15. Carbonyls, Carboxylic Acids and chirality IGCSE PDF

15. Carbonyls, Carboxylic Acids and chirality prefix / suffix homologous functional group example...

15. Carbonyls, Carboxylic Acids and chirality prefix / suffix homologous functional group example (* = usual use) series H H alkenes C C suffix -ene C C ethene H H suffix* -ol H H H alcohols C OH prefix hydroxy- H C C C O H propan-1-ol H H H H H H halogenoalkane C halogen prefix chloro- H C C C Cl bromo- 1-chloropropane H H H iodo- O H O suffix -al aldehydes C H H C C H prefix formyl- ethanal H H O H O suffix* -one ketones prefix oxo- H C C C H C propanone H H O H O suffix -oic acid carboxylic acids C OH H C C OH ethanoic acid H H H suffix -nitrile nitriles prefix cyano- H C C C N propanenitrile C N H H suffix* -amine H H H amines C NH2 prefix amino- propylamine H C C C NH2 Or propan-1-amine H H H H O H O esters H C C O C H methyl ethanoate C O -yl –oate H H O O acyl chloride CH 3 C ethanoyl chloride C -oyl chloride Cl Cl O O amide C -amide CH3 C ethanamide NH2 NH2 N Goalby chemrevise.org 1  H Aldehydes O Ketones H O H An aldehyde’s name ends in –al It always has the C=O bond on the H C C Ketones end in -one H C C C H first carbon of the chain so it does When ketones have 5C’s or more not need an extra number. It is by H H H in a chain then it needs a number default number one on the chain H propanone to show the position of the double ethanal bond. E.g. pentan-2-one H O H O H If two ketone groups then Carboxylic acids H H di is put before –one and H C C C C C H O These have the ending -oic an an e is added to the H H H acid but no number is stem H C C C necessary for the acid group pentane-2,4-dione as it must always be at the end of the chain. The H H O H O The prefix oxo- should be numbering always starts used for compounds that H3C C C from the carboxylic acid end propanoic acid contain a ketone group in O OH addition to a carboxylic acid If there are carboxylic acid groups on both ends of the or aldehyde 2-oxopropanoic acid chain then it is called a - dioic acid O O Ethanedioic acid C C HO OH Note the e in this name Nitriles H H H OH These end in –nitrile, but the C of the H3C CN group counts as the first carbon of H C C C CN C the chain. Note the stem of the name is C H3C N different : butanenitrile and not H H H butannitrile. butanenitrile 2-hydroxy-2-methylpropanenitrile Carboxylic acid derivatives Esters H H O H Esters have two parts to their names H C C C O C H O The bit ending in –yl comes from the alcohol that has formed it and is next to the single bonded oxygen. H H H The bit ending in –anoate comes from the carboxylic acid. O (This is the chain including the C=O bond) ethyl 3-methylbutanoate methyl propanoate Acyl Chlorides H3C O O O O add –oyl chloride to the stem CH3 C CH C C (CH2)3 C name Cl Cl H3C Cl Cl ethanoyl chloride 2-methylpropanoyl chloride Pentanedioyl dichloride Amides O Secondary and tertiary amides O are named differently to show the Add –amide to the stem CH3 C two (or three) carbon chains. H3C CH2 C NH CH3 name NH2 The smaller alkyl group is ethanamide preceded by an –N which plays N-methylpropanamide the same role as a number in positioning a side alkyl chain CH3 O CH3 O CH3 H3C CH C N CH3 H3C CH2 C N CH3 N,N,2-trimethylpropanamide N,N-dimethylpropanamide N Goalby chemrevise.org 2  15A Chirality A carbon atom that has Optical isomerism occurs in carbon compounds with 4 H H H H four different groups different groups of atoms attached to a carbon (called attached is called a chiral an asymmetric carbon). H C C C C H (asymmetric) carbon H H O H atom These four groups are arranged H tetrahedrally around the carbon. OH OH This causes two A mixture containing a 50/50 mixture of the different isomers that C C two isomers (enantiomers) is described as are not superimposable CH3 H3C being a racemate or racemic mixture. H5 C2 C2 H5 to be formed. They are H H mirror images Many naturally occurring molecules contain chiral C atoms, but are usually Two compounds that are optical isomers of found in nature as a pure enantiomer each other are called enantiomers. Different systems of nomenclature are is Optical isomers have similar physical and chemical properties, existence for optical isomers. D/L or +/- are but they rotate plane polarised light in different directions. commonly used, but both have