CHM121 COURSE COMPACT

Transcription

1 Course CHM 121 General Organic Chemistry (2 UNITS) Compulsory CHM121 COURSE COMPACT Course Duration Two hours per week for 15 weeks (30hours) Lecturer Data Name of the lecturer: Oluyori Abimbola Peter B.Sc, M.Sc., Ph.D (in view); Dr. Mrs Edewor-Kuponiyi Ph.D Department:of Physical Sciences College of Science and Engineering Office Location: A 132; 1 st College Building Consultation Hours: Mondays and Thursdays 11am-3pm Course Content : Introduction to and importance of Organic Chemistry. Qualitative analysis of organic compounds. Isolation and purification of organic compounds. Quantitative analysis of organic compounds. Determination of structure of organic compounds: Empirical, Molecular and Structural formulae. Hybridization: formation of sp 3, sp 2 and sp hybrid orbitals in carbon. Homologous series and functional groups. Isomerism: structural and stereoisomerism. Aliphatic hydrocarbon chemistry: alkanes alkenes and alkynes- IUPAC Nomenclature, physical properties, preparation and chemical reactions with simple mechanisms where applicable. Course Description Illustration below: This chemistry course teaches the introduction to organic chemistry and natural products. Some important terms are defined with examples. The chemistry of hydrocarbons will be discussed with emphasis on the chemistry of alkanes, alkenes and alkynes. The IUPAC nomenclature of organic compounds is also treated. Course Justification Illustration below: This aspect of chemistry exposes students to qualitative and quantitative organic chemistry, as well as well as the chemistry of hydrocarbons. The overall content of this course forms a prerequisite for important chemistry and biochemistry courses at higher levels.

2 Course objectives At the end of this course, students would: (i) Possess strong foundation in Organic Chemistry which forms the base for future building (ii) be equiped with the skills required for basic chemical transformations in synthetic organic chemistry Course Requirement Illustration below: To derive maximum benefits from this course and for quick understanding of the concepts, it is required that students be familiar with the fundamentals of Organic chemistry.. Method of Grading- An example below S/N Grading Score (%) 1. Test Test Assignment Final Examination 70 Total 100 Course Delivery Strategies: Lecturing method complimented with demonstrations will be adopted for efficient course delivery. LECTURE CONTENT Week 1: Introduction to and importance of Organic Chemistry/ Qualitative analysis of organic compounds Objectives At the end of the lectures for this week, the students should be able to state the importance of organic chemistry in the society carry out simple qualitative tests to determine the kind of compounds present in an unknown substance Description First hour: Organic chemistry is the is the aspect of chemistry which studies carbon and its special compounds excluding the carbon (iv) oxide, carbon (II) oxide, trioxocarbonates, and the

3 hydrogentrioxocarbonates(iv). Carbon is able to form these compounds because it 1. has the ability to cartenate (cartenation) 2. has the ability to form carbon-carbon single, double and triple covalent bonds 3. has the ability to combine with other non-metals such as nitrogen, sulphur, oxygen... The important Institutions/Organizations where organic chemistry has proved useful include but is no limited to 1. Classrooms in secondary schools and universities 2. University research laboratories 3. Pharmaceutical research laboratories/pharmaceutical Industries 4. Agrochemical Industries 5. The petroleum industry 6. Environmental Protection Agencies 7. Water treatment plants 8. Textile Industries 9. Food Industries etc. Second hour Qualitative analysis of organic compounds: Tests for phenols, reducing sugar, ketones, proteins, starch, carboxylic acids etc. Refer to CHM 129 Practical Manual for details. Study Questions: 1. Alcohols are insoluble in Lucas reagent. True or false? 2. When carboxyllic acid react with aqueous NaHCO 3, What would you observe? 3. Briefly describe the chemistry behind the silver mirror test. Reading List 3. Practical Chemistry for Schools and Colleges by G.O Ojokuku ISBN Week 2: Isolation and purification of organic compounds Objectives At the end of the lectures for this week, the students should be able to describe the principle behind the isolation of organic compounds from mixtures carry out simple purification techniques

