Oxford IBDP Chemistry Course Book 2014 Part 2

20 .1 T y p E s of or g a ni c r E a c T ion ● The curly arrow representing the iodine leaving originates at the bond between the hydrogen and iodine atoms. H CH H CH 3 C 3 C C + C ● A curly arrow goes from the lone pair or the negative charge on I to H H + Br δ+ C (the carbocation). Br Br ● The structural formula of the product 2-iodobutane is shown. Br δ- H CH 3 Electrophilic addition of halogens to alkenes C C Often used as a test for unsaturation in organic molecules, the Br Br electrophilic addition of a halogen (specically bromine) to an alkene follows the same mechanism as shown in gure 11. The halogen is ▲ Figure 11 The mechanism of the electrophilic a non-polar molecule with a net dipole of zero. It is polarized as it addition of bromine to propene forming approaches the electron-rich C =C of the alkene: electrons within the 1,2-dibromopropane halogen molecule are repelled, resulting in a temporary dipole. Electrophilic addition of interhalogens to alkenes δ+ δ- Br Cl Interhalogens are compounds in which two or more halogens are combined in a molecule. Differences in electronegativity between the δ+ δ- halogens will result in an electrophilic region of the molecule and this I Br determines which halogen will attack the pi bond (gure 12). ▲ Figure 12 Polarity in interhalogen molecules The addition reactions of halogens, interhalogens, and hydrogen halides to symmetrical alkenes all undergo the same mechanism as with H H unsymmetrical alkenes (gure 13). The difference is that Markovnikov’s rule does not apply. H H + C C C C H H H H Cl Br Br δ+ Electrophilic substitution reactions In sub-topic 10.2 it was stated that benzene does not readily undergo Cl addition reactions, preferring substitution reactions. The electrophilic δ- substitution mechanism of these reactions can be illustrated by the nitration of benzene. H H C C + The rst step requires the nitronium ion electrophile, NO to be generated. H H 2 Pure nitric acid contains only a small concentration of this electrophile, but Br Cl a nitrating mixture of sulfuric acid and nitric acid at 50 °C generates a ▲ Figure 13 The addition of the interhalogen BrCl to ethene forms 1-bromo-2-chloroethane higher concentration of nitronium ions, allowing the reaction to proceed at an acceptable rate. Sulfuric acid protonates nitric acid, which subsequently releases a water molecule to generate the electrophile: H O O + O + O + + N N H N H + H SO + HSO H O+ 4 2 2 4 + 2 O O nitric acid The nitronium ion is a strong electrophile. As the electrophile approaches the delocalized pi electrons of the benzene ring, the nitronium ion is attracted to the ring. Two electrons from the ring are donated and form a new C N bond. Additionally, a pi electron from the N =O bond of the nitronium ion moves onto the oxygen atom: + + H NO NO 2 2 445

20 ORGANIC CHEMISTRY (AHL) Molecular models are useful aids This is the rate-determining step of the mechanism. The addition of the for visualization of the shapes of nitronium ion to the C=C bond eliminates the aromaticity of the arene. molecules and can help with working out mechanisms and nomenclature. Water then acts as a base, deprotonating the carbocation intermediate and restoring the aromaticity of the system. + + + H H NO 2 NO 2 CH + HNO H SO C H NO + HO 3 _2___4 2 6 6 → 6 5 2 50 °C The product nitrobenzene is a yellow oil that can be isolated from the reaction mixture. Drawing mechanisms for electrophilic substitution reactions ▲ Figure 14 A molecular model of ● A curly arrow originates from delocalized electrons in benzene and 2,4,6-trinitrophenol (picric acid) + terminates at the NO electrophile. 2 ● Structural representation of the carbocation shows a partial, broken- line circle and a positive charge on the ring. ● A curly arrow representing the hydrogen ion leaving originates at the bond between the carbon and hydrogen atoms and terminates at the benzene ring cation. ● The structural formula of the organic product nitrobenzene is shown + along with the released hydrogen ion, H Reduction of carboxylic acids Carboxylic acids are reduced to aldehydes and eventually to primary alcohols while ketones are reduced to secondary alcohols, in reactions that are the reverse of the oxidation of alcohols (sub-topic 10.2). Two commonly used reducing agents are lithium aluminium hydride LiAlH and sodium borohydride NaBH . Lithium aluminium hydride is 4 4 regarded as a nucleophilic reducing agent that will reduce polar C =O bonds present in carboxylic acids, aldehydes, and ketones. LiAlH is the stronger reducing agent; it can reduce carboxylic acids to 4 primary alcohols while NaBH can reduce only aldehydes and ketones 4 to alcohols. The reduction equation is often represented in a simplied manner using the symbol [H] to represent the reducing agent. The reduction of an aldehyde to a primary alcohol can be represented by the general equation: R CHO + 2[H] → R CH OH 2 and the reduction of ketones to secondary alcohols by this general equation: R CO R′ + 2[H] → R CH(OH) R′ 446

20 .1 T y p E s of or g a ni c r E a c T ion Conversion of nitrobenzene to phenylamine re x (aniline) Reux is a process in which a reaction mixture is heated under controlled The nitration of benzene by electrophilic addition occurs when conditions for a period of time. A benzene is heated at 50 °C with a mixture of sulfuric acid and nitric condenser is used to cool the vapours acid. This initial step is described earlier in this topic: from volatile solvents and condense them back into the reaction mixture. The CH + HNO H SO C H NO + HO process ensures that the temperature 3 _2___4 2 remains constant over time and optimal 6 6 → 6 5 2 conditions for the reaction are achieved. 50 °C The subsequent conversion of nitrobenzene to phenylamine (aniline) (systematic name phenylamine) is described in two stages. Stage 1: Reduction of nitrobenzene Nitrobenzene is heated in a water bath under reux with a mixture of zinc and concentrated hydrochloric acid. The phenylammonium ion is formed and zinc is oxidized to zinc(II). + + 2+ water out 7H (aq) C H NO (l) + 3Zn(s) + → C H NH (aq) + 3Zn (aq) + 2H O(l) condenser 2 water in 6 5 2 6 5 3 round-bottom Stage 2: Formation of aniline clamps Aniline is formed by the deprotonation of the ammonium salt through heating mantle the addition of sodium hydroxide: with magnetic stirrer + C H NH (aq) + OH (aq) → C H NH (l) + H O(l) 6 5 3 6 5 2 2 H H N H ▲ Figure 16 Reux apparatus H C C H C C C C H H ▲ Figure 15 Molecular model of phenylamine (aniline) 447

20 organic cHEMisTry (aHL) 20.2 s thet te Understandings Applications and skills ➔ The synthesis of an organic compound stems ➔ Deduction of multi-step synthetic routes given from a readily available star ting material star ting reagents and the product(s). via a series of discrete steps. Functional group interconversions are the basis of such synthetic routes. Nature of science ➔ Retro-synthesis of organic compounds. ➔ Scientic method – in synthetic design, the thinking process of the organic chemist is one which invokes retro-synthesis and the ability to think in a reverse-like manner. ToK Background to designing a synthetic route Synthetic organic chemists Organic synthesis takes a starting material and converts it via a series of often use a method referred reactions into the desired product. Each step produces an intermediate to as et-the. in quantities less than the theoretical yield, so an efcient synthetic Star ting with knowledge of pathway will involve the smallest number of steps. For equilibrium the structure and proper ties reactions, conditions are selected that favour the products, thereby of the target compound, increasing the nal yield. they think “in reverse” to determine possible Reactions that convert one functional group to another, such as the synthetic pathways to oxidation of a primary alcohol to a carboxylic acid or the nucleophilic produce it. Imagination, substitution of a halogenoalkane, do not change the length of the intuition, and reasoning all carbon chain. Synthetic tools include controlled chain-lengthening and play their par t in scientic chain-shortening reactions while polymerization (topic 10 and option innovation. Imagination A) involves the formation of long molecules made up of repeating transcends the limitations monomer units. (Chain-lengthening and chain-shortening reactions are of acquired knowledge and not required for IB Chemistry.) opens up the possibility of new ideas. Retro-synthesis What are the roles of Knowledge of the types of reactions undergone by functional groups and these ways of thinking in their mechanisms allows chemists to determine possible steps in a synthetic solving synthetic pathway pathway, in both the forward and the reverse directions (gure 1). problems? Is retro- synthesis a combination Functional group interconversions, the conditions under which they occur, of understanding and and consideration of reaction rates form the background to this approach imagination? (sub-topics 10.1, 10.2, and 20.1). There is no single right or wrong way to solve synthetic problems; one methodology is outlined below. star ting compound product(s) ▲ Figure 1 Designing a synthetic pathway. The reversible arrows do not indicate equilibrium but rather a problem-solving approach that involves working both for wards from the reactant and backwards from the product (retro-synthesis). 448

20 . 2 s y n T H E T ic r o u T E s Step 1 ● Draw the structural formulae of both the starting compound and the desired product(s). ● Identify the functional group(s) present in the product. Step 2 ● List possible reactions that would produce the desired functional group(s). Step 3 ● Identify the functional group(s) present in the starting material and identify any relationship between the starting reagent and any intermediate compounds you have listed in step 2. Step 4 ● Design a reaction pathway that has the minimum number of steps. Include all the reaction conditions and reagents required. Worked example O H H H Design a synthetic route to produce ethyl H C + H C methanoate from chloromethane. O H Solution Step 1 H H The starting material, chloromethane, is a methanoic acid ethanol halogenoalkane while the product, ethyl methanoate, is an ester (gure 2). The condensation reaction is achieved by heating the carboxylic acid and alcohol in the presence of concentrated sulfuric acid, creating the sweet- smelling ester (sub-topic 10.2). Water is also formed in this reaction. chloromethane ethyl methanoate Step 3 Cl O C H H O H The starting material, chloromethane, needs C C to be converted to methanoic acid. Methanoic H H H acid is the product of the oxidation of a primary H H alcohol (sub-topic 10.2). Chloromethane, ▲ Figure 2 Structural formula of the starting compound, a halogenoalkane, is a reactive compound chloromethane and the ester product, ethyl methanoate that contains a polarized C Cl bond which is susceptible to attack by nucleophiles such as OH Step 2 (sub-topic 20.1). Ethyl methanoate is the product of a condensation (esterication) reaction between a carboxylic acid (methanoic acid, HCOOH) and an alcohol (ethanol, CH CH OH): 3 2 449

20 ORGANIC CHEMISTRY (AHL) Step 4 heat with dilute OH /S 2 reaction methanol: CH OH N 3 chloromethane: CH Cl 3 OS H/ 4 2 + H :noitcaer noitadixo ethyl methanoate: CH CH CHO heat with ethanol in conc. sulfuric acid/esterication methanoic acid: HCOOH 3 2 ▲ Figure 3 A reaction pathway showing how ethyl methanoate is produced from chloromethane, including reaction conditions and reagents Summary reaction pathways Examination questions may require you to deduce synthetic routes of up to four steps. Figures 4 to 6 show reaction pathways required for the SL/HL organic chemistry topics. When a mechanism is required, this is signied by “ M”. benzene nitrobenzene phenylamine (aniline) aldehyde carboxylic acid ▲ Figure 5 Reaction pathways for aromatic compounds ketone alcohol M alkene (eg ethene) polymer halogenoalkane ester dihalogenoalkane ▲ Figure 6 Reaction pathways for polymers M M alkane alkene ▲ Figure 4 Reaction pathways for aliphatic compounds 450

20 . 3 s T E r E oi s oM E r i s M 20.3 steemem Understandings Applications and skills ➔ Stereoisomers are subdivided into two ➔ Construction of 3-D models (real or vir tual) of a classes – conformational isomers, which wide range of stereoisomers. interconver t by rotation about a σ bond, and ➔ Explanation of stereoisomerism in non-cyclic congurational isomers that interconver t only alkenes and C and C cycloalkanes. by breaking and reforming a bond. 3 4 ➔ Comparison between the physical and ➔ Congurational isomers are fur ther subdivided chemical proper ties of enantiomers. into cis–trans and /Z isomers, and optical ➔ Description and explanation of optical isomers isomers. in simple organic molecules. ➔ Cis–trans isomers can occur in alkenes ➔ Distinction between optical isomers using a or cycloalkanes (or heteroanalogues) polarimeter. and dier in the positions of atoms (or groups) relative to a reference plane. According to IUPAC, E/Z isomers refer to alkenes Nature of science of the form R R C=CR R (R ≠ R , R ≠ R ) 1 2 3 4 1 2 3 4 where neither R nor R need be dierent ➔ Transdisciplinary – the three-dimensional 1 2 from R or R shape of an organic molecule is the foundation 3 4 pillar of its structure and often its proper ties. ➔ A chiral carbon is a carbon joined to four Much of the human body is chiral. dierent atoms or groups. ➔ An optically active compound can rotate the plane of polarized light as it passes through a solution of the compound. Optical isomers are enantiomers. Enantiomers are non-superimposable mirror images of each other. Diastereomers are not mirror images of each other. ➔ A racemic mixture (or racemate) is a mixture of two enantiomers in equal amounts and is optically inactive. isomerism Types of isomerism stereo- structural isomerism isomerism Stereoisomers have an identical molecular formula and bond multiplicity but show different spatial arrangements of the conformational congurational atoms. Stereoisomers can be subdivided into two major classes, isomerism isomerism conformational isomers and congurational isomers (gure 1). ▲ Figure 1 Classes of isomerism Conformational isomers can be interconverted by rotation about the σ bond, without breaking any bonds. Congurational isomers can 451

