Basic Organic Chemistry Lec (1) PDF

1 Basic Principle of Organic CHEMISTRY Lec (1) 3 ORGANIC CHEMISTRY  Bonding in Organic molecules  Atomic orbitals and hybridization 4 Octet Rule  A bond results from the attraction of nuclei for electrons  All atoms trying to achieve a stabl...

1 Basic Principle of Organic CHEMISTRY Lec (1) 3 ORGANIC CHEMISTRY  Bonding in Organic molecules  Atomic orbitals and hybridization 4 Octet Rule  A bond results from the attraction of nuclei for electrons  All atoms trying to achieve a stable octet  Octet Rule = atoms tend to gain, lose or share electrons to have 8 electrons makes an atom very stable  Orbital Theory and hybridization 5 Atoms are made of subatomic particles: protons, neutrons, & electrons 6  Electronegativity 7 Electronegativity : is a measure of the ability of an atom to attract a bonding pair of electrons. Fluorine (the most electronegative element) is assigned a value of 4.0.  Elements with a HIGH electronegativity have a STRONG pull-on electrons.  Elements with a LOW electronegativity have a WEAK pull-on electrons. 8 Electronegativity increase Electronegativity decrease 9 Bonding in Organic molecules When two atoms bond their DIFFERENCE in electronegativity determines the bond type Gain electrons Loss electron Share electron Two Major Types of Bonding 1- Ionic Bonding which is formed when two atoms of widely differing electronegativity 2-Covalent Bonding which is formed Between nonmetallic elements of similar electronegativity pairs of (e-) are shared between non-metal atoms 10 Ionic Bond  is formed between two atoms have large difference in electronegativity. ( EN more than 1.7).  which is formed by the transfer of one or more electrons from one atom to another  ionic compounds are made of ions.  ionic compounds are called Salts or Crystals 11 12  Covalent bond  Between nonmetallic elements of similar electronegativity  Formed by sharing electron pairs between non- metal atoms  there are two types of covalent bond:  Nonpolar covalent bond:  when electrons are shared equally (EN less than 0.5) as H2 ,O2,Cl2 Cl2 13  A polar covalent bond:  when electrons are shared between different atom as H2O and CO2 When the electronegativity of two bonding atoms is much higher and the atoms don’t pull equally (EN 0.5-1.7). 14  Orbitals theory and hybridization Atomic Orbitals  Electrons reside in general regions of space around the nucleus called shells.  Electrons restricted to specific regions of space called Atomic orbitals  There are four different kinds of orbitals for electrons s, p, d, and f.  The first shell, it is closest to the nucleus and is the smallest of the shells contain one orbital called the 1s orbital(spherical shape).  The second energy level contain one (s) orbital called 2s and three (p) (Px, Py, Pz) orbital called 2p orbitals (dumbbell shape).  The third energy level contain one (s) orbital called 3s , three (p) (Px, Py, Pz) orbital called 2p orbitals and five 3 (d) orbitals (dumbbell shape).  The s-orbital, which is spherical, is do not have more than two electrons  p-orbital, which is pairs of dumbbells, aligned along x, y and z-axis at 90° to each other  d-orbital, have complex shape, f-orbital: have very complex shape. 16 17 C 6 : 1s22s22p2 18  Hybridization  Hybridization is the mixing of atomic orbitals to form new orbitals with different energies and shapes than the original orbitals  Hybridization is the combining of different types of orbitals The hybridization of different types of orbitals allows the atom to form more bonds and bonds that are equal. This results in a more stable molecule. 1-SP3-hybridization or Tetrahedral (1S+ 3p) 2- SP2-hybridization or Trigonal (1S+ 2p) 3- SP-hybridization or Diagonal (1S+ 1p) In this process, the new orbitals come into existence and named as the hybrid orbitals. 19 1) sp3 – Hybridization  It occurs when one s-combine with three p- orbitals to form four equivalent  unsymmetrical tetrahedral orbitals (S + 3P = SP3).  