2007 ch158 a3 basicsorg update for 2009

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Warwick University Department of Chemistry Year 1, Course CH158: Foundations of Chemistry Section A3; Basics of Organic Chemistry Professor Martin Wills m.wills@warwick.ac.uk Important: Please bear in mind that organic chemistry ‘builds upon itself’ – you must make sure that you fully understand the earlier concepts before you move on to more challenging work Professor M Wills CH158 Year A3 Basics of Organic Chemistry Year Foundation course, Section A3; Nomenclature of Organic Compounds IUPAC has defined systematic rules for naming organic compounds These will have already been covered in detail at A-level and will only be mentioned briefly here The naming system (and the resulting names) can become very long with complex molecules, therefore this section will be restricted to simple compounds The IUPAC naming system involves the following components: - Identification of major chain or ring - Side chains and functional groups are added as appropriate, in alphabetical order - The sums of numbers for substituents are minimised Professor M Wills CH158 Year A3 Basics of Organic Chemistry Year Foundation course, Section A3; Nomenclature of Organic Compounds Examples: CH3 H2 C H2 C H3C C H2 CH C H2 CH3 C H2 is 3-methyloctane, not 5-methyloctane 2' major chain H3C H3C H3C C H2 H3C CH2 CH H2 C CH H2 C C C H2 1' CH CH CH3 CH H2 C C CH3 CH3 CH3 CH3 CH3 Is 5-(1’-methylethyl)-2,2,4-trimethyloctane CH3 CH3 H2C CH3 Is 4,5-diethyl-2,2-dimethylheptane It is NOT 3,4-diethyl-6,6-dimethylheptane! H3C H2 C CH CH3 H3C H2 C CH OH Butan-2-ol Professor M Wills CH3 Cl 2-chlorobutane CH158 Year A3 Basics of Organic Chemistry Year Foundation course, Section A3; Nomenclature of Organic Compounds Many common names persist in organic chemistry, despite IUPAC rules, e.g Compound ‘common’ name IUPAC name O H3C C CH3 Acetone Propanone Formaldehyde Methanal Acetic acid Ethanoic acid Dimethylether Methoxymethane O H C H O H3C H3C Professor M Wills C OH O CH3 CH158 Year A3 Basics of Organic Chemistry Year Foundation course, Section A3; Substitution level and functional groups The ‘substitution level’ of a carbon atom in an organic compound is determined by the number of attached hydrogen atoms: tertiary carbon (one H) H3C H3C CH H2 C CH3 H2 C CH C H2 Primary C (3 hs) CH CH3 CH3 CH3 Secondary C (2 Hs) CH3 C Quaternary C (0 Hs) The rules differ for certain functional compounds e.g alcohols: H3C H3C H2 C OH Primary alcohol (2Hs on C attached to O) Professor M Wills CH3 H3C CH OH C OH Secondary alcohol (1H on C attached to O) CH3 CH3 Tertiary alcohol (0Hs on C attached to O) CH158 Year A3 Basics of Organic Chemistry Year Foundation course, Section A3; Substitution level and functional groups In the case of AMINES, the rules are different: H3C H3C H H N N H CH3 Primary amine (2Hs on N) CH2 H3C N CH3 Secondary amine (1H on N) Tertiary amine (0Hs on N) Aromatic compounds: substitution position relative to group ‘X’ X Professor M Wills para meta ortho CH158 Year A3 Basics of Organic Chemistry Year Foundation course, Section A3; Substitution level and functional groups Functional groups will be dealt with as they arise, however the following should be committed to memory: R= alkyl group, joining at C atom, e.g CH3, c-C6H11, CH2CH2CH3 etc H R R R OH Alcohol NH2 C O Amide O Iodide Bromide Carboxylic acid Cl R C C O O Ketone R H R C C C C O O O N Acid Chloride Anhydride R R Ester O R Aldehyde OR R C I Br R OH R O Chloride Thiol R R Cl SH Amine C R R NH2 Imine O R N R O Nitro group A cyclic ester is called a lactone, a cyclic amide a lactam Professor M Wills CH158 Year A3 Basics of Organic Chemistry Year Foundation course, Section A3; Line drawing - the standard from this point in the course Line drawing represents an abbreviated ‘shorthand representation of organic structures: The rules are simple- Structures are written as a series of interconnected lines where each apex is the position of a carbon atom Heteroatoms (i.e not H or C) are shown H atoms are not shown with the exception of those on heteroatoms Examples Ethanol Ethanal Full structure H3C OH OH O O C H2 H3C C H Propene Abbreviated 'line-drawing' structure H3C CH2 N.b in some cases the H