Pharmaceutical Chemistry Lecture 1 PDF

PL1001 Pharmaceutical Chemistry CHEMISTRY OF AROMATIC COMPOUNDS Lecture 1 Dr Alberto Di Salvo Recommended reading Organic Chemistry by Clayden, Greeves, Warren and Wothers. Organic Chemistry by Loudon. Organic Chemistry by G Solomon & C Fryhle. All general chemistry...

PL1001 Pharmaceutical Chemistry CHEMISTRY OF AROMATIC COMPOUNDS Lecture 1 Dr Alberto Di Salvo Recommended reading Organic Chemistry by Clayden, Greeves, Warren and Wothers. Organic Chemistry by Loudon. Organic Chemistry by G Solomon & C Fryhle. All general chemistry and organic chemistry textbooks cover the chemistry of aromatic compounds Lecture 1 - content History and Importance of organic compounds Definition of aromatic compound: Hückel’s rule Classification of aromatic compounds Kekulé’s structure of Benzene and its limitations Resonance theory of benzene Molecular Orbital theory of benzene Comparison between cyclohexene and benzene Introduction Many drugs on the market are aromatics derivatives or contain an aromatic moiety Pharmaceutically important aromatics HO O OH O O H O N HO Aspirin Morphine An analgesic and antipyretic A narcotic analgesic Pharmaceutically important aromatics NH2 O N Cl N O O S O N N H Valium Sulfamethoxazole A tranquilliser An antimicrobial agent Historical background AROMATIC: fragrant substances. Later, benzene and its structural relatives were termed as aromatic. 1826 Michael Faraday discovered benzene, and he named it as ‘bicarburet of hydrogen’ because of the equal number of carbon and hydrogen atoms. 1834 Eilhardt Mitscherlich synthesised benzene by heating benzoic acid with calcium oxide. 1865 August Kekulé first noticed that all early aromatic compounds contained a six-carbon unit which is retained through most chemical transformations and degradation.  Definition: aromaticity criteria 1. Aromatic compounds contain one or more rings that have a cyclic arrangement of p orbitals. 2. Every atom of an aromatic ring has a p orbital. 3. Aromatic rings are planar. 4. The cyclic arrangement of p-orbitals in an aromatic compound must contain 4n + 2 π electrons, where n is any positive integer (0, 1, 2,....). In other words, an aromatic ring must contain 2, 6, 10,..... π electrons. These criteria are collectively called Hückel’s rule. (Erich Hückel 1931) Classification Benzene and its monocyclic derivatives (e.g. toluene) Polycyclic benzenoids (e.g. napthalene) Non-benzenoids (e.g. azulene) Macrocyclic (e.g. anulene) Heterocyclic (e.g. pyridine, pyrrole) anulene H N N pyrrole toluene azulene naphtalene pyridine Kekulé’s structure of benzene 1865: August Kekulé proposed the structure of benzene. In benzene H H C H C C or C C H C H H all six carbon atoms are in a ring all C atoms are bonded to each other by alternating single and double bonds one H atom is attached to each C atom all C and H atoms are equivalent  Limitations of Kekulé’s structure The Kekulé structure predicts that there should be two different 1,2-dibromobenzenes. In practice, only ONE 1,2-dibromobenze has ever been isolated. Kekulé proposed that these two forms are in equilibrium which is established so rapidly that it prevents isolation of the separate compounds. Later, this proposal was proved to be incorrect. No such equilibrium exists. Br Br Br Br  Resonance theory applied to benzene When two (or more) Lewis structures can be drawn for one molecule, the true structure lies somewhere in between the two (or more). The true structure of benzene lies in between the two Kekulé forms and is referred to as a RESONANCE HYBRID. Resonance theory applied to benzene The different structures contributing to the resonance hybrid are referred to as CANONICAL FORMS of the molecule. The more stable the canonical form, the more it contributes to the resonance hybrid. The resonance hybrid is more stable than any of the canonical forms. The difference in energy between the most stable canonical form and the hybrid is known as the RESONANCE ENERGY or DELOCALISATION ENERGY. Resonance theory applied to benzene + H2 H = -121 KJ mol-1 cyclohexene cyclohexane + 3H2 H = -209 KJ mol-1 benzene cyclohexane If benzene contained three individual C=C bonds then we would expect the enthalpy of hydrogenation to be 3 x -121 = - 363 kJ mol-1. The observed value of -209 kJ mol-1 is lower by 154 kJ mol-1. This amounts to the resonance energy of benzene and shows that the resonance hybrid is much more stable than either (contributing) canonical form. Molecular Orbital (MO) theory for the structure of benzene The bond angle for each C-C bond in benzene is 120o. This suggests that every carbon atom is sp2 hybridised. The bond length of 1.39Å lies in between that of a single bond (1.54Å) and a double bond (1.33Å). Each carbon forms one sigma bond to a hydrogen atom, and two individual sigma bonds to other carbons. This leaves 1 electron remaining in each non-hybridised pz orbital. The remaining pz electron on each carbon overlaps with each other to form a cloud of electron density above and below the plane of the molecule. MO theory applied to benzene bond angles 120o hybridisation sp2 MO theory applied to benzene According to MO theory, the six overlapping p-orbitals combine to form a set of six π molecular orbitals Antibonding MO’s E ψ6 ψ4 ψ5 six ‘parent’ p-orbitals ψ2 ψ3 ψ1 Bonding MO’s MO theory applied to benzene Three MO’s have energy lower than the isolated p orbitals and are termed bonding orbitals. Three MO’s have energy higher than the isolated p orbitals and are termed antibonding orbitals. Orbitals ψ2 and ψ3 are of the same energy and are termed degenerate. The closed bonding (fully occupied) shell accounts, in part, for the stability of benezene. Comparison of cyclohexene with benzene Cylcohexene is an alkene and is non-aromatic. It reacts like an alkene and undergoes electrophillic addition reactions. Benzene is aromatic and undergoes electrophillic substitution reactions. Cylcohexene Benzene KMnO KMnO COOH 4 4 No reaction HO HO 2 COOH 2 + + HO HO 3 3 No reaction OH HCl Ether HCl Cl No reaction Ether Worked examples Q1. Which one of the following molecules is aromatic? A B C D E Answer:E (naphthalene) It is planar Each carbon has an electron in a p-orbital Naphthalene does obey Hückel’s 4n + 2 rule (it has 10π electrons). Worked examples Q2. How many π–electrons does the following molecule possesses? A 8 B 10 C 12 D 14 E 16 Answer:D (14π electrons) 7 double bonds, 2 electrons in each C=C double bond does obey Hückel’s 4n + 2 rule (where n=3).