having a tetrahedral structure  bond angle 109.5 degrees  Ex: CH4 , H2O , NH3, BeCl2 20  The sp3 hybrid orbitals in CH4 21 H H sp3 σ 109.5° sp3 σ sp3 sp3 H H H sp3 sp3 sp3 σ sp3 H H 109.5° H 22 Types of Covalent Bonds A sigma (σ) bond is formed by end-to-end overlap of orbitals. All single bonds are σ bonds. The sp3 hybrid orbitals in NH3 O6 12 The sp3 hybrid orbitals in H2O  2) sp2 – Hybridization 25  Hybridization one s- and two p-orbitals are mixed form three sp2– hybrid orbitals.  having a planar triangular structure  Bond angle 120 degrees.  Ex: C2H2, BF3  The sp2 hybrid orbitals in C2H4 26 CH2=CH2 Ethene  A pi (p) bond is formed by sideways overlap of orbitals. 27 A p bond is weaker than a σ bond because sideways overlap is less effective than end-to- end overlap. A double bond consists of one σ bond and one p bond  3) SP – Hybridization 28  hybridization one s- and one p- orbital are mixed to form two sp – hybrid orbitals,  having a linear structure  Bond angle 180 degrees  29 The sp hybrid orbitals in C2H2 30 31 Hybridization NO. of σ bonds NO. of π Shape bonds SP3 S+ 3P Single bond 4 0 tetrahedral SP2 S+ 2P Double 3 1 planar bond triangular SP S+P Trible bond 2 2 Linear 32 Examples :  Empirical formula 33 Chemical formula Structural formula Molecular formula  1- Empirical Formula The empirical formula is the simplest whole-number ratio of the atoms in a compound. Examples: Benzene CH Glucose CH2O 34 C4H8 CH2 Factor =4 B2H6 BH3 Factor =2 35 2- Molecular Formula The molecular formula is the actual number of the toms of a compound. Examples : benzene C 6H 6 Glucose C6H12O6 CH2 C4H8 Factor =4 BH3 B2H6 Factor =2 36  Calculating Molecular Formulas for Compounds 37 Molecular formulas give the kind and number of atoms of each element present in the molecular compound the empirical formula of a compound of boron and hydrogen is BH3. Its molecular formula molar mass is 27.7g/mol. Determine the molecular formula of the compound. Knowing that the mass number of H=1, B=10  How to calculate the Empirical formula 38 The percent composition of a compound was found to be 63.5 % Silver, 8.2 % Nitrogen and 28.3 %Oxygen. Determine the Empirical formula of this compound ( Knowing that the mass number of Ag =107.87 , N=14, O= 16 Atoms Ag N O Mass 63.5 8.2 28.3 36.5 8.2 mass Mole = 0.585 28.3 = 1.769 Mole = 107.87 14 =0.585 16 Molar mass 0.585 0.585 Mole ratio 1.769 =1 0.585 = 1 =3 0.585 0.585 AgNO3 Empirical formula  How to calculate the Empirical & Molecular formulas 39 Find the Empirical & molecular formulas of a compound contains 38.7 g carbon , 9.62 g Hydrogen and 51.6 g Oxygen. Its molar mass is 62 g. ( Knowing that the mass number of C=12, H=1 , O=16 Atoms C O H Mass 38.7 51.6 9.62 Mole 38.7 51.6 9.62 12 = 3.22 =3.22 = 9.62 16 1 Mole ratio =1 = 1 =3 Empirical formula CH3O Empirical formula mass = (1x12) + (1x3) + ( 1x16) = 31.03 62 = 2 Molecular formula C2H6O2 Lecture 1 31.03  Problem1 The empirical formula of hexane is C3H7. Its molecular weight is 86.2. What is the molecular formula of hexane? (Ans:C6H14) Problem 2 Calculate the empirical formula of a compound whose composition is 65.68% C, 15.16% H, and 19.16% N? (Ans: C4H11N) Problem 3 The empirical formula of a compound is CH2O. 0.0835 moles of the compound contains 1.0 g of hydrogen. Molecular formula of the compound is Dr. Ghada Emad 41 Structural Classification of Carbon and Hydrogen Atoms Classification of Carbon Atoms:  A primary carbon atom (1o) is attached to only one other carbon atom.  A secondary carbon atom (2o) is attached to two other carbon atoms.  A tertiary carbon atom (3o) is attached to three other carbon atoms.  A quaternary carbon atom (4o) is attached to four other carbon atoms. Lectural 1  Classification of Hydrogen Atoms 42 Each Hydrogen atom is similarly classified and being given the same designation of 1o, 2o and 3o according to the carbon atom to which it is attached. 1 3 1 1 2 2 1 H H H H H H H H C C C H H C C C C H H 2 H H H H H H C H Lectural 1 H