4 Description First hour: Organic compounds are not usually found pure in nature. They are often found mixed with other chemical compounds. One important technique by which such separation is effected is called Chromatography. Thin Layer Chromatography TLC is an analytical technique which involves the use of a stationary phase and a mobile phase in an enclosed container/ the chromatographic tank. It can be used to help determine the number of components in a mixture, the identity of compounds, and the purity of a compound. Its ability to monitor the appearance of a product or the disappearance of a reactant makes it useful in the monitoring of the progress of a reaction. It takes about 5 minutes to complete a TLC procedure. The diagram below is illustrative. After identifying the number of spots by TLC, Preparative thin layer chromatography, a bigger version of TLC, is then used to isolate the various compounds. For more complex mixtures, the column chromatography may be employed before PTLC is used to purify the fractions. More sophisticated chromatographic techniques include the Gas Chromatography (GC), High Performance Liquid Chromatography (HPLC) - Observe pictures Second hour Recrystallization Recrystallization a process which is used to remove impurities from a crystalline material. The procedure involves dissolving the material in an amount of solvent that will produce a saturated solution at a temperature close to the boiling point of the

5 solvent. Insoluble impurities will not dissolve and so they are removed by the filtration (gravity) of the hot solution and the purified compound crystallizes as the filtrate cools. Suction filtration is used to isolate the purified crystals from the solution. A suitable solvent for recrystallization is one in which the desired compound is very soluble in the solvent at its boiling point, but is much less soluble in the solvent at room or icewater temperatures so that the compound comes out of solution as the mixture cools. The steps involved in recrystallization may be defined as follows: 1. Select the solvent. 2. Dissolve the material in the hot solvent. 3. Filter the solution if necessary. 4. Allow crystallization to take place. 5. Collect the crystals. 6. Wash the crystals. 7. Dry the crystals. Other methods which are also employed in purifying organic compounds are simple/fractional distillation separatory funnel filtration... Study Questions: 1. How many phases are usually involved in TLC? 2. Any solvent can be used for recrystallization. True or false? 3. If you follow the basic procedure for recrystallization and crystals do not appear, what steps would you take? 4. What is the significance of the R f value? Reading List Week 3: Determination of structure of organic compounds/ Empirical, Molecular and Structural formulae Objectives At the end of the lectures for this week, the students should be able to state the various spectroscopic techniques which are used in structural elucidation answer questions involving empirical, molecular and structural formulae

6 Description First hour: The structure of pure organic isolates can be elucidated by employing different combinations of the following: Infrared spectroscopy(ir)-functional group determination Gas Chromatography-Mass Spectroscopy (GC-MS)- Isolation and Molecular mass/structural pattern determination. Nuclear Magnetic Resonance Spectroscopy (NMR)- identification of the number and kinds of proton. UV-Visible Spectroscopy- Detection of Conjugation X-ray Crystallography- Determination of the arrangement of atoms/molecules in a crystal Activity: View pictures of each instrument on the internet Second hour The Empirical formula is the simplest formula of a compound e.g CH 2 The molecular formula shows the actual number of the different kind of atoms present i a compound. e.g the molecular formula of the compound which has the empirical formula CH 2 above may be C 6 H 12 The Structural formula is more detailed. It shows how the atoms in a molecule are connected in space. E.g the compound which has the molecular formula C 6 H 12 above may be cyclohexane, hex-1-ene, hex-2-ene, or hex-3-ene. The structural formula thus answers this question. cyclohexane Hex-1-ene Hex-2-ene Hex-3-ene Study Questions: 1. When g of an unknown carbohydrate was subjected to combustion analysis with excess oxygen, it produced g of CO 2 and g of H 2 O. What is the molecular formula? [MW of the unknown carbohydrate is g/mol] g sample of hydrocarbon undergoes complete combustion to produce 4.40g of CO 2 and 2.70g of H 2 O. What if the empirical formula of this compound? Reading List Week 4: Hybridization