20 ORGANIC CHEMISTRY (AHL) only be interconverted by the breaking of the σ or π bond or through rearrangement of the stereocentres. Conformational isomers Substituents and functional groups joined together by single σ-bonds can rotate freely, changing the three-dimensional arrangement of the atoms relative to one another. In contrast, a carbon–carbon double bond is composed of both a σ and a π bond and the arrangement of electron density above and below the internuclear axis means that no rotation is possible without breaking the π bond. Conformational isomers therefore differ from one another in the arrangement of atoms around a single bond. The rapid interconversion from one conformer to the other means that the separation of the individual isomers is virtually impossible. Ethane H In ethane, rotation about the carbon–carbon rear hydrogen H H bond results in two different conformations: eclipsed and staggered. In the eclipsed H conformation, the substituents (hydrogen atoms) on adjacent carbons are as close to one another H as is possible. When one half of the molecule rotates about the carbon–carbon bond, the rear carbon relative positions of the substituents change until the three hydrogen atoms on each carbon ▲ Figure 3 The Newman projection of ethane shows the relative are as far apart as possible – this is the staggered positions of the substituents conformation. The Newman representation of the eclipsed A Newman projection is a representation of conformation shows the hydrogen atoms slightly the three-dimensional structure which shows askew for clarity (gure 4), but the actual angle the conformation of the molecule by looking between hydrogens on the adjacent carbon along the carbon–carbon bond (gures 2 atoms is 0° and 3). The front carbon and its substituents are represented by lines projecting out of the The staggered conformation, with the hydrogens centre. The circle represents the carbon at the rear and bonds coming out of the circle show its on the adjacent carbon atoms positoned at substituents. 60° to each other, is more stable. The eclipsed conformation is of higher energy (less stable) due to repulsive interactions between the electrons of C H bonds. H H H H H H H H H H C H H H H H C H H H staggered eclipsed look along this bond ▲ Figure 2 Look along the carbon–carbon bond to construct the ▲ Figure 4 Newman projections of staggered and eclipsed Newman projection of ethane ethane conformers 452

20 . 3 s T E r E oi s oM E r i s M Conformational isomerism in cyclic hydrocarbons Cycloalkanes also show conformational isomerism. bent C C bond, conrmed by electron-density mapping: the bonding electron density is greatest The structural consequences arising from the bond outside the carbon–carbon internuclear axis. angles in C , C , and C cycloalkanes have been the Cyclobutane 3 4 5 The carbon ring bond angle increases from 60 ° in cyclopropane to 90° in cyclobutane; the molecule subject of extensive research. Torsional strain or of cyclobutane still experiences angle strain (90° < 109.5°) and torsional strain from the eclipsed torsional energy is the energy difference between arrangement of adjacent C H bonds. The strained four-membered ring in penicillin shows similar the staggered and eclipsed conformations. It is the angle strain; this is discussed in sub-topic D.2. result of the repulsion between bonding electrons. One way of minimizing the strain placed on this conformer is to pucker the ring. One of the four In an eclipsed conformation, pairs of bonding carbon atoms moves out of the plane of the ring, slightly increasing angle strain but signicantly electrons in the C H bonds will repel one another. decreasing torsional strain (gure 6). Cyclopropane The ring structure of cyclopropane lacks stability as the molecule experiences ring strain for two reasons. It exhibits torsional strain from repulsion of adjacent bonding electrons in C H bonds due to the ring rigidity (gure 5). It also 3 exhibits angle strain: the sp orbital angle is 109.5° (tetrahedral) but the internuclear bond angle in cyclopropane is only 60 ° resulting in a misalignment of the orbitals when they overlap end on to create the σ bond. The result is a (a) not quite eclipsed H H (a) cyclopropane H c H H 109.5° 60° c 3 H c H c c H 109.5° (b) typical alkane typical bent not quite eclipsed (b) bent bonding C–C bonds cyclopropane C–C bonds ▲ Figure 5 Cyclopropane shows (a) torsional strain; (b) angle ▲ Figure 6 (a) The planar ring structure of cyclobutane, with eclipsed C–H bonds, exhibits torsional and angle strain. (b) strain: the bond angle of 60° is much less than the ideal Puckering of the cyclobutane ring reduces torsional strain 3 109.5° for sp hybridized bonds Configurational isomers congurational isomerism Congurational isomers can be interconverted only by the breaking of bonds or through the rearrangement of the stereocentres. cis-trans and optical Congurational isomers are subdivided into cis–trans and E/Z isomers on /Z isomerism the one hand and optical isomers on the other (gure 7). ▲ Figure 7 Classes of congurational As mentioned above, congurational isomers exist due to the lack of isomerism rotation around the carbon–carbon double bond such as that present in aliphatic alkenes. Cis–trans isomers are determined by the positions of substituents relative to a reference plane. For alkenes this reference plane is the 453

20 ORGANIC CHEMISTRY (AHL) H H H CH carbon–carbon double bond. Cis-isomers have substituents on the same 3 side of the reference plane while in trans-isomers the substituents are on C C C opposite sides (gure 8). C Disubstituted cycloalkanes also exhibit cis–trans isomerism with the plane HC CH HC of symmetry being the ring (gure 9). 3 3 3 H mirror plane CH CH 3 3 CH 3 CH CH CH 3 3 3 cis-1,2-dimethylcyclobutane trans-1,2-dimethylcyclobutane (1S,2S)-isomer (1R,2R)-isomer ▲ Figure 8 Cis–trans congurational isomers of ▲ Figure 9 Cis–trans isomerism in 1,2-dimethylcyclobutane but-2-ene It is a relatively simple process to identify cis–trans isomers of disubstituted alkenes. However, a different naming convention is adopted for the naming of tri- and tetra-substituted alkenes. According to IUPAC, for isomers of the form R R C = CR R , where 1 2 3 4 R ≠R ,R ≠ R , and neither R nor R need be different from R or R , 1 2 3 4 1 2 3 4 the E/Z nomenclature rules outlined below can be applied. HC CH In the E/Z system relative priorities are assigned to the substituents on 3 3 each carbon of the carbon–carbon double bond. The Cahn–Ingold– Prelog (CIP system) rules for assigning the stereochemistry of substituted Br alkenes are named after the scientists who developed them in 1966. Their rules can be used to assign R or S conguration to each stereocentre ▲ Figure 10 3-bromopent-2-ene, and E or Z to a double bond. CH CH=C(Br)CH CH 3 2 3 In the examples given above, cis-isomers are the equivalent of Z isomers and trans-isomers are the equivalent of E isomers. Elemet Z highest The isomer 3-bromopent-2-ene has both bromine and ethyl substituents I 53 priority bonded to the same carbon (gure 10). To assign an E or Z conguration, 35 the priority of each atom bonded to the carbon atoms of the C =C bond Br 17 ↓ is rst established. This is achieved by ordering the atoms from highest to Cl lowest atomic number, Z. If both higher priority substituents are on the 9 same side of the double bond, the isomer is designated Z (comparable to F 8 cis). If they are on opposite sides it is designated E (comparable to trans). O In 3-bromopent-2-ene, one carbon of the double bond has methyl and hydrogen substituents and the other carbon has bromine and ethyl substituents. Table 1 lists substituents in order of priority and the two possible isomers are shown below. lowest higher priority higher priority higher priority lower priority priority N 7 HC Br HC C 6 3 C 3 CH CH C C 2 3 C H 1 H CH CH H Br lower priority 2 3 lower priority higher priority lower priority ▲ Table 1 To assign an E or Z conguration, (Z)-3-bromopent-2-ene ( )-3-bromopent-2-ene substituents are prioritized according to their atomic number, Z ▲ Figure 11 If both higher priority substituents are the same side of the double bond, the isomer is designated Z. If they are on opposite sides it is designated E 454

20 . 3 s T E r E oi s oM E r i s M Stereoisomerism in carotenoids Chemists have established that the enantiomers of many chemical Carotenoids are a large group of organic pigments (see sub- compounds have varying eects on the topic B.2) that display a wide range of stereochemical properties. human olfactory system (our sense Research into the effects carotenoids have on visual and motor of smell). Dierent enantiomers of the integration within the human body focuses on individual same compound have dierent odours. stereoisomers in isolation from their isomeric partners. Visual Recognition of this is of vital impor tance motor integration measures a child’s ability to make sense of visual to the perfume industry that has an information and then use it appropriately for a motor task such as annual income of billions of US dollars. writing, playing sports, or using tools and utensils. Natural products may exist as individual enantiomers or their mixtures. Precise Optical isomerism analysis is required to establish their olfactory proper ties in the development Optical isomerism is a type of congurational isomerism determined of new perfumes. by the presence of chiral carbon atoms. Also known as a stereocentre or asymmetric centre, a chiral carbon is bonded to four different atoms or groups of atoms. Optical isomers have the ability to rotate plane-polarized light and exist in pairs that are called enantiomers or diastereomers. O OH * OH NH * 2 H I I *N O NH 2 HO I N CH HO 3 OH I nicotine norepinephrine thyroxine ▲ Figure 12 Nicotine is naturally synthesized by the tobacco plant, norepinephrine is a neurotransmitter, and thyroxine is a hormone from the thyroid gland. All of these compounds are chiral: they each contain a stereocentre identied by * Enantiomers are non-superimposable mirror images of each other. They have no plane of symmetry and their optical activity is most readily assigned when the molecules are represented as three-dimensional images (gure 13). COOH COOH R R H C C H NH NH ▲ Figure 14 The ar t of making perfumes can be 2 2 traced back to ancient times in Egypt. Other ancient civilizations such as those of the ▲ Figure 13 Non-superimposable mirror images of the general formula of a 2-amino acid Arabs, Romans and Persians made perfume an impor tant aspect of their daily lives, promoting the development of scientic techniques used in the ex traction of scents from plants. Today, the markets in Fez, Morocco, oer a large variety of scents that can be mixed by ar tisans 455

20 ORGANIC CHEMISTRY (AHL) Optical isomers and plane-polarized light Under the same conditions, two optical isomers with the same general formula rotate the plane of polarized light by the same angle but in opposite directions (gure 15). One enantiomer rotates the plane of polarization in a clockwise direction; this is desig na te d the ( + ) enantiome r. The o the r e na nti om e r r o t a t e s th e plane of polarization in an anticlockwise direction and is designated the ( ) enantiome r. A 5 0 : 5 0 m i xt ur e of th e t wo e n a n t i om e rs i s called a racemic mixture (or racemate) and does not rotate plane- polarized light. light source ToK unpolarized light TOK identies eight specic ways of knowing (WOK). Individual learners polarized light undergo experiences which leads to the recognition of new information. The polarizer combination of this new information, pure enantiomer coupled with prior learning and understanding, enables the learner to analyser viewer determine what knowledge they will ▲ Figure 15 Rotation of plane-polarized light by a pure enantiomer construct and how this will integrate into their understanding of the world. A polarimeter can be used to determine the optical purity of the products of synthetic reactions (see sub-topic D.7). This technique is Optical isomers and their structure give commonplace in industry producing optically active products, examples indirect evidence of the existence of of which can be found throughout the pharmaceutical, fragrance food, a tetrahedrally bonded carbon atom. and chemical industries. The product’s effect on plane-polarized light can How do we use indirect evidence be compared with literature values to determine the purity of the desired to construct understanding? What enantiomer. ways of knowing help us to establish relationships between established theories and indirect evidence? The symbols d (for dextro) and l (for levo) steemem  mede Many therapeutic drugs are chiral molecules with only one enantiomer having the are now obsolete as stipulated by IUPAC desired pharmacokinetic and pharmacodynamic proper ties. phmket studies the body’s response to foreign compounds and changes caused by the and have been replaced by (+) and ( ). administered drug. It is associated with the absorption, distribution, metabolism and excretion of the drug by the body. phmdm studies the action of the drug on the systems of the body and how a drug binds to its target site. The separation of enantiomers can be a very expensive process and such drugs are often administered as racemic mixtures rather than as the pure active enantiomer. For example, synthetic compounds used as anaesthetics may be administered as a racemic mixture with one of the enantiomers having the intended therapeutic eects while the other may have undesired eects that can 456

