Basic Principles of Organic Chemistry Lecture Notes PDF

2 Basic Principle of Organic CHEMISTRY Chem 101 Lecture 2  Hydrocarbons 3 (Acyclic)Open Chain (Cyclic) Closed Chain Alkanes CnH2n+2...

2 Basic Principle of Organic CHEMISTRY Chem 101 Lecture 2  Hydrocarbons 3 (Acyclic)Open Chain (Cyclic) Closed Chain Alkanes CnH2n+2 Substituted Saturated cyclic Unsaturated cyclic Alkenes Alkynes hydrocarbon CnH2n CnH2n-2 C-OH C-CHO C-CO-C C-O-C C-NH2 C-COOH Alkanes (CnH2n+2) Nomenclature of alkanes (Paraffins) Saturated hydrocarbons or Alkanes (CnH2n+2) The nomenclature depend upon the number of C. atoms No. of Carbons Formula Name (CnH2n+2) 1 Methane CH4 2 Ethane C 2H 6 3 Propane C 3H 8 4 Butane C4H10 5 Pentane C5H12 6 Hexane C6H14 7 Heptane C7H16 8 Octane C8H18 9 Nonane C9H20 10 Decane C10H22  Common system 1- For unbranched chain, we put n (normal) before the name. 2- In presence of ONE CH3- branch in C2 position we put Iso-. 3- In presence of TWO CH3- branchs in C2 position we put Neo-. 7 8  The IUPAC Nomenclature rules 9 1-Identify the longest carbon chain. This chain is called the parent chain. 2-Identify all of the substituents (groups appending from the parent chain). 3-Number the carbons of the parent chain from the end that gives the substituents the lowest numbers 4-If the same substituent occurs more than once, the location of each point on which the substituent occurs is given. In addition, the number of times the substituent group occurs is indicated by a prefix (di, tri, tetra, etc.). 5-If there are two or more different substituents they are listed in alphabetical order in summary, the name of the compound is written out with the substituents in alphabetical order followed by the base name (derived from the number of carbons in the parent chain). Commas are used between numbers and dashes are used between letters and numbers. There are no spaces in the name   IUPAC Nomenclature 10 1-Find the longest continuous carbon chain 2-Number the C. chain from the side that gives the substituent the smallest number. d  Example: 11  3-Name the substituents as alkyl groups, 12 4- If the substituent repeated many times, we denote the number by adding Di, Tri, Tetra…..etc dimethyl 13. Identify and provide a name for the parent in each of the following compounds:  Haloalkanes CH3CH2Cl CH3CH2CH2F CH3CHBrCH3 Chloroethane 1-Fluoropropane 2-Bromopropane Ethyl chloride n-Propyl fluoride Isopropyl bromide. 1) When the parent chain has both a halo and an alkyl substituent attached to it, number the chain from the end nearer the first substituent 2-Chloro-4-methylpentane 2) Common names for simple haloalkanes are accepted by the IUPAC  alkyl halides (radicofunctional nomenclature). (CH3)3CBr CH3CH(CH3)CH2Cl (CH3)3CCH2Br 2-Bromo-2-methylpropane 1-Chloro-2-methylpropane 1-Bromo-2,2dimethylpropane tert-Butyl bromide Isobutyl chloride Neopentyl bromide 15 Q. Give an acceptable IUPAC name for each of the following alkanes: B) D)  Cycloalkanes (cycloparaffins) The general molecular formula for a cycloalkane is CnH2n. Alkane Cycloalkane 17 18 Cyclopropane Cyclobutane Cyclopentane Cyclohexane Cl H H3C C CH3 CH3 3 2 1 CH2CH3 Chlorocyclopentane Isopropylcyclohexane 1-Ethyl-3-methylcyclohexane  Methane (CH4) Preparation 1- Electric arc between carbon and hydrogen 2- Using water gas 3- Using CO2 and hydrogen 4- Using Aluminum carbide Reactions Generally, methane inert toward all types of reagents. 1- Combusion 2- Halogenation Chloronation of methane Mechanism The mechanism takes place in three steps: i- Initiation  ii- Propagation iii- Termination Generally, the halogenation could be arranged as in the following.  Preparation of alkane 1- Wurtz reaction - This reaction used for preparation of symmetrical alkanes with even number of C. atoms. - Using two different types of alkyl halides will results three different products. 2- From Grignard reagent - Treating Grignard with any acidic proton will gives an alkane. 3- Reduction of alkyl halide 4- Reduction of alkenes 5- Decarboxylation of salts of acids 6- Kolb electrolysis  What the percentage of the product I in the following 28 reaction,(the reactivity ratio between(1o:3oH) is (1:5)? No. of H Ratio Fraction yield % yield I 1H 9 1 9X1=9 (9/14)X100=64% II 3H 1 5 1X5=5 (5/14)X100=35.7% 9+5=14