CHE111_Lectures 1-2_872024.pdf

Field of Basic Sciences Organic Chemistry CHE 1 1 1 3 Credit hours: lecture (2 Cr. Hr) + laboratory (1 Cr. Hr) Lecture 1: Introduction Associate Prof. Gomaa A.M. Ali Sanad [email protected]...

Field of Basic Sciences Organic Chemistry CHE 1 1 1 3 Credit hours: lecture (2 Cr. Hr) + laboratory (1 Cr. Hr) Lecture 1: Introduction Associate Prof. Gomaa A.M. Ali Sanad [email protected] Summer 2024  Course Grading Grade Distribution Assessment Measures Weight Attendance 20 % Course activities (Quizzes, Assignments, Presentations, etc.) Lab 30 % Midterm Exam 10 % Final Exam 40 % Total 100 % Note: To pass a course: - Student must earn 60 % as a whole grade. - Student must earn at least 30% of its final exam mark (i.e. 12/40).  Grading System Marks % Collected Grade Points More than 97% A+ 4 93% to less than 97% A 4 89% to less than 93% A- 3.7 84% to less than 89% B+ 3.3 80% to less than 84% B 3.0 76% to less than 80% B- 2.7 73% to less than 76% C+ 2.3 70% to less than 73% C 2.0 67% to less than 70% C- 1.7 64% to less than 67% D+ 1.3 60% to less than 64% D 1.0 Less than 60% F 0.0  GPA Grade Point Average Calculations σ𝑪𝒐𝒖𝒓𝒔𝒆𝒔 (𝑮𝒓𝒂𝒅𝒆 𝑷𝒐𝒊𝒏𝒕𝒔∗𝑵𝒖𝒎𝒅𝒆𝒓 𝒐𝒇 𝑪𝒓𝒆𝒅𝒊𝒕 𝑯𝒐𝒖𝒓𝒔) Cumulative 𝑮𝑷𝑨 = σ𝑪𝒐𝒖𝒓𝒔𝒆 𝑪𝒓𝒆𝒅𝒊𝒕 𝑯𝒐𝒖𝒓𝒔 For a student who get the following marks in his first semester Course # Credit Total Grade Course 0∗4 + 3.0∗1 + 4∗2.0 + 3∗3.0 +(2∗3.7) Cumulative 𝐺𝑃𝐴 = (0+3+4+3+2) Hours marks points English 0 95 A 4.0 Math 3 60 D 1.0 27.4 Cumulative 𝐺𝑃𝐴 = = 2.28 Physics 4 70 C 2.0 12 Chemistry 3 80 B 3.0 Biology 2 90 A- 3.7  Course Content 1. Introduction 2. Formula weights, Avogadro’s number, empirical formulas from analyses 3. Chemical bonding and molecular structure 4. Lewis model of bonding 5. Resonance theory 6. Molecular orbitals (The case of H2 molecule) 7. Hybridization (sp3, sp2, sp) 8. Structural Formulas 9. Alkanes, Alkenes, Alkynes 10. Alcohols, Ethers, Aldehydes and Ketones 11. Carboxylic acids 12. Aromatic Compounds 13. Polymers  Learning Resources Organic Chemistry, 11th Edition, By Francis Carey and Robert Giuliano and Neil Allison and Susan Bane. McGraw Hill Higher Education, 2020. ISBN10: 1260148920, ISBN13: 9781260148923. Introduction to Organic Chemistry, 4th Edition by William Brown and Thomas Poon, John Wiley and Sons 2011. ISBN 978-0-470-38467-1 Organic Chemistry, 9th Edition by T. W. Graham Solomons and Craig B. Fryhle, John Wiley and Sons, 2008. ISBN 978-0-471-68496-1 Introduction What is Organic Chemistry? ▪ Literally, the term “organic” means “derived from living organisms”. ▪ Organic chemistry was originally the study of compounds extracted from living organisms and their natural products. ▪ It was believed that only living organisms possessed the “vital force” necessary to create organic compounds (vitalism). ▪ This concept started to change in 1828 after Friedrich Wöhler showed that it was possible to make urea (a constituent of urine) by evaporating an aqueous solution of the inorganic compound ammonium cyanate:  What is Organic Chemistry? ▪ The modern definition of organic chemistry is the chemistry of compounds containing carbon. ▪ Carbon Compounds are everywhere around us in food, flavors, fragrances, medicines, cosmetics, plastics, fibers, paints, adhesives, and of course in our bodies (DNA and proteins). 