. Example: Note: Full arrow is used to indicate the transfer of 2 electrons. Arrow head points to atom that receives the 2 electrons. 7.2 Types of attacking species (on organic compounds) There are 3 main types of attacking species in organic reactions. (i) Electrophile (“electron loving”): An electron-deficient species that can accept a pair of electrons. Usually positively charged, but also can be neutral. Examples: AlCl3, Br+, H+ Electrophiles are also Lewis acids. [RECALL: Theories of Acids & Bases] (ii) Nucleophile (“nucleus seeking”): A species that can donate electrons (has high electron density) as it has at least one lone pair of electrons. Usually neutral or negatively charged. Examples: NH3, H2O,OH−, CN− Nucleophiles are also Lewis bases. [RECALL: Theories of Acids & Bases] (iii) Free radical A highly reactive atom / molecule with an unpaired electron (usually produced by homolytic fission). Examples: Cl, CH3, NO Self-Practice 3: Which of the following can act as an electrophile? A Cl+ B CH2=CH2 C CN– D Na+ [A-Level 2001/P3/20(modified)] P a g e | 8-23 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 7.3 Types of organic compounds (one classification method) (i) Electron-rich functional groups (they contain  electrons) Examples: alkene, benzene (they do NOT contain polarised bonds) Such electron rich functional groups will attract electrophiles (“electron loving”) or Lewis acids. (ii) Electron-deficient functional groups (contains electron deficient carbon) The carbon atom is bonded to a more electronegative atom (hence they contain polarised bonds). Examples: halogenoalkanes, alcohols, aldehydes, ketones, carboxylic acids, acyl chlorides, esters + – + – + – + – + – + – Such functional groups that contain electron deficient carbon will attract nucleophiles (“nucleus seeking”) or Lewis bases. (iii) Neither electron rich nor electron deficient Alkanes contain only C–C and C–H bonds, which are very strong bonds Free radicals, which are very reactive, can break the C–H bonds P a g e | 8-24 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 7.4 Types of reactions (from point of view of organic compounds) (i) Substitution: involves replacing an atom (or a group of atoms) by another atom (or groups of atoms). This involves breaking and forming of single bonds. CH3CH2Cl + OH− → CH3CH2OH + Cl− (Cl atom is replaced by OH group) CH4 + Cl2 → CH3Cl + HCl (H atom is replaced by Cl atom) If attacking species is a free radical, it is called free radical substitution. If attacking species is an electrophile, it is called electrophilic substitution. If attacking species is a nucleophile, it is called nucleophilic substitution. (ii) Addition: involves two molecules combining together to form a single new product. This involves organic compounds having multiple bonds. Br2 + CH2=CH2 → BrCH2CH2Br CH2=CH2 + H2O → CH3CH2OH If attacking species is an electrophile, it is called electrophilic addition. If attacking species is a nucleophile, it is called nucleophilic addition. (iii) Elimination: involves the removal of atom or group of atoms from two adjacent atoms to form multiple bonds (C=C double bond). CH3CH2OH → CH2=CH2 + H2O (H2O is eliminated) CH3CH2Br → CH2=CH2 + HBr (iv) Hydrolysis: involves breaking covalent bonds by reaction with water. Water itself takes part in the reaction. Ethyl ethanoate (an ester) is hydrolysed to give ethanoic acid and ethanol. P a g e | 8-25 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 (v) Oxidation: gain of oxygen and/or loss of hydrogen atoms (increase in oxidation number for C) suitable oxidising agent CH3CH2OH + [O] CH3CO2H (vi) Reduction: loss of oxygen and/or gain of hydrogen atoms (decrease in oxidation number for C) suitable reducing agent CH3CO2H + 2[H] CH3CH2OH +H2O 7.5 Basic principles of mechanism drawing using curly arrows Electrons always flow from region of higher electron density (electron-rich site) to lower electron density (electron-deficient site). Example: electron-rich (higher electron density) functional group attracting electrophiles (lower electron density). P a g e | 8-26 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 Example: electron-deficient functional group (lower electron density) attracting nucleophiles (higher electron density). The movement of electron pair is illustrated using full curly arrows originating from electron-rich site to the electron-deficient site. Arrow head always point to atom