20 . 3 s T E r E oi s oM E r i s M interfere with the actions of the active component or, more seriously, have damaging eects on the body. CH CH 3 3 HO HO CH CH 3 3 O O (+)-ibuprofen (active) CH CH 3 3 ( )-ibuprofen (inactive) ▲ Figure 16 Ibuprofen is a chiral molecule; one enantiomer is active in the body while the other is not Physical and chemical proper ties of optical isomers ToK The two enantiomers of a particular substance have identical physical The visualization of three-dimensional properties such as boiling and freezing points, viscosity, density, and stereoisomers can be challenging. solubility. Many of their chemical properties are also identical except for Physical models and three- their reactions with other optical isomers, often in biological systems. dimensional images that have been Enzymes within the body are chiral and they can distinguish between created using computer software the enantiomers of their substrate (see sub-topic B.7). have been used to advance research into synthetic compounds. There are For example, limonene is a chiral molecule and the body can eight areas of knowledge, including mathematics, the natural sciences, distinguish between the two enantiomers in both taste and odour. the human sciences, the ar ts, history, ethics, religious, and indigenous Lemons and oranges contain the same isomer, ( +)-limonene. In knowledge systems. contrast, ( )-limonene is found in pine needles, star anise, peppermint, Discuss the role of physical and computer modelling in all areas of and spearmint and has no similarity in smell or taste to the ( +)-isomer knowledge. (gure 17). H * CH HC C CH 3 (+)-limonene 3 3 * 3 CH CH 2 2 ( )-limonene ▲ Figure 17 The (+)- and (–)-enantiomers of limonene Another example of a chiral molecule is the drug thalidomide (gure 18). (+)-enantiomer(eective isomer) It became available in the late 1950s and was prescribed by medical practitioners to pregnant women for the treatment of morning sickness, H which is nausea associated with pregnancy. One enantiomer of thalidomide is the effective drug, free of clinical side effects, with the O O intended therapeutic effect. The other enantiomer is a teratogen: it caused birth defects in babies born to some mothers who took the drug. H N In the human body the thalidomide enantiomers rapidly interconvert N O due to the relatively high acidity of the proton at the stereocentre. Regardless of the enantiomer administered, a racemic mixture is O stereocentre soon produced in the body. The drug is now used for the treatment of leprosy and cancer in males and those female patients who use ( )-enantiomer (teratogenic isomer) contraception and undergo regular pregnancy tests. H O O H N N O O ▲ Figure 18 The enantiomers of thalidomide 457

20 ORGANIC CHEMISTRY (AHL) Diastereomers ToK Diastereomers are different from the enantiomers of optical isomers. Chemists represent complex molecular Like enantiomers they are non-superimposable but they do not form structures as three-dimensional models. Modern-day models have mirror images. They have two or more stereocentres and differ in become sophisticated and detailed through advances in computing power. the conguration of at least one centre. In contrast to enantiomers, How do the scientists who elucidate diastereomers with the same general formula have different physical and complex structures in this way accurately represent them in two chemical properties. In option B.10 D-xylose and L-ribose are identied dimensions? What are the similarities and dierences in the two approaches as diastereomers due to the conguration at the C 2 and C 4 positions and what is the role of the dierent ways of knowing? (gure 19). O O H H 1 1 C C H 2 OH HO 2 H C C HO 3 H HO 3 H C C H 4 OH HO 4 H C C 5 5 CH OH CH OH 2 2 D-xylose L-ribose ▲ Figure 19 D-xylose and L-ribose are diastereomers due to the conguration at the C 2 and C 4 positions. They are non-superimposable but they are not mirror images 458

QuE sTions Questions 1 What is the correct order of reaction types in A. I and II only the following sequence? B. I and III only C H Br _I_ _I_I II_I_ C. II and III only → C H OH → C H COOH → C H COOC H 3 7 3 7 2 5 2 5 2 5 D. I, II and III [1] i ii iii IB, May 2009 A. substitution oxidation condensation B. addition substitution condensation C. oxidation 5 Which statement is correct about the substitution condensation enantiomers of a chiral compound? A. Their physical properties are different. D. substitution oxidation substitution B. All their chemical reactions are identical. IB, May 2011 C. A racemic mixture will rotate the plane of polarized light. 2 a) Identify the reagents used in the nitration D. They will rotate the plane of polarized light of benzene. [2] in opposite directions. [1] IB, May 2009 b) Write an equation or equations to show the + formation of the species NO from these 2 reagents. [1] 6 Which molecule has a chiral centre? c) Give the mechanism for the nitration of A. CH CH=CHCHO 3 benzene. Use curly arrows to represent the B. (CH ) C=CHCH OH movement of electron pairs. [2] 3 2 2 C. CH OCH CH 3 2 3 IB, May 2006 D. CH CHOHCH CH [1] 3 2 3 IB, May 2011 3 Which process can produce a polyester? A. Addition polymerization of a dicarboxylic acid 7 Which two molecules in gure 20 are cis–trans B. Condensation polymerization of a diol and isomers of each other? a dicarboxylic acid C. Addition polymerization of a diol and W HC H X H COOH 3 C C C dicarboxylic acid C CH HC CH HOOC 3 3 3 D. Condensation polymerization of a dicarboxylic acid IB, November 2010 Y HC H Z HC COOH 3 C 3 C C H C CH HOOC H 3 4 Which statements about substitution reactions ▲ Figure 20 are correct? I. The reaction between sodium hydroxide A. X and Z and 1-chloropentane predominantly follows B. X and Y an S 2 mechanism. N C. W and Y II. The reaction between sodium hydroxide D. W and Z [1] and 2-chloro-2-methylbutane predominantly follows an S 2 mechanism. IB, May 2011 N III. The reaction of sodium hydroxide with 1-chloropentane occurs at a slower rate than with 1-bromopentane. 459

20 ORGANIC CHEMISTRY (AHL) 8 Halogenoalkanes can undergo substitution ii) The reaction with 2-bromo-2- reactions with potassium hydroxide solution. methylbutane proceeds by an S 1 N mechanism. Describe this mechanism a) State an equation for the reaction of C H Cl 4 9 using structural formulas and curly with KOH. [1] arrows to represent the movement of b) Substitution reactions may occur by either electron pairs. [3] of two mechanisms namely S 1 or S 2. N N iii) Explain why 1-bromopentane reacts by Outline the meaning of the term S 1. [2] an S 2 mechanism whereas 2-bromo- N N 2-methylbutane reacts by an S 1 N c) Predict the mechanism (S 1 or S 2) N N mechanism. [3] expected for the reaction of the following iv) Explain whether the boiling point of halogenoalkanes with aqueous KOH. 1-bromopentane will be higher, lower, 1-chlorobutane to form butan-1-ol or the same as that of 2-bromo-2- methylbutane. [3] 2-chloro-2-methylpropane to form [2] v) The product C H OH formed from 2-methylpropan-2-ol. 5 11 the reaction with 1-bromopentane d) Explain the mechanism of each reaction in is warmed with ethanoic acid in part (c) using curly arrows to represent the the presence of a few drops of movement of electron pairs. [6] concentratedsulfuric acid. State the IB, November 2009 name of the type of reaction taking place and the structural formula of the organic product. [2] 9 There are several structural isomers with the IB, May 2011 molecular formula C H Br. 5 11 a) Deduce the name of one of the isomers which can exist as enantiomers and draw 10 Deduce a multi-step synthesis for each of thefollowing conversions. For each step three-dimensional representations of its state the structural formulae of the reactants andproducts and the conditions used for two enantiomers. [3] thereactions. b) All the isomers react when warmed with a dilute aqueous solution of sodium hydroxide according to the equation below. (i) 2-chlorobutane to butan-2-one (2 steps) CH Br + NaOH → C H OH + NaBr 5 11 5 11 (ii) propene to propyl ethanoate (2 steps) i) The reaction with 1-bromopentane (iii) benzene to aniline (phenylamine) (3 steps) proceeds by an S 2 mechanism. N Describe this mechanism using structural formulas and curly arrows to represent the movement of electron pairs. [3] 460

MEASUREMENT AND 21 A N A LY S I S ( A H L ) Introduction one technique results in a full structural identication of a molecule. This combined Although spectroscopic characterization approach forms the basis of this topic. techniques form the backbone of structural identication of compounds, typically no 21.1 Strs ntfiatn  ran mns Understandings Applications and skills ➔ Structural identication of compounds involves ➔ Explanation of the use of tetramethylsilane several dierent analytical techniques including (TMS) as the reference standard. 1 IR, H NMR and MS. ➔ Deduction of the structure of a compound ➔ 1 In a high resolution H NMR spectrum, single given information from a range of analytical peaks present in low resolution can split into characterization techniq u es (X- ray fur ther clusters of peaks. 1 crystallography, IR, H NMR, and MS). ➔ The structural technique of single crystal X-ray crystallography can be used to identify the bond lengths and bond angles of crystalline Nature of science compounds. ➔ Improvements in modern instrumentation – advances in spectroscopic techniques (IR, 1 H NMR and MS) have resulted in detailed knowledge of the structure of compounds. 461

21 M E A S U R E M E N T A N D A N A LY S I S ( A H L ) Advances in analytical techniques Improvements in modern instrumentation have led to advances in 1 spectroscopic techniques (IR, HNMR, and MS) resulting in detailed knowledge of the structure of compounds. Analytical techniques have a wide variety of applications such as: ● Testing for drug abuse by high-performance athletes. ● MS (in combination with other chromatographic techniques such as gas-chromatography (GC-MS) etc) can be used in forensic investigations for crimes. ● Protons in water molecules within human cells can be detected by magnetic resonance imaging (MRI), giving a three-dimensional view of organs in the human body (gure 1). In sub-topic 11.3, we discussed a number of the key analytical techniques used to identify the structure of an organic compound. The Figure 1 MRI scan of a healthy human brain. structural identication of compounds typically involves a combination With the advances made in the instrumentation used in MRI (due largely to the development of 1 more powerful magnets), MRI instruments can now even detect chemical changes in the brain of several different analytical techniques including IR, HNMR, and MS. stemming from external stimuli such as a ash of light. This has allowed neuroscientists 1 to pinpoint specic regions of the brain itself where brain activity is taking place and gain an H NMR spectroscopy understanding of the chemical principles underpinning our actual thought processes. In topic 11 we introduced the principles of proton nuclear magnetic Advances in MRI technology now allow advanced research to be carried out into the 1 dysfunctionality of the brain which is resonance spectroscopy ( H NMR). In this chapter we will now revisit this impor tant in deepening our understanding of neurological disorders and diseases such as 1 schizophrenia and Alzheimer ’s disease technique and look at some of the features of high-resolution H NMR. 1 High-resolution H NMR spectroscopy 1 In practice, most H NMR spectra do not consist of sets of single peaks, which may appear to be the case at low resolution. 1 A high-resolution H NMR spectrum can show further splitting of some absorptions. Splitting patterns result from spin–spin coupling. To understand spin–spin coupling, let us take the example of the 1 H NMR spectrum of 1,1,2-trichloroethane, whose structure is shown in gure 2(a). The molecule contains two types of hydrogen in different chemical environments. Let’s call these two different types of hydrogen H and H , respectively. a b Since protons have nuclear spin, they hence have a magnetic eld associated with them. Every proton can act as a tiny magnet. H can b adopt two alignments with respect to the applied magnetic eld, of H H H H H H a b a b a b H C C Cl H C C Cl H C C Cl a a a Cl Cl B Cl Cl Cl Cl o Figure 2(a) Structure of 1,1,2-trichloroethane, H spin aligned with B H spin opposed to B b o b o which has two types of hydrogen (H and H ) so hence deshields H so hence shields H a a a b Figure 2(b) Spin–spin coupling obser ved in 1,1,2-trichloroethane. Two combinations are in dierent chemical environments seen for the –CH (H ) proton b 462

21 . 1 S p e c T R o S c o p i c i d e N T i f i c A T i o N o f o R g A N i c c o M p o u N d S magnetic ux density, B . The magnetic moment of H aligns with B o b o forapproximately 50% of the molecules in the sample. The other 50 % Trmnl y ofthe molecules will have the magnetic moment of H opposing B . The terms ‘aks’ and ‘snals’ b o can be used interchangeably in 1 Therefore, the signal that will be observed for the methylene protons, H NMR spectroscopy. 1 CH (H ) will appear as a doublet, (d), in the high-resolution HNMR 2 a spectrum. This doublet consists of two lines of the same relative intensity. One of the two lines is located slightly upeld from the 1 original single peak observed in the low-resolution spectrum – this is due to 50% of the molecules having their H s shielded by H (H spin a b b 1 1 opposing B ) and the peak will appear at a lower chemical shift, δ. o 1 2 1 In the other case where the signal has moved slightly downeld from the original single peak in the low-resolution spectrum, 50 % of the 1 3 3 1 molecules will have their H s deshielded by H (spin of H aligning with a b b 1 4 6 4 1 B ) and the peak will appear at a higher chemical shift ,δ. The ratio of o 1 5 10 10 5 1 the intensities of the two lines of the doublet, (d), can be deduced using Pascal’s triangle and will be 1:1. Figure 2(c) Pascal’s triangle. This can be In the case of the methylene CH (H ) protons, four combinations used to deduce the splitting patterns in high- are possible: 1 2 a resolution H NMR spectra Combination 1: H and H magnetic moments aligned with B a1 a2 o (deshields H , so signal is shifted downeld to a higher δ). b Combination 2: H magnetic moment aligned with B and H a1 o a2 magnetic moment aligned against B o Combination 3: H magnetic moment aligned against B and H a1 o a2 magnetic moment aligned with B o B o Combination 4: H and H magnetic moments aligned against B a1 a2 o (shields H , so signal is shifted upeld to a lower δ). Figure 2(d) Spin–spin coupling b obser ved in 1,1,2-trichloroethane. Four combinations are seen for the In the case of combinations 2 and 3, the shielding effect of one cancels the deshielding effect of the other. Hence, the net effect is that there is methylene CH (H ) protons 2 a no change in the chemical shift of the single peak seen in the original 1 low-resolution H NMR spectrum. Therefore, the signal observed for the neighbouring CH (H ) proton will b split into three lines, which we call a triplet, (t). The ratio of the intensities of these lines again deduced from Pascal’s triangle will be 1:2:1. Ty  rtn Slttn 1 (d) Atal δ/m rm H NMR strm A: CH (t) 2 3.960 B: CH 5.762 Spin–spin coupling is actually transmitted through the electrons in the individual bonds. Therefore, spin–spin coupling depends on the way the hydrogens are related to each other in the bonding arrangements within the molecule. It is not necessary, however, to go through the above detailed treatment each time in order to determine the individual splitting patterns and associated intensities of lines resulting from spin–spin coupling. Deductions in fact can be made quite simply using two very simple rules. 463