11-cis-retinal VC=AA?? Absorbs light and allows vision  Organic vs. Inorganic Compounds C6H12O6 NaCl CO2 CH4 C2H6O C Ag Mine origins Plant or animal origins C H  Why Carbon? 1. Carbon can form strong covalent bonds to other carbon atoms and other elements as: hydrogen, nitrogen and oxygen. 2. Carbon can form chains and rings.  Principles of atomic structure Why do we study atomic structure? To answer, why there are many organic compounds, however, they only consist of only C and H. Pure substances are either Elements or Compounds.  Principles of atomic structure Atom ▪ It is the smallest unit of ordinary matter that forms a chemical element. ▪ An atom consists of a dense, positively charged nucleus containing protons and neutrons and a surrounding cloud of electrons. ▪ Each proton of the nucleus bears one positive charge (+Ve); electrons bear one negative charge (-Ve). Neutrons are electrically neutral; they bear no charge. Cp = - Ce ▪ Protons and neutrons have nearly equal masses (approximately 1atomic mass unit each) and are about 1800 times as heavy as electrons. Most atom mass comes from the mass of the nucleus; (electrons atomic mass is negligible). mp = mn  Principles of atomic structure Electronic Structure of Atom ▪ Electrostatic forces keep the electrons moving around the nucleus. These are the forces that pull negatively-charged and positively-charged particles towards each other. ▪ In nature, most atoms are stable. A stable atom has the same number of electrons as protons.  Principles of atomic structure ▪ Each element is distinguished by 2 numbers. ▪ The Atomic Number (Z) a number equal to the number of protons in its nucleus. Because an atom is electrically neutral, the atomic number also equals the number of electrons surrounding the nucleus. ▪ The Mass Number (A) is the total number of protons and neutrons in an atom.  Principles of atomic structure Volume of the atom ▪ Most of the volume of an atom comes from the electrons. ▪ The volume of an atom occupied by the electrons “electron cloud” is about 10,000 times larger than that of the nucleus. ▪ Electron cloud contributes very little to the mass of the atom, but make up most of the volume of the atom.  Principles of atomic structure Isotopes Atoms that have the same atomic number but differ in mass numbers due to different numbers of neutrons in their nuclei. Isotopes of Hydrogen Stable Stable Not Stable half-life of 12.32 years  Principles of atomic structure ❑ Energy levels of shells and subshells  Principles of atomic structure ❑ Atomic orbitals Atomic orbitals are mathematical functions that describe where an electron is likely to be found in an atom S Orbital: Spherical P Orbital: dump-bell shaped (spindle like) Principles of atomic structure The atom: Electrons (e-), Protons (+), Neutrons  Principles of atomic structure ❑ Valence Electrons ▪ They are the outermost electrons, the ones most likely to be involved in chemical bonding and reactions. ▪ In order to know the number of valence electrons of an element, Look at its group number in the periodic table. ▪ No of Valence electrons = Group No of Element ▪ For Example: No of Valence electrons of Na (group I-A) = 1 No of Valence electrons of C (group IV-A) = 4 Formula weights, Avogadro’s number, empirical formulas from analyses  Formula weights ❑ The formula