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Transcript
th Lecture Notes 4 Series: Organic Chemistry Nomenclature Due to the wide range of compounds that contain carbon, a systematic method has been introduced by IUPAC to name them. Alkanes Alkanes contain only carbon and hydrogen and are termed hydrocarbons. The important thing to remember about alkanes is that they contain only σ-bonds. The simplest alkane is methane. Question 4.1 (a) Predict the shape of the methane molecule. (b) What is the hybridisation at the carbon atom in methane? Methane contains four single carbon-hydrogen bonds. As it contains no β-bonds, methane is described as being saturated. In fact, all alkanes are saturated. The next member of the alkane family is ethane, C2H6. H H H C C H H H Now we have a carbon-carbon bond present. You will notice that we are still obeying the fundamental octet rule. Each carbon is forming four σ-covalent bonds and thus has eight electrons in its valence shell. Question 4.2 Draw a Lewis dot-cross diagram for ethane. The next alkane contains three carbons, the next one four carbons, the next one five carbons, and so on. The names arise directly from the number of carbons. The table below summarizes the first twelve members of the alkane family. The Alkanes No. of C atoms 1 2 3 4 5 6 7 8 9 10 11 12 Formula CH4 C2H6 C3H8 C4H10 C5H12 C6H14 C7H16 C8H18 C9H20 C10H22 C11H24 C12H26 48 Name Methane Ethane Propane Butane Pentane Hexane Heptane Octane Nonane Decane Undecane Dodecane A couple of important points arise from this table. • All the names end in ane - thus the alkanes. There is no other way around this-you must learn the names. It will make life a lot easier when naming other chemical species. • There is a definite trend in the formula of the alkanes. In fact they all have the general formula CnH2n+2. For example octane has eight carbon atoms and thus (2 x 8) +2 = 18 hydrogen atoms. We can thus recognise whether an organic chemical is an alkane or not. If it has the correct ratio of carbon:hydrogen of n:2n +2 it must be an alkane. Alkenes Alkenes also contain only carbon and hydrogen. However, this time the ratio of carbon:hydrogen is n:2n. The first member of the family is ethene, C2H4. What does the bonding look like in this molecule? Remember carbon must form four bonds to gain an octet, however this time there are not enough hydrogens to go around H H x x C C x x H H and thus the carbons are only forming three σ-bonds each. Can we do something to overcome this problem? Easy- form a π-bond between the carbons. Each carbon still has one electron it can donate to the other carbon and thus a bond can be formed. H H x x C C x x H H or H H C C H H Thus, there is now a double bond between the carbon atoms- one σ-bond and one π-bond. All alkenes contain a double bond and are thus termed unsaturated. The next member of the family is propene, C3H6. H H H H C C C H H Question 4.3 (a) The next member of the family contains four carbon atoms. Name it. (b) An alkene has the formula C8H16. What is its name? Alkynes Another family of compounds which contains only carbon and hydrogen exists. This family is called the alkynes. They have the general formula CnH2n-2. Carbon still obeys the octet rule and thus still forms four bonds. Carbon ALWAYS obeys the octet rule and thus ALWAYS forms 4 bonds. 49 Question 4.4 The first member of this family is ethyne, C2H2. Draw this molecule, clearly showing the orbitals involved in the bonding. The second member of the family is propyne, C3H4. The alkynes all contain a triple bond between two carbon atoms, and are therefore, like the alkenes, unsaturated. H H C C C H H Question 4.5 (a) Classify each of the following molecules as an alkane, alkene or an alkyne (i) C2H4 (ii) C7H12 (iii) C22H44 (iv) C10H18 (v) C18H38 (b) Consider the following molecule H 1 2 C C H H H C C C 3 4 H H H C C H H 5 H What is the hybridisation at each of the numbered carbons? (c) Complete the following table Molecule Name Ethane Hybridisation at carbon sp 3 C2H2 C2H4 (d) Name the following molecules (i) C4H6 (ii) C10H20 (iii) C7H14 Shorthand Organic chemists can get lazy at times. Instead of drawing out all the atoms in a molecule they sometimes use a form of shorthand. For instance when drawing out a propane molecule they might write C C C where it is taken as granted that there are actually hydrogen atoms at the end of the bonds. This shorthand can be taken a step further where we don’t even draw the carbon atoms! Propane can be drawn very simply as 50 where it is now recognised that there are CH3 groups at the points where the lines stop. CH2 = CH3 CH3 This comes in handy when drawing large molecules such as C10H22. Instead of laboriously drawing the molecule out we can draw – Question 5.6 What are the molecular formulae of the following compounds? (i) (ii) (iii) One last thing on shorthand is the use of “R”. When chemists write a molecule as R H H C C H the “R” simply signifies an organic group which they are not particularly interested in! That is to say that the “R” group will remain unchanged during a chemical reaction. Rings Alkanes, alkenes and alkynes are all linear molecules, where the carbon atoms are in a chain. There is however another way of joining carbons together and that is in rings. Consider for example the following molecule. H H H C H C C H H H H C = C H H C H H Here we have six carbon atoms joined in a ring. The naming of these compounds is still based upon the number of carbon atoms so this molecule is called cyclohexane. You will notice that the molecular formula for cyclohexane is C6H12. For all cycloalkanes the carbon:hydrogen ratio is Cn:H2n. This is exactly the same ratio as for alkenes. However, cycloalkanes are saturated molecules and they will have a very different chemistry from alkenes with the same molecular formula, so beware! 51 One can also get unsaturated rings such as the following molecule H C H C C H H C C H C H This pattern of a C-C single bond, C=C double bond, C-C single bond, C=C double bond, etc. is very important in organic chemistry. If this pattern exists, the molecule is said to be conjugated. If you have a conjugated ring, as in the above molecule, the molecule is described as being aromatic. We normally draw the above molecule as or The second structure is to show that the π-bonding in the molecule is delocalised, i.e., it is the average of the two resonance forms of the molecule. This particular molecule is very important in organic chemistry and is called benzene. Functional Groups Carbon has a rich chemistry. It can bond to many other atoms apart from hydrogen. These atoms are generally known as heteroatoms. Common heteroatoms are nitrogen, sulfur and the halides. The most important one however is oxygen which is not surprising as we live in an oxygen-rich planet. The chemical reactions that these compounds which contain heteroatoms undergo depend crucially on which groups of atoms are present. For instance, a compound which contains a C O group will have a very different chemistry from a compound which contains the group . C OH We can thus classify compounds into families which contain the same grouping of atoms or functional groups. (a) Alcohols (or Alkanols) Alcohols have the functional group C OH 52 The simplest alcohol is CH3OH, methanol. The suffix –ol is used to describe this functional group. The naming of the alcohols will come from the number of carbon atoms in the molecule. Thus C2H5OH is ethanol. The next one is: H H H H C C C H H H OH which is propanol and so on. (b) Aldehydes Aldehydes contain the functional group H C O The first three members of the family are