21 M E A S U R E M E N T A N D A N A LY S I S ( A H L ) O Rule 1: If a proton, H , has n protons as its nearest neighbours, that is a n × H , then the peak of H will be split into (n + 1) peaks. b a HC C CH CH Rule 2: The ratio of the intensities of the lines of the split peak can be 3 2 3 deduced from Pascal’s triangle. A B C 1 Let’s consider this in the case of the high-resolution H NMR spectrum of Figure 3 Structure of butan-2-one butan-2-one, whose structure is given in gure 3. ToK As can be seen, butan-2-one consists of hydrogens present in three different chemical environments. Let’s call these hydrogens, A-type, The intensity ratio of the lines in the B-type and C-type. The integration trace therefore is in the ratio 3:2:3. 1 Using section27 of the Data booklet and Pascal’s triangle, the peaks can be assigned as follows: high-resolution H NMR spectrum is given by the numbers in pasal’s Ty  rtn prt δ/m rm Slttn Atal δ/m rm tranl, a mathematical pattern. The stn 27  Data booklet numbers on the diagonals of Pascal’s 1 triangle add to the fbna sqn H NMR strm (0, 1, 1, 2, 3, 4, 8, 13, 21, 34, 55, 89, 144 ….), a set of numbers that A: CH – (s) 2.139 increases rapidly. Such sequences 3 2.2–2.7 (q) 2.449 are commonly found in nature, eg 0.9–1.0 1.058 the structures of leaves, fruitlets of a B: CH (t) pineapple, even the breeding of rabbits! 2 Find out what role the Fibonacci C: CH sequence plays in a molecule of DNA . 3 Why is mathematics such an eective 1 tool in science? Is mathematics the science of patterns? The following is the actual high-resolution H NMR spectrum for butan-2-one: 11 10 9 8 7 6 5 4 3 2 1 0 δ/ppm 1 Figure 4 High-resolution H NMR spectrum for butan-2-one (90 MHz in CDCl ), 3 Figure 5 An X-ray diractometer at a university consisting of a quar tet ( CH B), a singlet ( CH A) and a triplet ( CH C) research unit. X-ray crystallography is a 2 3 3 powerful structural technique used to determine the arrangement of atoms within a crystal Single-crystal X-ray crystallography using X-rays. The X-rays strike the crystal and are diracted into many specic directions The structural technique of single crystal X-ray crystallography can be depending on the location of electrons within the used to identify the bond lengths and bond angles of crystalline compounds. sample. From this, a 3D model of the electron density can be created and the mean position of The chemical community often shares chemical structural information on the atoms calculated the international stage. The Cambridge Crystallographic Database, ChemSpider developed by the Royal Society of Chemistry (RSC) and the Worldwide 464 Protein Data Bank (wwPDB) are examples that highlight the international dimension of the global scientic community.

21 . 1 S p e c T R o S c o p i c i d e N T i f i c A T i o N o f o R g A N i c c o M p o u N d S usl rsrs 1 ChemSpider, developed by the Royal Society of 2 The Worldwide Protein Data Bank (wwPDB) consists of Chemistry (RSC) is a free chemical structure database organizations that act as deposition, data processing which provides rapid text and structure search access and distribution centres for PDB data. The RCSB to over 29 million structures from hundreds of data Protein Data Bank is an information por tal to biological sources (http://www.chemspider.com/). macromolecular structures (http://www.rcsb.org/pdb/). Explanation of the use of tetramethylsilane 1 (TMS) as the reference standard in H NMR spectroscopy In IR spectroscopy (and also UV–Vis spectroscopy), the position of the absorption peaks can be associated with a wavelength, λ, or a 1 1 frequency, ν. In H NMR this is not possible, as the position of the H NMR signals depends on the strength of the externalmagnetic eld. Therefore, the frequencies can be variable, as no two magnets will be identical. In order to address this problem, a universal reference standard 1 has been agreed and hence the exact position of the H NMR signal can be found relative to the reference signal from the standard. This standard is TMS, tetramethylsilane, whose structure is shown in gure 6. TMS is used as a standard as it has the following advantages: 1 The 12 protons are in the same chemical environment, so there will be just one single peak, which will be strong. The chemical shift of CH 3 this signal for the TMS reference standard is assigned δ = 0 ppm. (All other chemical shifts are measured relative to this.) HC Si CH 3 3 2 TMS is inert (that is, it is fairly unreactive, so will not interfere with the sample being analysed). CH 3 3 It will absorb upeld (δ = 0 ppm), well removed from most other Figure 6 TMS is the universally protons involved in organic compounds which typically absorb agreed reference standard used in 1 H NMR spectroscopy downeld (the terms upeld and downeld have been discussed earlier in this topic). 4 It can be easily removed from the sample after measurement, as it is volatile, having a low boiling point of 26–27 °C. Worked example 1 a) Deduce the full structural formula of ethyl ethanoate. b) Using section 27 of the Data booklet, predict the high-resolution 1 HNMR spectrum of ethyl ethanoate. Your answer should refer to the integration trace on the spectrum, the approximate chemical shifts of the various protons, in ppm, any possible splitting patterns and the relative intensities of the lines of the splitting patterns. 465

21 M E A S U R E M E N T A N D A N A LY S I S ( A H L ) Solution a) H O H H H C C C H H H H A B C b) ● Three types of hydrogen atoms are present in different chemical environments. ● Integration trace showing ratio of hydrogen atoms: 3:2:3. Ty  prt Slttn Rlat v Atal δ/m rtn δ/m rm ntnsts 1 stn 27  Data booklet  lns  rm H NMR th slttn strm attrns A: CH 2.0–2.5 (s) – 2.038 3 3.7–4.8 (q) 0.9–1.0 B: CH (t) 1:3:3:1 4.119 2 C: CH 1:2:1 1.260 3 1 The actual high-resolution of H NMR spectrum of ethyl ethanoate recorded at 90 MHz in CDCl is shown in gure 7. 3 Sty t 11 10 9 8 7 6 5 4 3 2 1 0 Sometimes the chemical shifts δ/ppm that you predict from section 27 of the Data booklet dier 1 slightly from what you may Figure 7 H NMR spectrum of ethyl ethanoate 1 observe in the actual H NMR spectrum. In fact, chemical shifts may vary in dierent solvents and conditions. For example, some solvents have π-electron capabilities and/or can be involved in hydrogen bonding networks leading to variations in the chemical shifts depending on the solvents used. Notice this is the case with respect to the C type of protons in gure 7. 466

21 . 1 S p e c T R o S c o p i c i d e N T i f i c A T i o N o f o R g A N i c c o M p o u N d S Worked example 2 An unknown compound, X, of molecular formula, C H O, with a 3 6 1 characteristic fruity odour, has the following IR and H NMR spectra. 100 T% 50 0 3000 2000 1500 1000 500 4000 1 wavenumber/cm Figure 8 IR spectrum (in CCl solution) of X 4 10 9 8 7 6 5 4 3 2 1 0 δ/ppm 1 Figure 9 H NMR spectrum (300 MHz in CDCl ) of X 3 The MS of X showed peaks at m/z values = 58 and 29 (other peaks were also found). Deduce the structure of X using the information given and any other additional information from the Data booklet. For each spectrum assign as much spectroscopic information as possible based on the structure of X. Solution ● As the molecular formula of X is provided, it is worth deducing rst the IHD, which indicates the index of hydrogen deciency or degree of unsaturation For the generic molecular formula C H N O X : chn o x IHD = (0.5)(2c + 2 h x + n) 467

21 M E A S U R E M E N T A N D A N A LY S I S ( A H L ) Hence for C H O: 3 6 c = 3, h = 6, n = 0, o = 1, x = 0 IHD = (0.5)(6 + 2 6 0 + 0) = 1 Therefore, X contains either one double bond or onering. ● The compound has a typical fruity odour, which normally is characteristic of an ester. However, we note that, based on the molecular formula, X contains just one oxygen atom. This would rule out X being an ester ( COO ). An ether (C O C), a ketone (C CO C), an aldehyde (C CHO) or an alcohol (C OH) are possible classes, however, for X. ● Based on the above, we now examine the IR spectrum and see whether there is a strong IR absorption for C =O in the wavenumber range 1700–1750 cm 1 , based on section 26 of the Data booklet. Indeed, there is a strong peak at approximately 1740cm 1 , suggesting the presence of C =O. ● If C=O is present, then X might be either an aldehyde or a ketone. An aldehydic proton is quite characteristic in 1 the H NMR spectrum, with a chemical shift, δ, in the range 9.4–10.0 ppm, that is shifted considerably downeld, as seen from section 27 of the Data booklet. There does appear to be a single peak, in fact, at quite a large chemical shift value of δ = 9.8 ppm, suggesting the probable presence of an aldehydic proton. ● If X is an aldehyde, that means we now have identied a portion of the molecule, that is CHO. Since the remaining number of atoms must contain two carbons and ve hydrogens, this could only be the ethyl group, CH CH , suggesting that X is propanal, 2 3 CH CH CHO. 3 2 ● Let’s now te st this p r o po s e d s tr uc tur e , b a s ed o n t h e 1 spectroscopic data obtained from the H NMR spectrum. Three types of hydrogen atoms are present in different chemical environments, A, B, and C. Ty  prt Slttn Rlatv ntnsts Atal δ/m rtn δ/m rm  lns  th 1 rm H NMR stn 27  slttn attrns strm Data booklet A: CHO 9.4–10.0 (s) – 9.8 B: CH 2.2–2.7 (q) 1:3:3:1 2.5 2 0.9–1.0 (t) 1:2:1 1.1 C: CH 3 Notice that, for the C protons, the chemical shift observed on the spectrum is slightly outside the range predicted from section 27 of the Data booklet, but this is often the case as chemical shifts may vary in different solvents and conditions. 468

21 . 1 S p e c T R o S c o p i c i d e N T i f i c A T i o N o f o R g A N i c c o M p o u N d S O H H ● Having established all the spectroscopically assigned peaks in the 1 H NMR spectrum for the proposed structure of X, it is worth returning to the IR spectrum at this stage to conrm the additional H C C C H characteristic range for the infrared absorption due to the CH bonds in the wavenumber range 2850–3090 cm 1 H H . As can be seen from the IR spectrum, there are indeed absorptions occurring within Figure 10 Propanal this range. ● Finally, consider the MS. There should be a molecular ion peak at m/z = 58, corresponding to the relative molar mass of C H O, 3 6 calculated as 58.09. This indeed is present. In addition, the other dominant m/z value in the MS can be assigned as follows, using section 28 of the Data booklet: m/z = 29 ... signies presence of CH CH + indicating loss of CHO , 3 2 from molecule X, that is ( M 29). r ● This conrms compound X to be propanal. Note: The actual MS of propanal is as follows: 100 80 ytisnetni evitaler 60 40 20 0 15 20 25 30 35 40 45 50 55 60 65 70 75 10 m/z Figure 11 Source: SDBSWeb: http://sdbs.riodb.aist.go.jp (National Institute of Advanced Industrial Science and Technology) Qk qstn Can you explain the peaks on the MS of X greater than m/z = 58? 469

21 M E A S U R E M E N T A N D A N A LY S I S ( A H L ) Questions 1 An unknown compound, X, of molecular 2 An unknown compound, Y, of molecular formula, C H O , has the following IR and formula, C H O , has the following IR and 3 6 2 5 10 2 1 1 HNMR spectra. HNMR spectra. 100 100 T% 50 T% 50 0 3000 2000 1500 1000 500 0 3000 2000 1500 1000 500 4000 4000 1 1 wavenumber/cm wavenumber/cm Figure 12 IR spectrum of X (in CCl ) solution Figure 14 IR spectrum of Y (in CCl ) solution 4 4 12 10 8 6 4 2 0 2 16 14 12 10 8 6 4 2 0 δ/ppm δ/ppm 1 1 Figure 13 H NMR spectrum (90 MHz in CDCl ) of X Figure 15 H NMR spectrum (300 MHz in CDCl ) of Y 3 3 The MS of X showed peaks at m/z values = 74, The MS of, Y, showed peaks at m/z values = 102 45, and 29 (other peaks were also found). and 57 (other peaks were also found). Deduce the structure of X using the information Deduce the structure of Y using the information given and any other additional information given and any other additional information from the Data booklet. For each spectrum assign from the Data booklet. For each spectrum assign as much spectroscopic information as possible, as much spectroscopic information as possible, based on the structure of X. based on the structure of Y. 470

