2024 YISHUN INNOVA JUNIOR COLLEGE H2 Chemistry Past Paper - Alkenes PDF

Students’ Copy YISHUN INNOVA JUNIOR COLLEGE 2024 JC1 H2 CHEMISTRY HYDROCARBONS: ALKENES GUIDING QUESTIONS Which class of reagents do alkenes react with and why? What type of reactions do alkenes un...

Students’ Copy YISHUN INNOVA JUNIOR COLLEGE 2024 JC1 H2 CHEMISTRY HYDROCARBONS: ALKENES GUIDING QUESTIONS Which class of reagents do alkenes react with and why? What type of reactions do alkenes undergo and why? How do alkenes react with electrophiles? What is the major product obtained when an unsymmetrical alkene reacts with hydrogen halide and why? LEARNING OBJECTIVES At the end of the lesson, students should be able to: (a) explain the general reactivity of alkenes towards electrophilic reagents / electrophiles (b) describe the chemistry of alkenes as exemplified, where relevant, by the following reactions of ethene: (i) electrophilic addition of water / steam, hydrogen halides and halogens (ii) reduction via catalytic hydrogenation (catalytic addition of hydrogen) (iii) oxidation by cold, dilute manganate(VII) ions to form the diol (iv) oxidation by hot, acidified solution of manganate(VII) ions leading to the rupture of the carbon-to- carbon double bond in order to determine the position of alkene linkages in larger molecules (c) describe the mechanism of electrophilic addition in alkenes, using bromine with ethene as an example (d) apply Markovnikov’s rule to the addition of hydrogen halides to unsymmetrical alkenes, and explain the composition of products in terms of the stability of the carbocation intermediates. (e) recognise that the mechanisms of polar reactions involve the flow of electrons from electron-rich to electron- poor sites REFERENCES Peter Cann and Peter Hughes (2015), Cambridge International AS and A Level Chemistry Kim Seng Chan and Jeanne Tan (2017), Understanding Advanced Organic and Analytical Chemistry, The Learner’s Approach, Revised Edition Alkenes CONTENTS PAGE 1 Introduction 2 Nomenclature 3 Bonding and Hybridisation of Carbon in Alkenes 4 Isomerism 4.1 Constitutional Isomerism: Chain, Positional and Functional Group Isomerism 4.2 Stereoisomerism: Cis-Trans Isomerism 5 Physical Properties of Alkenes 5.1 Melting and Boiling Points 5.2 Solubility 6 Preparation of Alkenes 6.1 Elimination of Water, H2O, from Alcohols 6.2 Elimination of Hydrogen Halides, HX, from Halogenoalkanes 7 Chemical Properties of Alkenes 8 Reactions of Alkenes 8.1 Combustion 8.2 Reduction 8.3 Oxidation: Mild Oxidation and Oxidative Cleavage 8.4 Electrophilic Addition 8.4.1 Reaction with Halogens, X2, in Organic Solvent 8.4.2 Reaction with Aqueous Halogens, X2(aq) 8.4.3 Reaction with Hydrogen Halides, HX 8.4.4 Reaction with Water, H2O 9 Markovnikov’s Rule 10 Distinguishing Tests for Alkenes YIJC / JC1 H2 Chemistry / 2024 / Students’ Copy Pg 2 of 33 Alkenes 1 INTRODUCTION Alkenes, also known as olefins, are unsaturated hydrocarbons with at least one carbon-to-carbon double bond. Alkenes in general are used to manufacture a variety of polymers such as polyethylene, polyvinylchloride (PVC) and Teflon etc. Alkenes are also an important starting raw material for manufacturing of other organic chemicals. Lower alkenes (i.e. alkenes with lower number of carbon atoms) are commonly used as fuels. Acyclic alkenes with just one C=C double bond have the general formula of CnH2n. 2 NOMENCLATURE Alkenes have the suffix ‘–ene’. Steps in naming alkenes 1. Locate the longest continuous carbon chain containing the C=C bond (parent chain) 2. Any other substituents will be named as prefixes 3. Number the parent carbon chain such that the smallest set of numbers is used to identify the location of the first carbon atom of the C=C bond 4. Arrange substituents in alphabetical order, ignoring prefixes such as di-, tri-, tetra. EXAMPLE 1 alkene IUPAC name H H H H C C C H propene H H H H H H H C C C C C H pent-2-ene H H H H H C H H H H H 4-methylpent-2-ene H C C C C C H H H H cyclopentene H H H H buta-1,3-diene H C C C C H YIJC / JC1 H2 Chemistry / 2024 / Students’ Copy Pg 3 of 33 Alkenes 3 BONDING AND HYBRIDISATION OF CARBON IN ALKENES [Revisit 9A lecture notes, Introduction to Organic Chemistry, Section 5: Hybridisation and shapes of organic molecules.] In alkenes, each carbon