7 Objectives At the end of the lectures for this week, the students should be able to Apply the concept of hybridization to saturated and unsaturated hydrocarbons Define the pie and the sigma bond Description First hour: SP 3 Hybridization Hybridization While trying to explain the covalent bonding in more complicated in organic molecules with tetravalent carbon atoms e.g Methane, the need for the concept of hybridization (as explained by Linus Pauling, 1931) becomes obvious. As we know, carbon has four valence electrons (2s 2 2p 2 ) and forms four bonds. Because carbon uses two kinds of orbitals for bonding, 2s and 2p, we might expect methane to have two kinds of C-H bonds. However, all four C-H bonds in methane are identical and are spatially oriented toward the corners of a regular tetrahedron. Ground state C: 1s 2 2s 2 2p 2 Excited State C: 1s 2 2s 1 2p 3 Hybridization of one S and three P orbitals Hybrid orbitals: 1s 2 (SP 3 SP 3 SP 3 SP 3 ) Even if the excited state electronic configuration is used to form bonds at this stage, three of the bonds will be identical while the fourth will be different and this is not the case. Experiments show that all the four bonds are identical. What then is the answer? Hybridization!

8 Figure 3.1: SP 3 Hybridization Second hour SP 2 Hybridization This occurs when there is a double bond between two carbon atoms. As the name implies, the sp2 hybrid orbitals are formed from one S orbital and two P orbitals. The third P orbital is left unhybridized. Ground state C: 1s 2 2s 2 2p 2 Excited State C: 1s 2 2s 1 2p 3 Hybridization of one S and two P orbitals Three SP 2 hybrid orbitals and one unhybridized P orbital The hybridized orbitals overlap head to head to form a very strong covalent bond called the sigma bond (δ-bond) while the unhybridized p orbitals overlap side to side and forms the pie bond (π-bond) as shown in the following figure.

9 Figure 3.2: The formation of SP 2 orbitals in Ethene Study Questions: 1. Draw a suitable diagram to show the formation of the SP hybrid orbitals in ethyne. 2. State angle between two SP hybrid orbitals?. Reading List Week 5: Terminologies used in organic Chemistry Objectives At the end of the lectures for this week, the students should be able to define Homologous series Functional groups Alkyl group isomerism Description First hour: 1. Homologous series: A homologous series is a family of organic compounds which correspond to the same general molecular formula, have the same functional group and each member differ from the one preceeding it by a CH2 group. The first four members of four homologous series are shown in the following table

10 No. of C ALKANES ALKENES ALKYNES ALKANOL C1 Methane Methanol C2 Ethane Ethene Ethyne Ethanol C3 Propane Propene Propyne Propan-1-ol/Propan-2- ol C4 Butane But-1-ene/But-2- But-1-yne/But-2- Butan -1-ol... ene yne C5 Pent-1-ene... Pent-1-yne... Some homologous series and their fist four members 2. Functional groups: A functional group is a bond, an atom or a group of atoms which is common to a homologous series and is responsible for their chemical properties. Some functional groups are shown in the following table. Table 3.2: Functional Groups Second hour 3. Alkyl groups: An alkyl group is formed when one hydrogen atom is removed from a molecule of an alkane. They follow a general molecular formula C n H 2n+1 and they are usually attached e.g to a functional group. They are named by replacing

11 the ane ending in the name of the corresponding alkane with yl. Some examples are shown in the following table. Alkane Alkyl Methane (CH 4 ) Methyl (CH 3 ) Ethane (C 2 H 6 ) Ethyl (C 2 H 5 ) Propane (C 3 H 8 ) Propyl (C 3 H 7 ) Table 3.3: Alkyl Groups 4. Isomerism: This is a phenomenon in which two or more compounds have the same molecular formula but different structural formulae. There are TWO major types of isomerism: Structural and Stereo Structural Isomerism: this involves isomers that differ in the bonding arrangement and connectivity of atoms. Such isomers are of three types: (a) Chain isomerism: Pentane 2-methylbutane 2,2-dimethylpropane (b) Position isomerism: Br Br 1-Bromopentane 2-Bromopentane (c) Functional group isomerism: the isomers here have different functional groups. e.g

