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Basics of Organic Chemistry and Mechanism Revision for When You Are Lost and Confused Professor Martin Wills [email protected]

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Basics of Organic Chemistry and Mechanism Revision for When You Are Lost and Confused Professor Martin Wills M.Wills@Warwick.Ac.Uk Warwick University Department of Chemistry Basics of Organic Chemistry and Mechanism Revision for when you are Lost and Confused Professor Martin Wills [email protected] Unlike many other subjects, organic chemistry ‘builds upon itself’ – you must make sure that you fully understand the earlier concepts before you move on to more challenging work. In my experience, a lot of students struggle with organic chemistry because they ‘don’t get’ mechanisms. Anyone in this position will probably do badly in their degree course. You can’t succeed just by memorising, you have to understand the mechanisms. If you do feel stuck, lost and confused with organic chemistry then please read and work through this document and it should help you. Some is very basic – but none is meant to be patronising – so if you are 100% happy with a section then move on and work through it at your own pace. The secret to learning this is to be absolutely sure that you know and understand what is going on at each stage. Memorising is not good enough. Don’t move on until you are absolutely sure that you understand. 1 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 2 Nomenclature of Organic Compounds Examples: CH3 H2 H2 CH C C CH3 is 3-methyloctane, H3C C C C H2 H2 H2 not 5-methyloctane 2' major chain 1' H3C CH3 CH H H 8 2 2 2 C 6 CH 4 C CH3 H3C H3C C CH C 1 CH2 7 H2 5 3 H 2 2 CH3 CH3 CH3 H3C 6 CH 4 C CH3 C CH C 1 Is 5-(1’-methylethyl)-2,2,4-trimethyloctane 7 H2 5 3 H2C CH3 CH3 CH3 Is 4,5-diethyl-2,2-dimethylheptane H H 4 2 2 2 2 C CH3 4 C CH3 It is NOT CH H3C 1 H3C CH 1 3,4-diethyl-6,6-dimethylheptane! 3 3 OH Cl Butan-2-ol 2-chlorobutane Professor M. Wills 3 Nomenclature of Organic Compounds Many common names persist in organic chemistry, despite IUPAC rules, e.g. Compound ‘common’ name IUPAC name O C Acetone Propanone H3C CH3 O C Formaldehyde Methanal H H O C Acetic acid Ethanoic acid H3C OH O Dimethylether Methoxymethane H3C CH3 Professor M. Wills 4 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 CH3 CH Primary C (3 hs) H2 H2 C CH C CH3 H3C C CH C H2 CH3 CH3 CH3 Secondary C (2 Hs) Quaternary C (0 Hs) The rules differ for certain functional compounds e.g. alcohols: CH3 H3C H3C OH H2 CH C C H3C OH OH CH3 CH3 Primary alcohol Secondary alcohol Tertiary alcohol (2Hs on C attached to O) (1H on C attached to O) (0Hs on C attached to O) Professor M. Wills 5 Substitution level and functional groups In the case of AMINES, the rules are different: H3C H H3C H H3C CH2 N N N H CH3 CH3 Primary amine Secondary amine Tertiary amine (2Hs on N) (1H on N) (0Hs on N) Aromatic compounds: substitution position relative to group ‘X’ X para meta ortho Professor M. Wills 6 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.. R H C R R R R NH2 SH Cl OH O Alcohol Amine Thiol Chloride Aldehyde R OH R OR R R R R C C C Br I Bromide Iodide O OO Carboxylic acid Ester Ketone R NH2 R Cl R O R R H R O C C C C C N OOOO O N R Amide Acid Chloride Anhydride Imine Nitro group A cyclic ester is called a lactone, a cyclic amide a lactam Professor M. Wills 7 Line drawing 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 Full structure Abbreviated 'line-drawing' structure H3C OH OH Ethanol C H2 H C O O N.b. in some cases Ethanal 3 C the H atom of an H aldehyde may be H3C CH2 illustrated Propene C H H H Benzene C C H C C H CC H H Professor M. Wills 8 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. It is useful to know whether a reduction or oxidation takes place, because this allows a correct selection of reagent to be made. In some cases it is obvious that a reaction is an oxidation or reduction, in other cases they are not, for example: H3C OH C OH oxidation O H3C O H2 H3C (removal of two H atoms) C H CH H3C 2 reduction C H C CH3 H 3 (addition of two H atoms) C H2 CH2 H3C Oxidation or reduction? H2 C H C C H OH 3 (addition of O and C OH of two H atoms) H2 Professor M. Wills 9 Oxidation level To assign oxidation number (Nox), there are a number of methods you can use however this one is easy - 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 N = -3 (3 attached H atoms) OH ox H3C total -4 Ethanol C H 2 Nox =-1 (1 attached O atom, 2 attached H atoms) Nox = -3 (3 attached H atoms) H C O Ethanal 3 C total -2 H Nox = +1 (1 double bond to O atom, 1 attached H atoms) A change of ‘+2’ indicates an oxidation. A change of ‘-2’ indicates a reduction. note + 2 or -2 is the typical change in oxidation level. Professor M. Wills 10 Molecular Stability - covalent vs ionic bonding Many factors dictate the stability of atoms and ions. Hydrogen atoms gain stability if there are two electrons in their electron shell. For first and second row elements, significant stability is derived from an outer electronic configuration with 8 electrons. Atoms can achieve this by i) gaining or losing electrons or ii) sharing them. In the periodic table: Central elements cannot easily lose or gain electrons and prefer to bond covalently. These are H the most common atoms in organic compounds Li Be B C N O F Ne row 1 Na Mg Al Si P S Cl Ar row 2 Atoms at the extremes of 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 . : H H HH Two H atoms, covalent bond- electrons shared 1 electron each. Professor M. Wills 11 Molecular Stability - covalent vs ionic bonding Examples (nb the three dimensional shapes of the molecules will be discussed in a later section) Methane, CH4 H H Combine 4 H atoms (1 outer electron .. each) and 1 C atom (4 outer electrons) H ::C H HC H to form methane: .. H H Ethane C2H6 H H H H Combine 6 H atoms (1 outer electron .. .. each) and 2 C atom (4 outer electrons) H ::C C : H HC C H to form ethane .. .. H H H H Ethene C2H4 H Combine 4 H atoms (1 outer electron H H H : : each) and 2 C atom (4 outer electrons) C :: C CC to form ethen with double bond: .. .. H H H H Professor M. Wills 12 Molecular Stability - covalent vs ionic bonding Examples of covalent compounds – I realise that this is very basic however some people still struggle with it. It is not meant to be patronising so please move on rapidly if you are OK with this: Ethyne C2H2 Combine 2 H atoms (1 outer electron each) and 2 C atom (4 outer electrons) H: C ::: C : H C CH H to form ethyne with triple bond: Methoxymethane (dimethylether) H H H .. .. H.. .. H : C : O : C : H H C OC H Combine two C atoms (4 electrons), .. .. .. .. H one O atom (6 electrons) and six H H H H atoms (1 electron). Two pairs of electrons (lone pairs) reside on the oxygen atom and the molecule has a 'bent' structure. Methanal (formaldehyde) H .. H Combine one C atom, one O atom .. .. C ::O : C O and two H atoms with a C=O double .. .. bond. There are two lone pairs on O.
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