shown below. H C CH3 C O methanal (c) H H H CH3 O ethanal CH2 C O propanal Ketones Ketones are very similar to aldehydes except they have no hydrogen attached to the carbon which is attached to the oxygen. R C O R Two examples are shown below. O CH3 C O CH3 CH3CH2 C propanone (d) CH3 butanone Carboxylic acids Carboxylic acids contain the functional group O C OH The first two members of the family are methanoic acid and ethanoic acid. 53 O H C O OH methanoic acid CH3 C OH ethanoic acid Question 4.6 Calculate the oxidation number of carbon in the following compounds. (a) methane (b) methanol (c) methanol Question 4.7 Name the following molecules. (a) CH3CH2CH2OH (b) (e) CH3CH2CH2COOH (c) (d) methanoic acid CH3CH2CH2CH2CHO Esters Esters are formed when a carboxylic acid reacts with an alcohol. For example O CH3COOH + CH3CH2OH CH3 C O CH2CH3 + H2O ethyl ethanoate The naming of esters comes from the identity of the R′ and R″. If for instance R″ = CH3 then we have a methyl substituent. If R″ = C2H5 then we have an ethyl substituent. The fragment is named after the total number of carbon atoms in this fragment. Thus if we have CH3COO- , the fragment is called “ethanoate”. If instead we have CH3CH2COO-, we have propanoate and so on. The ester we formed in the above reaction is thus called ethyl ethanoate. Question 4.9 (a) Name the following substituent groups. (ii) (i) -C3H7 (b) Name the following esters. (i) CH3CH2CH2COOCH3 (ii) (f) -C4H9 CH3COOCH2CH2CH3 Ethers Ethers also contain an oxygen heteroatom. The functional group is R′-O-R″. If R′ = R″ = -C2H5 we will have diethyl ether. (When most people talk of “ether” they are actually talking about diethylether, which used to be employed as an anaesthetic during surgical procedures). (g) Amines and Nitriles These two functional groups both contain nitrogen but in very different forms as shown below. R R NH2 amine C N nitrile There are in fact different types of amines and they are differentiated by the number of carbons bonded to the nitrogen 54 H R N o 1 amine R R H R N o R H N R o 2 amine 3 amine o If for instance we have (CH3)3N we would name this 3 amine, trimethylamine. The naming of nitriles again comes from the number of carbons. Therefore CH3CN is ethanenitrile (which is better known to the older generation of chemists as acetonitrile). (h) Amides Not to be confused with amines! Amides contain the functional group O C N So for instance the compound: O H CH3CH2CH2CH2CH2C N H would be called hexanamide (as there are six carbons in the compound). (i) Arenes It might seem strange to include arenes in this list as they contain only carbon and hydrogen. Arenes are alkyl -substituted benzene rings. Examples are shown below CH3 CH2CH2CH2CH3 It is important to include arenes in this list of functional groups as the benzene ring substituent (known as the phenyl group) undergoes a wealth of chemistry. A table of all these functional groups (and a few more) is shown overleaf (don’t try and remember the extra ones!). Naming Compounds The title of this section might seem a little strange to you. Haven’t we just done nomenclature in the previous section? Yes we have, but knowing the vocabulary and putting it into action are two different skills. You might feel very comfortable in naming this compound- H H H H H H O C C C C C C H H H H H OH - as hexanoic acid. However, how would you name the following compound? 55 Cl H H H CH3 H O C C C C C C H CH3 H H H C H H C H OH H Gets a bit complicated doesn’t it? Luckily a set of rules have been put into place. So all we need to do is follow the rules and we should be able to name any organic compound. Branching in Alkanes Lets consider a compound with a molecular formula of C4H10. Its an alkane, so we must have butane. The compound must be H H H H H C C C C H H H H H However there is another molecule we can draw with the same molecular formula of C4H10. H H H H C C C H H H H C H H What we have done is introduce a branch in the carbon chain. Now we must be able to distinguish between these two compounds by giving them different names. In this case it is quite straightforward. We call the straight-chain molecule n-butane and the branched molecule iso-butane. However, when alkanes have more than one branch (as many do) we really do need a systematic way of naming them. Rule 1: Choose the longest carbon chain to base the name upon. Rule 2: Number the carbons in this longest chain beginning at the end nearest the first branch point. Rule 3: Identify and number the branches. Hydrocarbon branches -CH3 -CH2CH3 -CH2CH2CH3 -CH2CH2CH2CH3 Methyl ethyl propyl butyl CH3 CH3CHCH3 CH3CHCH2CH3 CH3 sec-butyl C CH2 CH3 C CH3CH2 CH3 C CH3 neopentyl tert-pentyl Example 4.1 56 CH2 CH2 isopentyl CH3 CH2 CH3 C H isobutyl tert-butyl CH3 CH3 H CH3 isopropyl pentyl CH3 CH3 C -CH2CH2CH2CH2CH3 Question: Name the following molecule. H H H H H C C C H H H C H H C H H H H C C C C H H H H H H Answer: When naming organic compounds use hyphens to separate numbers and letters, and commas to separate numbers. Example 4.2 57 Question: Name the following molecule. CH2 CH3 CH3 H H CH2CH3 C C C CH H H CH3 CH2CH2CH2CH3 Answer: When deciding which substituents to put first, simply put them in alphabetical order. Ethyl thus comes before methyl (Note: we don’t count the prefix, di- in this case, when deciding this order). Question 4.10 Name the following molecules. (a) C C C C C C C C C C C (b) C C C C C C C C C C c) 58 CH3 CH3 CH3 C C C C C CH2 CH3 Question 4.11 (a) Draw the structure of the following molecules (i) 3-methylhexane (ii) 3-ethyl-2-methylhexane (iii) 3-ethyl-4,7-dimethylnonane (b) Give a systematic name to isobutane. A result of branching in alkanes is that some carbons are now “different” from other carbons in the molecule. In the following molecule for instance – CH3 CH3 C o CH2 CH3 x CH o CH2 CH3 CH3 some carbons are only bonded to only one other carbon atom (those marked ). Other carbon atoms are o bonded to 2 carbon atoms (o). Those bonded to only one carbon atom are termed primary carbon atoms (1 ) o and those bonded to two are termed secondary carbon atoms (2 ). Both these types of carbon atoms are encountered in straight chain alkanes. However, with branching we now have two different carbon atoms, those bonded to three carbon atoms (x) and those bonded to four carbon atoms (•). These are termed o o tertiary carbon atoms (3 ) and quaternary carbon atoms (4 ) respectively. Naming Compounds with Functional Groups The names of most organic compounds can be viewed as having three main parts. Prefix-Parent-Suffix Prefix: identifies the where and what for the substituents. Parent: this is the backbone of the molecule, the name of which depends upon the number of carbon atoms. Suffix: this is particularly important as it describes the functional group present. Again, we establish the rules and follow them rigorously when naming these compounds. Rule 1: The parent hydrocarbon chain is the longest continuous carbon chain containing the highest priority functional group. Some compounds might contain more than one functional group and thus a priority list is required. For example, is the following compound named as an alcohol or as a carboxylic acid? O HO CH2 CH2 CH2 59 CH2 C OH The answer is as a carboxylic acid as the –COOH group has the highest priority. The full priority list is shown below. Functional Group -COOH -COOR -COX -CONH2 -CHO RCOR’ -OH -NH2 CnH2n-2 CnH2n CnH2n+2 Name carboxylic acid highest priority ester acid halide, X = Cl, Br amide aldehyde ketone alcohol amine alkyne alkene alkane An alkyl substituent (CnH2n+1) has the lowest priority. Halides are always named as substituents on the parent hydrocarbon. Common non-alkyl substituents are – Substituent -F -Cl -Br -I -NO2 Name Fluoro Chloro Bromo Iodo Nitro Rule 2: The chain is numbered in the direction that gives the highest priority functional group the lowest possible number. Of course for some functional groups by their very nature have to come at the end of a chain. Examples of these are C C O OH H aldehydes O carboxylic acids When numbering chains which include these functional groups start at the carbon attached to the heteroatoms. For example; CH3 CH2 CH2 COOH CH3 CH2 CH2 CH2 CH2 CHO 4 6 3 2 butanoic acid 1 5 4 3 hexanal 2 1 Rule 3: If the same number for the functional group is obtained in both directions the chain is numbered in the direction that gives the substituent the lowest number. For example the following molecule. 