CHM 102 Organic Chemistry I Lecture Notes PDF

FEDERAL UNIVERSITY DUTSE FACULTY OF SCIENCE DEPARTMENT OF CHEMISTRY CHM 102 ORGANIC CHEMISTRY I Lecture Notes Prepared by Dr. S. Habibu ([email protected]) Second Semester, 2019/2020 Session Federal University Dutse, CHM 102 Organic Chemistry I, 2019/...

FEDERAL UNIVERSITY DUTSE FACULTY OF SCIENCE DEPARTMENT OF CHEMISTRY CHM 102 ORGANIC CHEMISTRY I Lecture Notes Prepared by Dr. S. Habibu ([email protected]) Second Semester, 2019/2020 Session Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu Couse Outline  Historical survey of the development and importance of organic chemistry.  Nomenclature and classes of organic compounds, homologous series, functional groups.  Electronic theory in organic chemistry.  Saturated hydrocarbons and unsaturated hydrocarbons.  Isolation and purification of organic compounds,  Qualitative and quantitative organic chemistry, determination of structure of organic compounds. Couse Tutors Group Venue Programs Lecturers Name A GST Auditorium Chemistry, Botany & Zoology 1. Dr. M. N. DuruminIya 2. Abdulmalik Shehu B Student Center Main Hall Biology & Biochemistry Dr. M. B. Idris C Faculty of Science Main Hall Microbiology & Biotechnology Dr. S. Habibu D Student Center Hall-1 Physics and EMT Dr. I. S. Shina E Student Center Hall-2 Mathematics and Computing Dr. Hajarah Momoh F Faculty of Medicine Faculty of Medicine Dr. S. Nasir G Faculty of Agric Faculty of Agric Group A T. Abdulrahman H Faculty of Agric Faculty of Agric Group B S. L. Enesi Course Coordinator : Dr. S. Habibu |Page2 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu Relevant Textbooks * McMurry J. (2008) Organic chemistry (7th Edn) Brooks/Cole, Thomson Learning, Inc., * Solomons, T. G., & Fryhle, C. (2010). Organic chemistry (10th Edn). John Wiley & Sons, Inc. * Klein, D. R. (2012). Organic Chemistry as a Second Language (Vol. 2): John Wiley & Sons. * Hart, H., Hadad, C. M., Craine, L. E., & Hart, D. J. (2011). Organic chemistry: a short course: (13th Edn). Cengage Learning, USA. |Page3 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu DEVELOPMENT OF THE SCIENCE OF ORGANIC CHEMISTRY In the early nineteenth century, scientists classified all known compounds into two categories: organic and inorganic. Organic compounds were derived from living organisms (plants and animals), while inorganic compounds were derived from non-living sources (minerals and gases). This distinction was fuelled by the observation that organic compounds seemed to possess different properties than inorganic compounds. Organic compounds were often difficult to isolate and purify, and upon heating, they decomposed more readily than inorganic compounds. To explain these curious observations, many scientists subscribed to a belief that compounds obtained from living sources possessed a special “vital force” that inorganic compounds lacked. This notion, called vitalism, stipulated that it should be impossible to convert inorganic compounds into organic compounds without the introduction of an outside vital force. This idea discouraged chemists from trying to make organic compounds in the laboratory. But in 1828, the German chemist Friedrich Wöhler, then 28 years old, accidentally prepared urea, a well-known constituent of urine, by heating the inorganic substance ammonium cyanate. i.e., This experiment and others like it gradually discredited the vital-force theory and opened the way for modern synthetic organic chemistry. Over the decades that followed, other examples were found, and the concept of vitalism was gradually rejected. The downfall of vitalism shattered the original distinction between organic and inorganic compounds, and a new definition emerged. Specifically, organic compounds became defined as those compounds containing carbon atoms, while inorganic compounds generally were defined as those compounds lacking carbon atoms. Carbon is special due to its electronic structure and the position it occupy in the periodic table. As a group 4A element, carbon can share four valence electrons and form four strong covalent bonds. Furthermore, carbon atoms can bond to one another, forming long chains and rings. Carbon, alone of all elements, is able to form an immense diversity of compounds, from the simple to the staggeringly complex – from methane, with one carbon at om, to DNA, which can have more than 100 hundred million carbons. Despite the demise of vitalism in science, the word “organic” is still used today by some people to mean “coming from living organisms” as in the terms “organic vitamins” and “organic fertilizers.” The commonly used term “organic food” means that the food was |Page4 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu grown without the use of synthetic fertilizers and pesticides. An “organic vitamin” means to these people that the vitamin was isolated from a natural source and not synthesized by a chemist. In science today, the study of compounds from living organisms is called natural products chemistry. Importance of organic chemistry Organic chemistry occupies a central role in the world around us, as we are surrounded by organic compounds. The food that we eat and the clothes that we wear contain organic compounds. Our ability to smell odours or see colours results from the behaviour of organic compounds. The major constituents of living matter—proteins, carbohydrates, lipids (fats), nucleic acids (DNA and RNA), cell membranes, enzymes, hormones—are organic. The responses of our bodies to pharmaceuticals are the results of reactions guided by the principles of organic chemistry. A deep understanding of those principles enables the design of new drugs that fight disease and improve the overall quality of life and longevity. The growth of living things from microbes to elephants rests on organic reactions, and organic reactions provide the energy that drives our muscles and our thought processes. Other organic substances include the gasoline, oil, and tires for our cars; the clothing we wear; the wood for our furniture; the paper for our books; and plastic containers, camera film, perfume, carpeting, and fabrics. Pharmaceuticals, pesticides, paints, and adhesives are all made from organic compounds. Hardly a minute goes by when we’re not using something made of organic molecules, such as a pen, a computer keyboard, a music player, or a cellular phone. We view display screens made of organic liquid crystal arrays. In fact, there is not a single aspect of our lives that is not in some way dependent on organic chemistry. Hence, studying organic chemistry is a fascinating undertaking. |Page5 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu Drawing Chemical Structure Students should be able to know the different ways of presenting chemical structures of compound such as:  Lewis structure  Kékule structure  Condensed structure  Line-angle formula (skeletal structure). Exercises: 1. Carvone, a substance responsible for the odour of spearmint, has the following structure. Tell how many hydrogens are bonded to each carbon and give the molecular formula of carvone. 