A M AT E R I A LS Introduction to be put to many extraordinary uses. From metals to nanotechnology, research into the History has characterized civilizations by properties and uses of materials is sometimes the materials they use: Bronze Age, Stone serendipitous. Research often begins with Age, and Iron Age. Uses of materials were an end use in mind, in advance of specic developed based on observations of their knowledge about the composition and properties before an explanation of those construction of the material. properties had been proposed. Using the ideas of bonding and structure, materials are now classied and their properties manipulated A .1 Maa   Understandings Applications and skills ➔ Materials are classied based on their uses, ➔ Use of bond triangle diagrams for binary proper ties, or bonding and structure. compounds from electronegativity data. ➔ The proper ties of a material based on the ➔ Evaluation of various ways of classifying degree of covalent, ionic, or metallic character materials. in a compound can be deduced from its ➔ Relating physical characteristics (melting position on a bonding triangle. point, permeability, conductivity, elasticity, ➔ Composites are mixtures in which materials are brittleness) of a material to its bonding and composed of two distinct phases, a reinforcing structures (packing arrangements, electron phase that is embedded in a matrix phase. mobility, ability of atoms to slide relative to one another). Nature of science ➔ Improvements in technology – dierent materials ➔ Patterns in science – history has characterized were used for dierent purposes before the civilizations by the materials they used: Stone development of a scientic understanding of their Age, Bronze Age, and Iron Age. There are various proper ties. ways of classifying materials according to desired patterns. 471

A M AT E R I A L S Classifying materials Materials science involves understanding the structure and properties of a material and matching these to suitable applications. Type of bonding is one classication system employed by materials science, and four common types of material are metallics, ceramics, polymers, and composites. Each type is suitable for different end uses. Metallic substances exhibit metallic bonding (sub-topic 4.5). This makes them strong, malleable, and good conductors of heat and electricity. Approximately two-thirds of all elements are metals. The development of alloys has designed metallic materials suitable for many applications. At the atomic level, the freely moving electrons in metallic bonding confer ductility and strength, as well as conductivity. Ceramics are traditionally inorganic non-metallic solids formed between metals and non-metals. They have a crystalline structure and their ionic bonding (sub-topic 4.1) means they are brittle and usually show insulating properties, such as the ceramics familiar in plates and cups. However, a wide variety of ways of combining metals and non-metals leads to many ceramics with various uses. A compound of thallium, barium, calcium, copper, and oxygen forms a superconductor whereas glass is an amorphous insulating ceramic material. Bricks, tiles, electric capacitors, abrasives, and cement are other types of ceramics. Because of their numerous applications they are sometimes classied based on their uses. Polymers (also known as plastics) form a third classication of materials based on bonding. Plastics are covalently bonded long- chain molecules. (The formation of addition polymers is covered in sub-topic 10.2.) There are many uses and types of plastics and the industry is growing rapidly. ▲ Figure 1 Strain gauge and thermocouple In general, because of strong covalent bonds that exist throughout attached to a carbon–carbon composite polymer molecules these materials tend to be resistant to chemicals and sample for a stress test. Carbon–carbon is a do not corrode. With no free electrons or metal atoms they are generally composite material that consists of carbon good insulators of both heat and electricity and lighter than ceramics. bres in a matrix of graphite. It has been used Polymers can be engineered into many forms including thin exible in high temperature applications such as for bres, soft exible lms such as plastic wrap, or hardened plastics such the thermal tiles on NASA’s Space Shuttles as polyvinyl chloride (PVC) pipe. Pure and applied sciences not clear cut. Naturally occurring polymers such as rubber and silk have laid the foundations for the Pure science aims to establish a common plastics industry leading to many products from understanding of the universe; applied science engineering synthetic polymers. and engineering develop technologies that result in new processes and products. However, the boundaries between pure and applied sciences are Composites are mixtures composed of two distinct phases: a reinforcing phase embedded in a matrix. Each substance retains its own properties (as in any mixture, sub-topic 1.1); however the composite has specic properties not shown by either part of the mixture individually. Straw and clay formed an early composite used to build huts. In a composite 472

A .1 M At e r i A l s s c ie nc e in t r od u c t io n the reinforcing phase is made up of bres, particles, or a mesh which is embedded in a tough or ductile matrix, depending on the use. Each phase can be a metal, a ceramic, or a polymer. Examples of composites include breglass, carbon bre, and concrete. Aircraft wings, for example, can be made lighter and stronger by the use of composites. Identifying the desired properties of a material for a specic application, such as whether it needs to be re resistant, strong, porous or non- porous, a conductor or not, and then engineering materials to suit these properties forms the basis of the eld of materials science. Designer materials Understanding bonding and structure allows the packing arrangement provides pores that are large enough for individual water molecules to materials scientists to design and manufacture escape through but are too small for the passage of grouped hydrogen-bonded molecules in liquid new materials to desired specications. For water. This type of materials development needs to take into account any intermolecular forces example, waterproof breathable fabrics such as between the material and water, which might affect the movement of water vapour or liquid ® water. Bonding and structure are intrinsically linked with properties. Gore-Tex allow perspiration to evaporate while protecting the wearer from rain. Liquid water has extensive hydrogen bonding which produces large grouped particles, whereas water vapour exists as individual water molecules without hydrogen bonds. In Gore-Tex the material is layered so that Qk q 1 For the statement below identify the structural feature 2 Classifying materials according to the type of or proper ty from each pair that best serves the bonding is a useful system, but other classications application. have their place. For example, a metallurgist may be more interested in the grade of stainless steel A crucible for heating substances over a Bunsen in the alloy than in the type of bonding. Can you burner: high melting point/ low melting point; permeable to moisture / not permeable to moisture; suggest when the following classications may be loose packing arrangement / highly structured crystalline structure; metallic / ceramic. used? Nanomaterials, biomaterials, textiles, alloys, semiconductors. Bond triangle diagrams strongly ionic, eg CsF “The nature of the chemical bond is the problem at the heart of all chemistry.” Bryce Crawford Jr, 1953 Bonds between metals and non-metals vary from ionic to covalent strongly metallic, strongly covalent, in relation to the electronegativity difference between the two types eg Cs of atom. The greater the electronegativity difference, the higher eg F the ionic character. Strongly ionic compounds are crystalline, non- 2 conductors of electricity but moderate conductors of heat and have high melting points, whereas strongly covalent compounds have low ▲ Figure 2 A simple triangle of bonding. The melting and boiling points, are soft, and are poor conductors of both most metallic (least electronegative) element heat and electricity. A material with polar covalent bonds exhibits some is caesium, while the most electronegative ionic and some covalent character. Figure 2 shows how this can be is uorine illustrated simply in a triangle of bonding, while gure 3 gives a more comprehensive version. 473

A M AT E R I A L S CsF 3.6 3.4 3.2 3.0 2.8 2.6 2.4 ytivitagenortcele ∆ I 2.2 2.0 1.8 1.6 1.4 1.2 1.0 0.8 C 0.6 0.4 Figure 3 The diagram shows how the bond types ionic (I), metallic (M), covalent (C), and M semi-metallic (SM) depend on the dierence in electronegativity between the atoms involved. A 0.2 triangular bonding diagram is provided in section 29 of the Data booklet SM Cs F 2 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 3.2 3.4 3.6 3.8 4.0 average electronegativity Worked example Tin(II) chloride has a melting point of 247 °C while Electronegativity difference: SnCl = 3.2 2.0 = 1.2; lead(II) chloride has a melting point of 500 °C. Both are used in the production of aurene glass, PbCl = 3.2 1.8 = 1.4 an iridescent artwork glassware. One of these two substances exists as discrete molecules in the 2 vapour phase. Using electronegativity tables and the triangular bonding diagram in gure 3 identify Plotting (x, y) coordinates of (2.6, 1.2) on the which one, and explain your reasoning. bonding triangle diagram classies SnCl as mostly 2 covalent (60–70% covalent character). Plotting (x, y) coordinates of (2.5, 1.4) on the bonding triangle diagram classies PbCl as more 2 ionic (approximately 60 % ionic, 40% covalent). Solution Because SnCl shows more covalent property 2 Electronegativity values: Sn 2.0, Pb 1.8, Cl 3.2 than does PbCl it is more likely to exist as discrete 2 (2.0 + 3.2) molecules when vaporized. It will also have a lower ________ Average electronegativity: SnCl = = 2.6; 2 (1.8 + 3.2) melting point as covalent substances have lower ________ PbCl = = 2.5 2 2 melting and boiling points than do ionic substances. Questions 1 Magnesium oxide and manganese(II) oxide are strength at high temperatures. Explain your answer in terms of bonding and structure, both used in ceramics, but for different purposes. mentioning specically how the degree of ionic or covalent character inuences a) Using sections 8 and 29 in the Data booklet: packing arrangements and the ability of atoms to slide relative to one another. (i) state the average electronegativity for magnesium and oxygen and for manganese and oxygen 2 One method of sorting materials for recycling (ii) estimate the percentage covalent is to classify them as plastics, glass, paper, character for each compound. cardboard, and metals. Outline the advantages b) Deduce which of the two compounds is of using this classication rather than the more likely to be used to make kilns and metallic, ceramic, polymers, and composites crucibles because of its ability to retain its system. 474

A . 2 M e t A l s A n d i n d u c t i v e ly c o u p l e d p l A s M A ( i c p ) s p e c t r o s c o p y A .2 Ma a     ama (icp)   Understandings Applications and skills ➔ Reduction by coke (carbon), a more reactive ➔ Deduction of redox equations for the reduction metal, or electrolysis are means of obtaining of metals. some metals from their ores. ➔ Relating the method of extraction to the position ➔ The relationship between charge and the of a metal on the activity series. number of moles of electrons is given by ➔ Explanation of the production of aluminium by Faraday’s constant, F. the electrolysis of alumina in molten cryolite. ➔ Alloys are homogeneous mixtures of metals ➔ Explanation of how alloying alters proper ties of with other metals or non-metals. metals. ➔ Diamagnetic and paramagnetic compounds ➔ Solving stoichiometric problems using dier in electron spin pairing and their Faraday’s constant based on mass deposits in behaviour in magnetic elds. electrolysis. ➔ Trace amounts of metals can be identied and ➔ Discussion of paramagnetism and quantied by ionizing them with argon gas diamagnetism in relation to electron structure plasma in Inductively Coupled Plasma (ICP) of metals. Spectroscopy using Mass Spectroscopy ICP-MS ➔ Explanation of the plasma state and its and Optical Emission Spectroscopy ICP-OES. production in ICP-MS/OES. ➔ Identication of metals and abundances from simple data and calibration curves provided from ICP-MS and ICP-OES. ➔ Explanation of the separation and quantication of metallic ions by MS and OES. ➔ Uses of ICP-MS and ICP-OES. Nature of science ➔ Development of new instruments and techniques – ➔ Details of data – with the discovery that trace ICP spectroscopy, developed from an understanding amounts of cer tain materials can greatly enhance of scientic principles, can be used to identify and a metal’s performance, alloying was initially more quantify trace amounts of metals. of an ar t than a science. Reduction of metals Some metals such as gold can be mined directly as the element. However, most metals exist in nature in their oxidized states in compounds; for example, aluminium is found in bauxite as aluminium oxide, Al O . These metals can be extracted from their ores and are then 2 3 often alloyed to give them useful properties. 475

A M AT E R I A L S Because metals in ores are in an oxidized state, they need to be reduced to a zero oxidation state in the elemental form. Reduction by coke (carbon), a more reactive metal, or electrolysis are methods used to obtain metals from their ores. Reduction of iron ore in the blast furnace Reduction is carried out on a large scale industrially to obtain iron from iron ore. Most of the iron extracted is then processed further to produce steel. Iron ore is mainly the oxides Fe O and Fe O which are reduced 2 3 3 4 (sub-topic 9.1) by carbon in the form of coke in a blast furnace. ▲ Figure 1 Prospectors could pan for Coke is heated to form carbon dioxide, which reacts with more coke to elemental gold, Au, because this metal form carbon monoxide in the reducing furnace: is at the bottom of the activity series (section 25 in the Data booklet) and C(s) + O (g) → CO (g) therefore commonly found in its reduced form (zero oxidation state). It would be 2 2 impossible to nd elemental lithium as Li sits at the top of the activity series and CO (g) + C(s) → 2CO(g) can only be obtained by reduction 2 Carbon monoxide is a good reducing agent (it is easily oxidized) and reacts with the iron ore to produce molten iron, which is collected from the furnace: Fe O (s) + 3CO(g) → 2Fe(l) + 3CO (g) 2 3 2 At the very high temperatures in the blast furnace the coke can react directly with the iron ore and also act as a reducing agent itself: Fe O (s) + 3C(s) → 2Fe(l) + 3CO(g) 2 3 The carbon monoxide produced in this reaction can reduce more ore. Reduction by a more reactive metal A second means of obtaining elemental metals is reduction by a more active metal (sub-topic 9.1). Pure copper can be obtained from aqueous copper(II) sulfate by a single replacement reaction with solid zinc, for example: Zn(s) + CuSO (aq) → Cu(s) + ZnSO (aq) 4 4 Other redox reactions can be used to reduce the oxidized metal. For example, passing hydrogen gas over heated copper(II) oxide reduces the copper(II) oxide to elemental copper, while the hydrogen is oxidized to the +1 state: CuO(s) + H (g) → Cu(s) + H O(g) 2 2 Reduction by a more active metal or by carbon cannot be used to extract metals near the top of the activity series such as lithium, rubidium, or potassium. In this case electrolysis (sub-topic 9.2) allows us to obtain the pure metals. Once obtained the elemental metals must not be exposed to air or they will become oxidized again. Lithium is used in lithium batteries and obtained by electrolysis of molten lithium chloride to produce lithium metal and chlorine gas: 2LiCl(l) electrolysis 2Li(l) + Cl (g) ______ 2 → Δ 476