60 H H H OH Cl H C C C C C H H H H H H is 2,2-dichloropentan-3-ol (not 4,4-dichloropentan-3-ol). Example 4.3 Question: Name the following compound CH3 CH CH Cl CH3 CH2 C OH O Answer: The parent hydrocarbon is five carbons long 1 CH3 2 3 CH CH Cl CH3 4 CH2 5 C OH O The suffix will be –oic acid. The substituents are 3-methyl and 4-chloro. Thus we have 4-chloro-3-methylpentanoic acid. Example 4.4 Question: Name the following compound OH CH3 C O CH2 CH2 CH2 CH3 Answer: Example 4.5 61 CH Question: Name the following compound. OH CH3 CH CH2 CH CH3 Cl Answer: Question 4.12 Name the following compounds. (a) (b) CH3CH2 CCH3 CH3CHClCH2COOH (c) ClCH2COOH (d) ICH3 O Question 4.13 In the introduction to this section we mentioned it might be a bit complicated to name the following molecule. Cl CH2 CH CH2 C(CH3)2 CH2 COOH C2H5 Well, now you know the rules, show that is in fact quite simple by giving it a name! Question 4.14 o o o List the following alcohols as 1 , 2 or 3 alcohols. CH3 (a) CH3CH2OH (b) CH3CH2CH(OH)CH3 (c) CH3 C CH3 Isomers 62 OH Isomers are compounds that have the same numbers and kinds of atoms but differ in the way they are arranged. We have already seen that the molecular formula C4H10 can give rise to two different molecules- H3CCH2CH2CH3 H 3C CH CH3 CH3 n- butane isobutane These two molecules are isomers of each other. Question 4.15 In how many ways can you arrange the atoms for a molecular formula of C5H12 i.e. how many isomers can you draw? There are in fact many different types of isomerism that can occur in organic chemistry and this section will review them under two umbrella titles- constitutional isomers and stereoisomers. Constitutional Isomers This type of isomerism occurs when the atoms are connected differently. Butane and isobutane have a different way of connecting the carbon atoms and are thus constitutional isomers (as are all the isomers of C5H12- hope you got them all!). Another form of constitutional isomerism is when a molecular formula can give rise to different functional groups. For example if we have a molecular formula of C2H6O, how many ways can we connect the atoms together? The answer is two, as shown below. H3COCH3 or H3CCH2OH dimethyelther ethanol (Try this for yourselves: there is no other way of connecting these atoms while satisfying their valencies.) For larger molecules the number of different ways of combining the atoms together will increase, and another type of constitutional isomerism will arise:- that of the position of the functional group. For instance for C3H8O we have the following possibilities H3C O CH2CH3 H3C ethylmethyl ether CH2 CH2OH H3C CH CH3 OH propan-1-ol propan-2-ol Question 4.16 (a) Draw all the constitutional isomers for the molecular formula C4H10O. (b) Name all the compounds in part (a). [Hint: this question combines the three forms of constitutional isomerism mentioned- different carbon skeletons, different functional groups, and the different position of the functional groups.] The presence of unsaturation in a molecule also introduces the possibility of isomerism. This is really just an extension of the idea of functional groups having different positions. Consider the molecule butene. Given only this description we can actually draw two possible structures. CH3CH CHCH3 or CH2 CHCH2CH3 We then need to specify the exact position of the double bond. The two molecules would therefore be but-2ene and but-1-ene. 63 Question 4.17 Draw all the possible constitutional isomers for the molecular formula C6H12. [Hint: remember cycloalkanes have the same carbon:hydrogen ratio as alkenes and you can draw more than one cycloalkane here!]. Stereoisomers Stereoisomers are compounds where the atoms are connected in the same order but the molecules have different geometries. When talking of geometries, or shapes, we need to start thinking in three dimensions (3D). We will discuss three types of stereoisomer in this section: conformational isomers, geometric isomers and enantiomers. When thinking in 3D we need to remember the two very important concepts • There is rotation possible around a carbon-carbon single bond. • There is no rotation possible around a carbon-carbon double-bond. (a) Conformational Isomers Organic chemists have various ways of drawing 3D structures on a 2D page. One way is to show bonds coming out of a page and one going behind the page. Thus for ethane we can draw the following structure. H H H H C C H H If we now change our point of perspective when looking at the ethane molecule to an end-on view, so we’re looking directly along the C-C bond axis, then what we see is only one carbon atom. Attached to this carbon atom we see three hydrogen atoms. We can draw this as: H H H where the dot in the middle is the carbon atom that we can see. o The apparent HĈH angle is 120 C when we look at the molecule like this. We now draw a circle to represent the other carbon atom (the one we can’t see)H H back carbon front carbon H and then we put into the diagram the three hydrogens attached to this carbon. H H H H H H This type of representation is known as a Newman Projection. Newman projections are very useful when looking at what happens to the shape of a molecule when we rotate around a carbon-carbon bond. Upon rotation we can change the relative positions of the hydrogens on the back carbon with respect to the 64 hydrogens on the front carbon. There are an infinite number of possibilities here but two extremes stand out. These are shown below. H H H H H H H H H H H H eclipsed staggered In the eclipsed form one can only see the front carbon and the three hydrogens attached to it. The back hydrogens are hidden from view behind the front hydrogens. The back-hydrogens are said to be eclipsed. In the staggered conformation we can see all six hydrogen atoms as the back-hydrogens are staggered exactly ½ way between the front-hydrogens. It is important to note that the two molecules depicted above have different shapes and thus they are strictly speaking, stereoisomers of each other. However the difference in energy between the molecules is so small (the staggered conformation is only 12 kJ/mol more stable than the eclipsed conformation) that they cannot be separated from each other. The eclipsed and staggered forms of ethane are said to be conformational isomers or conformers. Why should the staggered confirmation be more stable? It is thought to be due to the fact the hydrogens on the front-carbon are at their furthest distance from the hydrogens on the back carbon. Thus the hydrogen electron clouds are at their furthest distance apart and thus (as electron clouds repel each other) they are at their lowest possible energy. The picture becomes slightly more complicated for larger alkanes. For instance if we look at butane, we now have to specify which carbon-carbon bond we have chosen to