Organic Chemistry Lecture Notes PDF

Prof. Dr. Hossieny Ibrahim Badr University in Assiut School of Biotechnology [email protected] Office number: Bio-326 1   Hybridization  The bonding in the hydrogen molecule is fairly straightforward, but the situation is more complicated in organic...

Prof. Dr. Hossieny Ibrahim Badr University in Assiut School of Biotechnology [email protected] Office number: Bio-326 1   Hybridization  The bonding in the hydrogen molecule is fairly straightforward, but the situation is more complicated in organic molecules with tetravalent carbon atoms.  Take methane, CH4, for instance. As we’ve seen, carbon has four valence electrons (2s2 2p2) 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.  In fact, though, all four C–H bonds in methane are identical and are spatially oriented toward the corners of a regular tetrahedron.  How can we explain this?  An answer was provided in 1931 by Linus Pauling, who showed mathematically how an s orbital and three p orbitals on an atom can combine, or hybridize, to form four equivalent hybrid orbitals with tetrahedral orientation.  When an s orbital hybridizes with three p orbitals, the resultant sp3 hybrid orbitals are unsymmetrical about the nucleus. 2   sp3 Hybridization and the Structure of Alkanes Carbon atom Hybridization (a) An sp3 orbital (b) Four tetrahedral sp3 orbitals 3  sp3 Hybridization and the Structure of Alkanes tetrahedral 109.5 25% s character 75% p character Four sp3 hybrid orbitals 4   sp3 Hybridization and the Structure of Alkanes Ex. Methane (CH4) Ex. Ethane (C2H6) s s s s C C C—H bond formed by sp3—s overlap sp3 sp3 s s C—C bond formed by sp3—sp3 overlap H H H No. of sp3 C atoms = 2 C C No. of σ-bond sp3–s = 6 H H No. of σ-bond sp3–sp3 = 1 H All bond angles 109.5° Ball stick model Sigma bonds (σ) 5   sp2 Hybridization and the Structure of Alkenes 2p 2p 2p 3sp2 2S 2S 3sp2 sp2 orbital sp2 orbital sp2 orbital p orbital Three sp2 orbitals sp2 orbital Trigonal planar Angle 120⁰ sp2 hybridized carbon atom Unhybridized p orbital 6 Ex. Ethylene (C2H4) No. of sp2 C atoms = 2 No. of σ-bond sp2–s = 4 2Pz 2Pz No. of σ-bond sp2–sp2 = 1 sp2 sp2 No. of π- bond = 1 C sp2 sp2 C Sigma bonds (σ) sp2 sp2 Pi bonds (π) H C H H C H Ball stick model 7   sp Hybridization and the Structure of Alkynes Ex. Ethyne (C2H2) Ethyne a linear shape No. of sp C atoms = 2 No. of σ-bond sp–s = 2 No. of σ-bond sp–sp = 1 No. of π- bond = 2 8   Covalent Bonding of Carbon Shape Example Tetrahedral Ethane Trigonal planar Ethene linear Ethyne No. of sp3 C atoms = 2 No. of sp2 C atoms = 2 No. of σ-bond sp3–s = 6 No. of σ-bond sp2–s = 4 No. of sp C atoms = 2 No. of σ-bond sp3–sp3 = 1 No. of σ-bond sp2–sp2 = 1 No. of σ-bond sp–s = 2 No. of π- bond = 1 No. of σ-bond sp–sp = 1 Ethane No. of π- bond = 2 Ethene Ethyne 9   Types of Organic Formulas  The Molecular formula of an organic compound simply shows the number of each type of atom present. It tells you nothing about the bonding within the compound.  The Empirical formula of an organic compound gives the simplest possible whole number ratio of the different types of atom within the compound.  The Condensed formula is also text-based; here, each carbon atom is listed separately, with atoms attached to it following.  A Displayed formula shows all of the atoms and all of the bonds present in an organic compound. The bonds are represented as lines.  Structural formula: Similar to displayed formula - not all bonds are shown, although all atoms are still indicated using subscript numbers. Carbon hydrogen bonds are often simplified.  