R AS hemistry A 032 for first assessment in 206 omplete Tutor Notes www.boomerchemistry.com Section: 4.. Basic concepts of organic chemistry Naming organic compounds page hemical formulae page Structural isomerism page 4..2 Alkanes Alkanes page Radical substitution page 205 Boomer Publications 23 3 35 38 40
Naming rganic ompounds Page 23 33 Naming rganic ompounds All organic compounds contain carbon. There are millions of organic compounds and the reason there are so many is that carbon can form single, double or triple bonds with other carbon atoms as well as bonding to oxygen, hydrogen, nitrogen, phosphorous and halogens: l arbon can form single, double or triple bonds with other carbons and bonds readily to other elements so there are almost an infinite number of organic compounds possible N Each carbon makes 4 bonds - always check your structures to make sure every carbon has 4 bonds The simplest organic compounds are the hydrocarbons. These compounds only contain hydrogen and carbon. ydrocarbons are compounds that contain only hydrogen and carbon. The simplest of the hydrocarbons are the alkanes. Alkanes have carbons with only single bonds. They are said to be saturated compounds as they are completely saturated with hydrogens - you cannot get any more hydrogens on! x methane 4 x butane 2 x ethane 5 x pentane 3 x propane You must learn the names of the first ten alkanes 6 x = hexane 7 x = heptane 8 x = octane 9 x = nonane 0 x = decane
Page 24 Naming rganic ompounds 33 Monkeys Eat Peanut Butter, may help you remember the first 4 compounds in the series. Remember the rest by reference to polygons - e.g. heptagon (7 sides) - heptane (7 carbons) etc. Saturated compounds have only single carbon-carbon bonds. Unsaturated ydrocarbons ydrocarbons with one or more carbon-carbon multiple bonds are called unsaturated compounds. Alkenes have at least one carbon-carbon double bond and the names of the alkene series are similar to alkanes except they have the suffix (ending) ene. A difference being you must state the position of the double bond if there is more than one possibility. For example: 2 3 4 Propene - double bond can only go in one position so no need to number but-2-ene You number the carbons to give the lowest number possible. For example, here you say but-2-ene rather than but-3-ene. 2 possibilities for butene but--ene 2 The names of the alkenes are ethene, propene, butene, pentene etc. Unsaturated compounds are compounds containing one or more carbon-carbon multiple bonds. 3 4
Naming rganic ompounds Page 25 33 Naming Branched Alkanes Using IUPA Rules rganic compound names are composed of the: PREFIX STEM SUFFIX Find the STEM by finding how many carbons in the longest carbon chain: 2 3 4 The longest carbon chain in this compound has 5 carbons. This compound will have the STEM pent 5 The STEM names for the first ten compounds are meth, eth, prop, but (pronounced bute), pent, hex, hept, oct, non and dec nce you have identified the longest carbon chain, identify the branches and count the number of carbons in the branches. For example here: 6 5 4 3 2 The longest chain in this branched alkane is 6 carbons and so the STEM is hex. There is a reason why this compound is numbered from right to left - see later There are two branches, at position 2 - a one carbon chain, and position 4 - a two carbon chain.