draw our Newman projection. If we choose the C1-C2 bond then our Newman projection will look like this: H H H H C2H5 H If instead we choose the C2-C3 bond then we will have: H H CH3 H CH3 H Now when we rotate the C2-C3 Newman projection we will have two different “forms” of eclipsed structures and two different staggered structures. The two eclipsed forms are shown below. CH3 H H CH3 H CH3 CH3 H H H 65 H H The two staggered forms are – H CH3 H H H H H CH3 H H CH3 CH3 Again each of the conformers can be converted into another by rotation around the C2-C3 bond. This is true for all conformers. Conformers are stereoisomers which can be made equivalent by rotation around carbon-carbon single bonds. (b) Geometric Isomers To return to the second point we listed at the start of the section- there is no rotation possible around a double bond. Ethene is a planar molecule. In fact the geometry immediately around an alkene functional group is always 2 planar due to the fact that the carbons are sp hybridised. If we now look at but-2-ene we can draw the molecule as H3C CH3 C C H H (I) However there is another possible structure. We could just as easily have drawn the molecule as H CH3 C C H H 3C (II) Note that we cannot form this molecule from the other one by simply rotating around the C=C bond, because no rotation is possible! The atoms are still connected in the same order but the shapes of the molecules are different. This means that we have another form of stereoisomerism. As the geometry around the double bond is different these are known as geometric isomers. How do we differentiate between them? Molecule (I) has the two methyl groups on the same side of the double bond. This is called the cis – isomer or the Z – isomer. The other isomer has the methyl groups on the opposite sides of the double bond. This is known as the trans – isomer or the E – isomer. H3C C H H CH3 CH3 C C H 3C H 66 C H cis- isomer (Z- isomer) trans- isomer (E- isomer) Question 4.18 Write down and name the geometric isomers possible for the following molecules. (a) 1,2-dichloroethene (b) 1-bromopropene (c) pent-2-ene Question 4.19 (a) Draw the two geometric isomers:- trans-1,2-dimethylcyclopropane and cis-1,2-dimethylcyclopropane. (b) Does the molecule 2-methylbut-2-ene exhibit geometric isomerism? If so, draw the isomers. (c) Is there a possibility of geometric isomerisation for the molecule 1,2-dichlorocyclohexane? As well as having the same carbon:hydrogen ratio cycloalkanes and alkenes also share the ability to exhibit geometric isomerism. Unlike conformers, geometric isomers can be separated from each other as they have quite distinct physical properties from each other. (c) Enantiomers 3 Another form of stereoisomerism arises from molecules which contain a chiral centre. A chiral centre is an sp hybridised carbon centre which is attached to four different substituents. An example would be the following molecule. CH3 H Cl I Believe it or not there is another way of drawing this molecule! If we imagine a mirror beside the molecule then we can envisage what the molecule will look like in the mirror (or through the mirror plane). CH3 I CH3 H H Cl Cl I mirror plane These two molecules are not the same! If you try and superimpose the two molecules on top of each other you will find it impossible to do! The two different isomers are called enantiomers. How do we differentiate between the two compounds? Once again we need to give each isomer a different name. The naming of enantiomers is quite a complicated business and not within the scope of this course. It is sufficient to recognise that when a chiral centre is present in a molecule, the isomers are known as the Risomer and the S-isomer. Lets now try the same thing for a molecule which does not contain a chiral centre. An example would be CHCl2I. CH3 CH3 Cl I Cl Cl 67 Cl I mirror plane Here we do not have four different substituents on the C atom. When we reflect through the mirror plane we will get a molecule that is superimposable upon the original. Therefore for this type of isomerism to occur we require the presence of a chiral centre. Chiral centres are also known as asymmetric centres or stereogenic centres. Enantiomers that occur in a 50:50 mixture with each other are called a racemic mixture or a racemate. Enantiomers however can be separated by various techniques. Indeed organic chemists spend an awful lot of time and energy trying to selectively synthesise one enantiomer. Why do they do this? Well, many biologically active compounds, although containing chiral centres, exist only in one enantiomeric form in nature. Therefore if one wants to mimic a naturally-occurring compound, say with a drug, only one enantiomer will be effective. Question 4.20 Consider the following molecule (a) (b) (c) (d) CH3CH(NH2)CH2CHCHCH(CH3)CH3 Draw an extended structure for the molecule, i.e. show the bonds. Identify the functional groups present in the molecule. Will this molecule display geometric isomerism? Explain your answer. Does this molecule have a chiral centre(s)? Indicate them clearly on your drawing of the structure. An Introduction to Reaction Mechanisms 68 Organic Chemistry is largely about making new carbon-containing compounds. To do this bonds in existing compounds must be broken and new bonds formed. If we understand the nature of how these two processes occur we will be well on our way to understanding reaction mechanisms, and perhaps even to predict how chemicals might react together. Bond Breaking and Bond Making Question 4.21 How many electrons are in the following bonds? (i) C (ii) C C (iii) C C C When a single covalent bond breaks, two different outcomes are possible. One way is for the two electrons in the bond to divide equally. This is known as homolytic fission and results in the formation of free radicals. X + Y → X• + Y• Each of X and Y now has an odd electron (as indicated by ). For instance if we homolytically split the carboncarbon bond in ethane then two methyl radicals will result. C2H6 → 2 CH3• methyl radical We denote the movement of single electrons with a single-headed arrow reaction mechanism for the homolytic splitting thus H3 C CH3 → . Thus we can draw the 2 CH3• Radicals are, in general, very reactive species. The other way to break a single covalent-bond is heterolytic fission, where the two electrons in the bond go to one of the atoms. The movement of two electrons is shown by a double-headed arrow X Y → X + + Y . - Now we have a cation and an anion forming. If a carbon-carbon bond breaks in this fashion then we have a carbocation and a carbanion resulting. The carbocation species is very important in organic chemistry and will be returned to later. Question 4.22 (a) How many electrons are around carbon in each of the following species + (i) CH3• (ii) CH3 (iii) CH3 (b) Would you expect these species to be reactive? Explain your answer. Bond making is as equally important as bond breaking. To form a single covalent bond we need two electrons. This can also happen in two ways. X Y 69 → X–Y or X+ Y- → X–Y i.e. the reverse of the dissociation mechanisms we have just talked about. If we look at the following bond formation mechanism: + Cl + CH3 → CH3Cl then we could draw the mechanism as Cl- + CH3 → CH3Cl where the chloride ion has donated two electrons to carbon. The chloride ion is termed a nucleophile and the carbocation is an electrophile. The formation of the bond comes from nucleophilic attack on an electrophile. A nucleophile is a “positive loving” species and will, itself, be negatively charged or electron rich. An electrophile is a “negative loving” species and will, itself, be positively