atom involved in the C=C bond is sp2 hybridised. hybridised unhybridised to form p orbital three sp2 hybrid orbitals 120o trigonal planar Bonding in ethene  bond formed when the sp2 hybridised orbital of C and the s orbital of H atom overlap head-on (sp2-s overlap). This is a single bond. In ethene, each of the carbon atom are sp2 hybridised. Each carbon atom forms two -bonds with hydrogen atoms through head-on overlap of two of its 2sp2 orbitals with the 1s orbitals of the hydrogen atoms. Another -bond is formed between the two  bond formed when the sp2 hybridised orbitals of  bond formed when carbon atoms when the remaining 2sp2 orbital the unhybridised p each C atom overlap of the each of the carbon atom head-on orbitals of each C head-on overlap with each other. (sp2-sp2 overlap) atom overlap side-way As the three 2sp2 orbitals are arranged in a trigonal planar shape, the three -bonds This is a double bond, as it consists formed also have a trigonal planar of a  bond and a  bond. arrangement and a bond angle of 120°. The remaining unhybridised p orbital of each From the Data Booklet, bond energy for of the two carbon atoms, which is (C=C) is 610 kJ mol–1 while the bond perpendicular to the plane of 2sp2 hybrid energy (C-C) is 350 kJ mol–1. orbitals, overlaps sideways to form a  bond. Note that the C=C bond energy is not Since all the C and H atoms lie on the same double of the C-C bond energy. This plane, the ethene molecule is planar. shows that a  bond is weaker than a  bond due to the less effective side-on overlapping between the p orbitals. YIJC / JC1 H2 Chemistry / 2024 / Students’ Copy Pg 4 of 33 Alkenes 4 ISOMERISM [Revisit 9B lecture notes, Introduction to Organic Chemistry, Section 2: Constitutional (structural) isomerism.] 4.1 Constitutional Isomerism Alkenes can demonstrate chain isomerism, positional isomerism and functional group isomerism. For example, H H H H H H H H H C C C C H H C C C C H positional isomerism H H H H but-1-ene but-2-ene H H H H H CH3 H H C C C C H H C C C H chain isomerism H H H but-1-ene 2-methylpropene H H H H H C C C C H functional group isomerism H H cyclobutane but-1-ene 4.2 Stereoisomerism: Cis-Trans Isomerism [Revisit 9B lecture notes, Introduction to Organic Chemistry, Section 3: Cis-trans isomerism.] For an alkene to exhibit cis-trans isomerism, it must fulfil two criteria: 1. restricted rotation of the carbon-carbon double bond (C=C) due to the presence of the  bonds, and 2. two different groups of atoms are bonded to each of the C atoms in the C=C bond. For example, but-2-ene exhibits cis-trans isomerism but but-1-ene does not. cis-but-2-ene trans-but-2-ene but-1-ene (there are two identical groups bonded to the same C atom involved in the C=C) YIJC / JC1 H2 Chemistry / 2024 / Students’ Copy Pg 5 of 33 Alkenes Since stereoisomers have the same structural formula but differs in the way the atoms are arranged in 3-dimensional space, the isomers must be drawn in a 3-dimensional manner in order to illustrate their differences. For cycloalkenes, [Revisit 9B lecture notes, Introduction to Organic Chemistry, Section 2: Constitutional (structural) isomerism – Extension of Learning.] Smaller ring cycloalkenes are invariably in the cis-arrangement. The trans-arrangement of these smaller ring cycloalkenes gives rise to ring-strain, rendering it unstable. Due to the geometrical constraint about the C=C bond, the smallest cycloalkene to exhibit cis-trans isomerism is cyclooctene. Differences Between Cis-Trans Isomers Cis-trans isomers generally have similar chemical properties – they usually react with the same reagents, but maybe at different rates. Cis-trans isomers have different physical properties (see Section 5). Cis-trans isomers may have different chemical reactivities due to the difference in the proximity of the functional groups. For instance, maleic acid (cis-butenedioic acid) melts at 140 oC and loses water to give maleic anhydride, while the trans isomer, fumaric acid, does not do this at its higher melting point of 300 oC. cis-butenedioic acid (maleic acid) trans-butenedioic acid (fumaric acid) YIJC / JC1 H2 Chemistry / 2024 / Students’ Copy Pg 6 of 33 Alkenes Cis isomers are generally less stable (more reactive) than trans isomers because the two bulky groups on the same