A . 2 M e t A l s A n d i n d u c t i v e ly c o u p l e d p l A s M A ( i c p ) s p e c t r o s c o p y The quantity of metal reduced at the cathode during electrolysis can be calculated using the current passed in the electrolysis, the time it is passed for, and the Faraday constant. This is the charge in coulombs (C) 1 on 1 mol of electrons and has the value 96500 C mol . For example, in the reduction of lithium from its ions: + Li +e → Li the equation shows that 1 mol of electrons are required to reduce 1 mol of lithium ions. Providing 1 mol of electrons requires 96500 C of charge from the electrolysis equipment. The amount of charge, Q, transferred can be calculated from the current I (in amperes, A) and time t (in s): Q = It 1 The SI unit, the ampere, is one coulomb per second; 1 A = 1 C s To reduce 1 mol of copper by electrolysis would take 2 mol of electrons: 2+ Cu + 2e → Cu Worked example Two electrolytic cells are connected in series so 2 mol e produce 1 mol Sn, therefore 0.031 mol e that the same current ows through both cells for the same length of time. One contains an aqueous produces 0.031/2 = 0.016 mol Sn solution of tin(II) sulfate and the other an aqueous solution of silver nitrate (gure 2). Calculate the 1 mass of each metal produced at their respective cathodes if a current of 2.5 A is allowed to run for 0.016 mol Sn × 118.71 g mol = 1.9 g Sn 20 minutes. + Ag +e → Ag 1 mol e produces 1 mol Ag, therefore 0.031 mol e produces 0.031 mol Ag 1 0.031 mol Ag × 107.87 g mol = 3.3 g Ag The production of aluminium Aluminium is one of society’s most useful metals. It is obtained from “alumina” or aluminium(III) oxide, Al O . Aluminium is a relatively active - + - + 2 3 metal so needs to be obtained from its ore by SnSO AgNO electrolysis which must take place in the molten 4 3 state. The exceedingly high melting point of Al O 2 3 (over 2000 °C) makes electrolysis of the native ore ▲ Figure 2 Two electrolytic cells in series economically unfeasible. Solution The Hall–Héroult process was developed to overcome this problem. Molten cryolite, Na AlF Q = It 3 6 is used as a solvent in the electrolysis allowing the 1 = 2.5 C s × 1200 s = 3000 C process to be carried out at lower temperatures. _300_0 C A large density difference between cryolite and = 0.031 mol e 1 96500 C mol molten aluminium also makes extraction of the Reduction equations: pure metal easier. The process is outlined on the next page and in gure 3. 2+ Sn + 2e → Sn 477

A M AT E R I A L S steel case graphite anode with the anodes and oxidize them. The anodes therefore need to be replaced periodically. The overall net equation is: 2Al O (l) + 3C(s) → 4Al(l) + 3CO (g) 2 3 2 + + If anodic oxidation can be avoided, then oxygen gas is produced. In this instance the reactions are: 3+ puried aluminium ore cathode: Al + 3e → Al(l) dissolved in molten cryolite 2 → O (g) + 4e anode: 2O 2 Molten aluminum has a density of 2.35 g cm 3 , which is approximately the same as that of graphite cathode molten aluminium molte n cryolite. Ho w e v e r, a m o lt e n mi xt u r e of cryolite saturated with alumina has a much ▲ Figure 3 The Hall–Héroult process for the ex traction of lower density (approximately 1.97 g cm 3 aluminium. Molten aluminium is more dense than the ). This cryolite–alumina mix ture so the product sinks below the reaction mix ture and can be run o allows the molten aluminium metal to sink to the bottom of the reaction vessel and be tapped off. If it remained in the reaction mixture it 1 Alumina is dissolved in molten cryolite. This could short-circuit the electrolysis apparatus. It has a melting point under 1000 °C so reduces is also important to keep the cryolite saturated the energy needed to create the molten ore for with alumina. For this reason the ore is continually electrolysis. fed into the vessel as the aluminium metal is drained off. 2 The steel case surrounding the molten substance is lined with graphite which serves The production of aluminium uses much more energy than recycling the metal as the cathode. Graphite anodes are inserted because the melting point of Al(s) is only 660 °C. The recycling process also does not into the electrolyte. require the additional electrical energy for electrolysis. 3 An electric current is passed which reduces the aluminium ions. The oxide ions react Alloys Alloys are homogeneous mixtures of metals with other metals or non- metals. By taking a readily available metal and adding small amounts of another material to it, certain desired properties can be greatly enhanced. For example, steel is stronger than iron, and stainless steel is produced in many grades of different composition depending on the purpose. Copper alloys such as bronze and brass have increased resistance to corrosion compared with pure copper. Trace amounts of titanium or scandium added to aluminium can greatly increase its strength and durability without compromising its low density for lightweight applications. Alloying aluminium allowed the development of technologies as diverse as compact hard disk drives and strong but light aircraft wings. The lanthanoids and actinoids (rare earth metals) are nding many alloying applications including in superconductors and lasers. Because the lanthanoids contain f-level electrons they have sharp 4f–4f electron transitions which makes them useful in optics applications such as for 478

A . 2 M e t A l s A n d i n d u c t i v e ly c o u p l e d p l A s M A ( i c p ) s p e c t r o s c o p y amplifying signals in optical bres and in phosphors in computer screens Alloying: Art and TVs. Materials such as ceramics can also be alloyed into metals, for example for use in dentistry. or science? Paramagnetic and diamagnetic materials With the discovery that trace amounts of One property of interest in a metal or alloy is its response to a magnetic certain materials can greatly enhance a metal's eld. Paramagnetic materials are attracted to a magnetic eld whereas performance, alloying was initially more of diamagnetic materials create a magnetic eld opposed to the applied an art than a science. Science has developed eld; and are therefore weakly repelled by an external magnetic eld. many ways to investigate matter indirectly, based In the atoms of a diamagnetic material the electrons are spin paired; upon established scientic foundations. 2 2 6 for example, neon is diamagnetic with electron conguration 1s 2s 2p (valence electron conguration shown below:) 2s 2p 2p 2p x y z so all 10 electrons exist in a paired state. Aluminium atoms, electron 2 2 6 2 1 conguration 1s 2s 2p 3s 3p have one unpaired p electron that is capable of being attracted to an external electric eld. Aluminium is paramagnetic: 3s 3p 3p 3p x y z A spinning electron creates a magnetic dipole. The spins of unpaired electrons can be temporarily aligned in an external eld, causing the material to be attracted to the applied magnetism. This is what happens in paramagnetic materials. In a ferromagnetic material the electron alignment induced by the magnetic eld can be retained, making a permanent magnet. For 2 6 example iron, electron conguration [Ar]4s 3d , can be heated and cooled in a magnetic eld and as the metal cools the unpaired Qk q electrons align themselves such that the magnetic eld created 2+ 1 Explain which ion Mn or by their spin is aligned with the applied eld. Banging or heating 2+ Zn will be attracted by a a permanent magnet can disrupt this alignment and weaken the magnetic eld and which magnet. Paramagnetic materials do not form permanent magnets will be repelled. in this way; their electrons are only temporarily aligned by the 2 Deduce how much copper externaleld. can be electroplated from As we have seen, in diamagnetic materials all the electrons are paired. an aqueous solution of In an external magnetic eld the paired electrons orientate themselves such that the eld created by their spin opposes the applied eld (Lenz’s copper(II) sulfate by a law, which is studied in IB Physics) and so the material will weakly repel the external eld. A superconductor exhibits perfect diamagnetism current of 2 A running for (sub-topic A.8). 20 minutes. Spectroscopic methods Trace concentrations of elements such as heavy metals in water are difcult to determine by chemical tests but can be detected by spectrophotometry techniques. Qualitative analysis showing which metals are present can be carried out by exciting electrons to higher 479

A M AT E R I A L S energy levels and detecting the characteristic wavelength of light emitted as these electrons return to lower energy levels; this is the process employed in atomic emission spectroscopy (AES) (gure 4). Concentrations (quantitative information) can be detected by the level of absorbance of this radiation in optical emission spectroscopy (OES). + e +energy + + p excitation p p e decay hν ▲ Figure 4 As excited electrons return from an excited-state to a lower energy level they emit a characteristic wavelength of light In mass spectrometry (MS) ions are introduced into a mass spectrometer and separated according to their mass-to-charge ratio. The detector receives a signal proportional to the concentration of the ion reaching it thus allowing both identication (qualitative analysis) and quantication. (Mass spectrometry is explained in more detail in topics11 and 21.) These spectroscopic techniques require that atoms are in the excited- state, or that they are fully ionized. Substances must also be atomized for spectroscopic analysis and this is usually accomplished by heating and/or electrical discharge, which bombards atoms with high-energy electrons to excite or ionize them. Plasma can also be used for the atomization and/or excitation of samples for spectroscopy. Plasma is one of the four states of matter and it consists of free electrons, positive ions, and neutral atoms or molecules. In plasma 1% or more of the electrons are dissociated from their atoms so plasma can conduct electricity and be inuenced by magnetic elds. Lightning, electric sparks, and the coloured neonlights used in advertising are all examples of matter in the plasma state. Argon is the gas that is ionized into plasma in inductively coupled plasma (ICP) discharges. Plasma can exist at temperatures much hotter than those reached in furnaces or other discharges (around 10 000 K), and can atomize and ionize any type of material. The emission region plasma discharge ionizes or excites the substance being analysed by plasma MS or OES. The argon is swirled through three concentric tubes of magnetic eld quartz in the torch. The tubes and the swirling action allow the high- temperature plasma to be contained in the centre so as to not melt induction the torch. A high frequency oscillating current is supplied to a coil coils surrounding the torch; this creates electromagnetic elds oscillating in resonance at a high frequency, approximately 30 MHz or more. An electric spark is passed to initiate the plasma by ionizing argon atoms, + knocking off electrons. These charged particles (Ar and e ) accelerate argon back and forth in the electromagnetic elds, occasionally colliding tangential ow + sample ow with other argon atoms and creating more Ar and e . The process + continues until Ar ions are being created at the same rate as electrons ▲ Figure 5 Schematic diagram of an ICP torch. are recombining with the ions to re-form argon atoms. The resulting The emission region is fur ther analysed by MS or OES “reball” of plasma reaches temperatures over 10000 K. This process 480

A . 2 M e t A l s A n d i n d u c t i v e ly c o u p l e d p l A s M A ( i c p ) s p e c t r o s c o p y of heating by magnetic induction is somewhat similar to the process used in induction hobs for cooking. The plasma is held in the centre of the torch until discharge by the swirling argon gas and strong magnetic eld surrounding it. The process of plasma formation is outlined in gure 6. a) b) c) e + Ar e + Ar Argon gas is swirled Oscillating radio frequency (RF) A spark produces some free through the torch. power is applied to the load coil. electrons in the argon. temperature/K excitation 6000 6200 + + 6500 M d) e) 6800 (ion) M * 8000 ionization 10 000 hν (atom) atomization excitation M M* hν + Ar e e + Ar + Ar e (gas) MX e + Ar vaporization (solid) (MX) n desolvation + (solution) M(H O) ,X 2 m The free electrons are accelerated A nebulizer sends in an aerosol carrying ▲ Figure 7 Metals and metallic compounds can by the RF elds causing fur ther the sample and this punches a hole in the be vaporized and ionized in an ICP discharge ionization and forming a plasma. plasma, creating an ICP discharge. ▲ Figure 6 Cross-section of an ICP torch and load coil depicting an ignition sequence A nebulizer sends a spray containing the sample that is to be analysed ▲ Figure 8 The sample is discharged in the into the plasma. The sample is ionized in the plasma ready for high-temperature plasma and viewed side-on spectroscopic analysis by OES or MS (gure 8). by instrumentation for wavelength detection and absorbance in optical (atomic) emission Figure 9 gives an overview of ICP-OES. The technique allows the spectroscopy (OES/AES) accurate detection of very small traces of many elements. The chart in gure 10 shows some detection limits which are constantly being 481 improved. The limiting factor in ICP-MS or ICP-OES is not the plasma but rather the quality of the samples and the accuracy of the calibration curve. Known standards are used for calibration which have uncertainties that must be allowed for. Once plotted the calibration curves can then be used to provide values for unknown concentrations. Creating a calibration curve that is accurate at very low concentrations requires a solution of known concentration to be prepared very accurately, and successive dilutions are then made to create the lower concentrations used in calibration.