charged or electron deficient. We have already seen electrophiles and nucleophiles in action in the formation of dative (coordinate) bonds in st the 1 semester (we just didn’t refer to them as nucleophiles and electrophiles!). Consider again, for example, the reaction between BCl3 and NH3. Cl3B NH3 → Cl3B – NH3 Question 4.23 (a) Identify the nucleophile and the electrophile in the previous reaction. (b) Identify each of the following species as an electrophile or a nucleophile. + (i) I (ii) HO (iii) CH3 (iv) CH3 (v) H + (iv) AlCl3 In real life it is very difficult to break carbon-carbon bonds or in fact any bonds in a molecule. Stable molecules will require an energy input to break bonds or a reactive centre that will encourage either nucleophilic or electrophilic attack. Ethane for instance will not spontaneously dissociate into two methyl radicals. Nor will it spontaneously + dissociate into CH3 and CH3 . Free radicals are commonly formed in photochemically induced reactions i.e. the interaction between molecules and light. The bromine molecule, for instance, can give rise to two bromine radicals thus Br2 + hυ → 2 Br• We show the photon as “hυ”. In this case the photon must have enough energy to break the bromine-bromine single bond in Br2. Question 4.24 -1 If the bond energy of Br2 is 229 kJ mol , what is the wavelength of the lowest energy photon which can dissociate Br2? Functional groups will give rise to reactive centres within molecules. For instance if we have a ketone containing the functional group O C 70 then we have a possible site for reaction. Consider the above functional group. The carbon is less electronegative than the oxygen and thus the polarity of the bond will be + - δ δ C O We now have a positive centre in the molecule which will be liable to attack from a nucleophile (Nu−). Nu C O However if carbon forms a bond to the nucleophile then it will be pentavalent and therefore break the octet rule. Instead carbon will also give two electrons away, in this case to oxygen. We thus have the reaction mechanism: Nu Nu C O C O This is called a concerted reaction mechanism where bonds are forming and breaking at the same time. Question 4.25 + Identify areas of positive charge and areas of negative charge by using δ and δ terminology in the following molecule. H 2N C C CH CH2 CH CH2 Cl Question 4.26 Will alkenes and alkynes be susceptible to nucleophilic or electrophilic attack? Explain your answer. Carbocations Consider the following reaction. CH3 CH3 H3C C H3C Cl C + Cl- CH3 CH3 The chloride ion “leaves” the molecule and is therefore simply called a leaving group. It leaves behind a carbocation. The carbon which used to be bonded to the chlorine has lost two electrons and thus has only six electrons around it. Six electrons are arranged into three electron pairs and thus the hybridisation at the 3 2 carbon has gone from sp in the parent molecule to sp in the carbocation. The carbocation formed in the above reaction will look like this. H3C + CH3 C o CHand 3 The central carbon has a vacant unhybridised p-orbital the angles between the C- C bonds are 120 . 71 Carbocations are very important intermediates in many organic reactions. They have a very short lifetime due to the fact that they are very powerful electrophiles i.e. the positive carbon centre wants to regain another two electrons to achieve a full valence shell. Question 4.27 Consider the following molecule. H H C Br H Suggest ways that this molecule might undergo bond breaking. Some carbocations are more stable than others. This comes from the so-called INDUCTIVE EFFECT. The carbon centre which is positively charged will try to pull electrons towards it from other parts of the molecule. Looked at another way the groups attached to the positively charged carbon can donate electron density through the σ-bonds to the electron-deficient centre. We draw this as follows C CH3 where the methyl substituent donates electron density towards the carbocation centre. Alkyl groups such as the methyl group are said to be electron-donating or to have a positive inductive effect. The inductive effect is the shifting of electron density within a σ-bond towards a positive centre in a molecule. If a substituent group is electron-donating it has a positive inductive effect and if it is electron withdrawing it is said to have a negative inductive effect. If electron density is “pushed” towards the carbocation centre then the carbocation will be stabilised. Therefore, the more electron-donating groups attached to the electron-deficient carbon the more stable it will be. Hydrogen atoms do not display any appreciable inductive effect. Generally, (i) the more alkyl groups attached to the electron-deficient carbon, the greater the inductive effect; and (ii) the larger the alkyl group substituent the more electron-donating it will be and therefore the greater the inductive effect. Question 4.28 (a) Arrange the following carbocations in order of increasing stability. + + + + CH3 CH3CH2 (CH3)3C (CH3)2CH (b) o o o Label each of the carbocations in the question above as 1 , 2 or 3 . Question 4.29 Which of the following carbocations is the more stable? + ClCH2CH2 or Explain your answer clearly. Question 4.30 72 + CH3CH2 Are carbon radicals more like carbocations or carbanions? Thus, does their stability sequence resemble that of carbocations or carbanions? Free Radical Substitution of Alkanes Alkanes are generally very unreactive (as they have no functional groups) but one particular reaction type is very important. This involves the substitution of a hydrogen within the alkane by a halide atom. The reaction between a halogen, such as Cl2, and an alkane, e.g. methane, proceeds via a free-radical mechanism. The reaction can be broken down into three steps; initiation, propagation and termination. Step1 Initiation: Generation of free radicals. Photochemically induced homolytic fission results in two Cl radicals. Cl2 + hυ → 2 Cl• H H Cl H C H Cl H C + H H H A chlorine radical abstracts a hydrogen atom from methane Hydrogen chloride and a methyl radical result Step 2 Propagation Cl Cl CH3 Cl + Cl CH3 Chloromethane is formed and another chlorine free radical forms which attracts another methane and the reaction continues The methyl radical the reacts with the chlorine molecule Step 3 Termination For a free radical substitution to stop, free radicals must react with each other. Cl Cl CH3 Cl Cl Cl CH3 Cl CH3 C2H6 CH3 Question 4.31 The previous reaction can also give rise to dichloromethane. Write a suitable reaction to form this product. When we consider the free radical substitution of larger alkanes then more than one free radical alkyl species is possible. Consider for example the reaction between a chlorine radical and n-butane. 