been superseded by the more useful and informative One enantiomer rotates it in one direction and the other enantiomer R/S system (this is not on the syllabus – for rotates it by the same amount in the opposite direction. information only). One optical isomer will rotate light clockwise (+)(called dextrorotatory). The other will rotate it anticlockwise(-)(called laevorotatory). Racemate A racemic mixture (a mixture of equal amounts of the two -ve enantiomer +ve enantiomer no rotation optical isomers) will not rotate plane-polarised light. Anticlockwise clockwise rotation rotation Chemical Reactions and Optical Isomers Formation of a racemate CH3 A racemate will be formed in a reaction H mechanism when a trigonal planar reactant or C intermediate is approached from both sides by NC: :CN an attacking species O Nucleophilic addition of HCN to aldehydes H and ketones (unsymmetrical) when the H H3 C CN trigonal planar carbonyl is approached from NC CH3 both sides by the HCN attacking species: There is an equal chance of C C results in the formation of a racemate either enantiomer forming so a racemate forms. No OH OH optical activity is seen N Goalby chemrevise.org 3  Formation of a racemate with SN1 mechanism H Br CH C H3C + CH2 3 H3C C CH2 CH3 C H3C C2H5 - OH H H :OH H The Br first breaks Because a The OH- ion can then attack C racemate forms away from the from either side resulting in there will be no haloalkane to form a H5C2 CH3 different enantiomers and a optical activity in planar carbocation racemate forms HO the products intermediate Comparison with SN2 mechanism In the SN2 mechanism no intermediates are formed and the reaction occurs via a transition state. - CH3 H Br OH H3C C CH2 CH3 HO C Br H3C C CH2 CH3 - H :OH CH 2CH 3 H If the reactant was chiral then during the reaction the opposite enantiomer would form. The product will rotate light in the opposite direction to the reactant A racemate can also be formed in the AS reaction of the electrophilic addition of HBr to an unsymmetrical alkene :Br - H H The bromide can H + attack this planar δ+ δ- C CH2 carbocation from C C H Br both sides leading to CH2 Br CH2 H3C CH3 a racemate H3C Br CH3 CH3 CH3 H2C CH CH2 CH3 :Br - Major product s 90% If the alkene is H H unsymmetrical, addition of + hydrogen bromide can C C CH2 CH3 CH2 CH2 CH2 CH3 Minor lead to isomeric products. H product H Br 10% N Goalby chemrevise.org 4 Compounds with C=O group Carbonyls are compounds with a C=O bond. 15B Carbonyls: Aldehydes and Ketones They can be either aldehydes or ketones H O H O H If the C=O is in the middle of If the C=O is on the end of the H. C C chain with an H attached it is an H C C C H the chain it is a ketone The name will end in -one aldehyde. H H H H The name will end in –al CH3COCH3 propanone CH3CHO ethanal Solubility in water CH3 The smaller carbonyls are soluble Pure carbonyls cannot hydrogen bond, but in water because they can form O H O C bond instead by permanent dipole bonding. hydrogen bonds with water. H CH3 δ- Reactions of carbonyls O In comparison to the C=C bond in The C=O bond is polarised because alkenes, the C=O is stronger and does O is more electronegative than δ+ not undergo addition reactions easily. C carbon. The positive carbon atom attracts nucleophiles. H3C CH3 This is in contrast to the electrophiles that are attracted to the C=C. nucleophile Oxidation Reactions Potassium dichromate K2Cr2O7 is Key point: Aldehydes Primary alcohol  aldehydes  carboxylic acid an oxidising agent that causes can be oxidised to alcohols and aldehydes to Secondary alcohol  ketones carboxylic acids, but oxidise. Tertiary alcohols do not oxidise ketones cannot be oxidised. Oxidation of Aldehydes Reaction: aldehyde  carboxylic acid H H O H H O Reagent: potassium dichromate (VI) solution and H C C C + [O]  H C C C dilute sulfuric acid. Conditions: heat under reflux H H H H H O H Full Equation for oxidation RCHO + [O]  RCO2H 3CH3CHO + Cr2O72- + 8H+  3 CH3CO2H + 4H2O + 2Cr3+ Aldehydes can also be oxidised using Fehling’s Observation: the orange dichromate ion solution or Tollen’s reagent. These are used as (Cr2O72-) reduces to the green Cr 3+ ion tests for the presence of aldehyde groups Tollen’s reagent Fehling’s solution Reagent: Tollen’s reagent formed by mixing aqueous ammonia and silver nitrate. The active substance is the complex