that receives the electrons (lower electron density or electron-deficient site). P a g e | 8-27 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 8. ISOMERISM Learning Outcome (i) Deduce the possible isomers for an organic molecule of known molecular formula. Isomerism occurs when 2 or more compounds have the same molecular formula but different arrangement of the atoms in the molecules. These compounds are known as isomers. ISOMERISM Constitutional Isomerism Stereoisomerism (Structural isomerism) - same molecular formula - same molecular formula - same structural formula but different - different structural formulae spatial arrangement of atoms (i.e. different arrangement of atoms) chain positional functional cis-trans enantiomerism group 8.1 Constitutional (structural) Isomerism Learning Outcome (a) Describe constitutional (structural) isomerism. Constitutional isomers have the same molecular formula but different structural formulae; i.e. different arrangement of atoms. There are three types of constitutional isomerism. 8.1.1 Chain isomerism Chain isomers have the same number of carbon atoms but different arrangement of carbon atoms in the carbon chain. e.g. C5H12 C C C C C CH3CH2CH2CH2CH3 pentane 2-methylbutane 2,2-dimethylpropane P a g e | 8-28 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 8.1.2 Positional isomerism Positional isomers have the same carbon chain, but different positions of a functional group(s). e.g. C3H7Cl CH3CH2CH2Cl 1-chloropropane 2-chloropropane 8.1.3 Functional group isomerism Functional group isomers have different functional groups. e.g. C3H6O C C C O propanal (aldehyde) C C C CH3 C CH3 propanone (ketone) O O e.g. C4H10O butan-1-ol (alcohol) ethoxyethane C C O C C (ether) e.g. C3H6O2 propanoic acid (carboxylic acid) O methyl ethanoate C C O C (ester) P a g e | 8-29 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 Constitutional (Structural) isomers with the same functional group have similar chemical properties but different physical properties. Constitutional (Structural) isomers with different functional groups have different chemical properties and different physical properties. 8.2 Stereoisomerism 8.2.1 Cis-trans isomerism Learning Outcome (b) Describe cis-trans isomerism in alkenes, and explain its origin in terms of restricted rotation due to the presence of π bonds [use of E, Z nomenclature is not required] (j) Identify cis-trans isomerism in a molecule of given structural formula. cis-trans isomerism usually occurs in alkenes due to:- 1. Restricted (or hindered) rotation that exists in C=C double bond due to the presence of a -bond. (Rotation about the C=C double bond cannot take place unless the -bond is broken.) 2. Each C atom of the C=C double bond has 2 different substituents attached to it. Example: ClCH=CHCl ❖ The cis isomer has two similar groups on the same side of the C=C bond. While the trans isomer has two similar groups on opposite sides of the C=C bond. ❖ Cis-trans isomers have similar chemical properties but differ in physical properties. ❖ The cis isomer usually has a higher boiling point compared to the trans isomer. P a g e | 8-30 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 ❖ Cis-trans isomerism is not exhibited for compounds with C=C in a ring. e.g. cyclohexene The ring structure effectively locks the C=C in a cis configuration as any attempt to form the trans isomer would cause a distortion of ring structure. ❖ Cis-trans isomerism can also occur in ring systems where each C atom of the single bond in a ring has 2 different substituents attached to it. (Rotation around a single bond is prevented by linkage in a ring) e.g. 1, 2-dichlorocyclopropane Lecture Practice 2: Do the following compounds exist as cis-trans isomers? (a) CH2=CH2 no _____________________ (b) CH3CH2CH=CHBr yes _____________________ (c) CH2ClCH2Cl no _____________________ Self-Practice 4: Do the following compounds exist as cis-trans isomers? (a) CH2=CF2 no (b) CH3CH2CH=CHCO2H yes (c) 1-bromobut-1-ene yes ❖ cis-trans isomers occur in pairs. A compound with n C=C bond that exhibits cis-trans isomerism, the total number of cis-trans isomers that can be formed is 2n. e.g. linoleic acid, CH3(CH2)4CH=CHCH2CH=CH(CH2)7COOH, gives 4 cis-trans isomers cis- cis- cis- trans- trans- cis- trans- trans- P a g e | 8-31 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 8.2.2 Enantiomerism (Optical Isomerism) Learning Outcome (c) Explain