side of the double bond in cis isomers cause steric hindrance. steric hindrance cis-but-2-ene trans-but-2-ene YIJC / JC1 H2 Chemistry / 2024 / Students’ Copy Pg 7 of 33 Alkenes 5 PHYSICAL PROPERTIES OF ALKENES 5.1 Melting and Boiling Points At room temperature and atmospheric pressure, the first 3 alkenes (C2 to C4) are gases; the next 11 alkenes (C5 to C15) are liquids while longer alkenes (C16 and longer) are solids. Recall in Chemical Bonding, alkenes have simple molecular structures and are non-polar. Hence, instantaneous dipole – induced dipole attractions exist between the alkene molecules. the difference in melting and boiling points of alkenes is influenced by the strength of the instantaneous dipole-induced dipole attractions between the molecules, which is dependent on: the electron cloud size of the molecule the surface area of contact between the molecules Differences in Physical Properties for Cis-Trans Isomers [Revisit 9B lecture notes, Introduction to Organic Chemistry, Section 3: Cis-trans isomerism.] For alkenes that exhibits cis-trans isomerism, in general, the cis isomer has a (i) higher boiling point but a (ii) lower melting point than the trans isomer. (i) The cis isomer has a higher boiling point than the trans isomer because the cis isomer is polar with permanent dipole – permanent dipole interactions between its molecules while the trans isomer is not polar with instantaneous dipole – induced dipole between its molecules. Hence, a larger amount of energy is required to overcome the stronger permanent dipole – permanent dipole interactions between the cis isomer molecules. An exception: In the case of butenedioic acid, the cis isomer (maleic acid) has a lower boiling point than the trans isomer (fumaric acid). This is because in the cis isomer, the close proximity of the two −COOH groups in the cis isomer allows for intramolecular hydrogen bonding. Hence the cis isomer exhibits less extensive intermolecular hydrogen bonding than the trans isomer. (ii) The cis isomer has a lower melting point than the trans isomer because the shape of the trans isomer allows for more efficient packing (i.e. the molecules can pack closer together) in the solid state, resulting in more molecules per unit volume. Hence, more extensive intermolecular forces of attraction formed. trans-isomer cis-isomer YIJC / JC1 H2 Chemistry / 2024 / Students’ Copy Pg 8 of 33 Alkenes 5.2 Solubility Alkenes, being generally non-polar, are insoluble in polar solvents such as water as they cannot form hydrogen bonds with water. Thus, the energy evolved when instantaneous dipole-induced dipole attractions is formed between the alkene and water molecules is insufficient to overcome the energy required to break instantaneous dipole-induced dipole attractions between the alkane molecules and the hydrogen bonds between the water molecules. However, they are soluble in non-polar solvents like tetrachloromethane, CCl4, and benzene. YIJC / JC1 H2 Chemistry / 2024 / Students’ Copy Pg 9 of 33 Alkenes 6 PREPARATION OF ALKENES 6.1 Elimination of Water, H2O, from Alcohols reagent: method 1: excess concentrated H2SO4 or method 2: Al2O3(s) conditions: heat type of reaction: elimination reaction general equation: elimination of H and OH from adjacent carbon atoms where R1, R2, R3 and R4 can be either a H atom or alkyl or aryl group example: CH3CH2OH → CH2=CH2 + H2O CH3CH(OH)CH2CH3 → CH3CH=CHCH3(major product) + H2O 6.2 Elimination of Hydrogen Halides, HX, from Halogenoalkanes reagent: KOH (or NaOH) in ethanol (ethanolic KOH) conditions: heat type of reaction: elimination reaction general equation: elimination of H and X from adjacent carbon atoms where R1, R2, R3 and R4 can be either a H atom or alkyl or aryl group and X represents a halogen example: CH3CH2Br → CH2=CH2 + HBr CH3CH(Cl)CH2CH3 → CH3CH=CHCH3 (major product) + HCl YIJC / JC1 H2 Chemistry / 2024 / Students’ Copy Pg 10 of 33 Alkenes [Revisit 9A lecture notes, Introduction to organic chemistry, Section 2.3: Degree of Substitution.] How to decide on the structure of the major product? In elimination reactions where multiple alkenes can form, the major product is typically the more stable, more highly substituted alkene. The degree of substitution in alkenes describes how many hydrocarbon groups (or