atom of an aldehyde may be illustrated C H H Benzene H C C H C C H C C H Professor M Wills H CH158 Year A3 Basics of Organic Chemistry Year Foundation course, Section A3; Oxidation level This is a useful tool for the understanding of organic reactions It is slightly different to the system used for the oxidation level of cations and anions In some cases it is obvious that a reaction is an oxidation or reduction, in other cases they are not, for example: H3C H3C H3C OH OH C H2 oxidation H3C O H3C (removal of two H atoms) CH2 C H reduction H3C (addition of two H atoms) CH2 C H Professor M Wills O C H CH3 C H2 Oxidation or reduction? (addition of O and of two H atoms) OH H3C H2 C C H2 OH CH158 Year A3 Basics of Organic Chemistry Year Foundation course, Section A3; Oxidation level To assign oxidation number (Nox), identify each each carbon atom that changes and assign oxidation numbers as follows: a) For each attached H assign ‘-1’ b) For each attached heteroatom (O, N, S, Br, Cl, F, I etc.) assign ‘+1’ c) Double or triple bonds to heteroatoms count double or triple respectively Then sum them for each molecule Example Ethanol Ethanal H3C H3C OH C H2 total -4 Nox =-1 (1 attached O atom, attached H atoms) O C H Nox = -3 (3 attached H atoms) Nox = -3 (3 attached H atoms) Nox = +1 (1 double bond to O atom, attached H atoms) total -2 A change of ‘+2’ indicates an oxidation A change of ‘-2’ indicates a reduction note + or -2 is the typical change in oxidation level Professor M Wills CH158 Year A3 Basics of Organic Chemistry 10 Year Foundation course section A3; Molecules in 3D Some things to be aware of: i) Symmetrical, tetrahedral, compounds have no overall dipole: H δ+ For example in the ammonium cation, NH4+ H δ+ H N δ+ The shape is still tetrahedral, and eacn N-H bond has a dipole, but there is no overall dipole, because they cancel out H δ+ H For the same reason, methane has no overall dipole either: ii) H C H H Molecules which are electron deficient, such as borane (BH3), retain a trigonal shape Why? – Well, without an electron pair, there is nothing to repel with!!! Borane, BH3: H H B (one B atom with three electrons and three H atoms with one electron each form borane, which has only electrons at the B atom: H Flat (trigonal), 120o bond angles no overall dipole H H B H H : :B : H H B In contrast, borohydride anion, NH4-, is tetrahedral, with no overall dipole: Professor M Wills H H H B H H H (3 sp2 orbitals and one p orbital, orthogonal to plan of BH3 atoms) CH158 Year A3 Basics of Organic Chemistry 26 Year Foundation course section A3; Molecules in 3D In the case of a carbon atom attached to three other groups (by two single bonds and one double bond) the single 2s and two2p orbitals mix (rehybridise) to form three sp orbitals These are all arranged at mutual 120 degree angles to each other and define a trigonal shape, the remaining p orbital projects out of the plane of the three sp2 orbitals and overlaps with an identical orbital on an adjacent atom to form the double bond: x 2s x 2p on a carbon atom combine to form x sp2 orbitals: C which lie at mutual 120 degrees in a molecule such as etheneC2H6, whilst the remaining p orbital forms the double bond: H H H H H C C H H H The resulting structure is rigid and cannot rotate about the C=C bond without breakage of the bond between the p-orbitals (the π bond) Professor M Wills CH158 Year A3 Basics of Organic Chemistry 27 Year Foundation course section A3; Molecules in 3D In the case of a carbon atom attached to two other groups (by one single bonds and one triple bond) the single 2s and one 2p orbitals mix (rehybridise) to form two sp orbitals These are all arranged at mutual 180 degree angles to each other and define a linear shape, the remaining p orbitals projecting out from the sp orbital to overlap with identical orbitals on an adjacent atom to form the triple bond: x 2s x 2p combine to form x sp2 orbitals: on a carbon atom C which lie at 180 degrees in a molecule such as ethyneC2H4, H whilst the remaining p orbitals form the triple bond: H H C C H Rehybridisation of orbitals of this type is not limited to carbon, of course Many other row and atoms (notably N) can rehybridise within organic molecules Professor M Wills CH158 Year A3 Basics of