ion of Reagent: Fehling’s solution containing blue Cu 2+ ions. [Ag(NH3)2]+. Conditions: heat gently Conditions: heat gently Reaction: aldehydes only are oxidised by Fehling’s Solution into a carboxylic acid and the copper ions Reaction: aldehydes only are oxidised by are reduced to copper(I) oxide.. Tollen’s reagent into a carboxylic acid and Observation: Aldehydes :Blue Cu 2+ ions in solution the silver(I) ions are reduced to silver atoms change to a red precipitate of Cu 2O. Ketones do Observation: with aldehydes, a silver mirror forms not react. coating the inside of the test tube. Ketones result in no change. CH3CHO + 2Cu2+ + 2H2O  CH3COOH + Cu2O + 4H+ CH3CHO + 2Ag+ + H2O  CH3COOH + 2Ag + 2H+ 5 N Goalby chemrevise.org Reduction of carbonyls Reducing agents such as NaBH4 (sodium tetrahydridoborate) Reagents: LiAlH4 in dry ether or LiAlH4 (lithium tetrahydridoaluminate) will reduce carbonyls Conditions: Room temperature and pressure to alcohols. Type of reaction: Reduction Role of reagent: Reducing agent Aldehydes will be reduced to primary alcohols Ketones will be reduced to secondary alcohols. H H O H H H H O H H H H H C C C + 2[H] H C C C O H H C C C H + 2[H]  H C C C H H H H H H H H O H H H propanal Propan-1-ol propanone H Propan-2-ol Addition of hydrogen cyanide to carbonyls to form hydroxynitriles R Reaction: carbonyl  hydroxynitrile When naming hydroxy Reagent: HCN in presence of KCN NC C OH nitriles the CN becomes Conditions: Room temperature and pressure part of the main chain Mechanism: nucleophilic addition H hydroxynitrile CH3COCH3+ HCN  CH3C(OH)(CN)CH3 CH3 NC C OH 2-hydroxy-2-methylpropanenitrile The extra KCN increases the CH3 concentration of the CN- ion CH3 CH3CHO + HCN  CH3CH(OH)CN nucleophile needed for the NC C OH first step of the mechanism 2-hydroxypropanenitrile H Nucleophilic Addition Mechanism δ- O O: - - H CN O H δ+ H3C C CH3 C H3C C CH3 H3C CH3 CN CN :CN- Reaction of carbonyls with iodine in presence of alkali Reagents: Iodine and sodium hydroxide Only carbonyls with a methyl group next to O Conditions: warm very gently the C=O bond will do this reaction. Ethanal is the only aldehyde that reacts. More H3C C H The product CHI3 is a yellow crystalline commonly is methyl ketones. precipitate with an antiseptic smell. This reaction is called the Iodoform test. CH3COCH3 + 3I2 + 4NaOH → CHI3 + CH3COONa + 3NaI +3H2O CH3COCH2CH3+ 3I2 + 4NaOH → CHI3 + CH3CH2COONa + 3NaI +3H2O N Goalby chemrevise.org 6 Reaction with 2,4-dinitro phenylhydrazine 2,4-DNP reacts with both aldehydes and ketones. The Use 2,4-DNP to identify if the compound is a product is an orange precipitate, It can be used as a test carbonyl. Then to differentiate an aldehyde for a carbonyl group in a compound. from a ketone use Tollen’s reagent. The melting point of the crystal formed can be used to help identify which carbonyl was used. Take the melting point of orange crystals product from 2,4-DNP. Compare melting point with known values in database H O 2N You don’t need to learn these equations H3C C.. H2N NH NO2 for the exam + O 2,4-DNP addition O 2N H O 2N H elimination of water H3C C NH NH NO2 H3C C N NH NO2 orange precipitate OH 15C Carboxylic Acids Solubility in Water  Acidity The smaller carboxylic (up to C4) H   The carboxylic acid are only weak acids acids dissolve in water in all O H O in water and only slightly dissociate, but proportions but after this the solubility  H3C C they are strong enough to displace rapidly reduces. They dissolve   H carbon dioxide from carbonates. because they can hydrogen bond to  O H O the water molecules. H CH3CO2H(aq) CH3CO2-(aq)+ H+(aq) Hydrogen bonding in solid ethanoic acid δ- Hydrogen bonding δ - between dimer in δ+ solid ethanoic acid O H O C CH3 H3C C δ - Solid ethanoic δ- δ+ O H O appears to have Mr of 120 Delocalisation The delocalised ion has equal C-O bond lengths. If The carboxylic acid salts are stabilised by delocalisation, delocalisation did not occur, the C=O bond would be which makes the dissociation more likely. shorter than the C-O bond. O O The pi charge cloud has O delocalised H3C C delocalised and spread out. The H3C C delocalisation makes the ion C H3C OH O more stable and therefore more likely to form. O N Goalby chemrevise.org 7 Strength of carboxylic acids O O Increasing chain length pushes delocalised electron density on to