2. Tell how many hydrogens are bonded to each carbon in the following compounds, and give the molecular formula of each substance: |Page6 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu NOMENCLATURE AND CLASSIFICATION OF ORGANIC COMPOUNDS There are two general ways of naming organic compounds: i. Trivial system ii. IUPAC system i). Trivial System of nomenclature: In trivial system, compounds derived their names from the source in which they are obtained or from the characteristic property of the compound. Examples: Table 1 Trivial system of Nomenclature Structure Name Source Formic acid Isolated from ants and named after the Latin word for ant, formica Urea Isolated from Urine Morphine A painkiller named after the Greek God of dreams, Morpheus Barbituric acid Named in honor of a woman named Barbara ii). IUPAC System: This is the most scientific and universally accepted system of nomenclature. The basic philosophy in the IUPAC (the International Union of Pure and Applied Chemistry) system is to name compounds as derivative of unbranched hydrocarbons. ALKANES The first four (n=1-4) unbranched chain saturated hydrocarbons are called methane, ethane, propane and butane. Higher members of the series are named from the Greek words |Page7 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu indicating the number of the carbon atoms present in the molecule, plus the suffix "-ane". The first twelve members are given in the following Table. Table 2. The names of the first 12 unbranched saturated hydrocarbons n Name Molecular Formula Constitutional Formula (CnH2n+2) 1 Methane CH4 CH4 2 Ethane C2H6 CH3CH3 3 Propane C3H8 CH3CH2CH3 4 Butane C4H10 CH3CH2CH2CH3 5 Pentane C5H12 CH3CH2CH2CH2CH3 6 Hexane C6H14 CH3CH2CH2CH2CH2CH3 7 Heptane C7H16 CH3CH2CH2CH2CH2CH2CH3 8 Octane C8H18 CH3CH2CH2CH2CH2CH2CH2CH3 9 Nonane C9H20 CH3CH2CH2CH2CH2CH2CH2CH2CH3 10 Decane C10H22 CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3 11 Undecane C11H24 CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2CH3 12 Dodecane C12H26 CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH3 IUPAC RULES FOR NAMING ORGANIC COMPOUNDS 1. Select the longest continuous chain of carbon atoms, the parent name is derived from it. 2. Groups attached to the main chain are called substituents, each substituent is given a name and a number to indicate the nature and the position of that substituent, respectively. Examples of substituents: (i) Alkyl groups: The group derived from one of the alkanes in Table 2 by the removal of a terminal hydrogen is called an alkyl group. The group name is found by removing "ane" from the alkane name and adding "yl". CH4 Methane becomes –CH3 Methyl CH3CH3 Ethane becomes –CH2CH3 Ethyl CH3CH2CH3 Propane becomes –CH2CH2CH3 Propyl CH3CH2CH2CH3 Butane becomes –CH2CH2CH2CH3 Butyl |Page8 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu If there are different alkyl groups attached to the parent chain, preference in numbering is given to the end bearing small alkyl group (see below). (ii) Halogens as substituents are called halo- e.g –Cl (chloro), –Br (bromo), –Fl (Fluoro), –I (Iodo) etc. 3. The parent chain is numbered from the side that gives the lowest position to the substituent. If there are more than one substituent, the parent chain is numbered from the side that gives the lowest combination of numbers. 4. When two or more identical groups are attached to the main chain, prefixes such as di-, tri-, tetra-, etc are used to indicate the number of occurrences of that substituent. The above compound is correctly named 2,3-dimethylpentane. The name tells us that there are two methyl substituents, one attached to carbon-2 and one attached to carbon-3 of a five-carbon saturated chain. 5. If two or more different types of substituents are present, they are listed alphabetically, except that prefixes such as di- and tri- are not considered when alphabetizing. B comes before C, hence, –Br should be given the lowest number. 6. Punctuation is important when writing IUPAC names. IUPAC names for hydrocarbons are written as one word. Numbers are separated from each other by commas and are separated from letters by hyphens. There is no space between the last named substituent and the name of the parent alkane that follows it. |Page9 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu Examples 1. 2. Give the IUPAC name of the following compounds: 3. Give the IUPAC name of the following compounds: CYCLOALKANES The prefix cyclo- is attached to the name. Examples Cyclopropane Cyclobutane Cyclopentane Cyclohexane Cycloheptane Cyclooctane Exercises 1. Give the IUPAC names of the following compounds: a. b. c. Br CH3 CH3 | P a g e 10 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu d. e. f. Br Br Br CH CH Br Cl Br 2. Provide the structural formulas for the following compounds: a. 2-methylhexane b. 2-bromo-4-methyloctane c. 4-ethyl-2,2-dimethylheptane d. 2-chlorobutane e. 1,1,3,3-tetrachlorocyclopropane 3. Write a structure for each of the compounds listed. Explain why the name given here is incorrect and give a correct name in each case. a. 2,3-fluoropropane b. 1-methylbutane c. 2-ethylbutane d. 1-methyl-2-ethylcyclopropane e. 1,1,3-trimethylhexane f. 4-bromo-3-methylbutane g. 1,3-dimethylcyclopropane ALKENES  Names end with – ene  Irrespective of the substituents the longest continuous chain must be chosen to include the maximum number of double bonds.  The double bonds must be numbered such that their sum is the lowest possible. | P a g e 11 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu Examples 1. Ehene Propene 2-methylpropene b. c. d. 2.. a. But-1-ene 2,5-Dimethylhex-2-ene 3-propylhex-1,4-diene 5-Methyl-hept-3-ene Exercises: Provide the structure of the following compounds a. 4,4-difluoro-3-methylbut-1-ene b. 1,1-difluoro-2-methylbuta-1,3-diene c. 5-methylcyclopenta-1,3-diene CYCLOALKENES Cycloalkanes are named by adding the prefix -cyclo and indicating the position of the double bond as well as the substituents where applicable. Examples: a. b. c. CH3 CH3 Br Br Cyclohexene CH3 CH3 2,5-Dimethyl-cyclohexa-1,3- 1,5-Dibromo-3,6-dimethyl-cyclohexa-1,3-diene diene | P a g e 12 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu ALKYNES  Names end with – yne Examples a. Propyne b. 4,4-dimethylpent-2-yne c. 4-ethyl-4-methylhept-5-ene-2-yne d. Pent-3-ene-1-yne e. 1-chlorobut-2-yne OTHER FUNCTIONAL GROUPS The parent compound is designated according to the following order of priority: Carboxylic Acids > Aldehydes > Ketones > Alcohols > Amines > Ethers > Hydrocarbons CARBOXYLIC ACIDS O  Has the general formula: R COOH or R C OH or RCO2H  Numbering is done such that carbon number 1 is a carbon to which the carboxylic acid group is attached.  – e in alkane is replaced by –oic acid for acyclic compounds.  –carboxylic acid is added (without removing – e) for cyclic compouds. Examples a. CH3CH2CH2CO2H b. c. Br COOH Butanoic acid HO O 3-bromo-2-methylbutanoic acid Cyclobutane carboxylic acid | P a g e 13 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu d. e. f. CO2H CO2H CH3 3-methylcyclohex-1-ene carboxylic acid cyclopent-3-ene carboxylic acid 2,2-dimethylpent-3-eneoic acid Where there is more than one carboxylic acid – di, – tri, etc are added before oic acid (for acyclic) or carboxylic acid (for cyclic). COOH Br COOH 3-bromocyclobutane-1,2-dicarboxylic acid Exercise Name the following: a. b. HO2C-CH=CH-CH2-CO2H CO2H Br ALDEHYDES O Aldehydes have the general form: R C H or RCHO Where R = alkyl group.  Numbering is done such that carbon number 1 is a carbon to which the aldehyde functional group is attached.  When R is acyclic, remove the ending e for alkane, alkene and add al. E.g. | P a g e 14 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu O H C H Methanal O H3C C H Ethanal O H3CH2C C H Propanal H O H2C C C H Prop-2-enal H C C CHO Prop-2-ynal  When R is cyclic, add carbaldehyde. E.g. O C H Cyclohexane carbaldehyde O H3C C H Name this compound. Br  Aldehydes occur as substituents are called: methanoyl or formyl. Example: O H H O H C C C C OH 3-methanoylpropanoic acid H H Or 3-formylpropanoic acid CHO 2,3-dimethanoylpentanoic acid CH3CH2 CH CH COOH CHO | P a g e 15 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu KETONES O Ketones have the general form: R C R' or RCOR’ Where R and R’ are alkyl groups.  – e in alkane is replaced by –one O H3C C CH3 Propanone O Butanone  Remember to indicate the position of ketonic functional group from five carbon atoms above.  Also, numbering is done such that carbon the carbon bearing the ketonic functional group is given a least position. O Pentan-3-one O 2,5-Dimethyl-hexan-3-one  When R is cyclic, it is regarded as a substituent: Example: O 1-Cyclopropyl-butan-1-one Br O 3-bromo-1-cyclohexyl-butan-1-one Br Br O 3,5-Dibromo-1-cyclopentyl-hexan-1-one  When both R’s are cyclic it is named as cycloalkanone. | P a g e 16 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu Example: O O Br Cyclohexanone 3-bromocyclopentanone  Ketones as substituents are called -oxo. Example: O O O Br O O HO OH C OH O O O 3-oxopentanoic acid 2,3-dioxohexanoic acid 4-Bromo-2,3-dioxo-heptanoic acid ALCOHOLS  Have the general form: R OH (where R = alkyl group)  Replace the ending -e by ol.  Numbering is done such that carbon number 1 is a carbon to which the alcahol functional group is attached. Example: OH OH CH3CH2OH HO CH3OH Ethanol Methanol Propan-2-ol But-3-ene-2-ol But-3-yne-2-ol Polyhydric alcohols  They are named by indicating the position of the -OH group and adding the prefix -di, -tri, -tetra, etc. for 2, 3, and 4 -OH’s respectively. Example: HO Ethan-1,2-diol OH OH Propan-1,2,3-triol HO OH | P a g e 17 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu  Alcohols as substituents are called hydroxy. Example: O OH 2-hydroxypropanoic acid HO THIOLS  They have the general form: R SH(where R is alkyl group)  Naming is done by adding thiol to name of alkanes without removing the ending -e. Example: CH3-SH Methanethiol CH3CH2-SH Ethanethiol HS Propan-2-thiol SH But-3-ene-2-thiol Cyclopentanethiol HS SH Propane-1,2,3-trithiol HS SH  Thiols as substituents are called mercapto. Example: O 3-mercaptopropanoic acid HS OH O SH O 3-mercaptopentan-1,5-dioic acid HO OH | P a g e 18 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu OH Name this compound HS COOH AMINES  Amines have the general form: R NH2 (where R = alkyl group).  Naming is done by replacing the -e in alkane with amine or  by taking the alkane as a substituent (i.e. alkyl) and adding amine to become akylamine. Example: CH3-NH2 Methanamine (or methyl amine) CH3CH2-NH2 Ethanamine (or ethyl amine) CH2=CH-CH2-NH2 Prop-2-enamine NH2 Cyclohexyl amine Classification of Amines  when the nitrogen atom is bonded to two H -atoms it is called primary amine,  when one of the H atoms in the amine functional group is replaced by an R – group, it is called secondary amine (or 2° amine), and  when both the two H atoms of the amine functional group are replaced by an R – group, it is called tertiary amine (or 3° amine). i.e. Primary amines (1° Secondary amines (2° amines) Tertiary amines (3° amines) amines) R NH2 R' R' NH2 R N H R N R'' H3C C H e.g. CH3 1- e.g. H3C N H dimethyl-amine e.g. N methylethanamine (or N-methyl methylamine) Dimethyl-propyl-amine or or (N-methyl methanamine) (N, N-dimethyl-propyl-amine) | P a g e 19 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu  amines as substituents are called amino. H2N O 2-aminopropanoic acid OH NH2 HO OH 2,3-diaminopentan-1,5-dioic acid O NH2 O CO2H 3-aminocyclohexane carboxylic acid NH2 O Cl CO2H Name this compound NH2 ETHERS R O R' (where R, R’ = alkyl group)  when R = R’, they are named as dialkyl ether e.g. CH3 O CH3 or (CH3)2 O Dimethyl ether Diethyl ether O or (CH3CH2)2 O  when R ǂ R’, they are named as alkyl alkyl ether e.g. Ethyl-propyl ether (or Ethoxy-propane) O NB: remember alphabetical order e before m  Ethers as substituents are called alkoxy e.g. O O OH 3-methoxypropanoic acid | P a g e 20 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu Exercises 1. Give the IUPAC names of the following compounds: a. (C2H5)2-CH-CH3 b. CH3-C(CH3)2-CH(CH3)-CH3 c. C2H5-CH(CH3)-CH2-CH(CH3)2 d. (CH3)2-CH-CH2-CH2-CH(CH3)2 e. CH3-CH2-CHClC(CH3)2-CHBrCH3 f. (CH3)2CHC=CH-CH2CH3 g. (CH3)2CHC=CH-CH2CH3 h. HCC-CH-CH2-CH3 2. Write the structural formula of the following compounds: a. 2,3-dimethylbutane b. 2,2,4-trimethylpentane c. 2-chloro-2,3-dimethylpentane d. 2,3-dibromo-1-chloro-3-methylhexane e. 2-chloro-2,3-dimethylbutane f. 3-propylpent-1,3-diene g. Oct-2-ene-5-yne h. hex-1,3-diene-5-yne i. 3-ethylpent-2-chloro-1,4-diene 3. Provide the IUPAC names of the following compounds. b. a. O OH d. NO2 CN c. OH OH e. CO2H OH f. CO2H OH HOOC HOOC O O | P a g e 21 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu CLASSIFICATION OF ORGANIC COMPOUNDS Organic compounds can be broadly classified as acyclic (open chain) or cyclic (closed chain). 