Organic Chemistry 28 Year Foundation course section A3; Molecules in 3D Conformation and configuration Configuration is a fixed stereochemical property of compounds Unlike conformation, a change in configuration requires bonds to be broken and formed Any molecule has a limited number of configurations in which it can exist Alkenes can exist in two configurations, for example but-2-ene may have the terminal methyl groups in a trans (across from each other) or cis (on the same side) position: H H3C C trans-but-2-ene H H3C cis but-2-ene C CH3 CH3 C H C H Changing trans butadiene into cis- butadiene (or vice versa) requires the breaking, and subsequent reforming, of the π bond This is a high- energy process and does not take place at room temperature At room temperature, but-2-ene (and other alkenes) can be physically separated into the two pure isomers Professor M Wills CH158 Year A3 Basics of Organic Chemistry 29 Year Foundation course section A3; Molecules in 3D - Conformation and configuration The configuration of an alkene can be obvious in some cases (such as but-2-ene) however in others it is not, for example is the molecule below a cis or trans alkene? H2C H3C C H CH3 C CH3 In order to provide an unambiguous means for assigning configuration to alkenes (and also to chiral centres as you will see later), organic chemists have adopted the ‘Cahn-Ingold-Prelog’ (CIP) rules for configurational assignment These are simple to use - first one assigns a ‘priority’ to each group attached to each carbon atom at each end of the alkene I will describe to priority rules in the next slide We then define the alkene as either Z (from the German zusammen, together) or E (from the German entgegen, across): High High C Z alkene: Low C E alkene: C Low High Low High C Low (relative priorities are of a group on a C atom to its partner on the same atom) Professor M Wills CH158 Year A3 Basics of Organic Chemistry 30 Year Foundation course section A3; Molecules in 3D Conformation and configuration The CIP priority rules are defined as follows, in their own order of priority: a) Atoms of higher atomic number have priority: CH3 H3C e.g C C C H H High High C H H In this molecule the attached carbon atoms at each end of the double bond have priority over the attached H atoms, hence this is a Z alkene b) When the attached atoms are identical on each side, isotopes of higher mass have priority D H3C e.g C H High High C C H H C H In this molecule the attached carbon and deuterium (deuterium is the H-2 isotope) atoms at each end of the double bond have priority over the attached H atoms, hence this is a Z alkene Professor M Wills CH158 Year A3 Basics of Organic Chemistry 31 Year Foundation course section A3; Molecules in 3D - Conformation and configuration The CIP priority rules are defined as follows, in their own order of priority: a) When the atoms and isotopes attached on each side are identical, move out until a point of difference is encountered and apply the following rules: a) Priority goes to the group with the element of highest atomic number at the point of difference H2C H e.g C H3C NH2 High Low C C C H3C CH3 CH3 C Low High E alkene b) Priority goes to the group with the highest sum of atomic numbers if the atoms are of the same types at the point of difference In the example below, the point of difference on the right hand side is two carbons away from the alkene carbon atom H2C H e.g C H3C CH2 H2C High Low CH3 C H2C CH3 CH C C Low High E alkene H2 C CH3 Professor M Wills CH158 Year A3 Basics of Organic Chemistry 32 Year Foundation course section A3; Molecules in 3D - Conformation and configuration This is how I worked out the last example (right hand side only): Imagine you are moving out from the Cα tothe adjacent atoms Going both up (to C1) and down (to C1') leads to a CH2 group, i.e no difference Moving to the next atom reveals that C2 (top branch) is attached to C,C,H but that C2' (lower branch) is attached to C,H,H The upper branch has the highest sum of attached atomic numbers and therefore has priority H2C CH H C H3C C α H2C 1' CH3 CH3 2' CH2 H2C Upper: Cα -> C1(C,H,H) -> C2(C,C,H) High Low C Lower: Cα -> C1(C,H,H) -> C2(C,H,H) H2 C C Low High E alkene CH3 There is one more rule: d) In the case of double