the COO- H3C CH2 C H3C CH2 C ion, making it more negative and OH O less stable. This make the acid less strong. Alkyl groups electron releasing Propanoic acid less acidic than ethanoic acid Cl O Cl O Electronegative chlorine atoms H C C delocalised withdraw electron density from H C C the COO- ion, making it less OH negative and more stable. This H O H make the acid more strong. Chlorine electron withdrawing chloroethanoic acid more acidic than ethanoic acid O H O H O OH In a dibasic acid the second HO2C- C C C C C C - group withdraws electron density O O H from the COO- ion, making it less H H H O H OH negative and more stable and weakens the O-H bond. This make HO2C- group electron withdrawing the acid more strong. Methods of preparing carboxylic acids Full Oxidation of Primary Alcohols Reaction: primary alcohol  carboxylic acid Observation: the Reagent: potassium dichromate(VI) solution and dilute sulfuric acid orange dichromate ion Conditions: use an excess of dichromate, and heat under reflux: (distil off product after (Cr2O72-) reduces to the reaction has finished) the green Cr 3+ ion H H H H H O + 2 [O] H C C C + H2O H C C C O H H H H H H O H CH3CH2CH2OH + 2[O]  CH3CH2COOH + H2O propan-1-ol Propanoic acid Oxidation of Aldehydes Reaction: aldehyde  carboxylic acid H H O H H O Reagent: potassium dichromate (VI) solution and dilute H C C C + [O]  H C C C sulfuric acid. Conditions: heat under reflux H H H H H O H Full equation for oxidation RCHO + [O]  RCOOH 3CH3CHO + Cr2O72- + 8H+  3 CH3COOH + 4H2O + 2Cr3+ Hydrolysis of Nitriles Reaction: Nitrile  carboxylic acid Reagent: dilute hydrochloric/ sulfuric acid. Conditions: heat under reflux CH3CH2CN + H+ + 2H2O  CH3CH2COOH + NH4+ N Goalby chemrevise.org 8  The Reactions of Carboxylic Acids Reduction of carboxylic acids to alcohols Lithium tetrahydridoaluminate (LiAlH4) is a Carboxylic acids will be reduced to primary alcohols strong reducing agent H H O H H H Reagents: LiAlH4 in dry ether Conditions: Room temperature and pressure H C C C + 4[H] H C C C O H + H2O Type of reaction: Reduction H H O H H H H Propanoic acid Propan-1-ol Role of reagent: Reducing agent Salt formation reactions of carboxylic acids Carboxylic acids can form salts with metals, alkalis and carbonates. acid + metal (Na)  salt + hydrogen The effervescence caused by production of CO2 2CH3CO2H + 2Na  2CH3CO2-Na+ + H2 with carboxylic acids with solid Na2CO3 or aqueous NaHCO3 can be used as a functional acid + alkali (NaOH)  salt + water group test for carboxylic acids CH3CO2H + NaOH  CH3CO2-Na+ + H2O acid + carbonate (Na2CO3)  salt + water + CO2 2CH3CO2H + Na2CO3  2CH3CO2-Na+ + H2O + CO2 Oxidation of methanoic acid O O It forms carbonic acid Carboxylic acids cannot be oxidised by using (H2 CO3 ) which can oxidising agents but methanoic acid is an H C + [O]  H O C decompose to give exception as its structure has effectively an O H O H CO2 aldehyde group Reaction of carboxylic acid with phosphorous (V) chloride Reaction: carboxylic acid  acyl chloride This reaction with PCl5 (phosphorous (V)chloride) is Reagent: PCl5 phosphorous(V) chloride used as a test for carboxylic acids. You would Conditions: room temp observe misty fumes of HCl produced. CH3COOH + PCl5  CH3COCl + POCl3 + HCl O O + PCl5  + POCl3 + HCl H3C C H3C C OH Cl N Goalby chemrevise.org 9 Esterification H H O H Esters have two parts Carboxylic acids react with alcohols, in the to their names, eg presence of a strong acid catalyst, to form H C C C O C H methyl propanoate. esters and water. H H H H+ The bit ending in –anoate The bit ending in –yl comes from Carboxylic Acid + Alcohol Ester + water comes from the carboxylic the alcohol that has formed it acid and includes the C in and is next to the single bonded the C=O bond. oxygen. H H H O H H O The reaction is reversible. The H+ reaction is quite slow and needs + H C C O H H C C O C C H + H2O H3C C heating under reflux, (often for OH H H H H H several hours or days). Low yields (50% ish) are achieved. An acid CH3CO2H + CH3CH2OH CH3CO2CH2CH3 + H2O catalyst (H2SO4) is needed. ethanoic acid ethanol ethyl ethanoate Uses of Esters Esters can have pleasant smells Esters are sweet smelling For use in perfumes they need to be non toxic, soluble in solvent such as compounds that can be used in ethanol, volatile (turns into gas easily), and not react with