A M AT E R I A L S ) 01x( stnuoc noissime 6 1.0 mg dm 3 photons from 5 5 standard element of interest 4 photons from photometer other elements 3 sample emission 2 0.5 mg dm 3 standard 1 sample concentration diraction grafting (separates light of 0 blank dierent wavelengths) 0 0.2 0.4 0.6 0.8 1.0 3 concentration/mg dm slit computer light ▲ Figure 11 An example calibration cur ve plasma ame circulating magnetic eld aspirated sample sample radio frequency generator pump aspirator argon gas sample ▲ Figure 9 Conceptual diagram of ICP-OES 3 ICP-OES detection limits/µg dm ▲ Figure 12 ICP torch in an ICP Atomic Li Be B C N Emission Spectrometer, photos cour tesy of Brian Young, Virginia Tech Chemistry Dept. 0.3 0.1 1 40 na Na Mg Al Si P S Cl 3 0.1 3 4 30 30 ns K Ca Sc V Ti Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br 20 0.02 0.3 0.5 0.5 2 0.4 2 1 5 0.4 1 4 20 20 50 na Rb Sr Y Nb Zr Mo Ru Th Pd Ag Cd In Sn Sb Te I 30 0.06 0.3 5 0.8 3 6 5 3 1 1 9 30 10 10 na Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Ti Pb Bi 10 0.1 1 4 15 8 5 0.4 5 10 4 1 30 10 20 Ce Pr Nd Sm Eu Gd Tb Dy Ho Er Tm Y b Lu 5 1 1 2 0.1 1 2 2 0.4 1 0.5 0.3 0.2 Th U 70 15 ▲ Figure 10 Detection limits for dierent elements by ICP-OES ▲ Figure 13 ICP torch emission Some obvious limits apply to ICP-MS and ICP-OES. Argon cannot be 482 analysed, and neither can carbon dioxide as any argon supply usually contains this. Using water or organic solvents prevents H, O, or C atoms from being analysed because of the quantity of these elements in the solvent. At the high temperatures used in ICP-MS and ICP-OES, any solvent sprayed from the nebulizer not only vaporizes but also atomizes as any covalent bonds are broken. Advantages of ICP-MS and ICP-OES over other analytical techniques include a larger linear calibration, and the ability to detect multiple elements at low concentrations.

A . 2 M e t A l s A n d i n d u c t i v e ly c o u p l e d p l A s M A ( i c p ) s p e c t r o s c o p y Questions 1 Magnesium is an essential component of b) For each of the processes named in a), chlorophyll and traces of it can be found outline how the process occurs. [3] in various uids extracted from plants. IB, November 2005 Its concentration may be estimated using 3 Some vaporized magnesium is introduced into a inductively coupled plasma optical emission mass spectrometer. One of the ions that reaches spectroscopy (ICP-OES). 25 + the detector is Mg a) Describe the plasma state as used in ICP. a) Identify the number of protons, neutrons, b) The calibration curve shown in gure 14 25 + and electrons in the Mg ion. [1] was set up using ICP-OES. b) State how this ion is accelerated in the mass spectrometer. [1] 900 25 2+ 800 700 c) The Mg ion is also detected in this mass 600 500 spectrometer by changing the magnetic 400 300 eld. Deduce and explain, by reference 200 100 to the m/z values of these two ions of 01 0 25 2+ ▲ Figure 14 magnesium, which of the ions Mg s stnuock/langis 25 + and Mg is detected using a stronger magnetic eld. [2] IB, November 2006 4 Which ion would undergo the greatest deection in a mass spectrometer? 16 + A. O 16 2+ B. O 18 2+ 10 20 60 80 C. O 3 concentration of Mg /mol dm 16 18 + D. ( O O) [1] IB, November 2004 5 a) Traditionally, the r a w m a te r ia l s for t h e i) Outline how such a calibration curve production of iron are iron ore, coke, might have been produced. limestone, and pr e he a te d a i r. I r o n ii) Comment on the use of this calibration oxidesare reduced in a blast furnace curve for detecting magnesium in by both carbon and carbon monoxide plants which has an approximate to form iron. Give the equation for the 3 concentration of 1000 mg dm in reduction of iron(III) oxide by carbon solution. monoxide. [1] iii) Explain how ICP-MS or ICP-OES/ b) In many modern blast furnaces, AES is used in determining whether the metal taken up by the plant was hydrocarbons (such as methane) are magnesium or manganese, based on the separation techniques of MS and also adde d to the p r e he a te d a ir. T hi s OES/AES used. produces carbon monoxide and hydrogen. The hydrogen formed can also act as a reducing agent. Give the equation for 2 A sample of germanium is analysed in a mass the reduction of magnetite, Fe O , by 3 4 spectrometer. The rst and last processes hydrogen. [1] in mass spectrometry are vaporization and IB, November 2003 detection. a) State the names of the other three processes in the order in which they occur in a mass spectrometer. [2] 483

A M At e r i A l s A .3 caa Understandings Applications and skills ➔ Reactants adsorb onto heterogeneous ➔ Explanation of factors involved in choosing a catalysts at active sites and the products catalyst for a process. desorb. ➔ Description of how metals work as ➔ Homogeneous catalysts chemically combine heterogeneous catalysts. with the reactants to form a temporary ➔ Description of the benets of nanocatalysts in activated complex or a reaction intermediate. industry. ➔ Transition metal catalytic proper ties depend on the adsorption/absorption proper ties of the metal and the variable oxidation states. ➔ Zeolites act as selective catalysts because of their cage structure. ➔ Catalytic par ticles are nearly always nanopar ticles that have large surface areas per unit mass. Nature of science ➔ Use of models – catalysts were used to increase reaction rates before the development of an understanding of how they work . This led to models that are constantly being tested and improved. Models of catalysis many catalysts are toxic and their disposal can be difcult. Can new theories or advances in Models of how catalysts work have been areas such as nanotechnology nd catalysts that developed based on observations and theories, are even more effective and environmentally and these are constantly being tested and sound? reworked. The use of catalysts has had tremendous benets, but is not without risk as Homogeneous and heterogeneous catalysts A catalyst increases the rate of a reaction and is left unchanged at the end of the reaction. A homogeneous catalyst is in the same phase as the reactants, takes the part of a reactant, and is reformed as a product at the end of the reaction. A heterogeneous catalyst is in a different phase than that of the reactants. Catalysts work by providing an alternative reaction pathway for the reaction that lowers the activation energy, as shown in a potential energy prole diagram (gure 1; see also sub-topic 6.1). A higher proportion of reactant particles therefore achieve the required activation energy as a result (see the Maxwell–Boltzmann distribution curve in sub-topic 6.1). 484

A . 3 c A t A ly s t s potential energy of the activated complex for the uncatalysed reaction uncatalysed reaction A + B → AB a E for AC* + B → AB + C tsylatac tuohtiw  a ygrene in catalysed reaction  for A + C → AC* catalysed reaction A +B + C → AB + C, a where C represents the catalyst in catalysed reaction reactants ∆H reaction progress In an aa mx Figure 1 The activation energy for the catalysed reaction is lower than that for the (marked * in gure 1) bonds uncatalysed reaction are both forming and breaking and the reaction could fall Mechanisms of catalysis either side of the hill. The “valley” on the blue line shows The mechanism by which the activation energy is lowered varies a stable intermediate. This between homogeneous and heterogeneous catalysts. A homogeneous intermediate then reacts in catalyst forms bonds with one or more of the reactants resulting in either step 2 forming the products a reaction intermediate , which then further reacts, or an activated and regenerating the catalyst. complex, which is a temporary transition state. In either case the A one-step reaction has just an energy needed for the reactant molecules to complete the reaction is activated complex without an reduced as the reaction occurs between reactant and catalyst rather than intermediate. between one reactant and another. For example, in the general reaction: A + B →AB A and B need to collide and overcome the activation energy for the reaction. In the catalysed reaction: A + C → AC* AC* + B → AB + C the collision between A and the catalyst C to form the intermediate AC* requires less energy than does the collision between A and B to form AB. The intermediate (or activated complex) then reacts with B to form the product AB and regenerate the catalyst. In this way the homogeneous catalyst enters the reaction, but is left unchanged at the end. You will recall from topic 10 that esterication is a reversible reaction between a carboxylic acid and an alcohol. The reactants are heated in the presence of a catalyst, usually concentrated sulfuric acid: CH CH COOH + CH OH + CH CH COOCH + H O 3 3 2 __H__ 3 2 3 2 ↽⇀ Δ 485

A M AT E R I A L S + The H ion from sulfuric acid forms an intermediate with the reactants which allows the water molecule to leave and the ester to form at a much lower activation energy than is the case without the acid catalyst. The catalyst is regenerated in the reaction. A heterogeneous catalyst is in a different phase from the reactants, usually a solid catalyst for a gaseous reaction or a reaction in solution. Transition metals are common heterogeneous catalysts. These solids bring reactant molecules together in an orientation that enables them to react readily, thus reducing factors that inhibit the reaction. t heterogeneous catalyst A c catalyst u reactant A adsorbs d onto surface of solid o catalyst at active site r p product desorbs from catalyst tsylatac tsylatac A tcudorp The most common primary reaction occurs catalyst B catalysts are platinum, on catalyst palladium, and rhodium. reactant B adsorbs onto surface Copper is sometimes used, but of solid catalyst at active site is less common in industrial catalytic processes. B A Figure 2 The action of a heterogeneous catalyst Reactant molecules adsorb onto the heterogeneous catalyst at active sites. The process of adsorbing onto the surface of a solid catalyst affects the bonds in the reactants so that they are stretched, weakened, and sometimes broken. The reaction occurs on the surface of the catalyst, in one or several steps, and the products desorb from the surface of the catalyst. Homogeneous and heterogeneous catalysts compared Homogeneous catalysts are in the same phase as the reactants resulting in close contact between reactant and catalyst molecules. They work under mild conditions and have good selectivity for the desired products. A disadvantage of being in the same phase is that the catalyst needs to be removed after the reaction. This is usually accomplished by distillation, which might destroy the catalyst if a high temperature is required to distil off either the product or the catalyst. In industrial processes it is generally easier to separate large quantities of product from a heterogeneous catalyst than from a homogeneous catalyst. With a heterogeneous catalyst there is a lower effective concentration of catalyst because the reaction can only occur on the surface of the solid. 486

A . 3 c A t A ly s t s Forming the catalyst into a mesh is one way of increasing the effective Figure 3 Molecular graphic showing how a surface area. catalytic conver ter conver ts pollutants into harmless gases. The yellow framework is a A distinct disadvantage of heterogeneous catalysis is that catalytic lattice of copper atoms – the catalyst. Atoms of poisoning can occur when other compounds react with the surface the pollutant gases carbon monoxide, CO and of the catalyst. These might be products that remain, or foreign nitrogen monoxide, NO are shown adsorbed substances in the reaction mixture. In either case accumulation on the onto the copper surface (oxygen = red; catalyst surface will reduce its effectiveness. For example, the majority carbon = green; nitrogen = blue). The reaction of cars have a catalytic converter (gure 3) to reduce the emission of produces the harmless gases nitrogen, N harmful pollutants such as nitrogen oxides and incomplete combustion products including carbon monoxide. The catalyst converts carbon 2 monoxide to carbon dioxide and nitrogen oxides to nitrogen. “Leaded” (right) and carbon dioxide, CO (upper centre). petrol cannot be used in a car with a catalytic converter because the lead strongly adsorbs onto the surface of the catalyst and prevents the 2 adsorption of carbon monoxide. The blue and white zones show molecular orbitals involved in breaking apar t an NO Nanocatalysts molecule over a cluster of 10 copper atoms (brown) The use of nanoparticles has bridged the gap between homogeneous and heterogeneous catalysts. Due to their small size nanocatalysts naa  are par ticles are sometimes referred to as catalytic particles. Most heterogeneous that have dimensions less than nanocatalysts are metal nanoparticles which supply catalytically 100 nm and exhibit proper ties active sites. Catalytic nanoparticles have a large surface area per unit that dier from those of the bulk mass. They can provide a large contact area and can be introduced to material. Individual molecules a reaction mixture in the same way as homogeneous catalysts while are usually not considered to be providing adsorption/desorption sites as a heterogeneous catalyst. nanopar ticles but small clusters of them may be. Common properties that are considered when selecting a catalyst include: selectivity for only the desired product; conversion efciency; the ability to work in the conditions necessary for the reaction; environmental impact; cost; lifetime; and susceptibility to poisoning. Nanocatalysts generally have a high conversion efciency because of their small size and largesurface area. They can be engineered for maximum selectivity which reduces catalytic poisoning by unwanted substances. Enzymes,for example, can achieve greater than 99.99 % selectivity, me a ni ng l e ss t ha n 1 in 100 0 0 c o n ve r s io n s gives an unwanted by-product. This level of selectivity is rarely found in synthetic catalysts. Nanocatalysts can also provide low energy consumption and a longlifetime. Many nanocatalysts contain various forms of carbon, including graphite, carbon nanotubes, fullerenes, and graphene. Transition metal catalysts Ceramics provide some useful catalysts, but the most widely used inorganic catalysts are transition metals due to their variable oxidation states and high adsorption capacity. The variable oxidation states of transition metals allow them to enter many reactions as homogeneous catalysts. They can form complexes (topic 13), allowing them to form bonds with many substances. This frequently involves a change in oxidation state, which is returned to the original state when the reaction is over. 487