1. Acyclic or open chain compounds: These are compounds also known as aliphatic compounds, they have branched or straight chains. Examples: CH3 CH3CH3 HO CH3 C CH3 H O O Ethane Acetaldehyde Isobutane Acetic acid 2. Alicyclic or closed chain or ring compounds: These are cyclic compounds which contain carbon atoms connected to each other in a ring (homocyclic). When atoms other than carbon are also present then it is called as heterocyclic. They exhibit some properties similar to aliphatic compounds. Examples: O Cyclopropane Cyclohexane Cyclohexene Tetrahydrofuran 3. Aromatic compounds They are compounds containing benzene and other ring related compounds. Similar to alicyclic, they can also have heteroatoms in the ring. Such compounds are called as heterocyclic aromatic compounds. Examples: N S Benzene Anthracene Pyridine Thiophene (Aromatic) (Heterocyclic aromatic) Homologous Series and Functional Groups Organic compounds can also be classified into families according to their structural features and the members of a given family often have similar chemical behavior. The structural features that make it possible to classify compounds in to families are called functional groups. | P a g e 22 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu A functional group is a group of atoms in a compound that is responsible for the characteristic chemical behavior in every molecule where it occurs. The table below contains the structures of some common functional groups: Name General Structure Example Alkene (double bond) H H C C C C H H Ethene Alkyne (triple bond) C C H C C H Ethyne Arene (aromatic ring) Benzene Alcohol R OH CH3 OH Methanol (where R = alkyl group) Ether R O R' CH3 O CH3 (where R, R’ = alkyl group) Dimethyl ether O Ester O R C O R' (where R, R’ = alkyl group) H3C C O CH3 Methylethanoate Amine R NH2 H 3C NH2 (where R = alkyl group) Methylamine Nitrile R C N H 3C C N (where R = alkyl group) Ethanenitrile Nitro R NO 2 H3C NO 2 (where R = alkyl group) Nitromethane Thiol R SH H3 C SH (where R = alkyl group) Methanethiol O O Aldehyde R C H (where R = alkyl group) H3C C H Ethanal O O Ketone R C R' H3C C CH3 (where R = alkyl group) Propanone | P a g e 23 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu O O Carboxylic acid H3C C OH R C OH (where R = alkyl group) Ethanoic acid O O Acid halide R C X H3C C Cl (where X = F, Cl, Br, etc.) Ethanoyl chloride O O Amide H3C C NH2 R C NH2 (where R = alkyl group) Ethanamide O O O O Anhydride R C O O C R' H3 C C O O C CH3 (where R, R’ = alkyl group) Ethanoic anhydride Homologous series is a series of organic compounds with the same functional group in which each member differ from its adjacent member by a fixed unit (CH2). Any two successive members of a homologous series are called homologs. General Properties of homologous series i). All members have the same functional groups and hence, possess similar chemical properties. ii). Members differ from adjacent members by CH2. iii). All members can be represented by a general formula e.g. alkanes (CnH2n+2), alcohol (CnH2n+1 OH) etc. iv). Different members of the series can be prepared using the same general method of preparation. v) As the molecular weight increases, members of a homologous series show gradual increase in physical properties such as melting point, boiling point, etc. Exercise Identify the functional groups in each of the following molecules: a. Methionine, an amino acid b. Ibuprofen, a pain reliever. | P a g e 24 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu c. Capsaicin, the pungent substance in chili peppers: ELECTRONIC CONCEPT IN ORGANIC CHEMISTRY Atomic Orbitals and Electron Configuration  An orbital is a region of space where there is a high probability of finding an electron.  All s orbitals are spheres. The relative energies of atomic orbitals in the first and second principal shells are as follows:  Electrons in 1s orbitals have the lowest energy because they are closest to the positive nucleus.  Electrons in 2s orbitals are next lowest in energy.  Electrons of the three 2p orbitals have equal but higher energy than the 2s orbital.  Orbitals of equal energy (such as the three 2p orbitals) are called degenerate orbitals  An atomic orbital represents the region of space where one or two electrons of an isolated atom are likely to be found.  molecular orbital (MO) represents the region of space where one or two electrons  of a molecule are likely to be found.  When atomic orbitals combine to form molecular orbitals, the number of molecular  orbitals that result always equals the number of atomic orbitals that combine. HYBRIDIZATION In order for atoms to combine to form molecules the process has to be thermodynamically favorable. The energy must be sufficiently to cause an unpairing of electrons, and to allow atom to exhibit its most common valency which depends on the number of unpaired electrons. Atoms which have more unpaired electrons after acquiring an energy are said to be in an excited state. According to quantum mechanics, the electronic configuration of a carbon atom (6 electrons) in its lowest energy state—called the ground state—is that given as follows: | P a g e 25 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu 2 p2 px py pz Energy 2 S2 1 S2 Ground state electronic configuration of carbon Carbon forms four covalent bonds. The valence electrons of carbon (those used in bonding) are those of the outer level, that is, the 2s and 2p electrons. However, these electrons are at different energy levels. Hence to achieve bond formation, these electrons must be at the same energy level. Therefore, an electron has to move from s orbital to p orbital (excitation). That is: The four unpaired electrons occupy four different orbitals which are unequal in energy and other characteristics. In order to form four bonds which are equal in every respect, the four orbitals have to undergo hybridization. Hybridization, therefore, is the mixing of two or more unequal orbitals to form two or more new equal hybrid orbitals. SP3 HYBRIDIZATION The ground state electron configuration of carbon cannot satisfactorily describe the bonding structure of methane (CH4), in which the carbon atom has four separate C – H bonds, because the electron configuration shows only two atomic orbitals capable of forming bonds (each of these orbitals has one unpaired electron). This would imply that the carbon atom will form only two bonds, but we know that it forms four bonds. We can solve this problem by imagining an excited state of carbon in which a 2s electron has been promoted to a higher energy 2p orbital. | P a g e 26 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu 2p 2p px py pz px py pz Excitation Energy 2S 2S 1S 1S Excitation of an electron in C atom Now the carbon atom has four atomic orbitals capable of forming bonds, however, these orbitals have different geometries. Therefore, the excited state electron configuration cannot explain the observed tetrahedral geometry and the angle of 109.5° in methane. To solve this problem the orbitals, have to undergo hybridization. That is: 2p px py pz four degenerate sp3 orbitals hybridize Energy 2s 1s 1s sp3 hybridization This procedure gives us four orbitals that were produced by averaging one s orbital and three p orbitals, and therefore we refer to these atomic orbitals as sp3-hybridized orbitals. Thus, sp3 hybridization results in to: i. Formation of four (sp3– hybrid) orbitals of equal energy and shape. ii. The four orbitals spread as widely as possible and are therefore