12 HO Ethanol O Methoxymethane Stereoisomerism: This involves isomers that have the same connectivity, but differ in the spatial arrangement of atoms. There are two classes: (a) Geometric isomerism: Double bond containing isomers which differ in their relative orientations. (b) Optical isomers (Enantiomers): These are isomers which are chiral, nonsuperimposable mirror images of each other. A chiral molecule is one which contains a carbon atom which has four different groups attached to it. Study Questions: 1. Define the following terms: (a) Homologous series (b) Functional group (c) Isomerism 2. How many chain isomers are there in the ninth member of the alkane homologous series?. References/Reading List 3. Schaum s Outline Organic Chemistry by Herbert Meislich, Howard Nechamkin and Jacob Sharefkin ISBN

13 Week 6 Topic: Computer Based Test Objectives: To examine the students on all that has been taught during the first five weeks of the semester. References/Reading List: 3. Schaum s Outline Organic Chemistry by Herbert Meislich, Howard Nechamkin and Jacob Sharefkin ISBN Week 7: Stereochemistry Objectives At the end of the lectures for this week, the students should be able to Differenciate between R and S configurations Identify chiral, achiral and meso compounds Description First hour: Stereoisomerism While Stereochemistry is the study of the three-dimensional structure of molecules, Isomerism is the phenomenon in which 2 or more compounds have the same molecular formula but different structural formula Stereoisomers have the same molecular formula, same bonding sequence but different spatial orientation. Stereochemistry has its own language and terms that need to be learnt in order to fully communicate and understand the concepts. Some important terms Stereoisomers are compounds with the same connectivity but different arrangement or orientation in space. Enantiomers are stereoisomers that are non-superimposable mirror images; they differ only in direction (+ or -) of optical rotation

14 Diastereomers are stereoisomers that are not mirror images of each other. They are different compounds having different physical properties An Asymmetric center/chiral center is an sp 3 carbon with 4 different groups attached to it. Optical activity refers to ther ability of a compound to rotate plane polarized light. Chiral compound refers to a compound that is optically active (achiral compound will not rotate light) A Polarimeter is a device that measures the optical rotation of the chiral compound. In general: if there is no asymmetric carbon, the compound is usually achiral; if there is 1 asymmetric carbon, the compound is always chiral; if there are 2 or more asymmetric carbons, the compound may or may not be chiral. Question: Cis-1,2-dichlorocyclopentane contains two asymmetric carbons but is achiral. Why is this so? Answer: because it contains an internal plane of symmetry.it is a meso compound. Second hour Absolute Configuration/(R) and (S) Nomenclature In order to assign R or S configuration to a molecule, the substituents around the chiral carbon are usually assigned priorities by using the cahn-ingold-prelog priorities. (C.I.P. Priorities) For example: I > Br > Cl > S > F > O > N > 13 C > 12 C > 3 H > 2 H > 1 H Using a 3-D drawing or model, put the 4th priority group in back. Look at the molecule along the bond between the asymmetric carbon and the 4th priority group.

15 Draw a curly arrow from the 1 st priority group to the 2 nd group to the 3 rd group and check if the direction of the arrow is clockwise or anticlockswise. Clockwise arrow (R) configuration Anti-clockwise arrow (S) configuration Treat double and triple bonds as if both atoms in the bond were duplicated or triplicated e.g Study Questions: 1. Define the term Meso compound 2. Draw a 3-dimensional formula for (R)-2-chloropentane. 3. Give the name of the following structure, taking the stereochemistry into consideration. O OH OH Reading List 3. Practical Chemistry for Schools and Colleges by G.O Ojokuku ISBN Week 8: Classification of Hydrocarbons and Nomenclature Objectives At the end of the lectures for this week, the students should be able to Describe the classification of hydrocarbons Name hydrocarbons and other organic compounds using the IUPAC nomenclature system Description First hour:

16 Hydrocarbons are compounds which consist of only carbon and hydrogen in their molecule. IUPAC NOMENCLATURE OF HYDROCARBONS Important steps towards naming Organic Compounds Step 1 Find the parent hydrocarbon. (a) Find the longest continuous chain of carbon atoms in the molecule, and use the name of that chain as the parent name. The longest chain may not always be easy to identify; you may have to turn corners. Answer is nonane