Page 26 Naming rganic ompounds 33 The branches form the PREFIX (start) of the name. Branches are named according to how many carbons are present as follows: Number of carbons 2 3 4 5 6 7 8 9 0 PREFIX methyl ethyl propyl butyl pentyl hexyl heptyl octyl nonyl decyl ethyl branch on position 4 6 5 4 3 2 methyl branch on position 2 The full IUPA name of the compound above is: hyphens separate numbers and letters 4-ethyl-2-methylhexane compound is an alkane so has suffix ane ethyl comes before methyl alphabetically PREFIX STEM SUFFIX Numbering of the arbon hain Always number the carbon chain to give the lowest numbers. For example, the compound above could be numbered as follows: 2 3 4 5 6 Numbering this way would give the branches at positions 3 and 5 (adds to 8) Numbering from R to L gives branches at positions 2 and 4 (adds to 6) - use way that gives lowest total
33 Some more examples: Naming rganic ompounds Page 27 branch has only one carbon so prefix is methyl position of branch 4 3 2 2-methylbutane carbon chain numbered to give branch on the lowest number compound is an alkane so has suffix ane longest carbon chain has 4 carbons so STEM is but 5 4 3 2 three branches on positions 2, 2 and 4. Note if we numbered from L to R, branches would be on 2, 4 and 4 -number to give the lowest total longest carbon chain is 5 carbons long so STEM is pent 2,2,4-trimethylpentane compound is an alkane so has SUFFIX ane 2 commas separate numbers 3 4 all three branches have one carbon and so prefix will be trimethyl Naming Alkenes Naming alkenes is similar to alkanes except they have the suffix ene and you must state the position of the double bond, if there is more than one possible place it can go. carbon branch on position 2 so PREFIX 2-methyl The compound is an alkene so has SUFFIX, ene. Double bond is on position 2-methylbut--ene 4 carbons in longest carbon chain so has STEM but
Page 28 Naming rganic ompounds 33 Naming aloalkanes - compounds that contain halogens F, l, Br or I bonded to a carbon. Treat these as you would branches. Br 2 3 Br has 2 bromines on positions and 2 so PREFIX is,2-dibromo,2-dibromopropane compound is a haloalkane so has the SUFFIX ane chain has 3 carbons so has STEM prop Br 2 3 4 2-bromo-3-methylbutane bromine and methyl groups in positions 2 and 3 respectively so PREFIX is 2-bromo-3-methyl (bromine before methyl alphabetically) 4 carbons in chain so STEM is but compound is a haloalkane so SUFFIX is ane Functional Groups The functional group is part of the molecule responsible for its reactions. These change the suffix. therwise, the naming is the same as for alkanes. alkene suffix ene aldehyde suffix anal carboxylic acid suffix anoic acid ketone suffix anone alcohol suffix anol
Naming rganic ompounds Page 29 33 Examples of each: has a methyl group on position 2 so has the PREFIX 2-methyl 2-methylpropanal 3 carbons in carbon chain so STEM is prop compound is an aldehyde so has SUFFIX anal Br 5 4 3 2 3-bromo-3-ethylpentanoic acid bromine and ethyl group on position 3 so PREFIX is 3-bromo-3-ethyl (bromine is placed before the ethyl alphabetically) 5 carbons in chain so STEM is pent carboxylic acid so SUFFIX is anoic acid 5 4 3 2 3-methylpentan-2-one has a methyl group on position 3 so PREFIX is 3-methyl the chain has 5 carbons so STEM is pent 3 carbons in chain so has STEM prop compound is ketone so has the suffix anone. The = is on position 2 so has the SUFFIX an-2-one Br 2 3 I 2-bromo--chloro-3-iodopropan-2-ol l bromine, chlorine and iodine on positions 2, and 3 respectively so has PREFIX 2-bromo--chloro-3- iodo (placed in alphabetical order, starting with chloro on position - comes alphabetically first) compound is an alcohol so has the suffix anol. The is on carbon 2 so SUFFIX is an-2-ol
Page 30 Naming rganic ompounds 33 Multiples of the Same Group When you have more than one of the same group attached to a carbon chain, you use the following: di : 2 groups the same tri : 3 groups the same tetra : 4 groups the same For example: l l Br Br 5 4 3 2 Br l l,,3-tribromo-4,4,5,5-tetrachloro-2,3-dimethylpentane