water. perfumes and flavourings. Esters can be used as solvents Although polar, they do not form hydrogen bonds (reason: there is no for polar organic substances hydrogen bonded to a highly electronegative atom) thus, they have much lower b.p. than the hydrogen-bonded carboxylic Ethyl ethanoate is used as a acids they came from. They are also almost insoluble in water solvent in glues and printing inks Hydrolysis of esters Esters can be hydrolysed and split up by either heating with acid or with sodium hydroxide. i) with acid This reaction is the reverse reaction of ester formation. When an reagents: dilute acid (HCl) ester is hydrolysed a carboxylic acid and an alcohol are formed. conditions: heat under reflux H+ This reaction is reversible and does CH3CH2CO2CH2CH3 + H2O CH3CH2CO2H + CH3CH2OH not give a good yield of the products. ethyl propanoate ii) with sodium hydroxide reagents: dilute sodium hydroxide CH3CH2CO2CH3 + NaOH  CH3CH2CO2- Na+ + CH3OH conditions: heat under reflux methyl propanoate sodium propanoate methanol The carboxylic acid salt product is the anion of the carboxylic acid. This reaction goes to completion. The anion is resistant to attack by weak nucleophiles such as alcohols, so the reaction is not reversible. 10 N Goalby chemrevise.org Carboxylic acid derivatives: Acyl Chlorides Acyl Chlorides O Acyl chlorides are The Cl group is classed as a good leaving much more reactive groups (to do with less effective delocalisation.) CH3 C than carboxylic acids This makes acyl chlorides and acid anhydrides Cl much more reactive than carboxylic acids and ethanoyl chloride esters Reaction with water Reaction with alcohol Change in functional group: acyl chloride  Change in functional group: acyl chloride  ester carboxylic acid Reagent: alcohol Reagent: water Conditions: room temp. Conditions: room temp. RCOCl (l) + CH3CH2OH  RCO2CH2CH3 + HCl (g) RCOCl (l) + H2O  RCO2H + HCl (g) O H O H H O O CH 3 C + CH3CH2OH  H C C O C C H + HCl CH3 C + H2O  CH3 C + HCl (g) Cl H H H Cl OH Observation: Steamy white fumes of HCl are Observation: Steamy white fumes of HCl given off are given off This reaction for making esters is much better than using carboxylic acids as the reaction is much quicker and it is not a reversible reaction Reaction with ammonia Reaction with primary amines Change in functional group: acyl chloride  Change in functional group: acyl chloride  primary amide secondary amide Reagent: ammonia Reagent: primary amine Conditions: room temp. Conditions: room temp. RCOCl (l) +2NH3  RCONH2 + NH4Cl (s) RCOCl +2CH3NH2  RCONHCH3 + CH3NH3+Cl- O O O O + CH3NH3+Cl- CH3 C + 2NH3  CH3 C + NH4Cl (s) CH3 C + 2CH3NH2  CH3 C NH CH3 Cl NH2 Cl N-methylethanamide Observation: white smoke of NH4Cl is given off N Goalby chemrevise.org 11 Polyesters There are two types of polymerisation: addition and condensation Condensation Polymerisation In condensation polymerisation there are two different monomers The two most common types of that add together and a small molecule is usually given off as a condensation polymers are side-product e.g. H2O or HCl. polyesters and polyamides which involve the formation of an ester The monomers usually have the same functional group on both ends linkage or an amide linkage. of the molecule e.g. di-amine, di carboxylic acid, diol, diacyl chloride. Forming polyesters uses these reactions we met earlier in the course Carboxylic Acid + Alcohol  Ester + Water Acyl chloride + Alcohol  Ester + HCl If we have the same functional group on each end of molecule we can make polymers so we have the analogous equations: dicarboxylic acid + diol  poly(ester) + water diacyl dichloride + diol  poly(ester) + HCl Using the carboxylic acid to make the ester or amide would need an acid catalyst and would only give an equilibrium mixture. The more reactive acyl chloride goes to completion and does not need a catalyst but does produce hazardous HCl fumes. Terylene- a common polyester O O O O n + n HO CH CH OH C C O CH2 CH2 O + 2n-1 H2O C C 2 2 n HO OH ethane-1,2-diol The -1 here is because at benzene-1,4-dicarboxylic acid each end of the chain the H Terylene fabric is used in clothing, tire cords and OH are still present O O O O n C (CH2)3 C + n HO OH + 2n-1 HCl C (CH2)3 C O O Cl Cl n pentanedioyl dichloride benzene-1,4-diol N Goalby chemrevise.org 12  It is also possible for