R-groups) are directly bonded to the carbon atoms of the C=C bond. An alkene with more alkyl groups attached to the carbons of the C=C bond is considered more substituted. The prediction of the major product in the elimination reaction by the stability of the alkenes formed is known as the Saytzeff’s Rule. For example, when 2-bromobutane is bubbled into hot ethanolic KOH, three alkenes are formed in different proportions. Note that the trans-isomer is more stable than cis-isomer (hence trans-isomer present in greater quantity than cis-isomer) due to reduced steric hindrance when the R groups are on opposite sides of the double bond. YIJC / JC1 H2 Chemistry / 2024 / Students’ Copy Pg 11 of 33 Alkenes CHECKPOINT 1 Draw the product that is formed when the alkenes are treated with H2(g) in the presence of Pt catalyst at room temperature. reagents and Product Alkene conditions ethanolic KOH, heat excess concentrated H2SO4 YIJC / JC1 H2 Chemistry / 2024 / Students’ Copy Pg 12 of 33 Alkenes 7 CHEMICAL PROPERTIES OF ALKENES Learning outcomes (a) explain the general reactivity of alkenes towards electrophilic reagents / electrophiles (b) describe the chemistry of alkenes as exemplified, where relevant, by the following reactions of ethene: (i) addition of water/steam, hydrogen halides and halogens (ii) reduction via catalytic hydrogenation (iii) oxidation by cold, dilute manganate(VII) ions to form the diol (iv) oxidation by hot, concentrated manganate(VII) ions leading to the rupture of the carbon-to-carbon double bond in order to determine the position of alkene linkages in larger molecules Alkenes, though generally non-polar, are more reactive in comparison to alkanes. Recall in Introduction to Organic Chemistry, you learnt that the functional group of alkenes is the C=C bond, which determines the characteristic reactions that alkenes undergo. The reactions that alkenes undergo are: combustion reduction oxidation electrophilic addition reactions Why do alkenes undergo electrophilic addition? [Revisit 9C lecture notes, Introduction to Organic Chemistry Section 1.1: Types of Organic Species.] Alkenes are attracted to electrophiles (electron-deficient species). This is due to the high electron density of C=C double bond as  electrons being not directly involved in holding the carbon nuclei in place thus alkenes are able to attract electrophiles. Alkenes are unsaturated and thus they undergo addition reactions, in particular electrophilic addition reactions. During the reaction, the weaker  bond is broken instead of the  bond. In place two strong  bonds are formed in the product. YIJC / JC1 H2 Chemistry / 2024 / Students’ Copy Pg 13 of 33 Alkenes 8 REACTIONS OF ALKENES Learning outcomes (d) describe the chemistry of alkenes as exemplified, where relevant, by the following reactions of ethene: (i) addition of water/steam, hydrogen halides and halogens (ii) reduction via catalytic hydrogenation (iii) oxidation by cold, dilute manganate(VII) ions to form the diol (iv) oxidation by hot, concentrated manganate(VII) ions leading to the rupture of the carbon-to-carbon double bond in order to determine the position of alkene linkages in larger molecules 8.1 Combustion Like alkanes, alkenes undergo complete combustion to produce carbon dioxide and water. 3n CnH2n + 2 O2 → nCO2 + nH2O 8.2 Reduction Alkenes are reduced to alkanes by hydrogen. Commercially, the reaction is important in the manufacture of solid or semi-solid fats (margarine) from vegetable oils. This process is called catalytic hydrogenation and is a form of heterogeneous catalysis, where the catalyst and reactants are in different phases. reagent: H2(g) conditions: nickel, Ni, heat or platinum, Pt, at room temperature type of reaction: reduction (addition) general equation: where R1, R2, R3 and R4 can be a H atom or alkyl or aryl group example: CH2=CH2 + H2 → CH3CH3 Other reducing agents such as sodium borohydride, NaBH4, and lithium aluminium hydride, LiAlH4, cannot be used for reduction of alkene. Both of these reducing agents can be considered to be sources of the nucleophilic hydride anion (:H−). Since the C=C of the alkene is electron-rich, it repels the approach of an electron rich nucleophile. YIJC / JC1 H2 Chemistry / 2024 / Students’ Copy Pg 14 of 33 Alkenes 8.3 Oxidation