73 (i) Cl H CH2 CH2 CH2 CH2 CH3 CH2 CH2 CH3 + HCl CH CH2 CH3 CH3 (ii) Cl H CH CH2 CH2 CH3 CH3 o + HCl o Reaction (i) gives rise to a 1 free radical and reaction (ii) gives rise to a 2 free radical. Which free radical is more stable? Free radicals, like carbocations are electron-deficient (free radicals require one more electron to obey the octet rule, carbocations require two more electrons). Therefore free radicals will have the same o o o order of stability as carbocations – 1 <2 <3 . Thus reaction (ii) will be favoured over reaction (i). The favoured product of this reaction will be 2-chlorobutane. Question 4.32 o o o (a) What order of stability would you predict for 1 , 2 and 3 carbanions? Justify your answer. (b) What would be the favoured product of the reaction between isobutane and Cl2? Again, justify your answer. (c) The reaction between Cl2 and butane also gives rise to 3,4-dimethylhexane as one of the minor products. Suggest a reaction mechanism which will give rise to this product. Electrophilic Addition Reactions of Alkenes An addition reaction is one in which the bond order between two carbon atoms decreases. For instance, we can add a species across a double bond. C C XY + X Y C C In the above reaction the bond order between the two carbon atoms has decreased from two to one. Addition reactions of alkenes proceed via a carbocation intermediate. Alkenes are nucleophilic, that is, they have a centre of electron density in the double bond. We therefore require a positive centre to react with it. If we take a polar molecule such as HBr then we will have a positive center on the H atom. δ+ δ- H – Br Thus the first step of the reaction will be a nucleophilic attack by the alkene at the hydrogen end of the HBr molecule. Taking ethane as an example: C C H Br As the carbon-hydrogen bond forms the hydrogen-bromine bond will simultaneously break to give – H H C C H H H 74 + Br The bromide ion will now attack the carbocation thus H H C C H H Br- H H3C CH2Br The sum total of the reaction is that we have converted an alkene into an alkyl halide. We need to be careful when dealing with larger alkenes. Consider for example the reaction between propene and HBr. Two carbocation intermediates are now possible depending on which carbon atom of the double bond uses the pi electrons to form a bond to the hydrogen atom. H3C Br- CH CH2 + CH2 CH2 CH2 H Br CH3 CH CH3 + Br- Question 4.33 (a) Which is the more stable carbocation in the previous example, A or B? (b) Which will be the preferred product of this reaction, 1-bromopropane or 2-bromopropane? [Note we will get both products but one will be the major product and the other the minor product] (c) 1-bromopropane and 2-bromopropane are isomers. Which type of isomers are they? For even larger alkenes we’ve got another variable to consider, that of the possibility of the formation of a different kind of isomers during the reaction. Consider the reaction between cis-but-2-ene and HBr H3C CH3 C H3C C H + H Br H C H H CH3 C H + Br Only one carbocation intermediate is possible in this case. However, the subsequent nucleophilic attack from 2 the bromide ion can now come from two possible directions. Remember a carbocation is sp -hybridised and trigonal planar. Thus the bromide ion can attack either from above or below. Above H3 C H C H C CH3 H H3 C + Br H C* C H H Below H 3C H C H C CH3 H H 3C + Br- 75 H C H C* CH3 Br H Br CH3 At first sight these products might not look different from each other. However, if we take a closer look we will see that the carbon marked * is in fact a chiral centre, and the two products formed are in fact enantiomers of each other, i.e. non-superimposable mirror images. As the carbocation intermediate is planar, the attack from the nucleophile is equally likely to attack from above or below the plane. Therefore we will end up with a 50:50 mixture of both enantiomers or a racemate (racemic mixture). You will note that a racemic mixture of enantiomers is only possible when a chiral centre results at the reaction centre. Question 4.34 How many products will be observed in the following reactions? Justify your answer by drawing suitable reaction mechanisms for the formation of each product. If more than one constitutional isomer is predicted indicate which will be the major product. CH3CH2 CH3 (a) 2-methylbut-2-ene + HBr (b) trans - but-2-ene + HBr (c) C C + HBr H3C CH3 Ozonolysis One particularly important oxidation reaction of alkenes is their reaction with ozone, O3. This reaction cleaves the double bond in the alkene to form a carbonyl species i.e. a species containing the C=O group. The reaction can be summarised as shown below. R R C O C + C O R R R R R R O3 C The exact identity of the products of this reaction will depend upon the nature of the R groups. If for instance we start with the following alkene H H3C x C* C CH3 H3C then the products will be H3C * C O + x O H C CH3 H3C Why is this reaction so important? Well it allows us to identify the alkene starting material if its structure is unknown. If for instance we identify the products of ozonlysis as ethanal and butanone then we would know the identity of the starting alkene by reverse logic. H CH3 C O O C CH2CH3 H 3C 76 O3 C C CH2CH3 H CH3 H or C C Note we cannot say which of the two geometric isomers is the starting material. Question 4.35 Ozonolysis of alkene A gave the following aldehydes. Specify the reaction conditions and deduce the structure of alkene A. A → CH3CH2C(CH3)2CHO and (CH3)3CCHO Organic Redox Chemistry Carbon is in group 14 of the periodic table, with four valence electrons. The rules of assigning formal oxidation numbers dictates that when carbon combines with oxygen it will exhibit positive oxidation numbers. However when carbon combines with hydrogen it will exhibit negative oxidation numbers. The two extremes of this arrangement are CO2, with C in the +4 oxidation state, and CH4, with C in the -4 oxidation state. The oxidation states in between +4 and –4 will be encountered for carbon compounds which contain both oxygen and hydrogen. Question 4.36 Assign an oxidation number to carbon in the following compounds (i) CO (ii) CH3OH (iii) CH2O (iv) HCOOH Therefore if we want to convert an alcohol into an aldehyde then we would consider using an oxidising agent. If instead we want to convert a carboxylic acid to an aldehyde we would use a reducing agent and so on. Another example of thinking of organic reactions in terms of oxidation numbers is if we go from an alkene to an alkane. An example would be – C2H4 → C2H6 ethane ethane If we calculate the change in formal oxidation number of carbon in this reaction we should see that carbon is going from –2 to –3 i.e. it is being reduced. Thus we would add a reducing agent to ethene to obtain ethane. Of course we also need to add hydrogen atoms, but that isn’t a problem because if we add H2(g) to ethene we are adding a source of hydrogen and a reducing agent in one. Thus organic chemists talk about the addition of hydrogen to a compound as a reduction. If hydrogen is removed from a compound then they will have effected an oxidation. Similarly if they add oxygen to a compound they term it an oxidation (as the formal oxidation number of carbon will increase). If oxygen is removed then we will have reduced the compound. Inter-Conversion of Alcohols/Aldehydes/Ketones/Carboxylic Acids o If we oxidise a 1 alcohol then we will get an aldehyde 77 H R C OH H oxidizing agent C H O R Question: What would we use as an oxidising agent? Answer: A species which has a high oxidation state. In this particular case we use CrO3 in acidic media. Question 4.37 + Name the product when propan-1-ol is treated with CrO3/H . o If we oxidise a 2 alcohol then we will get a ketone H R C OH CrO3 / H+ R C R' O R' + Again we can use CrO3/H as our oxidising agent. Question 4.38 + If we react a tertiary alcohol with CrO3/H , what product would you predict (use tert butylalcohol as an example)? Hint: be careful! Question 4.39 o + In fact, if we treat a 1 alcohol with excess CrO3/H then it will be oxidised all the way through to the o + corresponding carboxylic acid. However, it we treat a 2 alcohol with excess CrO3/H the reaction stops at the formation of the ketone. Explain. If, instead, we wish to go from an aldehyde or a ketone to an alcohol we should recognise that this reaction is a reduction and we should therefore treat the aldehyde/ketone with a reducing agent. A very common reducing agent in organic chemistry is LiAlH4, (lithium aluminium hydride). LiAlH4 is a