polyamides and polyesters to form from one monomer, if that monomer contains both the functional groups needed to react 3 repeating units H O H O H O HO O CH C HO C C O C C O C C OH H 3C OH CH3 CH3 CH3 2-hydroxypropanoic acid (lactic acid) 3 repeating units poly(lactic acid) OH O O O OH 1 repeating unit 4-hydroxypentanoic acid O O It is possible for some of these compounds to form O O O various cyclic esters under different conditions from forming the polymer. You do not need to learn these but may be asked to O deduce structures from information given Chemical reactivity of condensation polymers The reactivity can be explained by the presence of polyesters can be broken down by hydrolysis and polar bonds which can attract attacking species are, therefore, biodegradable such as nucleophiles and acids Polyesters can be hydrolysed by acid and alkali With HCl a polyester will be hydrolysed and split up in to the original dicarboxylic acid and diol With NaOH an polyester will be hydrolysed and split up into the diol and dicarboxylic acid salt. N Goalby chemrevise.org 13  15E. Spectroscopy and chromatography The effect of different types of radiation on molecules i infrared in analysis – infra red energy causes bonds to vibrate. This can be used to identify the types of bond in a molecule ii microwaves for heating- certain molecules absorb the microwaves causing them to rotate iii radio waves in NMR – can cause the hydrogen nucleus to change its spin state. This can give us information about the arrangements of hydrogens in a molecule. iv ultraviolet in initiation of reactions – UV energy can break bonds such as the Cl-Cl bond or C-Cl bond NMR spectroscopy NMR spectroscopy involves interaction of materials with the low- The radio waves used in energy radiowave region of the electromagnetic spectrum proton NMR cause the hydrogen nucleus to change NMR spectroscopy is the same technology as that used in ‘magnetic its spin state. resonance imaging’ (MRI) to obtain diagnostic information about internal structures in body scanners e.g. scanning for brain disorders Equivalent Hydrogen atoms. In addition the intensity (integration value) In an H NMR spectrum, there is one of each signal is proportional to the number signal for each set of equivalent H atoms. of equivalent H atoms it represents. Ha H b 3 Ha Hb Hc H d a a H Hd H C C O C C C C c aH Hc H H Hb H H C Ha d a 2 a Ha 4sets of equivalent H’s: Ethanol has 3 ratio 6:1:2:3 1 groups of different hydrogen atoms O H3C C CH3 1 signal a a a b O c CH3c H3C CH2 C O CH3 O a b c H3C CH2 C O C CH3 3 sets of equivalent H’s: ratio 3:2:3 c CH3 Br 3 sets of equivalent H’s: ratio 3:2:9 a b c H3C CH CH2 CH3 CH CH2 CH3 a b c d 3 sets of equivalent 4 sets of equivalent H’s: ratio 3:1:2:3 H’s: ratio 3:1:2 Solvents This means that in the H NMR the Samples are dissolved in solvents without any 1H atoms, e.g. CCl4, CDCl3. solvent will not give any peaks CCl4 is a non-polar compound that is a good solvent for non-polar organic molecules CDCl3 is a polar covalent molecule that is a good solvent for polar organic molecules N Goalby chemrevise.org 14 Calibration and shift A small amount of TMS (tetramethylsilane) is added to the sample to calibrate the spectrum CH3 The spectra are recorded on a TMS is used because: scale known as the chemical it only gives one signal H3C Si CH3 shift (δ), which is how much the its signal is away from all the other H signals field has shifted away from the gives strong signal so only a small amount needed field for TMS.. it is non-toxic CH3 it is inert tetramethylsilane it has a low boiling point and so can be removed from sample easily 10 9 8 7 6 5 4 3 2 1 0 The δ is a measure in parts per million (ppm) is a relative δ chemical shift (ppm) scale of how far the frequency of the proton signal has shifted away from that for TMS. H NMR shift The δ depends on what other atoms/groups are near the H – more electronegative groups gives a greater shift. 