what is meant by a chiral centre (d) Deduce whether a given molecule is optically active based on the presence or absence of chiral centres and/or a plane of symmetry (j) Identify chiral centres in a molecule of given structural formula A carbon atom with four different substituents attached to it is said to be chiral. It is usually marked with an asterisk, *. e.g. Enantiomerism arises when compounds have at least one chiral centre (carbon atom with 4 different substituents attached to it) and do not have an internal plane of symmetry, and the mirror images formed are non-superimposable. e.g. 2-hydroxypropanoic acid, CH3CH(OH)CO2H CH3 These isomers appear in a pair. Individually they are called enantiomers. For a compound with n chiral centres, the total number of enantiomers that can be formed is 2n. Additional reading: Every object has a mirror image. Objects can be divided into two classes: those who can match exactly on their mirror images (can be superimposed) and those that cannot. Objects that cannot be superimposed on their mirror image: example your left hand cannot superimpose on your right hand, though both are mirror images of each other Objects that can be superimposed: example tennis balls, conical flask etc P a g e | 8-32 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 Learning Outcome (f) Recognise that optical isomers have identical physical properties except in the direction in which they rotate plane-polarised light. (g) Recognise that optical isomers have identical chemical properties except in their interactions with another chiral molecule. (h) Recognise that different stereoisomers exhibit different biological properties, for example in drug action. How to recognise whether a compound displays enantiomerism? See whether compound has at least one carbon atom with 4 different substituents attached to it. The particular carbon atom is said to be chiral. It is usually marked with an asterisk, *. Below is a generic way to represent enantiomerism. Example: butan-2-ol CH3 The pair of ENANTIOMERS ❖ have identical physical properties (e.g. melting and boiling points, density), except the direction of rotation of the plane polarised light; ❖ have identical chemical properties, except the reaction with other chiral molecules ❖ have different biological properties (e.g. effects of drugs); P a g e | 8-33 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 One enantiomer will rotate plane-polarised light to the right or clockwise, while the other will rotate the polarised light to the left or anti-clockwise. Enantiomers are said to be optically active. It is important to recognise that an optically active sample rotates plane-polarised light and contains chiral molecules. Optical Activity If a substance can rotate the plane of polarised light, it is said to be optically active. Below is a simple illustration on how to determine whether a substance is optically active: Ordinary light vibrates in all direction. When light is passed through the polarising filter, the light will vibrate in a single direction only. After the plane polarised light passes through an optically active sample, the light will rotate and vibrate at an angle from the original. Lecture Practice 3 Draw the pair of enantiomers of the following molecule. P a g e | 8-34 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 Lecture Practice 4 State the type/s of isomerism shown by CH3CH=CHCHClCH3 and the total number of stereoisomers. cis-trans isomerism and enantiomerism Number of stereoisomers = 2n = 21+1 = 4 (n refers to the number of chiral carbon + no of C=C that can exhibit cis-trans isomerism) CH3 Cl Cl CH3 CH3 Cl Cl CH3 CH3 Racemic mixture Enantiomers rotate plane-polarised light by the same angle but in the opposite direction. A 50:50 mixture of the two enantiomers therefore has no effect on plane polarised light because the direction of rotation of plane polarised light by one enantiomer is cancelled out by the other enantiomer. This equimolar mixture of the two enantiomers is called a racemic mixture and is optically inactive. P a g e | 8-35 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 Meso compounds Does it mean that all molecules with chiral centres are optically active? The presence of chiral centres is a useful but not a sufficient condition for the molecule to be optically active. For example, meso compounds contain chiral centres but also an internal plane of