weight (FW) of a substance is the sum of the atomic weights of the atoms in its chemical formula. ❑ For an element such as Na: Formula weight = Atomic weight = 23 amu (atomic mass unit) ❑ For a molecule such as C2H6O: Formula weight = Molecular weight = 2×(12 amu) + 6×(1 amu) + 1×(16 amu) = 46 amu  Molar Mass ❑ The molar mass of a substance is the mass in grams of one mole of this substance ❑ The molar mass in grams per mole (g/mol) of any substance is numerically equal to its formula weight in atomic mass units (amu). ❑ For example: ❖ The formula (molecular) weight of CO2 is 44 amu The molar mass of CO2 is 44 g/mol.  Molar Mass Q) What is the molar mass of glucose, C6H12O6? ❑ Our first step is to determine the molecular weight (MW) of glucose: MW of glucose = 6×(12 amu) + 12×(1 amu) + 6×(16 amu) = 180 amu Because glucose has a molecular weight of 180 amu, 1 mol of this substance (6.022 × 1023 molecules) has a mass of 180 g. In other words, C6H12O6 has a molar mass of 180 g/mol.  Avogadro’s Number and the Mole ❑ One mole is the amount of matter that contains as many objects (atoms, molecules, or whatever other objects we are considering) as the number of atoms in exactly 12 g of isotopically pure 12C. ❑ From experiments, scientists have determined this number to be 6.022 × 1023 and called it Avogadro’s number (NA). ❑ Therefore, a mole of atoms, a mole of molecules, or a mole of anything else all contain Avogadro’s number of objects: 1 mol of C atoms = 6.022 × 1023 C atoms 1 mol of H2O molecules = 6.022 × 1023 H2O molecules Q) How many molecules are in 3 moles of water?  Interconverting Masses and Moles 𝒎𝒂𝒔𝒔 𝒐𝒇 𝒔𝒖𝒃𝒔𝒕𝒂𝒏𝒄𝒆 (𝒈) Number of moles = 𝒎𝒐𝒍𝒂𝒓 𝒎𝒂𝒔𝒔 (𝒈/𝒎𝒐𝒍) Q) Calculate the number of moles and molecules of glucose (C6H12O6) in a 5.38 g sample. Since the molar mass of glucose = 180 g/mol (see previous slide) 5.38 (𝑔) Then, the number of moles = = 0.02989 mol 180 (𝑔/𝑚𝑜𝑙) The number of molecules = Number of moles ×Avogadro’s number (NA) = 0.02989 × 6.022 × 1023 = 1.8 × 1022 molecules  Empirical and Molecular Formulas ❑ The empirical formula (the simplest formula): A formula with the smallest whole-number ratio of the elements that make up a compound. So, it tells us only the relative amounts of each element in a compound. However, it is not necessary to show the actual numbers of atoms in the compound. ❑ The molecular formula (the true formula): A formula that specifies the actual number of atoms of each element in one molecule of a compound. It may be the same as the empirical formula or multiple of it. ❑ Example: The molecular formula of ethylene is C2H4, but the empirical formula is CH2. In both formulas, the ratio between C and H is 1: 2, but only the molecular formula shows the actual number of atoms in the molecule.  Chemical Formulas Molecular formula: It indicates the actual number of atoms of each elements in one molecule of a substance H2O CH4 Cl2 Empirical formula: It simply gives the ratio of the combination of the constituent elements MF H2O2 EF HO MF C2H4 EF CH2 Structural Formula: It shows the way of attachment of each atom to the other in molecule. CH4 Molecular Weight: It is the atomic masses of each component present in the molecule. C= 12 g/mol H2O = 18 g/mol  Empirical and Molecular Formulas ❑ We can obtain the molecular formula for any compound