hemical Formulae Page 3 34 Empirical Formulae and Molecular Formulae Empirical formula gives the simplest ratio of each atom of each element in a compound. Molecular formula gives the actual number of atoms of each element. Example: A hydrocarbon has the relative molecular mass of 56.0 and a % composition by mass of 85.7% carbon and 4.3% hydrogen. alculate the emperical formula. We can say that 00g of material would have 85.7 g of carbon and 4.3 g of hydrogen. mass of material in grams m / g 85.7 4.3 M / g mol- 2.0.0 divide mass, m, by molar mass, M, to get the number of moles Number of moles 7.4 7.4 The simplest whole number ratio 2 Molar mass M m 4.3 divide number of moles by the 7.4 smallest value (7.4) to get simplest ratio n number of moles The empirical formula is therefore 2 The compound has a relative molecular mass of 56.0 so we need to work out how many units of 2 there are in 56 ne unit of 2 has a relative mass of 4.0 so to find out how many of these are in 56.0: 56 4 = 4 units 4 x 2 = 4 8 This is the molecular formula
Page 32 hemical Formulae 34 Displayed Formula, Structural Formula and General Formula The displayed formula is when all the atoms and all the bonds are shown. For example, 2-methylbutane has the displayed formula: Its molecular formula would be 5 2. In this case, the empirical formula (simplest ratio of each atom of each element) would also be 5 2 The structural formula would look like this: 3 ( 3 ) 2 3 Note how the branched methyl group is shown in brackets to show it is a branch at position 2 on the carbon chain. ere are some more examples: Displayed Formula Structural Formula 3 () 3 Molecular Formula 3 8 propan-2-ol 2-methylpropanoic acid 3 ( 3 ) 4 8 2
hemical Formulae Page 33 34 Displayed Formula Structural Formula Molecular Formula Br 2 (Br) 2 3 5 Br 3-bromopropanal Skeletal Formula This is where carbon chains are represented by a zig zag line and the carbon to hydrogen bonds are not shown. For example: = ethane = propane = 2-methylpentane
Page 34 hemical Formulae 34 For bonds other than the carbon to hydrogen, these bonds are shown along with the chemical symbol to represent the element l l l = l 2,2-dichloropropan--ol = propene = cyclopentane
Structural isomerism Page 35 35 Isomerism There are different types of isomerism. In AS chemistry you need to know about structural isomerism and Z/E isomerism which is a form of stereoisomerism. You will learn about Z/E isomerism in the Alkenes section. Structural isomers have the same molecular formula but a different structural formula. Examples of structural isomers: butane and 2-methylpropane 4 0 the same molecular formula 4 0 3 2 2 3 different structural formula 3 ( 3 ) 3 Another example: propan--ol and propan-2-ol 3 8 the same molecular formula 3 8 3 2 2 different structural formula 3 () 3
Page 36 Structural isomerism 35 When trying to find different structural isomers for the same molecular formula, it is often easiest to use skeletal formulae. For example, finding the alkane structural isomers of 6 4 start with a straight chain first next, move a methyl group around 2 3 X next, move two methyl groups around these two are the same compound so are not isomers 4 5 If you are unsure whether two structures are different structural isomers, name the two structures using IUPA rules. If the two structures come out with the same name, they are the same compound. In the example above, both structures give the name 2-methylpentane.
Structural isomerism Page 37 35 General Formula General formula is a mathematical type expression that describes a homologous series (see later). Alkanes have the general formula n 2n+2 It is therefore possible to work out the molecular formula for an alkane with any number of carbons. For example, an alkane with 24 carbons (n=24), would have the molecular formula would be 24 50 Alkenes have the general formula n 2n omologous Series omologous series have the same functional group but each successive member of the series differs by 2 For example, the first 3 compounds in the homologous series of carboxylic acids: methanoic acid ethanoic acid propanoic acid yclic ompounds ere are some cyclic compounds which have the general formula n 2n cyclopropane cyclobutane cyclohexane
Page 38 Alkanes 36 Why Do the Alkanes ave Different Boiling Points? Alkanes are nonpolar, covalently bonded compounds with a simple molecular structure. The forces that need to be overcome therefore to boil them are induced dipole-dipole interactions. Long chain hydrocarbons have more surface contact (NT GREATER SURFAE AREA) than short chain hydrocarbons. The induced dipole-dipole interactions for long chain hydrocarbons are greater and so boiling points are higher as more energy is required to overcome these forces. Branched hydrocarbons have less surface contact than their equivalent straight chain isomers and so increased branching leads to a lowering of boiling points. For example, the structural isomers of 7 6 eptane Boiling point = 98 2,3-dimethylpentane Boiling point = 89.5 2,2,3-trimethylpentane Boiling point = 80.9 INREASED BRANING leads to... LESS SURFAE NTAT leads to... DEREASE IN INDUED DIPLE-DIPLE INTERATINS leads to... DEREASE IN BILING PINTS