Depending on the condition (temperature), alkenes can undergo mild or strong oxidation when treated with KMnO4 in acidic or alkaline medium. (A) Mild Oxidation When treated with cold (acidic/alkaline) KMnO4, alkenes undergo mild oxidation to form a diol through a partial C=C bond cleavage. reagent: KMnO4 (aq) in dilute H2SO4 (aq) (or in dilute NaOH(aq) ) conditions: cold type of reaction: oxidation general equation: diol where R1, R2, R3 and R4 can be an H atom or alkyl or aryl group example: CH2=CH2 + H2O + [O] → CH2(OH)CH2(OH) observation: in acidic medium, in alkaline medium, decolourisation of purple KMnO4 decolourisation of purple KMnO4 and formation of a brown precipitate of MnO2. (B) Oxidative Cleavage In hot KMnO4, alkenes undergo oxidative cleavage to form ketones and/or carboxylic acids. The C=C double bond completely cleaves to form C=O bond. reagent: KMnO4 (aq) in dilute H2SO4 (aq) (or in dilute NaOH(aq) ) conditions: heat type of reaction: oxidation observation: in acidic medium, in alkaline medium, decolourisation of purple KMnO4 decolourisation of purple KMnO4 and formation of a brown precipitate of MnO2. Potassium dichromate, K2Cr2O7, is a weaker oxidising agent compared to KMnO4. Hence it is unable to oxidise alkenes, be it mild or strong oxidation. YIJC / JC1 H2 Chemistry / 2024 / Students’ Copy Pg 15 of 33 Alkenes Depending on the groups that are bonded to the carbon atoms involved in the C=C bond, different products are formed upon the oxidation of alkenes. when two alkyl or aryl groups are bonded to the carbon atom, a ketone is formed. ketone when one alkyl or aryl group and one hydrogen atom are bonded to the carbon atom, a carboxylic acid is formed. carboxylic acid when two hydrogen atoms are bonded to the carbon atom, carbon dioxide gas and water are formed. The oxidative cleavage reaction is useful in 1. distinguishing between alkenes with and without terminal C=C double bond (C=C bond between first and second carbon atoms of the alkene). For alkenes with terminal C=C double bond, effervescence of CO2 is observed upon oxidation. Bubbling the gas into Ca(OH)2(aq) forms white precipitate. 2. identifying the position of the C=C double bond(s) in the alkene by analysing the structure of the cleavage products. 3. shortening the carbon chain length. For example, CH3CH=CH2 + 5[O] → CH3COOH + CO2 + H2O (3 carbon atoms) (2 carbon atoms) YIJC / JC1 H2 Chemistry / 2024 / Students’ Copy Pg 16 of 33 Alkenes CHECKPOINT 2 Draw the structural formula of the organic products that is formed when the alkenes are treated with KMnO4 in different conditions. reagents and Alkene structural formula of product(s) conditions KMnO4, NaOH(aq) cold KMnO4, H2SO4(aq) heat concentrated acidic KMnO4, heat CHECKPOINT 3 An alkene with the molecular formula of C7H14 was found to yield propanoic acid, CH3CH2COOH, and butanone, CH3COCH2CH3, on oxidation with hot acidified potassium manganate(VII). Suggest the structure of this alkene. YIJC / JC1 H2 Chemistry / 2024 / Students’ Copy Pg 17 of 33 Alkenes 8.4 Electrophilic Addition Learning outcomes (d) describe the mechanism of electrophilic addition in alkenes, using bromine with ethene as an example (e) apply Markovnikov’s rule to the addition of hydrogen halides to unsymmetrical alkenes, and explain the composition of products in terms of the stability of carbocation intermediates A reaction mechanism shows the sequence of reaction steps by which the reaction occurs. A mechanism describes in detail which bonds are broken and formed, and when and how they are broken and formed. Most mechanisms in organic chemistry can be represented by a series of labelled diagrams with possible electron movements in the particles represented by curly arrows. [Revisit 9C lecture notes, Section 2.1 Heterolytic fission, Section 2.2 Homolytic fission, Section 2.3 Curly Arrows – Illustrating Electron Transfer] A half curly arrow, “ ”, is used to denote the movement of a single electron. A full curly arrow, “ ”, is used to denote the movement of a pair of electrons. The  electrons in alkenes attract electrophiles (electron-deficient species) and since alkenes are unsaturated, they undergo addition reactions. The breaking and forming of all the bonds in this mechanism are heterolytic. A heterolytic fission of a covalent bond always result in the formation of a cation and an anion. 