directed towards the corners of a regular tetrahedron with an angle of 109.5 °. | P a g e 27 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu sp3 109.5o sp3 sp3 sp3 Tetrahedral arrangement in sp3 orbitals This is the most stable configuration. The orbitals are oriented in such a way that electrons are as far apart as possible, hence repulsive forces are reduced to minimum. SP2 HYBRIDIZATION Depending on the compound to be formed, the carbon atom may have three of its unpaired orbitals hybridized. un hybridized 2p orbital px py pz 2pz three degenerate sp2 hybridize orbitals Energy 2s 1s 1s sp2 hybridization Sp2 hybridization produces: i. Three sp2 hybrid orbitals of the same energy and shape ii. one pz unhybridized (un affected) orbital. The three sp2 hybrid orbitals assume the shape of trigonal planar with an angle of 120 °. sp2 120o sp2 sp2 Trigonal planar arrangement in sp2 orbitals | P a g e 28 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu SP HYBRIDIZATION Sometimes, the carbon can have only two of its unpaired electrons hybridized. un hybridized 2p orbitals (py & pz) px py pz 2py 2pz hybridize Energy two degenerate sp 2s orbitals 1s 1s sp hybridization Sp – hybridization results in: i. two sp – hybrid orbitals of equal energy and shape. ii. two unhybridized p – orbitals of ( py and pz). The two sp orbitals will assume a linear shape (angle of 180 °). 180o sp sp Linear arrangement in sp orbitals In the formation of covalent bonds, electrons are shared between atoms and these atoms must approach sufficiently closely enough for the orbitals of one atom to overlap with the orbital of the other. Since each orbital can only contain a maximum of two electrons, therefore, an orbital of one atom containing the two electrons is used and these two bonding electrons are shared between the two atoms. MOLECULAR ORBITALS The overlap of two atomic orbitals results in the formation of two molecular orbitals. That is: i. bonding molecular orbital (lower in energy), and ii. anti-bonding molecular orbital (higher in energy). At this level, only the bonding orbital will be discussed. The greater the degree of overlapping between two atomic orbitals the stronger the bond formed. | P a g e 29 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu Type of bonds a. Sigma bond (δ - bond): – This is formed as a result of overlapping of two orbitals to form a single bond. There are there possible ways in which s and p orbitals may overlap. 1. Two s orbitals overlapping. 2. Two p orbitals overlapping. 3. One s and one p orbitals overlapping together. All the above three cases result in δ – bond (single bond) formation and the bonding electrons are likely to be located near an imaginary line connecting the two nuclei of the atoms involved. b. pi bond (π - bond): This is formed as a result of lateral overlapping of two parallel p – orbitals of adjacent atoms. pi - bond The degree of overlapping in π – bond is less than in δ – bond and the bonding electrons are located away from the imaginary line joining the two nuclei. Hence, π – bonds are weaker than δ – bonds. | P a g e 30 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu Illustration of the bond types 1. Carbon – carbon single bonds: Examples:- a. Methane (CH4) 1s2 2s1 2p3 C (exited state) sp3 H H H H 1s 1s 1s 1s H H 109.5o sp3 H H 1s H H H H In each case there is an overlapping between s – orbital of H – atom and sp3 hybrid orbital of the central carbon atom. b. Ethane (CH3 – CH3) Each of the two carbon atoms in ethane undergoes sp3 hybridization and will therefore have four sp3 hybrid orbitals of equal energy and shape. These orbitals will arranged themselves in tetrahedral manner. | P a g e 31 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu H H H H H 109.5o sp3 H H C C sp3 109.5o H H H H H In the case of ethane above, there are : i. overlapping between s – orbitals of six H – atoms and six sp3 hybrid orbitals of two C – atoms. 2. Carbon – carbon double bonds: e.g. ethene (CH2 = CH2). In carbon – carbon double bods the hybridization is sp2. 1s2 2s1 2p3 C (exited state) sp2 unhybridized i. each carbon atom has three sp2 hybrid orbitals of equal energy and shape. qii. each of the three sp2 hybrid orbitals will arrange themselves in trigonal planar. iii. Each carbon will have one unhybridized pz orbital. The two unhybridized pz orbitals of the two carbon atoms will lie parallel to each other and will thus overlap laterally to form an additional π – bond.  - bond  - bond H H H sp 2 H sp2 sp2 C sp2 sp2 C C C H sp2 H H H In this case there is: i. overlap between two sp2 – hybrid orbitals of the two carbon atoms. | P a g e 32 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu ii. overlap between s – orbitals of four hydrogen atoms and four sp2 – hybrid orbitals of the two carbon atoms. iii. lateral overlapping between two unhybridized pz orbitals to form π – bond. 3. Carbon – carbon triple bonds: e.g. ethyne (HC CH ). In the formation of carbon – carbon triple bod the hybridization is sp. 1s2 2s1 2p3 C (exited state) sp2 unhybridized Here, each carbon will have: i. Two sp hybrid orbitals of equal energy and shape, arranged in linear passion. ii. two unhybridized (py and pz) orbitals lying parallel to each other, which can overlap laterally to form additional π – bonds.   -bond sp sp C sp C sp H C C H   -bond There is: i. overlap between two sp – hybrid orbitals of the two carbon atoms. ii. overlap between s – orbitals of two hydrogen atoms and two sp – hybrid orbitals of the two carbon atoms. iii. lateral overlapping between four unhybridized py and pz orbitals to form two π – bonds. | P a g e 33 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu Summary sp3 H H H2 CH3 sp3 sp2 sp2 C C2 C3 C i. H3C sp2 C sp C sp C sp H H sp CH3 sp 3 H sp sp ii. H3C 3 C C CH2 iii. H3C CH3 sp H H3C CH3 sp3 Questions 1a.Define the term hybridization. b. Fill in the blank spaces in the table below: Type of carbon – carbon bond CH3 – CH3 CH2 = CH2 HC CH Hybridization Average bond angle Geometry 2a. show with an example each of the orbitals involved in the formation of π and δ – bonds. b. Explain why the C – H of ethyne is more acidic than ethene or ethane. 