symmetry, which causes the compound to be achiral and hence optically inactive. The plane of symmetry makes one half of the molecule a mirror image of the other half. A molecule is achiral if it is superimposable on its mirror image. Most achiral molecules do have an internal plane of symmetry. Example: tartaric acid Since tartaric acid has 2 chiral carbons, it exhibits enantiomerism. However, molecules 1 and 2 are actually the same molecule (i.e. superimposable on each other) as there is a plane of symmetry within the molecule. Hence it is a meso compound. This plane of symmetry is not present in molecule 3 or 4 and they are a pair of enantiomers (mirror images of each other). They can be distinguished by the direction to which each isomer rotates plane polarised light. If molecule 3 rotates plane polarised light clockwise by 10⁰, molecule 4 would rotate plane polarised light anticlockwise by the same magnitude. Hence only 3 (2n – 1) stereoisomers (2 enantiomers and 1 meso compound) are present. P a g e | 8-36 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 Self-Practice 5: 1. [A-Level 2003/P1/Q20] Many drugs exhibit enantiomerism. The diagrams show the structure of three drugs. What is the total number of chiral carbon centres in these three structures? A 1 B 2 C 3 D 4 2. Compound G is optically active. How many chiral centre(s) does compound G have? A 1 B 2 C 3 D 4 3. [A-Level 2005/P1/Q21] Which molecules have isomers that exhibit cis-trans isomerism? I II III IV C3H6BrI C3H5I C3H4I2 C3H4BrI A I, II and III only B II, III and IV only C II and IV only D III and IV only Still need a little more explanation about Enantiomerism? Watch the video here: http://youtu.be/RBtgAz70_JY (2min:24s) P a g e | 8-37 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 Answers Thinking Time 1 Since the 2  bonds are perpendicular to each other, the two methylene groups (=CH2) are perpendicular to each other too and the molecule is non-planar. Self-Practice 1: Q1: B Q2: 1 & 2 Self-Practice 2: Q1: 2-bromo-3-chloropentane / 2-methylpropene Q2: (a) hex-2-ene (b) 3-bromo-2-chloropentane CH3CHClCHBrCH2CH3 Self-Practice 3: A Self-Practice 4: a) no b) yes c) yes Self-Practice 5: Q1: A Q2: C Q3: B P a g e | 8-38 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 APPENDIX: FULL CLASSIFICATION As previously mentioned, as the complexity of the compound increase, it becomes more convenient and important to identify the compound by its functional groups. In this section, you would find a useful summary of all the functional groups as well as examples of each of these functional groups. While they may look overwhelming at this point, they will become less daunting after you have completed the entire series of lectures on Organic Chemistry. Simplest homologous series: alkanes General formula: CnH2n+2, n = 1, 2, 3, … Members: CH4 C2H6 C3H8 CnH2n+2 (except for cyclic alkanes) Several other homologous series can be considered to be derived from alkanes (by replacing a H atom with a functional group.) P a g e | 8-39 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 In the table below, R, R’ represent alkyl groups of general formula CnH2n+1. organic structure of homologous series general formula examples compounds functional group CnH2n+2 ethane (C2H6) alkane n = 1, 2, … --- butane (CH3CH2CH2CH3) CnH2n aIkenes hydrocarbons n = 2, 3, … ethene (C2H4) arene _ CH3 benzene (C6H6) methylbenzene(C6H5CH3) CnH2n+1X bromoethane (CH3CH2Br) halogenoalkane n = 1, 2, … C X X = F, Cl, Br, I halogen derivatives halogenoarene _ X chlorobenzene (C6H5Cl) H H hydroxy CnH2n+1OH alcohols H C C OH compounds n = 1, 2, … H H ethanol (CH3CH2OH) P a g e | 8-40 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 OH phenols _ OH phenol (C6H5OH) H CnH2n+1CHO aldehyde C O n = 0, 1, 2, … R carbonyl ethanal (CH3CHO) compounds R' CnH2n+2CO ketone C O n = 2, 3, … R propanone CH3COCH3 O CnH2n+1CO2H carboxylic acid C n = 0, 1, 2, … OH ethanoic acid (CH3CO2H) O carboxylic ester R−CO2R’ R C acids and O R' derivatives ethyl ethanoate (CH3CO2CH2CH3) O acyl chloride R−COCl C Cl propanoyl chloride (CH3CH2COCl) P a g e | 8-41 Organic Chemistry: Introduction & Isomerism 2024 Semester 2 H H H C C C N nitrile R−CN H H propanenitrile (CH3CH2CN) H H H H C N amine R−NH2 C N H H H methylamine (CH3NH2) nitrogen compounds O amide R−CONH2 C NH2 ethanamide (CH3CONH2) R' amino acid H2N C CO2H H2N C CO2H R aminoethanoic acid (H2NCH2CO2H) P a g e | 8-42