8.4.1a Reaction with Halogens, X2, in Organic Solvent reagent: Cl2 or Br2 in CCl4 conditions: room temperature, dark type of reaction: electrophilic addition reaction general equation: where R1, R2, R3 and R4 can be an H atom or alkyl or aryl group example: CH2=CH2 + Br2 → CH2BrCH2Br observation: decolourisation of pale yellow Cl2 or decolourisation orange-red Br2 YIJC / JC1 H2 Chemistry / 2024 / Students’ Copy Pg 18 of 33 Alkenes The reaction is carried out in the dark and room temperature because this is to prevent the alkyl group from undergoing free radical substitution reaction in the presence of light and heat with the halogen (Recall in Alkanes). 8.4.1b Mechanism for Reaction with Halogens, X2, in Organic Solvent Mechanism: Electrophilic Addition Step 1: The alkene  electrons are attracted to Br+ causing a heterolytic fission of the Br-Br bond to form the carbocation intermediate and Br−. Step 2: The Br− is attracted to the C+ to form the product. In Step 1, the electron-rich -bond of alkene will polarise and induce a dipole in an approaching Br2 molecule forming Br+−Br−, by repelling the electrons of Br–Br bond. (a) A pair of electrons flows from the -bond of alkene to the electron deficient bromine atom, Br+. A bond is formed between one of the C atoms and the Br atom. The movement of electrons away from the other C atom results in it being positively charged i.e. a carbocation. (b) The Br2 molecule undergoes heterolytic fission to form a Br– ion. In Step 2, a pair of electrons flows from Br– ion to positively charged C atom in the carbocation, forming a C–Br bond. YIJC / JC1 H2 Chemistry / 2024 / Students’ Copy Pg 19 of 33 Alkenes 8.4.2a Reaction with Aqueous Halogens, X2(aq) reagent: Cl2(aq) or Br2(aq) conditions: room temperature, dark type of reaction: electrophilic addition reaction general equation: for symmetrical alkenes where R can be the same alkyl or aryl group or H atom for unsymmetrical alkenes (Markovnikov’s rule applies) example: CH2=CH2 + Br2(aq) → CH2BrCH2OH (major product) observation: decolourisation of pale yellow Cl2(aq) or decolourisation of orange Br2(aq) The decolourisation of Br2 in CCl4 or Br2 (aq) can be used as a test for the presence of alkenes. 8.4.2b Mechanism for Reaction with Aqueous Halogens, X2(aq) When there is more than one type of nucleophile (i.e. electron-pair donors) present, a mixture of products can be formed in Step 2 of the mechanism. As observed from the electrophilic addition mechanism shown above, a carbocation intermediate is formed in Step 1. YIJC / JC1 H2 Chemistry / 2024 / Students’ Copy Pg 20 of 33 Alkenes There are 2 possibilities in Step 2, either: An electron pair flows from Br– ion to positively charged C atom in the carbocation, forming a C–Br bond. OR An electron pair flows from O atom in H2O to positively charged C atom in the carbocation, forming a C–O bond. This is followed by an electron pair flows from the O–H bond to the positively charged O atom i.e. heterolytic fission of O–H bond to give the bromoalcohol and HBr. In the reaction with aqueous halogens, the bromoalcohol is formed as the major product since water is present in excess since it is used as a solvent. It is also observed that there could be three possible products for unsymmetrical alkenes (See section 9). YIJC / JC1 H2 Chemistry / 2024 / Students’ Copy Pg 21 of 33 Alkenes Markovnikov’s rule applies. Why is that so? Step 1: The alkene  electrons are attracted to X (+) causing a heterolytic fission of the X–X bond to form two possible carbocation intermediates and X−. Step 2: The nucleophile X− as well as OH− are attracted to the carbocation intermediates to form 3 possible products. YIJC / JC1 H2 Chemistry / 2024 / Students’ Copy Pg 22 of 33 Alkenes If the reaction requires you to draw the mechanism for the reaction with aqueous halogens, you may show the mechanism that forms the major product, or as required by the question. CHECKPOINT 4 Ethene is shaken with an aqueous mixture of bromine and sodium chloride. Which of the following compounds will not be formed? A CH2BrCH2OH B CH2BrCH2Cl C CH2BrCH2Br D CH2(OH)CH2Cl YIJC / JC1 H2 Chemistry / 2024 / Students’ Copy Pg 23 of 33 Alkenes When the electrophilic addition is