3. Indicate the type of hybridization of each carbon atom in the following compound: H2 C H2C C CH 4. Give the hybridization state of each of the carbon atoms (a – e ) in the compound below: e CH2 b a c d | P a g e 34 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu SATURATED AND UNSATURATED HYDROCARBONS Hydrocarbons A hydrocarbon is an organic compound made of only carbon and hydrogen atoms joined by covalent bonds. It is possible for double or triple bonds to form between carbon atoms and even for structures, such as rings, to be formed. Classification Hydrocarbons are classified into two major classes namely saturated and unsaturated hydrocarbons. Saturated hydrocarbons are those that contain only single covalent bonds and have as many hydrogen atoms as possible attached to every carbon. They are also known as aliphatic hydrocarbons. Unsaturated hydrocarbons are those that contain double or triple covalent bonds between adjacent carbon atoms. There are three types of unsaturated hydrocarbons: alkenes, alkynes and aromatic hydrocarbons. Alkenes have one or more carbon-carbon double bond. Alkynes have one or more carbon-carbon triple bond, and aromatic hydrocarbons have a benzene or aromatic ring in their structure. ALKANES Alkanes are saturated hydrocarbons having carbon – carbon single bonds in their structures. All the carbon atoms are sp3 – hybridized. They have a general formula of CnH2n+2 where n is the number of carbon atoms. Structural isomerism in alkanes This is a process by which two or more compounds have the same molecular formula but different structural formula due to different ways in which the atoms are linked together. Methane (CH4), ethane (C2H6) and (C3H8) have no structural isomers. Isomerism in alkanes start from butane (C4H10) which has two isomers. That is, (C4H10) Butane i. CH3CH2CH2CH3 n -butane ii. CH3 2-methylpropane (isobutane) CH H3C CH3 Pentane (C5H12) has three isomers which are n-pentane, 2-methylbutane and 2,2- dimethylpropane. | P a g e 35 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu CLASSIFICATION OF CARBON ATOMS i. primary carbon atom (1°): carbon atom in which there is only one other carbon atom attached to it. e.g. C CH2 CH3 primary C - atom ii. secondary carbon atom (2°): carbon atom in which there are two other carbon atoms attached to it. e.g. C CH2 CH3 secondary C - atom iii. Tertiary carbon atom (3°): carbon atom in which there are three other carbon atoms attached to it. e.g. C C CH CH3 Tertiary C - atom PHYSICAL PROPERTIES OF ALKANES For the straight chain alkanes at room temperature, C1 – C4 are gases, C5 – C17 are liquids, while higher members are all solids. Boiling points The boiling points of alkanes increases with an increase in molecular mass. This is because of attractive intermolecular forces (Van der Waals forces), the bigger the alkane molecule, the greater the attractive intermolecular forces between the alkane molecules and therefore are more difficult to separate into the gas phase. Branching on the other hand, lowers the boiling points. | P a g e 36 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu Melting points The trends for boiling points in alkanes are very similar for melting points as well, the larger the molecule, the higher the melting point. However, odd-numbered alkanes (odd number of carbons) have lower melting points than even-numbered alkanes. This is because the even- numbered alkanes pack closer together more easily, making it more difficult to break down the solid material. Density The density of alkanes increases with increase in molecular mass. However, all alkanes are less dense than water. Solubility Alkanes are almost completely insoluble in water. They are, however, soluble in organic solvents like benzene and ether. Liquid alkanes tend to be good solvents themselves for other organic molecules. CHEMICAL PROPERTIES (CHEMICAL REACTIONS) OF ALKANES Alkanes are not very reactive because they are stable i.e. all the C–C and C–H bonds are very strong and not easily broken to allow the alkane molecule to undergo a chemical change. However, some of the reactions that alkanes undergo include combustion, halogenation, nitration, and cracking. Combustion This is a rapid oxidation of alkanes at high temperatures in the presence of excess oxygen, to produce carbon dioxide, water and energy. The general equation for the combustion of alkanes is: CnH2n+2 + Excess O2 nCO2 + (n + 1)H2O + Energy Example: CH4 + 2O2 CO2 + 2H2O + Energy Halogenation Alkanes undergo halogenation in the presence of light or heat (250 – 400 °C), yielding a mixture of halogenated products and hydrogen halides. Example: CH4 + Cl2 CH3Cl + HCl Nitration Alkanes react with nitric acid vapor at high temperature to yield nitroalkanes. i.e.  R-H + HNO3 R-NO2 + H2O e.g. | P a g e 37 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu  CH3-H + HNO3 CH3-NO2 + H2 O PREPARATION OF ALKANES 1. From CO and H2 Alkanes are obtained when CO is hydrogenated over suitable metal catalyst. The general equation is: nCO + (2n + 1)H2 cat. CnH2n+2 + nH O 2 e.g. CO pd + 3H2 CH4 + H2O 2. From CO2 and H2 Alkanes are also obtained when CO2 is hydrogenated over suitable metal catalyst. The general equation is: nCO2 + (3n + 1)H2 cat. CnH2n+2 + 2nH2O e.g. pd 5CO2 + 16H2 C5H12 + 10H2O 3. By Hydrogenation (Reduction of Alkenes and Alkynes) Alkenes or alkynes can be hydrogenated in the presence of metal catalyst to produce the corresponding alkanes. Examples: CH2 CH2 + H2 pt CH3-CH3 CH + 2H2 pt CH3-CH3 CH 4. From Alkyl halides and Grignard reagent Alkyl halides react with Grignard reagent to form alkanes. Grignard reagent has the following general formula: R – MgX and is generally called alkyl magnesium halide. Examples: CH3CH2Br + CH3CH2MgBr CH3CH2CH2CH3 + MgBr2 CH3Cl + CH3CH2MgCl CH3CH2CH3 + MgCl2 5. Crude oil as a main source of hydrocarbons Crude oil (petroleum) acts as the key source of hydrocarbon (alkanes). The crude oil is separated into its various fractions, each containing a mixture of alkanes, by fractional distillation in an oil refinery. Below is a table showing the products obtained from a typical petroleum fractional distillation. Name of fraction No. of C atom Boiling point range Natural gas C1 – C5 < 40 °C Petrol (gasoline) C5 – C10 40 °C – 175 °C | P a g e 38 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu Kerosine (paraffin) C10 – C14 175 °C – 225 °C Diesel oil C14 – C18 > 275 °C Lubricating oil, waxes, e.t.c. > C18 – Asphalt, bitumen > C40 – PETROLEUM CRACKING Fractional distillation of crude oil gives a lot of lubricating oil but not enough petrol to meet the increasing demand. To meet the required yield of petrol (C5 – C10), molecule with large numbers of atoms are broken down to smaller ones. Again, the petrol obtained from fractional distillation gives large proportion of straight – chain hydrocarbons which do not burn so smoothly as the branched – chain. Definition: Cracking is the breakdown of large molecules in fuel oils in to smaller ones needed in petrol. Types of cracking i. Thermal cracking: this is carried out at high temperature (of about 700 °C) and a high pressure of about 30 atm. ii. Catalytic cracking: occurs at about 500 °C and atmospheric pressure. This process uses a mixture of silicon(iv)oxide and aluminum as catalyst. It is easier to control and is more preferred to thermal. Examples: C16H34 C8H18 + C8H16 (diesel oil) C8H18 C6H14 + C2H4 (kerosine) (petrol) (ethene) Generally, Alkane Smaller alkane + Smaller alkene Uses of cracking i. Highly demanded petrol can be obtained from larger, less demanded petroleum fractions. ii. Branched – chain alkanes, which burn more smoothly in an internal combustion engine, can be obtained from straight – chain alkanes. iii. The bye – products of cracking can be used as raw materials in Petro-chemicals industries. For example, ethene (and other alkenes) is polymerized to give useful polymer. Also, H2 when evolved is trapped and can be used in the manufacture of ammonia (Harber’s process) or in the process of hardening fat to margarine. | P a g e 39 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu ALKENES (OLEFINS)  They form a homologous series of unsaturated hydrocarbons containing one or more C – C double bonds.  