In a Bond-line (Zig-zag) formula, most hydrogen atoms are omitted, and line ends or vertices represent carbons. Functional groups and atoms other than carbon or hydrogen are still shown. Easiest to draw & commonly used. 10  Types of Organic Formulas 11 Examples Structural formula Bond-line formula Structural formula Bond-line formula Structural formula Bond-line formula Structural formula Bond-line formula 12 Problem  Convert the following condensed structures into Bond-line structures: ❶ (CH3)3CCH(CH3)CH2CH3 ❷ (CH3)2CHCH(CH3)CH(CH3)2 ❸ (CH3)2CHC(CH3)2CH2CH3 ❹ (CH3)2CHCH2CH3 ❺ (CH3)2CHCH(CH3)2 ❻ (CH3)2CH(CH2)3CH3 ❼ BrCH2(CH2)3CH(CH2CH3)2 ❽ (CH3)3CCH2CH(Br)COCH3 ❾ (CH3)3COC(CH3)3 ❿ (CH3)3COH ⓫ (CH3)2CHCH(CH2CH3)CH2CH(CH3)2 ⓬ (CH3)2C(Br)CH(OH)CH2CH(CH3)2 ⓭ CH3C(OH)2CH(CH3)CH2C(CH3)3 13 Problem A C B A B 14 Problem C4 N1  What is the hybridization of C1: ……  What is the hybridization of C2: ……  What is the hybridization of C3: ……  What is the hybridization of C4: ……  What is the hybridization of N1: ….. C1 C2  What is the hybridization of N2: ….. C3  What is the hybridization of N3: ….. N3 N2 A B  Identify the hybridization of the carbon atoms (A,B,C,D and E) E labeled in the following molecule. C D 15   Isomerism  Isomers are different compounds with the same Molecular formula, but their atoms are arranged in a different order (i.e. the atoms are bonded in different ways).  There are two primary types of isomerism, which can be further categorized into different subtypes. These primary types are: 1- Structural (Constitutional) Isomerism 2- Stereoisomerism. Structural (Constitutional) Stereoisomerism ❶ ❷ ❸ ❹ 16   Structural (Constitutional) Isomerism ❶Chain Isomerism  For methane (CH4), ethane (CH3—CH3)and propane (CH3—CH2—CH3), there is only one type of carbon arrangement.  As the number of carbon increases to 4, there are two ways for the carbon atoms to be connected: as a straight-chain (blue structure below) and as a branch on the chain (red structure below). Butane 2-methylpropane 17   Structural Isomerism ❶Chain Isomerism Constitutional isomers of C5H12 isomer I isomer II isomer III For alkanes with 6 carbons, there are a total Constitutional isomers of C6H14 of Five constitutional isomers. 18   Structural Isomerism ❷Position Isomerism C4H9OH C3H9N C6H4Cl2 19   Structural Isomerism ❸ Functional Isomerism C2H6O C4H8O2 C3H6O 20   Structural Isomerism ❹ Ring-chain Isomerism C4H8 C4H6 1,3-butadiene Cyclobutene Methyl cyclopropene 21   Stereoisomerism  Stereoisomerism, or spatial isomerism, is a form of isomerism in which molecules have the same molecular formula and sequence of bonded atoms (constitution), but differ in the three-dimensional orientations of their atoms in space ❶ Geometric Isomerism  There is one major requirement for geometric isomerism in compounds containing a carbon-carbon double bond. Each double- bonded carbon atom must be attached to two different atoms or functional groups. H H H CH3 H H H3C H C C* C C * C C* C C H CH3 H H H3C CH3 H CH3 Not Isomers (Same compound) Geometric Isomers 22  ❶ Geometric Isomerism (a) Cis-Trans Notation for Geometric Isomers (b) E-Z Notation for Geometric Isomers (a) Cis-Trans Notation: A B A A  The Latin prefixes cis and C C C C trans translate to “on this A B B B side of” and “on the other side of,” respectively. trans isomer cis isomer  cis isomer has functional groups on the same side of A D A A the double bond. C C C C  trans isomer has functional B A D B groups on opposite sides of the double bond. trans isomer cis isomer 23   Examples H H H3C H H H Cl H *C C * C C H3C CH3 C C C C H CH3 Cl Cl H Cl Cis-2-Butene trans-2-Butene Cis-1,2-dichloroethene trans-1,2-dichloroethene H H H H3C CH2 H H H H C C C C H3C CH2 CH2 CH3 H CH2 CH3 cis-3-hexene trans-3-hexene 24  ❶ Geometric Isomerism (b) E-Z Notation:  Although the terms cis and trans are useful when identifying geometric isomers for disubstituted alkenes, they do not apply to alkenes with three or four substituents.  