A M AT E R I A L S Transition metals also make good heterogeneous catalysts: many gases will adsorb to their surface. Weak bonds form between the reactant gases and the catalyst surface, locally increasing the reactant concentration at adsorption sites. Reactant bonds are weakened as described earlier, resulting in an increased rate of reaction. Some transition metals are toxic and should be used only if a suitable alternative is not available. Figure 4 Pumice, a porous rock Zeolites with a density less than water, is a naturally occurring zeolite Zeolites are microporous substances made of alumina silicate which has a cage-like structure providing a large surface area. Zeolites are cheap, plentiful, and occur naturally in many forms including pumice (gure 4). There are over 100 different structures of cages, cavities, channels, and other types of framework. A zeolite can act as a microscopic sieve, allowing only certain molecules through depending on their size and structure. Zeolites work by both adsorption and cation exchange. Figure 5 Molecular graphic of a zeolite structure. Zeolites are used to remove heavy metal ions from water supplies. Water Zeolites have a cage-like structure composed molecules pass through the “molecular sieve” while larger complex ions of channels, cavities, and various frameworks. of the metals are trapped by the sieve. Zeolites are also used as catalysts This structure acts as a molecular sieve as well for cracking in the petroleum industry. Their cation exchange properties as providing a large surface area for catalysis are used to remove the “hard water” ions of calcium and magnesium: these ions stay on the zeolite while potassium and sodium ions exchange out. Many washing powders contain zeolites to make washing in hard- water locations more effective. Questions 1 Catalysts may be homogeneous or other factors which should be considered heterogeneous. when choosing a suitable catalyst for an industrial process. [2] a) Distinguish between homogeneous and IB, May 2010 heterogeneous catalysts. [1] b) Discuss two factors which need to be 3 Compare the modes of action of homogeneous considered when selecting a catalyst for and heterogeneous catalysts. State one example aparticular chemical process. [2] of each type of catalysis using a chemical equation and include state symbols. [4] c) Identify the catalyst used in the catalytic cracking of long chain hydrocarbons and IB, May 2009 state one other condition needed. [2] 4 Carbon nanotubes can be used as catalysts. IB, May 2011 a) Suggest two reasons why they are 2 a) State one advantage and one disadvantage effective heterogeneous catalysts. that homogeneous catalysts have over b) State one potential concern associated heterogeneous catalysts. [2] with the use of carbon nanotubes. b) Apart from their selectivity to form the required product and their cost, discuss two 488

A . 4 li Q u id c r ys tA l s A .4 lq a Understandings Applications and skills ➔ Liquid crystals are uids that have physical ➔ Discussion of the proper ties needed for a proper ties (electrical, optical, and elasticity) substance to be used in liquid-crystal that are dependent on molecular orientation to displays (LCD). some xed axis in the material. ➔ Explanation of liquid-crystal behaviour on a ➔ Thermotropic liquid-crystal materials are pure molecular level. substances that show liquid-crystal behaviour over a temperature range. ➔ Lyotropic liquid crystals are solutions that show the liquid-crystal state over a (cer tain) range of concentrations. ➔ Nematic liquid crystal phase is characterized by rod-shaped molecules which are randomly distributed but on average align in the same direction. Nature of science ➔ Serendipity and scientic discoveries – Friedrich Reinitzer accidently discovered owing liquid crystals in 1888 while experimenting on cholesterol. The discovery of liquid crystals The observation of two separate melting H points for the substance cholesteryl benzoate (gure 1) led to the serendipitous discovery of O H H liquid crystals by Friedrich Reinitzer in 1888. O Continued experimentation and improvements in instrumentation over the years have developed Figure 1 The molecular structure of cholesteryl benzoate. this eld into an industry of ultra-high-denition Friedrich Reinitzer, an Austrian botanist, noticed two melting liquid-crystal displays. points for this compound in 1888. It turned cloudy at one temperature and clear at another, which led to fur ther research into liquid-crystal behaviour 489

A M AT E R I A L S The properties of liquid crystals Liquid crystals are a state of matter intermediate between crystalline and liquid. They are uids whose physical properties are dependent on molecular orientation relative to some xed axis in the material: a liquid crystal has molecules that can ow like a liquid but line themselves up in a crystalline order. Liquid crystal molecules maintain this orientational order, aligning mostly the same way, but not their positional order; they can slide over each other as in a liquid. The molecular shape of many liquid crystals is linear or at with very little branching. They frequently contain long-chain alkyl groups which form long, thin, rigid, rod-shaped molecules, or linear chains of aromatic rings that form at disc shapes, or a combination of both. The ability of these chains to align when a weak electric eld is applied is what forms the liquid-crystal state. Liquid crystals are often polar molecules so they change orientation when an electric eld is applied. They normally show a fairly rapid switching speed, changing in orientation when the eld is reversed. Molecular lengths of 1.3nm are typically needed for visual displays. HC 9 4 O CH 4 9 O O O O HC O O 9 4 O O CH O 4 9 CH O O 4 9 CN CH 4 9 Figure 2 Liquid crystal molecules have an alkyl group and a polar end, as in 4-cyano-4’- pentylbiphenyl, shown on the left. A series of aromatic benzene rings can form at disk shapes as in the benzene alkanoate derivative pictured on the right To form useful liquid crystals a substance needs to be chemically stable and have a liquid-crystal phase that is stable over a suitable temperature range. Worked example eld. The nitrile group (C ≡N) confers a degree ofpolarity so that the molecules align in a The structure of 4-cyano-4’-pentylbiphenyl, a common direction when a weak electric eld commercially available crystalline material used isapplied. in electrical liquid-crystal display (LCD) devices, is shown in gure 3. CH CN The biphenyl group makes the molecules more rigid and rod-shaped. The benzene ring is also 5 11 chemically stable and will not decompose under stress such as UV radiation, pressure, or at slightly Figure 3 elevated temperatures. Explain how the three different parts of the molecule – CN, C H , and the biphenyl group – 5 11 The long alkyl chain C H ensures that the contribute to the properties of the compound used 5 11 molecules cannot pack together too closely and so in LCD electrical display devices. helps maintain the liquid-crystal state. Its length gives a rod-like shape and as it is an alkyl chain it Solution is chemically stable. It is essential for liquid crystals to be polar so that they can be inuenced by a weak electric 490

A . 4 li Q u id c r ys tA l s Transmitting light: Forming LCD displays Figure 4 Nematic liquid crystal molecules align in an electric eld Liquid-crystal displays are used in many lightweight applications including digital watches, calculators, and laptops because of their low energy consumption. The ability of liquid crystal molecules to transmit light depends on the orientation of the molecules. This can be controlled by the application of a small voltage across a thin lm of the material, forming light and dark areas of the display. Thermotropic liquid-crystal materials are pure substances that show liquid-crystal behaviour over a temperature range between the solid and liquid states. Lyotropic liquid crystals are solutions that show the liquid-crystal state at certain concentrations. The nematic liquid-crystal phase is characterized by rod-shaped molecules that are randomly distributed but on average align in the same direction. In an electric eld the molecules of a nematic liquid crystal become orientated as shown in gure 4(b). The molecules can still slide over each other but in general they maintain their alignment. In LCD displays nematic liquid crystals are often placed in layers at Figure 5 A liquid-crystal strip thermometer right angles to each other with each pixel containing liquid crystal being used to measure body temperature. sandwiched between two polarized glass plates. These plates each have a The dierent areas of the strip contain set of grooves and the two sets are at 90 ° to each other. The liquid crystal thermotropic liquid crystals that are designed molecules in contact with the glass line up with the grooves and the to respond at dierent temperatures molecules in between form a twisted arrangement between the plates that is held by intermolecular bonds. Light passing through the rst lter becomes polarized. When the polar nematic liquid crystal molecules are aligned with the grooves they allow the polarized light to pass through the lm and the pixel appears bright. As a voltage is applied across the lm the polar molecules all align with the eld rather than with the grooves. The plane-polarized light is no longer aligned with the orientation of the liquid crystal molecules and so the pixel appears dark. Thermotropic liquid crystals change behaviour over a range of O temperatures. Biphenyl nitriles are thermotropic liquid crystals that naturally exist in the nematic phase as shown in gure 4(b), their rod- S shaped molecules distributed randomly but on average pointing in the same direction. Increased thermal agitation disrupts this directional order O - until it is lost, as in gure 4(a), when the normal liquid phase isformed. + Lyotropic liquid crystals have a hydrophilic end that is polar and easily N attracted to polar molecules such as water, and a hydrophobic end that a is non-polar and repelled by polar molecules. These substances take on liquid-crystal arrangements as the Figure 6 Soap contains a polar end that is concentrationincreases. Soap is an example soluble in water and a non-polar end capable of a lyotropic liquid crystal (gure 6). of dissolving oils and fats. This enables soapy water to wash greasy dishes Soap forms spherical micelles (gure7). micelle bilayer sheet The polar ends on the outside of the sphere are surrounded by water molecules with non-polar oil or grease encapsulated in the centre. At high concentrations rod-like micelles are formed that have liquid-crystal properties and can form bilayer sheets. Figure 7 Spherical and bilayer sheets are formed by soap at increasing concentrations 491

A M At e r i A l s In lyotropic liquid crystals rigid structures occur at higher concentrations. Micelles group together into hexagonal layers and then rod-shaped liquid-crystal structures (gures 8 and 9). Figure 8 Rod-like micelle structure of a lyotropic liquid crystal. The entire micelle will ow like a liquid but will retain its orientation uid behaviour erutarepmet rod-like liquid-crystal structure forms micelles form micelles begin to form hexagonal liquid crystal concentration Figure 9 Lyotropic liquid crystals restructure according to the concentration. At cer tain concentrations they exhibit liquid-crystal proper ties; at other concentrations they do not Kevlar (gure 10) is a lyotropic liquid crystal. The linked benzene rings make the rod-shaped molecules rigid. The alignment of these molecules depends on the concentration of the solution. There are strong intermolecular hydrogen bonds between the chains giving a very ordered and strong structure. These bonds can be broken with concentrated sulfuric acid, as oxygen and nitrogen atoms become protonated, breaking the hydrogen bonds. Kevlar is discussed in sub-topic A.9. t Figure 10 Molecules of Kevlar. Strong hydrogen bonding occurs between the chains. Can you identify where the hydrogen bonding occurs? 492

A . 4 li Q u id c r ys tA l s Questions 1 Liquid-crystal displays are used in digital 7 Distinguish between thermotropic and watches, calculators, and laptops. lyotropic liquid crystals and state one example of each type. [4] Describe the liquid-crystal state, in terms of molecular arrangement, and explain what IB, May 2009 happens as temperature increases. [3] 8 The structure of 4-pentyl-4-cyanobiphenyl, IB, May 2011 a commercially available nematic crystalline material used in electrical display devices, is 2 Discuss three properties a substance should have shown in gure 12. if it is to be used in liquid-crystal displays. [3] IB, May 2011 CH CN 5 11 3 Kevlar is an example of a lyotropic liquid crystal. Outline what is meant by lyotropic Figure 12 Describe and explain in molecular terms the liquid crystal. [2] workings of a twisted nematic liquid crystal. [4] IB, May 2010 IB, November 2009 (part) 4 Detergents are one example of lyotropic liquid crystals. 9 a) Compare the positional and directional order in a crystalline solid, a nematic State one other example of a lyotropic liquid phase liquid crystal, and a pure liquid. crystal and describe the difference between Show your answer by stating yes or lyotropic and thermotropic liquid crystals. [3] no in a copy of table 1. [2] IB, May 2010 ca nma p q  ha q 5 a) Name a thermotropic liquid crystal. [1] a b) Explain the liquid-crystal behaviour of the thermotropic liquid crystal named in part (a), on the molecular level. [4] Positional order IB, May 2010 6 Describe the meaning of the term liquid Directional order crystals. State and explain which diagram Table 1 in gure 11, a or b, represents molecules that are in a liquid crystalline phase. [2] b) Outline any two principles of a liquid- crystal display device. [2] IB, May 2009 Figure 11 IB, May 2009 493

A M At e r i A l s A .5 pm Understandings Applications and skills ➔ Thermoplastics soften when heated and harden ➔ Description of the use of plasticizers in polyvinyl when cooled. chloride and volatile hydrocarbons in the ➔ A thermosetting polymer is a prepolymer formation of expanded polystyrene. in a soft solid or viscous state that changes ➔ Solving problems and evaluating atom economy irreversibly into a hardened thermoset by in synthesis reactions. curing. ➔ Description of how the proper ties of polymers ➔ Elastomers are exible and can be deformed depend on their structural features. under force but will return to nearly their ➔ Deduction of structures of polymers formed from original shape once the stress is released. polymerizing 2-methylpropene. ➔ High density polyethene (HDPE) has no branching allowing chains to be packed together. ➔ Low density polyethene (LDPE) has some branching and is more exible. ➔ Plasticizers added to a polymer increase the exibility by weakening the intermolecular forces between the polymer chains. ➔ Atom economy is a measure of eciency applied in green chemistry. ➔ Isotactic addition polymers have substituents on the same side. ➔ Atactic addition polymers have the substituents randomly placed. Nature of science ➔ As a result of advances in technology (X-ray repeating units was integral in the development of polymer science. diraction, scanning tunnelling electron microscopes, etc.), scientists have been able to ➔ Ethics and risk assessment – polymer understand what occurs on the molecular level development and use has grown quicker than and manipulate matter in new ways. This allows an understanding of the risks involved, such as new polymers to be developed. recycling or possible carcinogenic proper ties. ➔ Theories can be superseded – Staudinger ’s proposal of macromolecules made of many 494