Have general formula CnH2n.  The unsaturated carbon atoms are sp2 hybridized.  There is a presence of π – bond in addition to δ – bonds. PHYSICAL PROPERTIES OF ALKENES  They are virtually insoluble in water.  They are soluble in non – polar solvents e.g. benzene, however, unsymmetrical alkenes are slightly polar due to electron – donating properties of the alkyl groups.  From ethene to butene are gases. Structural isomerism in alkenes Ethene and propene have no structural isomers. Isomerism start from butene (C4H8). (C4H8) Butene i. CH3CH2CH=CH2 But-1-ene ii. CH3CH=CHCH3 But-2-ene ii. CH3 2-methylpropene H3C C CH2 (i) and (ii) above are called positional isomers because they differ only in the position of double bond. (iii) is a branched chain isomer. Geometrical isomerism in alkenes Restriction of free rotation about C – C double bond creates the phenomenon of geometrical isomerism. An alkene having a formula: R – CH=CH – R can have two isomers depending on the position of the two alkyl groups. R R H R C C C C H H R H cis - isomer trans - isomer R - on the same side R - on opposite sides Examples: H3C CH3 H CH3 C C C C H H H3 C H cis - but-2-ene trans - but-2-ene | P a g e 40 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu PREPARATION OF ALKENES 1. By dehydration (removal of water) of alcohols - H2O C C C C OH H alkene alcohol The reagent used for the dehydration is concentrated H2SO4 and the ease of dehydration of alcohol is in the following order: 3° > 2° > 1°. Example: H H H H conc. H2SO4 + H2O H C C H H C C H 170 °C H OH ethene ethanol 2. By dehydrohalogenation (removal hydrogen halide) of haloalkanes C C + HX C C H X alkene haloalkane The reagent used for the dehydrohalogenation is a strong base called “alcohol potash” which is a combination of ethanol and potassium hydroxide (C2H5OH/KOH) and the ease of dehydrohalogenation is in the following order: 3° > 2° > 1°. Example: H H H H C2H5OH/KOH + HCl C C H H C C H H  H Cl ethene chloroethane Formation of mixed products (Saytzeff orientation) Weather in the dehydration of alcohols or dehydrohalogenation of haloalkanes, mixtures of alkenes are formed when an unsymmetrical compound is used. In the mixtures of the alkene formed is alkene which is always at higher yield and is called “major product” while the other at a lower yield is called “minor product”. Examples: | P a g e 41 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu H H H H H H H H H H H H conc. H2SO4 H C C C C H H C C C C H + H C C C C H  H H OH H H H H H 81% (major product) 19 % (minor product) but-2-ene but-1-ene Reason: stability of alkenes increases with substitution. i.e. R R R R H R H R H H R C C R >H C C R > H C C R >H C C > H C C H H The more stable the alkene, the more readily it is formed. Question: Give the complete equation for the dehydrohalogenation of 2-bromopentane, showing the major and the minor product formed. 3. By dehalogenation of vicinal dihalides H H H H Zn(dust) + ZnBr2 H C C H H C C H Br Br ethene 4. Petroleum cracking Already discussed under alkanes. Chemical properties (Chemical reactions) of alkenes Alkenes are very reactive compounds compared to alkanes. They undergo addition reactions rather than substitution. Their double bond containing π – electrons actas a source of electrons, therefore functioning as a base. Hence, their reactions are described as electrophilic additions. 1. Halogenation ( addition of halogens) C C + X2 C C X X This act as the best method for preparing vicinal di halides. Example: H H H H CCl4 H C C H + Br2 H C C H Br Br 2. Addition of hydrogen halide (HX) (hydrohalogenation) Alkenes react with hydrogen halides (HX) to form alkyl halides. i.e. | P a g e 42 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu C C + HX C C (HX = HCl, HBr, HF) H X Example: H H H H H C C H + HBr H C C H H Br Addition of HX across unsymmetrical double bonds (Markovnikov orientation) When a haloacid (HX) is added across unsymmetrical double bond, there are two possibilities of forming two different haloalkanes. Out of these two possibilities only one is found to form practically. For example, H H H H3C C CH2 + HBr H3C C C Br (Not formed) H H 1-bromopropane But, H H H H3C C CH2 + HBr H3C C C H (formed) Br H 2-bromopropane “Markovnikov rule” when a halo acid(hydrogen halide) is added across unsymmetrical double bond, the hydrogen from the acid becomes attached to the carbon atom bearing the higher number of hydrogens while the halogen goes to the carbon atom having the smaller number of hydrogens. Reason: The reaction proceeds via the formation of carbonium ion in which the stability decreases in the following order: 3° > 2° > 1°. More examples: H H Cl i. + HCl H CH3 CH3 Br ii. + HBr | P a g e 43 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu Addition of HX across unsymmetrical double bonds in the presence of peroxides e.g. H2O2 (Anti Markovnikov orientation) When a haloacid (HX) is added across unsymmetrical double bond, in the presence of peroxides, the hydrogen becomes attached to the carbon atom bearing the smaller number of hydrogen atom(s) while the halogen goes to the carbon atom bearing the higher number of hydrogens. Example: H H H HBr H3C C CH2 H3C C C Br H2O2 H H 1-bromopropane Also, H H H HCl i. H2O2 Cl CH3 CH3 HBr ii. H2O2 H Br Question: The addition of hydrogen halides to unsymmetrical aliphatic unsaturated hydrocarbons proceeds according to Markovnikov’s rule. (a). State the rule, (b) using a named aliphatic unsaturated hydrocarbon, give the name and structure of the Markovnikov’s and anti- markovnikov’s products, in its reaction with hydrogen bromide. 3. Oxidation reactions a. Oxidation with KMnO4 (hydroxylation): alkene readily decolorizes alkaline solution of purple KMnO4 producing diol (glycol). i.e. aq. KMnO4/OH C C C C OH OH b. Oxidation with Osmium tetra oxide (OsO4): Also, diol is produced. i. aq. OsO4/Et2O C C C C ii. aq. NaSO2 OH OH | P a g e 44 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu c. Oxidation with ozone (Ozonolysis): H H O O i. O3/CCl4 R C C R' R C H + H C R' ii. Zn dust/H2O Examples: H H O O i. O3/CCl4 i. H3C C C H H3 C C H + H C H ii. Zn dust/H2O O O ii. i. O3/CCl4 ii. Zn dust/H2O + H ketone aldehyde 4. Hydrogenation Alkenes are hydrogenated by passing them over a finely divided Pt, Pd, or Ni catalyst in a hydrogen atmosphere to give alkanes. i.e. Pt, Pd, or Ni C C + H2 C C H H H H H H Pt, Pd, or Ni Example: H C C H + H2 H C C H H H 5. Polymerization The process by which many simple molecules (monomers) join together to form very large family of long chain is known as polymerization. The long chain molecules are called polymers. Example: n H2C CH2 CH2 CH2 CH2 CH2 n Ethene (monomer) Polyethene (polymer) ALKYNES  They contain C – C triple bonds.  