Alkanes Page 39 36 Alkanes as Fuels You must be able to write equations for the complete and incomplete combustion of these compounds and state the conditions under which they occur. Incomplete ombustion This occurs when there is limited supply of oxygen. ne of the products is carbon monoxide which is toxic. People that have gas fired boilers are advised to have carbon monoxide detectors. Tips for balancing combustion equations: 3 8 (g) +? 2 (g) 3(g) + 4 2 (l) start with just one molecule of the thing you are burning, then work out how many molecules of carbon monoxide it will make (3 carbons in propane so will make 3 x ) and how many molecules of water (8 hydrogens in propane will make 4 x 2 ) You then work out how many molecules of oxygen you need to balance. We can see we have 7 atoms of oxygen of the right hand side so we will need 3½ molecules of oxygen (it is perfectly acceptable to use half molar quantities) 3 8 (g) + 3½ 2 (g) 3(g) + 4 2 (l) omplete ombustion When the oxygen supply is plentiful, complete combustion occurs and carbon dioxide is produced rather than carbon monoxide. When balancing the equation, do as before, work out how many molecules of carbon dioxide and water will be made and then work out how many molecules of oxygen will be needed: 3 8 (g) + 5 2 (g) 3 2 (g) + 4 2 (l)
Page 40 Radical Substitution 36 Radical Substitution Mechanisms show the steps for how we believe a reaction takes place at a molecular level. When writing mechanisms, we draw curly arrows to show the movement of a pair of electrons. For example, when a covalent bond breaks by heterolytic fission we show the shaired pair of electrons moving from the bond to one of the atoms to form a negatively charged ion: A X A + + X - When bonds break by homolytic fission, the shaired pair of bonding electrons are split equally resulting in two species with a single, unpaired electron which we call radicals. When we want to show a single electron moving we use an arrow with half an arrow head which looks like a fish hook: A X A. + X. radicals Radical Substitution of Alkanes to form aloalkanes The overall equation for the radical substitution of an alkane to form a haloalkane is as follows: u. v. light 2 6 (g) + l 2 (g) 2 5 l(g) + l(g) We will now go through the mechanism to show the individual steps for how we believe this reaction takes place. st Stage - Initiation l l l. + l. chlorine molecule u. v. light chlorine radicals The initiation step involves the halogen, in this case chlorine, undergoing homolytic fission due to the presence of ultraviolet light to form two chlorine radicals. Radicals are very reactive.
Radical Substitution Page 4 36 2nd Stage - Propagation The propagation stage involves two repeated steps that build up the chain reaction. This is the stage where the two products are formed as well as further radicals. The first propagation step involves the reaction of the alkane, in this case ethane with the chlorine radical. ne of the products, hydrogen chloride, is formed along with an alkyl radical:.l. + l alkyl radical The second propagation step involves the alkyl radical formed in the previous step reacting with a chlorine molecule to produce the other product, in this case chloroethane, and a further chlorine radical.. l l l +.l chloroethane chlorine radical 3rd Stage - Termination The final stage is the termination stage, which as the name suggests, brings the chain reaction to a close. This is where two radicals come together to form a molecule. From the chlorine and alkyl radicals we have made so far, there are 3 possibilities:.l.l l l Two chlorine radicals can pair up to form a chlorine molecule.
Page 42 Radical Substitution 36.. Two alkyl radicals can come together to form a longer chain alkane, in this case butane is produced...l l And chlorine and alkyl radicals can come together to produce a haloalkane. Radical reactions are difficult to control and will inevitably produce a plethora of products in addition to the desired products. The mechanism for the reaction above can be written in condensed form as follows. It does not matter which of the ways you learn to describe the mechanism. st Stage - Initiation u l 2 (g). v. light.l(g) +.l(g) 2nd Stage - propagation Step.l(g) + 2 6 (g). 2 5 (g) + l(g) Step 2. 2 5 (g) + l 2 (g) 2 5 l(g) +.l(g) 3rd Stage - Termination.l(g) +.l(g). 2 5 (g) +. 2 5 (g) l 2 (g) 4 0 (g). 2 5 (g) +.l(g) 2 5 l(g)