from its empirical formula if we know the molecular weight (or the molar mass) of the compound. ❑ The subscripts in the molecular formula of a compound are always whole- number multiples of the subscripts in its empirical formula. ❑ This multiple can be found by dividing the molecular weight by the empirical formula weight: 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒂𝒓 𝒘𝒆𝒊𝒈𝒉𝒕 Whole-number multiple (n) = 𝒆𝒎𝒑𝒊𝒓𝒊𝒄𝒂𝒍 𝒇𝒐𝒓𝒎𝒖𝒍𝒂 𝒘𝒆𝒊𝒈𝒉𝒕 Molecular formula = n × Empirical formula ❖ If n = 1, then the molecular and empirical formulas are the same.  Empirical and Molecular Formulas ❑ Exercise: An organic compound has an empirical formula of C3H4 and an experimentally determined molecular weight of 121 amu. What is its molecular formula? (Atomic masses are C = 12 and H = 1 amu) SOLUTION The empirical formula weight = 3 × 12 + 4 × 1 = 40 amu 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 121 Thus, Whole-number multiple (n) = = = 3.03 ≈ 3 𝑒𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑤𝑒𝑖𝑔ℎ𝑡 40 We then multiply each subscript in the empirical formula by 3 to give the molecular formula: C9H12.  Empirical and Molecular Formulas ❑ See the following examples: Molecule name Molecular formula Empirical formula (lowest ratio of elements) Glucose C6H12O6 CH2O Benzene C6H6 CH Acetic acid C2H4O2 CH2O Ethanol C2H6O C2H6O  Empirical Formula from Combustion Analysis ❑ When an organic compound containing carbon and hydrogen is completely combusted, the carbon is converted to CO2 and the hydrogen is converted to H2O. From the masses of CO2 and H2O we can calculate the number of moles of C and H in the original sample and thereby the empirical formula. ❑ If oxygen is present, its mass percentage can be determined by subtracting the measured mass percentages of C and H from 100.  Empirical Formula from Combustion Analysis ❑ Exercise: Isopropyl alcohol is composed of C, H, and O. Combustion of 0.255 g of isopropyl alcohol produces 0.561 g of CO2 and 0.306 g of H2O. Determine the empirical formula of isopropyl alcohol. SOLUTION 12 Mass of C = 0.561 (g) × = 0.153 𝑔 (1 mol of CO2 (44 g) contains1 mol of C (12 g)) 44 𝐶 𝑚𝑎𝑠𝑠 0.153 % C in the sample = × 100 = × 100 = 60 % 𝑠𝑎𝑚𝑝𝑙𝑒 𝑚𝑎𝑠𝑠 0.255 2 Mass of H = 0.306 (g) × = 0.0343 𝑔 (1 mol of H2O (18 g) contains 2 mol of H (2 g)) 18 𝐻 𝑚𝑎𝑠𝑠 0.0343 % H in the sample = × 100 = × 100 = 13.45 % 𝑠𝑎𝑚𝑝𝑙𝑒 𝑚𝑎𝑠𝑠 0.255 % O in the sample = 100 – [% C + % H] = 26.55 %  Empirical Formula from Combustion Analysis ❑ Continue solution: C H O Atomic percentages (%) 60 % 13.45 % 26.55 % Assume a 100 g sample 60 g 13.45 g 26.55 g Divide by the atomic weight to 60 13.45 26.55 = 𝟓 mol = 𝟏𝟑. 𝟒𝟓 mol = 𝟏. 𝟔𝟔 mol 12 1 16 convert masses to moles Divide by the smallest number 3 8.1≈ 𝟖 1 (1.66) Remember: You should obtain whole numbers. If you get fractions, you should multiply by a small whole number until you have whole numbers. Therefore, the empirical formula is C3H8O Quiz ❑ Complete the following table Molecular formula Empirical formula Molecular weight C6H6 C4H8O2 C2H6O C3H6O (Atomic masses: C = 12, O = 16, H = 1 amu)  Quiz ❑ Could the empirical formula determined from chemical analysis be used to tell the difference between acetylene, C2H2, and benzene, C6H6? Answer: No, it couldn’t because both compounds have the same empirical formula CH. Thank You