The E/Z system, on the other hand, applies to disubstituted, trisubstituted, and tetrasubstituted alkenes.  The isomer with configuration Z (from the German word zusammen, meaning “together”) has substituents with higher priority on the same side (Zame Zide) of the double bond.  The isomer with configuration E (from the German word entgegen, meaning “opposite”) has substituents with higher priority on opposite sides of the double bond. 25  E-Z Notation: Low priority Low priority LP LP High priority High priority HP HP Z - isomer High priority Low priority HP LP Low priority High priority LP HP E - isomer 26   Assigning Group Priorities: The Cahn, Ingold, Prelog rules:  Look at the atoms directly attached to each carbon of the double bond.  Rank them according to decreasing atomic number.  Priority of common atoms: 53I > 35Br > 17Cl > 16S > 9F > 8O > 7N > 6C > 1H  CH3CH2─ > CH3─ > H Cl H Br  Examples C C C Br CH3 Cl Br H Z - isomer C C Cl CH3 Cl CH2 CH3 E - isomer C C H2 N H Cl CH2 CH3 Br CH3 C C C C E - isomer HO CH3 H3C Br Z - isomer E - isomer 27   Stereoisomerism ❷ Optical Isomerism  Optical isomers are molecules that are non-superimposable mirror images of each other and they are not identical.  Molecules with a chiral carbon show this kind of isomerism.  Chiral carbons, also known as asymmetric carbons, refer to a carbon atom that is bonded to four different functional groups. Mirror Chiral Chiral carbon carbon L-amino acid D-amino acid Enantiomers Enantiomers 28   Chirality and Biological Properties  In human body, the biological functions are modulated by a lot of enzymes and receptors.  Enzymes and receptors are essentially proteins, and proteins are made up of amino acids.  Amino acids are examples of naturally exist chiral substances.  Because amino acids are chiral, proteins are chiral so enzymes and receptors are chiral as well.  The binding site of enzyme or receptor is chiral, so it only binds with the enantiomer whose groups are in the proper positions to fit into the binding site. As shown in the diagram, only one enantiomer binds with Binding site of enzyme Binding site of enzyme the site, but not the other enantiomer. 29 Problem Identify all chiral carbons in the following molecules O OH HO Br OH HO OH OH Cl OH OH O O O H C N OH OH N O NH2 O 30  Classification of Carbon Atoms  Primary (1°) bonds to One carbon atom.  Secondary (2°) bonds to Two carbon atoms.  Tertiary (3°) bonds to Three carbon atoms.  Quaternary (4°) bonds to Four carbon atoms. 2° hydrogen 3° hydrogen 1° hydrogen 1° 1° Example: Identify the carbon 2° 1° atoms in the following molecule 1° 2° as primary, secondary, tertiary, or 3° 3° quaternary. 2° 4° 1° 1° 1° 31  Functional Groups in Organic Chemistry Cl X Cl Cl Alkyl halide 1° 2° 3° Alkene Alkyne Aromatic X= F,Cl, Br, I) OH OH OH H3C C N O NO 2 1° 2° 3° Alcohol Ether Nitrile Nitro H H NH2 O N N O H O 1° 2° H 3° Aldehyde Aldehyde Ketone Note: C=O is usually called as Amine Carbonyl group 32   Functional Groups in Organic Chemistry O O O O O H N N N OH O H H Carboxylic acid Ester Amide O O 2 H O H O 1 3 H3C Anhydride O CH3 O 11 10 4 Br OH 5 H C O O 8 3 6 H N NO 2 9 2 7 CH CH3 33 Problem Identify the functional groups HO O H OH O O O Br O O H3C O NO 2 H2N CH NH2 34