3 O H3C CH2 C O CH3 3 2 δ ppm 3 H H O H C C C H H O H 2 1 δ ppm N Goalby chemrevise.org 15 Spin-Spin coupling in H NMR Nuclei in identical chemical In high resolution H NMR each signal in the spectrum can be split into environments do not show further lines due to inequivalent H’s on neighbouring C atoms. coupling amongst themselves! Splitting of peak = number of inequivalent H’s on neighbouring C atoms + 1 a b O c signal singlet doublet triplet quartet quintet H3C CH2 C O CH3 The peak due to group a will be a triplet as it is next to b appearance (a carbon with 2 H’s) The peak due to group b will be a quartet as it is next to a Split number (a carbon with 3H’s) 1 2 3 4 5 of peaks The peak due to group c will number of be a singlet as it is next to a neighbouring carbon with no H’s) 0 1 2 3 4 inequivalent H atoms For 6 split peaks use the relative size 1:1 1:2:1 1:3:3:1 1:4:6:4:1 term hextet or multiplet H O H H The peak due to group c will a be a triplet as it is next to a H C C O C C H carbon with 2 H’s a H Shift 0.7-1.2 b H H c Integration trace 3 c The peak due to group a will The peak due to group b will b be a singlet as it is next to be a quartet as it is next to a carbon with 0 H’s a carbon with 3 H’s Shift 2.1-2.6 Shift 3.7 -4.1 Integration trace 3 Integration trace 2 ppm N Goalby chemrevise.org 16 Mass spectrometry Measuring the Mr of an organic molecule Spectra for C4H10 If a molecule is put through a mass spectrometer Mass spectrum for butane it will often break up and give a series of peaks caused by the fragments. The peak with the 43 Molecular ion largest m/z, however, will be due to the complete C4H10+ molecule and will be equal to the Mr of the molecule. This peak is called the parent ion or 29 molecular ion 58 Fragmentation Molecular ion formed: M → [M]+. + e– When organic molecules are passed through a mass spectrometer, it detects both the whole molecule and The molecule loses an electron and fragments of the molecule. becomes both an ion and a free radical Several peaks in the mass spectrum occur due to fragmentation. This process produces an ion The molecular ion fragments due to covalent bonds breaking: [M]+. → X+ + Y. and a free radical. The ion is responsible for the peak Relatively stable ions such as carbocations R+ such as CH3CH2+ and acylium ions [R-C=O]+ are common. The more stable the ion, the greater the peak intensity. The peak with the highest mass/charge ratio will be normally due to the original molecule that hasn’t fragmented (called the molecular ion). As the charge of the ion is +1 the mass/ charge ratio is equal to Mr. Equation for formation molecular ion Mass spectrum for butane C4H10 [C4H10]+. + e– m/z 58 43 Equations for formation of fragment ions from molecular ions [C4H10]+. [CH3CH2CH2]+ +.CH3 m/z 43 29 C4H10 = 58 [C4H10]+. [CH3CH2]+ +.CH2CH3 m/z 29 Equation for formation molecular ion Mass spectrum for butanone CH3CH2COCH3 [CH3CH2COCH3]+. + e– m/z 72 The high peak 43 at 43 due to [CH3CO]+ Equations for formation of fragment ions from molecular ions stability of acyl group [CH3CH2COCH3]+. [CH3CH2CO]+ +.CH3 m/z 57 29 [CH3CH2CO]+ [CH3CH2COCH3]+. [CH3CO]+ +.CH2CH3 m/z 43 +. [CH3CH2]+ 57 [CH3CH2COCH3] [CH3CH2COCH3]+. [CH3CH2]+ +.COCH3 m/z 29 72 N Goalby chemrevise.org 17  Chromatography The mobile phase may be a liquid or a gas. The stationary phase may be a solid (as in thin- Chromatography is an analytical technique that separates layer chromatography, TLC) or either a liquid or components in a mixture between a mobile phase and a solid on a solid support (as in gas stationary phase. chromatography, GC) Separation by column chromatography depends on the balance between solubility in the moving phase If the stationary phase was polar and the moving and retention in the stationary phase. phase was non- polar e.g. Hexane. Then non- polar compounds would pass through the column more quickly than polar compounds as they would A solid stationary phase separates by adsorption, have a greater solubility in the non-polar moving A liquid stationary phase separates by relative solubility phase. (Think about intermolecular forces) HPLC stands for high performance liquid In gas-liquid chromatography GC the mobile phase is chromatography. a inert gas such as nitrogen, helium, argon. HPLC: stationary phase is a solid silica The Stationary phase is a liquid on an inert solid. HPLC: mobile phase a liquid Gas-Liquid Chromatography Gas-liquid chromatography can be used to separate In gas-liquid chromatography, the mobile mixtures of volatile liquids. phase is a gas such as helium and the stationary phase is a high boiling point The time taken for a particular compound to liquid absorbed onto a solid. travel from the injection of the