17 (b) If two different chains of equal length are present, choose the one with the larger number of branch points as the parent. 3- branch points 2-branch points A B Hence we choose the parent chain (red ink) in A Step 2 Number the atoms in the longest chain. (a) Beginning at the end nearer the first branch point, number each carbon atom in the parent chain. (b) If there is branching an equal distance away from both ends of the parent chain, begin numbering at the end nearer the second branch point Step 3 Identify and number the substituents. (a) Assign a number, or locant, to each substituent to locate its point of attachment to the parent chain. 2-methyl, 4-methyl, 4-methyl = 2,4,4-trimethyl 3-ethyl (b) If there are two substituents on the same carbon, give both the same number. There must be as many numbers in the name as there are substituents. E.g...4,4- in a

18 Step 4 Write the name as a single word. Use hyphens to separate the different prefixes, and use commas to separate numbers. If two or more different substituents are present, cite them in alphabetical order. If two or more identical substituents are present on the parent chain, use one of the multiplier prefixes di-, tri-, tetra-, and so forth, but don t use these prefixes to determine the alphabetical order. The name of the compound is: 3-ethyl-2,4,4-trimethylnonane Step 5 Name a complex substituent as though it were itself a compound. In some particularly complex cases, this fifth step is necessary whenevert a substituent on the main chain has sub-branching. E.g Number the branched substituent beginning at its point of its attachment to the main chain, and identify it in this case, a 2-methylpropyl group. The substituent is treated as a whole and is alphabetized according to the first letter of its complete name, including any numerical prefix. It is set off in parentheses when naming the entire molecule. For historical reasons, some of the simpler branched-chain alkyl groups also have nonsystematic, common names, as noted

19 earlier. The common names of these simple alkyl groups are so well entrenched in the chemical literature that IUPAC rules make allowance for them. Hence, the following compound is properly named either 4-(1-methylethyl)heptane or 4-isopropylheptane. When writing an alkane name, the nonhyphenated prefix iso- is considered part of the alkyl-group name for alphabetizing purposes, but the hyphenated and italicized prefixes sec- and tert- are not. Thus, isopropyl and isobutyl are listed alphabetically under i, but sec-butyl and tert-butyl are listed under b. Note: We will mostly make use of the stick diagram (line structure) for all the structures in this course. Hence, it is important that you get acquainted with this method. C-C is a line, there must be four bonds around each carbon atom and the hydrogens are omitted. Second hour The entire class practices the study questions and more! Study Questions 1. Draw the structure of the following compounds: (a) 2,2,5-trimethyl Undecane (b) 3-isopropyl-4-methyloctane

20 2. Give the names of the following structures: (a) Cl (b) I Cl References/Reading List Organic Chemistry by R.T. Morrison and R.N. Boyd Organic Chemistry by John McMurry Schaum s Outline Organic Chemistry by Herbert Meislich et al Week 9: Alkanes Objectives At the end of the lectures for this week, the students should be able to Describe the basic chemistry of alkanes State the physical and chemical properties of alkanes Describe the preparation of alkanes Description First hour: Alkanes : Saturated hydrocarbons; C n H 2n+2, C-C functional group Sources: 1. The main source is petroleum, accompanied by natural gas 2. the potential second source is Coal Petroleum : a mixture of alkanes C1-C40. Naphthenes(cycloalkanes) are also in the mixture Natural gas: a mixture of Methane, ethane and propane in ratio 12:2:1 Higher alkanes only make up 3%