and triple bonds, ‘dummy’ atoms should be added and counted in the determination of priority See next slide Professor M Wills CH158 Year A3 Basics of Organic Chemistry 33 Year Foundation course section A3; Molecules in 3D - Conformation and configuration Here is an example of the determination of configuration for an alkene attached to a double bond: C HC H C α C H3C H2C 1' H CH2 2' CH2 H2C CH C C α H2C H3C H2 C C CH2 2' CH2 1' H2C CH3 < dummy atoms Upper: Cα -> C1(C,C,H) H2 Lower: Cα -> C1(C,H,H) C CH3 High Low C C Low High E alkene Professor M Wills CH158 Year A3 Basics of Organic Chemistry 34 Year Foundation course section A3; Molecules in 3D - Conformation and configuration CIP priority rules are also applied to the determination of configuration at chiral centres (a chiral molecule is one which is not superimposable on its mirror image, rather like your hands) The simplest form of a chiral centre is one with a carbon atom attached to four different groups E.g mirror CBrClFH is a chiral molecule the two forms (known as enantiomers) can be illustrated thus: F Cl F C C Br H H Cl Br To assign a configuration to a chiral molecule such as the one shown above we first assign CIP priorities to all four groups using the same rules: e.g Cl Br Professor M Wills F C H C = highest priority to = lowest priority CH158 Year A3 Basics of Organic Chemistry 35 Year Foundation course section A3; Molecules in 3D - Conformation and configuration We then view the molecule, with the assigned priorities, along the C-4 bond (with the behind the central carbon atom Finally, draw an arrow from atom with priority to priority to priority in turn: 3 atom is behind C C C In this case the arrow is clockwise; this is therefore referred to as a R isomer (R comes from the Latin rectus, for ‘right’) Isomers of this type are sometimes called ‘enantiomers’ F Cl Br R enantiomer C H The mirror image of the molecule above is the S enantiomer (from the Latin sinister for ‘left’) F C H Professor M Wills Cl Br C C S enantiomer CH158 Year A3 Basics of Organic Chemistry 36 Year Foundation course section A3; Molecules in 3D - Conformation and configuration Here are a couple of examples - can you see the derivation of the configuration? H2N SH R configuration SH 1 H CH3 H2N H CH3 S configuration 2 One carbon atom makes all the difference! Professor M Wills CH158 Year A3 Basics of Organic Chemistry 37 Year Foundation course section A3; Molecules in 3D - Conformation and configuration This is quite important – please read it Some other conventions are used to defined the configuration at chiral centres e.g l - molecule with a negative optical rotation (from the Greek for levorotatory; left) d - molecule with a positive optical rotation (from the Greek for dextrarotatory; right) The D/L notation ( a very old convention) is derived from the signs of optical rotation of R and S glyceraldehyde respectively: HO H H OH HO CHO D (+)-glyceraldehyde (also R) HO CHO L (-)-glyceraldehyde (also S) The trivial convention for the absolute configurations of sugars derives from the D/L notation above D-glucose is the natural enantiomer (costs £20/kg) whilst L-glucose is very rare (£31/g!) Professor M Wills CH158 Year A3 Basics of Organic Chemistry 38 Year Foundation course section A3; Molecules in 3D - Conformation and configuration Amino acids are classified into L- (natural) and D- (unnatural) R H H R H2N CO2H L - amino acid H2N CO2H D- amino acid Most L-amino acids are of S- configuration Despite all the different notations, R and S is the one YOU should learn how to use Professor M Wills CH158 Year A3 Basics of Organic Chemistry 39 Year Foundation course section A3; Molecules in 3D - Conformation and configuration The two mirror-images of chiral compounds can have dramatically different physical properties That is because we ourselves are made up of molecules of one ‘handedness’ Try assigning R/S to these: O NHiPr O OH O NHiPr CO2H OH CO2H NH2 HS Et O Cl Et O Cl N O Et O NH N O O NH Thalidomide R is mutagenic S is anti-emetic N Professor M Wills O Penicillamine R is an antiarthritic S is a highly toxic Propranolol: R is a heart drug S is a contraceptive Limonene: R has a lemon odour S has an orange odour N NH2 HS O CO2H CO2H O Et 'Dual' R is an herbicide S is a pesticide Ibuprofen R is anti-inflammatory S is inactive CH158 Year A3 Basics of Organic Chemistry 40 [...]