source of hydride ions, H in solution. Addition Reactions of Alkenes An addition reaction of an alkene involves the following transformation C X XY C Y C C We have already seen that hydrogen halide species, HX, can add across the double bond via a carbocation intermediate. However other species can add across the double bond to give a variety of products. In some cases the mechanism might be different but the overall reaction involves transforming an unsaturated species into a saturated species. Thus the overall effect is a reduction. Below are some examples of addition reactions (i) H3C CH3 C H3C C H H2O / H+ 78 H3C OH H C C CH3 CH3 H Effectively we have added the elements of water across the double bond. This is called HYDRATION and is a method for preparing alcohols (ii) H H H H C H2 C H H catalyst H C C H H H Here we have added hydrogen across the double bond to form an alkane. The presence of the ‘catalyst’ enables the HYDROGENATION reaction to proceed at a faster rate. (iii) Br Br H H C Br2 C H H H C C H H H Bromination of the double bond again results in a saturated species. Elimination Reactions of Alkyl Halides If addition reactions of alkenes are considered to be reductions then there must be an opposite reaction that will be considered an oxidation. The opposite of addition is elimination, where a small molecule is eliminated leaving behind an unsaturated alkene. For example: H H Br C C H CH3 CH3 H HBr CH3 C H C CH3 In the above reaction HBr is eliminated to convert an alkyl halide into an alkene. This type of reaction is catalysed by a very strong base, namely tert-butoxide: CH3 H3C O- C CH3 If we look at the alkyl halide we will notice that the bond most likely to be broken is the C-Br bond as it is the most polar, and Br will be the best leaving group. H H C H x Br C CH 3 CH3 If this happens of course then the carbon marked with an X will be electron deficient. Therefore we require to move an electron pair towards it. H H C Br C H' CH3 79 CH3 All we’ve got to do therefore is “encourage” the carbon-hydrogen, C-H’ bond to break! How do we do this? By + recognizing that if the H’ atom gives up its two electrons then it will become H , or an acid. What better therefore than to add a base such as our tert-butoxide ion! The full reaction mechanism can then be drawn as H H Br C C CH3 CH3 H H CH3 C - O H3C C C + Br- + (CH3)3C OH CH3 H CH3 CH3 Therefore we end up eliminating H and Br from the alkyl halide and form an alkene. You’ll note that we require a very strong base to do this as the hydrogen bonded to the carbon is very weakly acidic. Nucleophilic Substitution Reactions of Alkyl Halides A substitution reaction occurs when a substituent at a carbon centre is replaced with another, different, substituent. If we have nucleophilic substitution then we can outline the general reaction as Nu- C + - X C Nu + X- - where Nu is the nucleophile and X is the halide leaving group. Two different modes of reaction mechanism can take place and we will deal with each in turn. SN1 Reactions The diagram below shows what SN1 stands for. SN1 only one reactant is involved in the formation of the reaction intermediate Substitution Nucleophilic How do SN1 reactions proceed? The answer is via a carbocation intermediate. The first step is therefore the breaking of the carbon-halogen bond. R' R'' R' C R'' X R''' C + X- R''' The carbocation thus formed then undergoes nucleophilic attack. R' R' R'' C R''' + Nu 80 R'' C R''' Nu An example of SN1 would be the following reaction. H3C C CH3 CH3 CH3 H3C Cl C + OH- H3C C CH3 CH3 CH3 OH SN1 reactions go via a carbocation intermediate. Once again we need to consider the consequences of going through a carbocation intermediate. Consider the following reaction in a basic medium. CH3 C2H5 sp CH3 3 C Br C2H5 sp 2 C + Br- C3H7 C3H7 2 Once again we are forming an sp hybridised carbon centre which is trigonal planar. Again (as was the case in the addition reactions of alkenes) we must recognise that the nucleophile, OH , now can either attack from below or above. H5C2 OH C CH3 C3H7 CH3 C2H5 - C OH C3H7 H3C OH C C2H5 C3H7 Two enantiomers will thus result and we will have a racemic mixture of products. If the carbocation centre has three different substituents attached to it then a racemic mixture of the two enantiomers will result. Question: Answer: What type of alkyl halides will undergo SN1 reactions? Those that can form stable carbocation intermediates. Question: Answer: Which alkyl halides will form the most stable carbocation intermediates? o Tertiary alkyl halides as they will form 3 carbocations. Tertiary alkyl halides undergo substitution via an SN1 reaction mechanism. SN2 Reactions Would we expect CH3Br to undergo an SN1 type reaction? If it did then the first step would be 81 H H H C Br H C + Br- H H is the least stable carbocation that can form as it has no stabilising inductive effect from alkyl substituents. + CH3 Question 4.40 o o o Designate what type of carbocation (1 , 2 or 3 ) the following alkyl halides will give rise to if they undergo an SN1 type of reaction. CH3CH2I CH3C(CH3)2CH2I CH3CH2C(CH3)2I CH3CH2CH(CH3)I o o Generally primary alkyl halides will give rise to 1 carbocations which are not stable. Therefore 1 alkyl halides do not undergo the SN1 reaction. Rather they undergo an SN2 route. Question 4.41 How many reactants are involved in the formation of the reaction intermediate in an SN2 type reaction? o Instead of forming carbocations 1 alkyl halides form a five-coordinate intermediate. If, for example, we look at the reaction of iodoethane with bromide ions: Br- H3C C + H I H Firstly bromide approaches the central carbon atom from the opposite side to the iodide substituent. What then happens is the carbon simultaneously starts to form a bond with the bromine and starts to break the bond with the iodine. We thus have a very short-lived pentavalent-carbon intermediate. CH3 Br C I H H signifies bond forming or breaking The iodide ion then leaves the intermediate to give: CH3 + C Br H I- H At no stage has a carbocation intermediate formed as in SN1. How many products form in SN2 reactions? Well, again it will depend upon the starting alkyl halide. If we start with an alkyl halide that has three different substituents on the carbon attached to the halide (i.e a chiral centre) then we need to specify whether the starting material is a pure enantiomer or a racemic mixture. This 82 is important, as during an SN2 reaction the attack by the nucleophile is unidirectional i.e. it attacks from one direction, the direction from opposite the leaving group. This will result in inversion at the chiral C-atom. To help explain this, consider the following alkyl halide. H C H5C2 I CH3 - When this is attacked by a nucleophile, say OH , then the following intermediate will form: H5C2 H C OH I CH3 and finally we will get: H HO C C2H5 CH3 If we therefore started with, say, the R-isomer of the alkyl halide we will get the S-isomer of the final alcohol, and only the S-isomer. This is called inversion. If instead we started with a racemate of the starting material then because all the starting material will be inverted to the other enantiomer a racemic mixture of the product will result. o 1 alkyl halides undergo SN2 substitution pathways o 3 alkyl halides undergo SN1 substitution pathways o 2 alkyl halides undergo either SN1 or SN2 substitution pathways Acids & Bases in Organic Chemistry Some organic acids are very easy to spot – they’re not called the carboxylic acids for nothing! Others, however, are less obvious. Terminal alkynes for instance are considered to be acids, albeit very weak ones. Carboxylic Acids Carboxylic