carried out in the presence of other anions, the anions can also act as a nucleophile. For example, when propene reacts with Br2(aq) in the presence of NaCl(aq), three different products will be formed. Step 1: Step 2: There are 3 different nucleophiles (Br−, H2O and Cl−) that can be attracted to the carbocation intermediate. YIJC / JC1 H2 Chemistry / 2024 / Students’ Copy Pg 24 of 33 Alkenes 8.4.3a Reaction with Hydrogen Halides, HX reagent: HX(g) where X = Cl, Br or I conditions: room temperature type of reaction: electrophilic addition reaction general equation: for symmetrical alkenes where R can be the same alkyl or aryl group or H atom for unsymmetrical alkenes (Markovnikov’s rule applies) where R can be an alkyl or aryl group example: CH2=CH2 + HBr → CH3CH2Br CH2=CHCH3 + HBr → CH3CHBrCH3 (major product) HCl(g) and HBr(g) can also be generated in-situ, by reacting NaCl(s) or NaBr(s) with concentrated H2SO4. NaX + H2SO4 → NaHSO4 + HX (where X is Cl or Br) The order of reactivity of HX is HI > HBr > HCl. The weaker the H-X bond is, the greater the reactivity with alkenes. YIJC / JC1 H2 Chemistry / 2024 / Students’ Copy Pg 25 of 33 Alkenes CHECKPOINT 5 Draw the mechanism to show the reaction between ethene and HBr(g) to form bromoethane. Show all charges and relevant lone pairs and show the movement of the electron pairs by using curly arrows. 8.4.3b Mechanism for Reaction with Hydrogen Halides, HX Step 1: The alkene  electrons are attracted to H (+) causing a heterolytic fission of the H-Br bond to form the carbocation intermediate and Br−. Step 2: The Br− is attracted to the carbocation to form the product. For unsymmetrical alkenes, apply Markovnikov’s rule (See section 9), Step 1: The alkene  electrons are attracted to H (+) causing a heterolytic fission of the H-Br bond to form two possible carbocation intermediates and Br−. YIJC / JC1 H2 Chemistry / 2024 / Students’ Copy Pg 26 of 33 Alkenes Step 2: The nucleophile Br− is attracted to the carbocation intermediates to form 2 possible products. Since the secondary carbocation is more stable than the primary carbocation as it has more electron-donating alkyl groups, the positive charge on the carbon atom is more dispersed. It will be formed in greater proportion and subsequently forming the major product. YIJC / JC1 H2 Chemistry / 2024 / Students’ Copy Pg 27 of 33 Alkenes 8.4.4a Reaction with Water, H2O reagents and laboratory method industrial method conditions: (i) concentrated H2SO4, cold H2O(g), H3PO4 (catalyst), high (ii) H2O(l), heat temperature and high pressure type of reaction: electrophilic addition reaction general equation: for symmetrical alkenes where R can be the same alkyl or aryl group or H atom for unsymmetrical alkenes (Markovnikov’s rule applies) example: CH2=CH2 + H2O → CH3CH2OH CH2=CHCH3 + H2O → CH3CH(OH)CH3 (major product) Electrophilic addition reaction may result in the formation of a racemic mixture that does not exhibit optical activity. This is because the carbocation intermediate formed after Step 1 is trigonal planar and thus the nucleophile can be attracted to the C+ from either side of the plane with equal probabilities. If the carbon become chiral, the resulting mixture is racemic, as equal amount of the 2 enantiomers are formed, like in the reaction of propene with hydrogen bromide. The optical activity of each of the enantiomers exactly cancels out each other, and hence the product does not rotate plane polarised light. YIJC / JC1 H2 Chemistry / 2024 / Students’ Copy Pg 28 of 33 Alkenes Step 2: When drawing the mechanism, only 1 arrow has to be shown in Step 2. Key points to include when describing electrophilic addition mechanism name of mechanism: electrophilic addition construct balanced equations for each elementary step with the following details o use of full curly arrows to denote movement of electron pairs o show dipoles (+ and - charges) whenever appropriate o show lone pair of electrons o indicate the rate-determining step/ slow step o draw the correct structural formula of the intermediate and final product. YIJC / JC1 H2 Chemistry / 2024 / Students’ Copy Pg 29 of 33 Alkenes 9 MARKOVNIKOV’S RULE Learning outcome (f) apply Markovnikov’s rule to the addition of hydrogen halides to unsymmetrical alkenes, and explain the composition