21 Physical Properties The larger the alkane molecule, the larger its surface area, the stronger its intermolecular forces (Van der Waal s forces) and therefore the higher the melting and the boiling points The first 4 alkanes are gases, the next thirteen are liquids and C 18 and above are solids. An effect of branching is observed for all homologous series: a branched chain isomer has a lower boiling point than a straight chain isomer. n-butane (b.pt 0 o C) and isobutene (b.pt -12 o C) Alkanes Are soluble in non-polar solvents and they are good solvents for compounds of low polarity. They are less dense than water Second hour Preparation Hydrogenation of alkenes Reduction of alkyl halides Coupling of alkyl halides with organometallic compounds Reactions Halogenation/ Mechanism: Free radical substitution reaction. Combustion Pyrolysis Study Questions 1. The chemistry of alkanes mainly involve...reaction. 2. In an attempt to prepare an alkane, Tiyati coupled Lithium propyl copper with n- nonyl bromide. What alkane will she obtain? 3. Show the mechanism for the chlorination of methane. References/Reading List 3. Schaum s Outline Organic Chemistry by Herbert Meislich, Howard Nechamkin and Jacob Sharefkin ISBN

22 Week 10: Alkenes Objectives At the end of the lectures for this week, the students should be able to Describe various methods of preparation of alkenes Describe the important reactions of alkenes Description First hour: Alkenes are unsaturated hydrocarbons; C n H 2n Preparation Dehydrohalogenation of alkyl halides Dehydration of alcohols Dehalogenation of vicinal dihalides Reduction of alkynes Second hour Alkenes react with the use of their C=C functional group; they undergo addition reactions. The loosely held π electrons are available to electron seeking reagents (electrophiles) The mechanism is shown in the hydrohalogenation shown below. Stage 1: Formation of the carbocation Stage 2: The halide ion attacks the carbocation to yield a neutral addition product Other reactions which follow the same mechanism include: Catalytic hydrogenation Halogenation Addition of sulphuric acid Hydration Halohydrin formation Study Questions 1. Explain the mechanism behind electrophilic addition reaction 2. Draw the structure and state the name of the product of the reaction of propene with chlorine in water.

23 References/Reading List 3. Schaum s Outline Organic Chemistry by Herbert Meislich, Howard Nechamkin and Jacob Sharefkin ISBN Week 11: Alkynes Objectives At the end of the lectures for this week, the students should be able to Describe various methods of preparing alkynes Describe the reactions of alkynes Differenciate between alkenes and terminal alkynes Description First hour: Alkynes are unsaturated hydrocarbons; C n H 2n-2 ; C C func onal group; SP hybridiza on. Sources of Acetylene: Alkanes Preparation Dehydrohalogenation of alkyldihalides The reaction of metal acetylides with primary alkyl halides. Second hour Reactions of Alkynes Alkynes undergo electrophilic addition because of the same reason as in alkenesanvailability of electrons. The intermediate is however, different. A vinyl cation is formed here as opposed to a carbocation in alkenes. The mechanism is shown below. The following reactions follow the mechanism Hydrogenation Halogenations Hydration

24 Acidity of alkynes: they are very weak acids... Study Questions 1. What are tautomers? 2. During the hydration of ethyne, ethanol is not obtained as expected. Rather, Ethanal is obtained. Give an explanation for this observation. References/Reading List 3. Schaum s Outline Organic Chemistry by Herbert Meislich, Howard Nechamkin and Jacob Sharefkin ISBN Week 12 Topic: Computer based Test 2 Objectives: To examine the students on all that has been taught since the 7 th week of the semester. References/Reading List 3. Schaum s Outline Organic Chemistry by Herbert Meislich, Howard Nechamkin and Jacob Sharefkin ISBN Week 13 Topic: Revision Objectives: To strengthen the understanding of the students on all that has been taught during the semester. References/Reading List 3. Schaum s Outline Organic Chemistry by Herbert Meislich, Howard Nechamkin and Jacob Sharefkin ISBN Practical Chemistry for Schools and Colleges by G.O Ojokuku

25 ISBN Week 14 Topic: Revision Objectives: To answer all the questions which the students would have generated in the place of reading. Reading List: 3. Schaum s Outline Organic Chemistry by Herbert Meislich, Howard Nechamkin and Jacob Sharefkin ISBN Practical Chemistry for Schools and Colleges by G.O Ojokuku ISBN Week 15 Topic: Examination Objectives: To examine the students on all that has been taught during the semester. Reading List: List of reference material