Have a general formula of CnH2n-2  The unsaturated carbon atoms are sp – hybridized and are attached to each other by one δ – bond and two π – bonds. PHYSICAL PROPERTIES OF ALKYNES  C2 – C4 are gases.  C5 – C8 are liquids. | P a g e 45  Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu  Boiling points increase with increase in relative molecular mass.  They are insoluble in water but soluble in non-polar organic solvents. PREPARATION OF ALKYNES 1. By dehydrohalogenation of vicinal alkyl halides H H H C2H5OH/KOH NONH2 C C C C C C alkyne X X X Chemical Properties (reactions) of alkynes 1. Acetylide formation Terminal alkynes (i.e. alkynes having triple bonds at the end of the chain) reacts with: (i). Cu+ ions in ammoniacal copper (I) chloride solution. (ii). Ag+ ions in AgNO3 to form insoluble heavy metals dicarbides (acetylides). C H 2Cu+ NH3. H2O H C + Cu C C Cu + 2H+ R.T red ppt. copper(I)dicarbide (copper(I)acetylide) This reaction is used to test for terminal triple bond because, alkynes of the type R C C R do not react. 2. Formation of sodium acetylide (Ethyl sodium) Ethyne reacts with sodium metal in liquid ammonia to form ethyl sodium (sodium acetylide). C C H + 2Na NH3 (aq) 2 H 2H C C Na + H2 ethylsodium (sodium acetylide) 3. Hydration of Ethyne Ethyne is hydrated by bubbling the gas through dilute H2SO4 at 60 °C in the presence of mercury(II)sulphate as a catalyst. That is, H dil. H2SO4 C H H C C H H C C H H C HgSO4 / 60°C H OH H O alcohol aldehyde | P a g e 46 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu An einol alcohol is formed which is unstable and therefore rearranges itself to form the more stable aldehyde. The process is in dynamic equilibrium but the equilibrium lying very much towards the right – hand – side. This phenomenon is called “tautomerism”. 4. Addition of HCN This reaction takes place under pressure and in the presence of barium cyanide as catalyst. Example: H H Ba(CN)2 H C C H + HCN C C C N H propenenitrile 5. Hydrogenation Hydrogenation takes place in the presence of metal catalysts such as Pd, Pt, or Ni. Example: H H H H H2 C H Pt H C C H H C C H H C + H2 Pt H H 6. Polymerization (conversion to benzene) Ethyne polymerizes to benzene when passed through a heated tube containing complex organo – nickel catalyst. That is, C H heated tube, 60-70°C 3 H C complex Ni - cat. benzene TEST FOR HYDROCARBONS 1. Alkenes and alkynes decolorize alkaline solution of purple potassium tetraoxomanganese (vii) (KMnO4) while alkanes do not. 2. Alkenes and alkynes decolorize bromine water (a solution of bromine in water) while alkanes do not. 3. Alkenes and alkynes decolorize Br2 / CCl4 while alkanes do not. The above three tests are called “test for unsaturation” and are used to test the presence of double or triple bonds in hydrocarbons or other organic functional groups. Hence, any of the test can be used to distinguish between alkanes on one hand, and alkenes and alkynes on the | P a g e 47 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu other hand. However, these reactions can not differentiate between alkenes and alkynes since both undergo the same reactions with the reagent. 4. Ethyne (and other alkynes with terminal triple bonds) form a reddish – brown precipitate with a solution of copper chloride in liquid ammonia, while alkanes and alkenes do not. That is, 2 H C C H + 2 Cu(NH3)2 Cl 2 Cu C C Cu + 2HCl + 4NH3 The above reaction can be used to distinguish between alkenes and alkynes. | P a g e 48 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu ISOLATION AND PURIFICATION OF ORGANIC COMPOUNDS Before investigating the structure and properties of an organic compound, the compound must be pure. Common methods of purification and identification of organic compounds include: 1. Melting and boiling points determination. 2. Crystallization and recrystallization. 3. Distillation: a. Fractional distillation, b. Steam distillation, c. Vacuum distillation. 4. Liquid extraction. 5. Sublimation. 6. Chromatographic separations: a. Paper chromatography, b. Thin – layer chromatography. Determination of melting point A known compound may usually be identified (or characterized) by determining its melting point or boiling point (if liquid). The melting point of a compound is defined by physical chemist as the temperature at which the solid and liquid phases are at equilibrium. The common method of melting point determination used in organic chemistry, is a range of temperature over which a small amount of the solid in a thin – walled capillary tube first visibly soften and finally completely liquifies. For the sake of simplicity, such melting point stages are called melting points. The melting point of an organic compound if very nearly pure will be small (0.5 – 1.0 °C) and the substance is said to melt sharply. The presence of impurities even in minute quantities, usually depresses, or lowers the melting point and widens the range. Thus, the melting point is not only useful as a means of identification but also as a criterion for purity. Depression of the melting point of a pure compound by an impurity is also useful as an aid in identification. Consider for example, a substance A thought to be identical with a compound B because A and B have the same melting points. If a small amount of A is mixed with small amount of B and the melting point is measured (mixture melting point), and the mixture melting point is the same as the melting point of B, then A and B are probably identical. However, if the mixture melting point is depressed below the melting point of B (and the range broadens), then A acts as an impurity to B and the two compounds cannot be identical. | P a g e 49 Federal University Dutse, CHM 102 Organic Chemistry I, 2019/2020 Session, Dr. S. Habibu Crystallization and Recrystallization Organic compounds (and many other inorganic compounds) are more soluble in hot solvents than in cold. An impure solid organic compound when dissolve in a proper amount (usually small amount) of an appropriate solvent at an elevated temperature, will precipitate after cooling. If the hot solution is filtered, before being allowed to cool, dirt and other insoluble impurities will be removed from the solution. The liquid filtrate containing the pure solid is then allowed to cool and fine crystals of the solid will form. The leftover liquid after recovery of the pure solid is called “mother liquor