sample to where it leaves the column to the detector is known as Sample in its retention time. This can be used to identify Flow a substance. control oven Some compounds have similar retention times so will not be distinguished. display Waste column outlet Basic gas-liquid chromatography will tell us how many components there are in the mixture by the Carrier gas detector number of peaks. It will also tell us the abundance of each substance. The area under each peak will be proportional to the abundance of that component. It is also possible for gas-liquid chromatography machine to be connected to a mass Most commonly a mass spectrometer is combined spectrometer, IR or NMR machine, enabling all with GC to generate a mass spectra which can be the components in a mixture to be identified. analysed or compared with a spectral database by computer for positive identification of each component in the mixture. GC-MS is used in analysis, in forensics, environmental analysis, airport security and space probes. N Goalby chemrevise.org 18 TLC Chromatography (thin-layer A mixture can be separated by chromatography and chromatography) identified from the amount they have moved. (Can be used with mixtures of amino acids) Method: Thin-layer chromatography Wear plastic gloves to prevent contamination a) Wearing gloves, draw a pencil line 1 cm above the from the hands to the plate bottom of a TLC plate and mark spots for each sample, equally spaced along line. pencil line –will not dissolve in the solvent b) Use a capillary tube to add a tiny drop of each solution to a tiny drop – too big a drop will cause different different spot and allow the plate to air dry. spots to merge c) Add solvent to a chamber or large beaker with a lid so that is no more than 1cm in depth Depth of solvent– if the solvent is too deep it d) Place the TLC plate into the chamber, making sure that will dissolve the sample spots from the plate the level of the solvent is below the pencil line. Replace lid– to prevent evaporation of toxic solvent the lid to get a tight seal. e) When the level of the solvent reaches about 1 cm from Will get more accurate results if the solvent is the top of the plate, remove the plate and mark the solvent allowed to rise to near the top of the plate but level with a pencil. Allow the plate to dry in the fume the Rf value can be calculated if the solvent cupboard. front does not reach the top of the plate f) Place the plate under a UV lamp in order to see the spots. Draw around them lightly in pencil. dry in a fume cupboard as the solvent is toxic g) Calculate the Rf values of the observed spots. UV lamp used if the spots are colourless and not visible If using amino acids then ninhydrin spray can be used instead of UV lamp to locate the spots Rf value = distance moved by amino acid distance moved by the solvent Each substance has its own Rf value Measure how far each spot travels relative to the solvent front and calculate the Rf value. Compare Rf values to those for known substances. Some substances won't separate because similar compounds have similar Rf values. So some spots may contain more than one compound N Goalby chemrevise.org 19 Bringing it all together C H O 1. Work out empirical formula 66.63/12 11.18/1 22.19/16 Elemental analysis C 66.63% H 11.18% O 22.19% =5.5525 =11.18 =1.386875 =4 =8 =1 2. Using molecular ion peak m/z value from mass spectrum calculate molecular formula Mr empirical formula C4H8O = 72 molecular ion peak m/z value= 144 If Mr molecular formula 144 then compound is C8H16O2 3. Use IR spectra to identify main C8H16O2 could be an ester, carboxylic acid or combination of bonds/functional group alcohol and carbonyl. Look for IR spectra for C=O and O-H bonds There is a C=O but no O-H absorptions, so must be an ester. C-H C=O CH3 4. Use NMR spectra to give details of carbon chain singlet of area 9 H3C C H At δ =0.9 CH3 4 peaks – only 4 different environments. Means 3 CH3 groups 9 Peak at δ 4 shows H–C–O Peak at δ 1.2 Peak at δ 2.2 shows H–C=O shows R-CH3 Area 2 suggests CH2 Area 2 suggests CH2 Area 3 means CH3 Quartet means next to a Singlet means adjacent to Triplet means next CH3 C with no hydrogens to a CH2 H H3C CH3 H O H3C O C H H3C O C C H 2 2 3 H H 5 4 δ ppm 3 2 1 Put all together to give final structure CH3 O H3C C CH2 C O CH2 CH3 CH3 N Goalby chemrevise.org 20