of products in terms of the stability of carbocation intermediates This idea of forming major and minor products in the electrophilic addition of alkenes is summarised in Markovnikov’s rule. Markovnikov’s rule states that, when a molecule HA (where HA can be H-Cl, H-Br, H-I, H-OH) is added to a C=C bond in an unsymmetrical alkene, the H atom is added to the carbon atom which has the greater number of H atoms resulting in the major product. The basis of Markovnikov’s rule is the stability of the carbocations. A carbocation is a positively charged ion, in which the carbon atom is bearing the positive charge. where R can be an H atom, alkyl or aryl group 1. Carbocations can be classified as primary (1o), secondary (2o) or tertiary (3o) carbocations based on the number of alkyl (R) groups bonded to the carbon bearing the positive charge. 2. The driving force of a reaction is the formation of a stable product. Hence, a more stable carbocation will be formed faster since its formation is favoured over a less stable carbocation. The stability of a carbocation, and its rate of formation, increases in the following order: YIJC / JC1 H2 Chemistry / 2024 / Students’ Copy Pg 30 of 33 Alkenes Reasons for the Difference in Stability of the Carbocations Recall in Chapter 9C Introduction to Organic Chemistry Section 1.2 (B): Electronic effects, we learnt that the alkyl group is known to be an electron-donating group (EDG). Therefore, the electron-deficiency on the carbon atom bearing the positive charge (C+) would be dispersed by a greater extent when it is bonded to more of these alkyl groups. The more electron-donating alkyl groups bonded to the carbon, the positive charge would more effectively dispersed. Thus, the more stable is the carbocation. On the other hand, substituents such as halogens, –COOH or –NO2 groups are electron- withdrawing groups (EWGs) which intensify the positive charge on the carbon and destabilise the carbocation. If G is an electron-donating group (e.g. alkyl If G is an electron-withdrawing group (e.g. group), the positive charge on the carbon is halogens), the positive charge on the carbon dispersed, hence the carbocation is is intensified, hence the carbocation is stabilised. destabilised. EDG: electron-donating group EWG: electron-withdrawing group CHECKPOINT 6 Draw the skeletal formula of the major products that are formed when 3-methylpent-2-ene reacts with hydrogen chloride. YIJC / JC1 H2 Chemistry / 2024 / Students’ Copy Pg 31 of 33 Alkenes CHECKPOINT 7 Draw the displayed formula of the major product formed when 4-methylpenta-1,3-diene is reacted with an excess of (a) bromine in hexane; (b) bromine water; (c) hydrogen bromide; and (d) steam and phosphoric acid at high temperature and high pressure. (a) (b) (c) (d) YIJC / JC1 H2 Chemistry / 2024 / Students’ Copy Pg 32 of 33 Alkenes 10 DISTINGUISHING TESTS FOR ALKENES Distinguishing tests are simple chemical tests used to distinguish different organic compounds based on the different functional group they contain and are commonly done in a test-tube. These tests involved reactions that give the following observations: (i) colour change of the reaction mixture; (ii) formation of precipitate; and/or (iii) gas produced. To distinguish alkenes from the other functional groups: reagents and Br2(l) or Br2 in CCl4 or Br2(aq) conditions: room temperature, dark observation: alkenes other functional groups decolourisation of reddish no decolourisation of reddish brown Br2(l); brown Br2(l); decolourisation orange-red Br2 no decolourisation orange-red in CCl4; Br2 in CCl4; decolourisation orange Br2(aq) no decolourisation orange Br2(aq) reagents and KMnO4 in dilute H2SO4 (or dilute NaOH) conditions: cold observation: alkenes other functional groups decolourisation of purple KMnO4 no decolourisation of purple (acidic) KMnO4 decolourisation of purple KMnO4 and formation of a brown precipitate of MnO2 (alkaline) To identify alkenes with terminal C=C bond (=CH2): reagents and KMnO4 in dilute H2SO4 conditions: heat observation: alkenes with =CH2 alkenes without =CH2 decolourisation of purple decolourisation of purple KMnO4 with effervescence. KMnO4 Gas evolved gives a white precipitate with limewater. CHEM~IS~TRY YIJC / JC1 H2 Chemistry / 2024 / Students’ Copy Pg 33 of 33