... FC(C)=1 -(1)-2(0) = 0 Hence the formal charge on each atom in ethane is zero N.b - use a atomic group number of ‘1’ for hydrogen * i.e count from 1 to 8 across the row Professor M Wills CH158 Year 1 A3 Basics of Organic Chemistry 19 Year 1 Foundation course, Section A3; Formal Charge Further examples: Ethene - H C H for each (equivalent) C atom, FC(C)=4 -(4)-2(0) = 0 for each (equivalent) H atom, FC(H)=1... atom, FC(H)=1 -(1)-2(0) = 0 for the N atom, FC(O)=5 -(4)-2(0) = +1 Hence the formal charge on the atoms in the molecule is: H H N H H CH158 Year 1 A3 Basics of Organic Chemistry 20 Year 1 Foundation course, Section A3; Formal Charge Further examples: Protonated water (overall charge of +1 and a lone pair on O) H O H H for each (equivalent) H atom, FC(H)=1 -(1)-2(0) = 0 for the O atom, FC(O)=6 -(3)-2(1)... later section) Professor M Wills CH158 Year 1 A3 Basics of Organic Chemistry 18 Year 1 Foundation course, Section A3; Formal Charge Formal charge is a method for assigning charge to individual atoms in molecules Although it does not always give a ‘perfect’ picture of true charge distribution, it is very helpful when reaction mechanisms are being illustrated The definition of formal charge on a given (row... Hence the formal charge on each atom in ethene is zero C H Methoxymethane (remember this moelcule has two lone pairs on O) H H C H O C H H for each (equivalent) C atom, FC(C)=4 -(4)-2(0) = 0 for each (equivalent) H atom, FC(H)=1 -(1)-2(0) = 0 for the O atom, FC(O)=6 -(2)-2(2) = 0 H Hence the formal charge on each atom is zero Ammonium cation (overall charge of +1) H H N H Professor M Wills H for each... Borane-ammonia complex for each (equivalent) H atom, H H FC(H)=1 -(1)-2(0) = 0 H N B H for the N atom, FC(O)=5 -(4)-2(0) = +1 H H for the B atom FC(B)=3-(4)-2(0) =-1 Methyl cation (only 6 electrons around C): Hence the formal charge on the atoms in the molecule is: Hence the formal charge on the atoms in the molecule is: C H H H H H N B H H H F F H O Tetrafluoroborate anion: H H H B F F Use the formal charge... orbital low energy, σ The electrons 'drop' into a lower energy position, which provides a driving force for the reaction, and stability Always bear this in mind when thinking about molecular orbital structure Professor M Wills CH158 Year 1 A3 Basics of Organic Chemistry 16 Year 1 Foundation course, Section A3; Bond Polarity Covalency suggests equal sharing, but this is rarely the case because atoms differ... the rows can gain or lose an electron to form an ionic lattice (e.g NaCl) The simplest example is where two hydrogen atoms combine to form H2, with a covalent bond between the atoms: H + H Two H atoms, 1 electron each Professor M Wills H : H H H covalent bond- electrons shared CH158 Year 1 A3 Basics of Organic Chemistry 11 Year 1 Foundation course, Section A3; Molecular Stability - covalent vs ionic... atom (4 outer electrons) to form methane: Ethane C2H6 Combine 6 H atoms (1 outer electron each) and 2 C atom (4 outer electrons) to form ethane H H C H H H H H H C C H H H : C :C H H : H H H H Ethene C2H4 Combine 4 H atoms (1 outer electron each) and 2 C atom (4 outer electrons) to form ethen with double bond: Professor M Wills H H : : C :: C H H H C H H C H CH158 Year 1 A3 Basics of Organic Chemistry... filled orbitals, which contain negatively charged electrons, and therefore repel each other This is known as the ‘valence shell electron pair repulsion’ (or VSEPR), and often dominates the shape of molecules Professor M Wills CH158 Year 1 A3 Basics of Organic Chemistry 24 Year 1 Foundation course section A3; Molecules in 3D The VSEPR model for the structure of molecules also explains why molecules such as... dipole lone pair CH158 Year 1 A3 Basics of Organic Chemistry 25 Year 1 Foundation course section A3; Molecules in 3D Some things to be aware of: i) Symmetrical, tetrahedral, compounds have no overall dipole: H δ+ For example in the ammonium cation, NH4+ H δ+ H N δ+ The shape is still tetrahedral, and eacn N-H bond has a dipole, but there is no overall dipole, because they cancel out H δ+ H For the same
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