acids are weak acids i.e. they do not fully ionise in solution. Rather an equilibrium is established. - RCOOH + H2O ⇌ RCOO + H3O + Although carboxylic acids are technically weak acids they’re still, in the realm of the organic chemist at least, considered to be a good source of protons or hydronium ions. This is due to the stability of the carboxylate ion. 83 O R C O- Is this the only way we can draw the structure or is there the possibility of another resonance form? How about this? O- O R C R O- C O The presence of another resonance structure stabilizes the carboxylate ion as now the negative charge is delocalised and the average structure can be drawn as O C R O Alcohols Alcohols are strange species. They can act either as an acid or as a base. Alcohols are less acidic than water but they can act as proton donors. + R – O – H + base → R – O + base – H An example would be the reaction of phenol with sodium amide, NaNH2, a strong base. H O- O + NH2- NH3 + Compounds resulting from substitution of a benzene ring are usually known by their nonsystematic names e.g. phenol. Some other examples are CH3 COOH toluene benzoic acid NH2 aniline However alcohols can also accept protons and therefore exhibit basic properties. This arises because of the lone pairs on the oxygen atom of the alcohol. H R H+ O R O H H oxonium ion 84 We shouldn’t be too surprised by this behaviour as it is exactly analogous to the behaviour of water, another amphiprotic species. H O + OH- OH- H2O + H H O H+ + H3O+ H Would we expect alcohols to react with carboxylic acids? The answer is yes – acids react with bases! An example would be the following reaction. CH3CH2COOH acid + CH3CH2OH base → CH3CH2COOCH2CH3 + H2O ethyl propanoate The exact mechanism of the ester formation is quite complicated and we shouldn’t push the acid-base story too far. Organic chemists in fact call this type of reaction a CONDENSATION as a small molecule, in this case water, is produced in the reaction. Terminal Alkynes Terminal alkynes contain the following group R – C ≡ C – H The terminal hydrogen is very weakly acidic and therefore we would expect terminal alkynes to react with strong bases such as sodium amide. R C C H + - NH2 R C C + NH3 This is a very important reaction in organic chemistry as it gives a route to form carbon-carbon bonds. Question 4.42 (a) Is the anion R – C ≡ C an electrophile or a nucleophile? (b) Is the anion R – C ≡ C an acid or a base? - The anion, R – C ≡ C is looking to attack a positive centre on another molecule. What type of molecules contain positive centres? Easy – compounds which contain carbon attached to a more electronegative atom, such as an halogen (X) R R C C C H X R C C CH2 R + X- H If we think about this reaction type very carefully we should be able to recognise it as a nucleophilic substitution. Question 4.43 Would you predict the above reaction to go via an SN1 or an SN2 mechanism? Explain your answer. 85 We could also react this powerful nucleophile with CO2 in acidic media. O R C C O C O H + R C C C OH So we have another route possible to prepare carboxylic acids. Reagents and Reaction Conditions Now we’ve got a general picture (however small!) of the reactions that can occur, an important issue to consider is how to actually go about doing these reactions in the lab or in the workplace. This is important because if one chooses the wrong conditions then the reaction might not occur at all, or even worse the reaction will proceed but to the wrong products! Most organic reactions need to have the following specified. (a) solvent (b) temperature (c) catalyst (d) reagents to be used The presence of a catalyst is of particular importance. Catalysts are added to speed things up. However catalysts can speed some reactions up faster than others and therefore one catalyst might give a different product from another catalyst with the same reactants! An example of this would be the reactions of alkynes with hydrogen gas. What would we expect to get? R – C ≡ C – R’ + H2 → ? Here we have got an unsaturated molecule reacting with a reducing agent. We would therefore expect to get ADDITION across the triple bond. Possible products therefore are: H H C C R R' H C R' R C R H H H C C H H R' The actual product formed depends upon the experimental conditions. (i) H H R C C Lindlar catalyst R' H2 C C R R' cis- alkene (ii) R C C R' Pd/C 2 H2 R H H C C H H R' alkane The so-called Lindlar catalyst stops the reaction at the alkene (note it is also specifically the cis-alkene) while the use of a palladium on carbon catalyst takes the reaction further. Therefore when we react H2 with an alkyne we must specify the catalyst used. 86 o Another example of the importance of specifying the reaction conditions is when oxidising 1 alcohols. R CH2 CrO3 H+ OH R CrO3 H+ CHO R COOH + The use of CrO3/H will oxidise the alcohol all the way through to the carboxylic acid. However the use of pyridinium chlorochromate (PCC) in dichloromethane allows the isolation of the aldehyde. R CH2 OH PCC CH2Cl2 R CHO Sometimes we can use our knowledge of the reaction conditions to yield structural information as in the following example. Example 4.6 Question: A hydrocarbon of unknown structure has the molecular formula C8H10. On catalytic hydrogenation over the Lindlar catalyst, 1 equivalent of H2 is absorbed. On hydrogenation over a Pd/C catalyst 3 equivalents of H2 react. Draw a structure for the hydrocarbon that fits the data. Answer: 87 Question 4.44 Predict the products from reaction of hex-1-yne with H2 over (a) Lindlar catalyst (b) Pd/C catalyst Question 4.45 Predict the products from reaction of dec-5-yne with (b) H2, Pd/C (a) H2, Lindlar catalyst Question 4.46 Predict the products from reaction of pentan-1-ol with (a) CrO3 (b) PCC in CH2Cl2 Putting it All Together! The process of learning at the tertiary level can be viewed as having three stages. • Knowing the facts • Understanding the facts • Applying the knowledge gained It is the last point above that is the most important in chemistry. What can you do with what you have at your fingertips? We will therefore end this tutorial with some problem-solving exercises in organic chemistry. The best way to learn how to solve problems is to attempt to solve problems and therefore we will take you through some worked examples and then leave the rest up to you! Example 4.7 Question: An unknown hydrocarbon, A, has a formula of C6H10. On catalytic hydrogenation over Pd/C it reacts with only 1 equivalent of H2. A also undergoes reaction with ozone to give a single product a dialdehyde, B. Suggest a structure for A and B Answer: 88 Example 4.8 Question: How could you prepare 2-methylhexane from an alkyl halide and an alkyne? Answer: 89 Question 4.47 In Example 4.8 we broke the C4-C5 bond to choose our starting materials. (a) Still by breaking the C4-C5 bond, choose another alkyl halide and another alkyne which will give the same product. (b) Do a retrosynthetic analysis to make 2-methylhexane by breaking the C5-C6 bond. Example 4.9 Question: An aldehyde A (C5H10O) can be reduced by LiAlH4 to B. Treatment of B with concentrated H2SO4 gave C. Ozonolysis of C affords methanal and D (C4H8O). D gave a positive iodoform test which identifies methylketones. Identify A, B, C and D. Answer: 90 Example 4.10 Question: A hydrocarbon A adds one equivalent of hydrogen in the presence of a Pd/C catalyst to form n-hexane. When A is reacted vigorously with KMnO4 a single carboxylic acid containing three carbon atoms is isolated. Give the structure and name of A. Answer: Question 4.48 If we start with hex-2-ene, what would be the products upon vigorous treatment with KMnO4? 91